From c0f00b0a3aec0125331bf3b8f85ed74ae34fb44b Mon Sep 17 00:00:00 2001 From: krahets Date: Thu, 10 Nov 2022 04:00:10 +0800 Subject: [PATCH 01/23] Update LICENSE and README. --- LICENSE.md | 488 ++++++++++++++++++++++++++++++++++++++++------------- README.md | 10 +- 2 files changed, 377 insertions(+), 121 deletions(-) diff --git a/LICENSE.md b/LICENSE.md index d379d94..3b7b82d 100644 --- a/LICENSE.md +++ b/LICENSE.md @@ -1,173 +1,427 @@ -# Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International +Attribution-ShareAlike 4.0 International + +======================================================================= + +Creative Commons Corporation ("Creative Commons") is not a law firm and +does not provide legal services or legal advice. Distribution of +Creative Commons public licenses does not create a lawyer-client or +other relationship. Creative Commons makes its licenses and related +information available on an "as-is" basis. Creative Commons gives no +warranties regarding its licenses, any material licensed under their +terms and conditions, or any related information. Creative Commons +disclaims all liability for damages resulting from their use to the +fullest extent possible. + +Using Creative Commons Public Licenses + +Creative Commons public licenses provide a standard set of terms and +conditions that creators and other rights holders may use to share +original works of authorship and other material subject to copyright +and certain other rights specified in the public license below. The +following considerations are for informational purposes only, are not +exhaustive, and do not form part of our licenses. + + Considerations for licensors: Our public licenses are + intended for use by those authorized to give the public + permission to use material in ways otherwise restricted by + copyright and certain other rights. Our licenses are + irrevocable. Licensors should read and understand the terms + and conditions of the license they choose before applying it. + Licensors should also secure all rights necessary before + applying our licenses so that the public can reuse the + material as expected. Licensors should clearly mark any + material not subject to the license. This includes other CC- + licensed material, or material used under an exception or + limitation to copyright. More considerations for licensors: + wiki.creativecommons.org/Considerations_for_licensors + + Considerations for the public: By using one of our public + licenses, a licensor grants the public permission to use the + licensed material under specified terms and conditions. If + the licensor's permission is not necessary for any reason--for + example, because of any applicable exception or limitation to + copyright--then that use is not regulated by the license. Our + licenses grant only permissions under copyright and certain + other rights that a licensor has authority to grant. Use of + the licensed material may still be restricted for other + reasons, including because others have copyright or other + rights in the material. A licensor may make special requests, + such as asking that all changes be marked or described. + Although not required by our licenses, you are encouraged to + respect those requests where reasonable. More_considerations + for the public: + wiki.creativecommons.org/Considerations_for_licensees + +======================================================================= + +Creative Commons Attribution-ShareAlike 4.0 International Public +License + +By exercising the Licensed Rights (defined below), You accept and agree +to be bound by the terms and conditions of this Creative Commons +Attribution-ShareAlike 4.0 International Public License ("Public +License"). To the extent this Public License may be interpreted as a +contract, You are granted the Licensed Rights in consideration of Your +acceptance of these terms and conditions, and the Licensor grants You +such rights in consideration of benefits the Licensor receives from +making the Licensed Material available under these terms and +conditions. + + +Section 1 -- Definitions. + + a. Adapted Material means material subject to Copyright and Similar + Rights that is derived from or based upon the Licensed Material + and in which the Licensed Material is translated, altered, + arranged, transformed, or otherwise modified in a manner requiring + permission under the Copyright and Similar Rights held by the + Licensor. For purposes of this Public License, where the Licensed + Material is a musical work, performance, or sound recording, + Adapted Material is always produced where the Licensed Material is + synched in timed relation with a moving image. + + b. Adapter's License means the license You apply to Your Copyright + and Similar Rights in Your contributions to Adapted Material in + accordance with the terms and conditions of this Public License. + + c. BY-SA Compatible License means a license listed at + creativecommons.org/compatiblelicenses, approved by Creative + Commons as essentially the equivalent of this Public License. + + d. Copyright and Similar Rights means copyright and/or similar rights + closely related to copyright including, without limitation, + performance, broadcast, sound recording, and Sui Generis Database + Rights, without regard to how the rights are labeled or + categorized. For purposes of this Public License, the rights + specified in Section 2(b)(1)-(2) are not Copyright and Similar + Rights. + + e. Effective Technological Measures means those measures that, in the + absence of proper authority, may not be circumvented under laws + fulfilling obligations under Article 11 of the WIPO Copyright + Treaty adopted on December 20, 1996, and/or similar international + agreements. + + f. Exceptions and Limitations means fair use, fair dealing, and/or + any other exception or limitation to Copyright and Similar Rights + that applies to Your use of the Licensed Material. + + g. License Elements means the license attributes listed in the name + of a Creative Commons Public License. The License Elements of this + Public License are Attribution and ShareAlike. + + h. Licensed Material means the artistic or literary work, database, + or other material to which the Licensor applied this Public + License. + + i. 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Sui Generis Database Rights means rights other than copyright + resulting from Directive 96/9/EC of the European Parliament and of + the Council of 11 March 1996 on the legal protection of databases, + as amended and/or succeeded, as well as other essentially + equivalent rights anywhere in the world. + + m. You means the individual or entity exercising the Licensed Rights + under this Public License. Your has a corresponding meaning. + + +Section 2 -- Scope. + + a. License grant. + + 1. Subject to the terms and conditions of this Public License, + the Licensor hereby grants You a worldwide, royalty-free, + non-sublicensable, non-exclusive, irrevocable license to + exercise the Licensed Rights in the Licensed Material to: + + a. reproduce and Share the Licensed Material, in whole or + in part; and + + b. produce, reproduce, and Share Adapted Material. + + 2. Exceptions and Limitations. For the avoidance of doubt, where + Exceptions and Limitations apply to Your use, this Public + License does not apply, and You do not need to comply with + its terms and conditions. + + 3. Term. The term of this Public License is specified in Section + 6(a). + + 4. Media and formats; technical modifications allowed. The + Licensor authorizes You to exercise the Licensed Rights in + all media and formats whether now known or hereafter created, + and to make technical modifications necessary to do so. The + Licensor waives and/or agrees not to assert any right or + authority to forbid You from making technical modifications + necessary to exercise the Licensed Rights, including + technical modifications necessary to circumvent Effective + Technological Measures. For purposes of this Public License, + simply making modifications authorized by this Section 2(a) + (4) never produces Adapted Material. + + 5. Downstream recipients. + + a. Offer from the Licensor -- Licensed Material. Every + recipient of the Licensed Material automatically + receives an offer from the Licensor to exercise the + Licensed Rights under the terms and conditions of this + Public License. + + b. Additional offer from the Licensor -- Adapted Material. + Every recipient of Adapted Material from You + automatically receives an offer from the Licensor to + exercise the Licensed Rights in the Adapted Material + under the conditions of the Adapter's License You apply. + + c. No downstream restrictions. You may not offer or impose + any additional or different terms or conditions on, or + apply any Effective Technological Measures to, the + Licensed Material if doing so restricts exercise of the + Licensed Rights by any recipient of the Licensed + Material. -Creative Commons Corporation (“Creative Commons”) is not a law firm and does not provide legal services or legal advice. Distribution of Creative Commons public licenses does not create a lawyer-client or other relationship. Creative Commons makes its licenses and related information available on an “as-is” basis. Creative Commons gives no warranties regarding its licenses, any material licensed under their terms and conditions, or any related information. Creative Commons disclaims all liability for damages resulting from their use to the fullest extent possible. + 6. No endorsement. Nothing in this Public License constitutes or + may be construed as permission to assert or imply that You + are, or that Your use of the Licensed Material is, connected + with, or sponsored, endorsed, or granted official status by, + the Licensor or others designated to receive attribution as + provided in Section 3(a)(1)(A)(i). -### Using Creative Commons Public Licenses + b. Other rights. -Creative Commons public licenses provide a standard set of terms and conditions that creators and other rights holders may use to share original works of authorship and other material subject to copyright and certain other rights specified in the public license below. The following considerations are for informational purposes only, are not exhaustive, and do not form part of our licenses. + 1. Moral rights, such as the right of integrity, are not + licensed under this Public License, nor are publicity, + privacy, and/or other similar personality rights; however, to + the extent possible, the Licensor waives and/or agrees not to + assert any such rights held by the Licensor to the limited + extent necessary to allow You to exercise the Licensed + Rights, but not otherwise. -* __Considerations for licensors:__ Our public licenses are intended for use by those authorized to give the public permission to use material in ways otherwise restricted by copyright and certain other rights. Our licenses are irrevocable. Licensors should read and understand the terms and conditions of the license they choose before applying it. Licensors should also secure all rights necessary before applying our licenses so that the public can reuse the material as expected. Licensors should clearly mark any material not subject to the license. This includes other CC-licensed material, or material used under an exception or limitation to copyright. [More considerations for licensors](http://wiki.creativecommons.org/Considerations_for_licensors_and_licensees#Considerations_for_licensors). + 2. Patent and trademark rights are not licensed under this + Public License. -* __Considerations for the public:__ By using one of our public licenses, a licensor grants the public permission to use the licensed material under specified terms and conditions. If the licensor’s permission is not necessary for any reason–for example, because of any applicable exception or limitation to copyright–then that use is not regulated by the license. Our licenses grant only permissions under copyright and certain other rights that a licensor has authority to grant. Use of the licensed material may still be restricted for other reasons, including because others have copyright or other rights in the material. A licensor may make special requests, such as asking that all changes be marked or described. Although not required by our licenses, you are encouraged to respect those requests where reasonable. [More considerations for the public](http://wiki.creativecommons.org/Considerations_for_licensors_and_licensees#Considerations_for_licensees). + 3. To the extent possible, the Licensor waives any right to + collect royalties from You for the exercise of the Licensed + Rights, whether directly or through a collecting society + under any voluntary or waivable statutory or compulsory + licensing scheme. In all other cases the Licensor expressly + reserves any right to collect such royalties. -## Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International Public License -By exercising the Licensed Rights (defined below), You accept and agree to be bound by the terms and conditions of this Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International Public License ("Public License"). To the extent this Public License may be interpreted as a contract, You are granted the Licensed Rights in consideration of Your acceptance of these terms and conditions, and the Licensor grants You such rights in consideration of benefits the Licensor receives from making the Licensed Material available under these terms and conditions. +Section 3 -- License Conditions. -### Section 1 – Definitions. +Your exercise of the Licensed Rights is expressly made subject to the +following conditions. -a. __Adapted Material__ means material subject to Copyright and Similar Rights that is derived from or based upon the Licensed Material and in which the Licensed Material is translated, altered, arranged, transformed, or otherwise modified in a manner requiring permission under the Copyright and Similar Rights held by the Licensor. For purposes of this Public License, where the Licensed Material is a musical work, performance, or sound recording, Adapted Material is always produced where the Licensed Material is synched in timed relation with a moving image. + a. Attribution. -b. __Adapter's License__ means the license You apply to Your Copyright and Similar Rights in Your contributions to Adapted Material in accordance with the terms and conditions of this Public License. + 1. If You Share the Licensed Material (including in modified + form), You must: -c. __BY-NC-SA Compatible License__ means a license listed at [creativecommons.org/compatiblelicenses](http://creativecommons.org/compatiblelicenses), approved by Creative Commons as essentially the equivalent of this Public License. + a. retain the following if it is supplied by the Licensor + with the Licensed Material: -d. __Copyright and Similar Rights__ means copyright and/or similar rights closely related to copyright including, without limitation, performance, broadcast, sound recording, and Sui Generis Database Rights, without regard to how the rights are labeled or categorized. For purposes of this Public License, the rights specified in Section 2(b)(1)-(2) are not Copyright and Similar Rights. + i. identification of the creator(s) of the Licensed + Material and any others designated to receive + attribution, in any reasonable manner requested by + the Licensor (including by pseudonym if + designated); -e. __Effective Technological Measures__ means those measures that, in the absence of proper authority, may not be circumvented under laws fulfilling obligations under Article 11 of the WIPO Copyright Treaty adopted on December 20, 1996, and/or similar international agreements. + ii. a copyright notice; -f. __Exceptions and Limitations__ means fair use, fair dealing, and/or any other exception or limitation to Copyright and Similar Rights that applies to Your use of the Licensed Material. + iii. a notice that refers to this Public License; + + iv. a notice that refers to the disclaimer of + warranties; -g. __License Elements__ means the license attributes listed in the name of a Creative Commons Public License. The License Elements of this Public License are Attribution, NonCommercial, and ShareAlike. + v. a URI or hyperlink to the Licensed Material to the + extent reasonably practicable; -h. __Licensed Material__ means the artistic or literary work, database, or other material to which the Licensor applied this Public License. + b. indicate if You modified the Licensed Material and + retain an indication of any previous modifications; and + + c. indicate the Licensed Material is licensed under this + Public License, and include the text of, or the URI or + hyperlink to, this Public License. -i. __Licensed Rights__ means the rights granted to You subject to the terms and conditions of this Public License, which are limited to all Copyright and Similar Rights that apply to Your use of the Licensed Material and that the Licensor has authority to license. + 2. You may satisfy the conditions in Section 3(a)(1) in any + reasonable manner based on the medium, means, and context in + which You Share the Licensed Material. For example, it may be + reasonable to satisfy the conditions by providing a URI or + hyperlink to a resource that includes the required + information. + + 3. If requested by the Licensor, You must remove any of the + information required by Section 3(a)(1)(A) to the extent + reasonably practicable. -j. __Licensor__ means the individual(s) or entity(ies) granting rights under this Public License. + b. ShareAlike. -k. __NonCommercial__ means not primarily intended for or directed towards commercial advantage or monetary compensation. For purposes of this Public License, the exchange of the Licensed Material for other material subject to Copyright and Similar Rights by digital file-sharing or similar means is NonCommercial provided there is no payment of monetary compensation in connection with the exchange. + In addition to the conditions in Section 3(a), if You Share + Adapted Material You produce, the following conditions also apply. -l. __Share__ means to provide material to the public by any means or process that requires permission under the Licensed Rights, such as reproduction, public display, public performance, distribution, dissemination, communication, or importation, and to make material available to the public including in ways that members of the public may access the material from a place and at a time individually chosen by them. + 1. The Adapter's License You apply must be a Creative Commons + license with the same License Elements, this version or + later, or a BY-SA Compatible License. -m. __Sui Generis Database Rights__ means rights other than copyright resulting from Directive 96/9/EC of the European Parliament and of the Council of 11 March 1996 on the legal protection of databases, as amended and/or succeeded, as well as other essentially equivalent rights anywhere in the world. + 2. You must include the text of, or the URI or hyperlink to, the + Adapter's License You apply. You may satisfy this condition + in any reasonable manner based on the medium, means, and + context in which You Share Adapted Material. -n. __You__ means the individual or entity exercising the Licensed Rights under this Public License. __Your__ has a corresponding meaning. + 3. You may not offer or impose any additional or different terms + or conditions on, or apply any Effective Technological + Measures to, Adapted Material that restrict exercise of the + rights granted under the Adapter's License You apply. -### Section 2 – Scope. -a. ___License grant.___ +Section 4 -- Sui Generis Database Rights. - 1. Subject to the terms and conditions of this Public License, the Licensor hereby grants You a worldwide, royalty-free, non-sublicensable, non-exclusive, irrevocable license to exercise the Licensed Rights in the Licensed Material to: +Where the Licensed Rights include Sui Generis Database Rights that +apply to Your use of the Licensed Material: - A. reproduce and Share the Licensed Material, in whole or in part, for NonCommercial purposes only; and + a. for the avoidance of doubt, Section 2(a)(1) grants You the right + to extract, reuse, reproduce, and Share all or a substantial + portion of the contents of the database; - B. produce, reproduce, and Share Adapted Material for NonCommercial purposes only. + b. if You include all or a substantial portion of the database + contents in a database in which You have Sui Generis Database + Rights, then the database in which You have Sui Generis Database + Rights (but not its individual contents) is Adapted Material, - 2. __Exceptions and Limitations.__ For the avoidance of doubt, where Exceptions and Limitations apply to Your use, this Public License does not apply, and You do not need to comply with its terms and conditions. + including for purposes of Section 3(b); and + c. You must comply with the conditions in Section 3(a) if You Share + all or a substantial portion of the contents of the database. - 3. __Term.__ The term of this Public License is specified in Section 6(a). +For the avoidance of doubt, this Section 4 supplements and does not +replace Your obligations under this Public License where the Licensed +Rights include other Copyright and Similar Rights. - 4. __Media and formats; technical modifications allowed.__ The Licensor authorizes You to exercise the Licensed Rights in all media and formats whether now known or hereafter created, and to make technical modifications necessary to do so. The Licensor waives and/or agrees not to assert any right or authority to forbid You from making technical modifications necessary to exercise the Licensed Rights, including technical modifications necessary to circumvent Effective Technological Measures. For purposes of this Public License, simply making modifications authorized by this Section 2(a)(4) never produces Adapted Material. - 5. __Downstream recipients.__ +Section 5 -- Disclaimer of Warranties and Limitation of Liability. - A. __Offer from the Licensor – Licensed Material.__ Every recipient of the Licensed Material automatically receives an offer from the Licensor to exercise the Licensed Rights under the terms and conditions of this Public License. + a. UNLESS OTHERWISE SEPARATELY UNDERTAKEN BY THE LICENSOR, TO THE + EXTENT POSSIBLE, THE LICENSOR OFFERS THE LICENSED MATERIAL AS-IS + AND AS-AVAILABLE, AND MAKES NO REPRESENTATIONS OR WARRANTIES OF + ANY KIND CONCERNING THE LICENSED MATERIAL, WHETHER EXPRESS, + IMPLIED, STATUTORY, OR OTHER. THIS INCLUDES, WITHOUT LIMITATION, + WARRANTIES OF TITLE, MERCHANTABILITY, FITNESS FOR A PARTICULAR + PURPOSE, NON-INFRINGEMENT, ABSENCE OF LATENT OR OTHER DEFECTS, + ACCURACY, OR THE PRESENCE OR ABSENCE OF ERRORS, WHETHER OR NOT + KNOWN OR DISCOVERABLE. WHERE DISCLAIMERS OF WARRANTIES ARE NOT + ALLOWED IN FULL OR IN PART, THIS DISCLAIMER MAY NOT APPLY TO YOU. - B. __Additional offer from the Licensor – Adapted Material.__ Every recipient of Adapted Material from You automatically receives an offer from the Licensor to exercise the Licensed Rights in the Adapted Material under the conditions of the Adapter’s License You apply. + b. TO THE EXTENT POSSIBLE, IN NO EVENT WILL THE LICENSOR BE LIABLE + TO YOU ON ANY LEGAL THEORY (INCLUDING, WITHOUT LIMITATION, + NEGLIGENCE) OR OTHERWISE FOR ANY DIRECT, SPECIAL, INDIRECT, + INCIDENTAL, CONSEQUENTIAL, PUNITIVE, EXEMPLARY, OR OTHER LOSSES, + COSTS, EXPENSES, OR DAMAGES ARISING OUT OF THIS PUBLIC LICENSE OR + USE OF THE LICENSED MATERIAL, EVEN IF THE LICENSOR HAS BEEN + ADVISED OF THE POSSIBILITY OF SUCH LOSSES, COSTS, EXPENSES, OR + DAMAGES. WHERE A LIMITATION OF LIABILITY IS NOT ALLOWED IN FULL OR + IN PART, THIS LIMITATION MAY NOT APPLY TO YOU. - C. __No downstream restrictions.__ You may not offer or impose any additional or different terms or conditions on, or apply any Effective Technological Measures to, the Licensed Material if doing so restricts exercise of the Licensed Rights by any recipient of the Licensed Material. + c. The disclaimer of warranties and limitation of liability provided + above shall be interpreted in a manner that, to the extent + possible, most closely approximates an absolute disclaimer and + waiver of all liability. - 6. __No endorsement.__ Nothing in this Public License constitutes or may be construed as permission to assert or imply that You are, or that Your use of the Licensed Material is, connected with, or sponsored, endorsed, or granted official status by, the Licensor or others designated to receive attribution as provided in Section 3(a)(1)(A)(i). -b. ___Other rights.___ +Section 6 -- Term and Termination. - 1. Moral rights, such as the right of integrity, are not licensed under this Public License, nor are publicity, privacy, and/or other similar personality rights; however, to the extent possible, the Licensor waives and/or agrees not to assert any such rights held by the Licensor to the limited extent necessary to allow You to exercise the Licensed Rights, but not otherwise. + a. This Public License applies for the term of the Copyright and + Similar Rights licensed here. However, if You fail to comply with + this Public License, then Your rights under this Public License + terminate automatically. - 2. Patent and trademark rights are not licensed under this Public License. + b. Where Your right to use the Licensed Material has terminated under + Section 6(a), it reinstates: - 3. To the extent possible, the Licensor waives any right to collect royalties from You for the exercise of the Licensed Rights, whether directly or through a collecting society under any voluntary or waivable statutory or compulsory licensing scheme. In all other cases the Licensor expressly reserves any right to collect such royalties, including when the Licensed Material is used other than for NonCommercial purposes. + 1. automatically as of the date the violation is cured, provided + it is cured within 30 days of Your discovery of the + violation; or -### Section 3 – License Conditions. + 2. upon express reinstatement by the Licensor. -Your exercise of the Licensed Rights is expressly made subject to the following conditions. + For the avoidance of doubt, this Section 6(b) does not affect any + right the Licensor may have to seek remedies for Your violations + of this Public License. -a. ___Attribution.___ + c. For the avoidance of doubt, the Licensor may also offer the + Licensed Material under separate terms or conditions or stop + distributing the Licensed Material at any time; however, doing so + will not terminate this Public License. - 1. If You Share the Licensed Material (including in modified form), You must: + d. Sections 1, 5, 6, 7, and 8 survive termination of this Public + License. - A. retain the following if it is supplied by the Licensor with the Licensed Material: - i. identification of the creator(s) of the Licensed Material and any others designated to receive attribution, in any reasonable manner requested by the Licensor (including by pseudonym if designated); +Section 7 -- Other Terms and Conditions. - ii. a copyright notice; + a. The Licensor shall not be bound by any additional or different + terms or conditions communicated by You unless expressly agreed. - iii. a notice that refers to this Public License; + b. Any arrangements, understandings, or agreements regarding the + Licensed Material not stated herein are separate from and + independent of the terms and conditions of this Public License. - iv. a notice that refers to the disclaimer of warranties; - v. a URI or hyperlink to the Licensed Material to the extent reasonably practicable; +Section 8 -- Interpretation. - B. indicate if You modified the Licensed Material and retain an indication of any previous modifications; and + a. For the avoidance of doubt, this Public License does not, and + shall not be interpreted to, reduce, limit, restrict, or impose + conditions on any use of the Licensed Material that could lawfully + be made without permission under this Public License. - C. indicate the Licensed Material is licensed under this Public License, and include the text of, or the URI or hyperlink to, this Public License. + b. To the extent possible, if any provision of this Public License is + deemed unenforceable, it shall be automatically reformed to the + minimum extent necessary to make it enforceable. If the provision + cannot be reformed, it shall be severed from this Public License + without affecting the enforceability of the remaining terms and + conditions. - 2. You may satisfy the conditions in Section 3(a)(1) in any reasonable manner based on the medium, means, and context in which You Share the Licensed Material. For example, it may be reasonable to satisfy the conditions by providing a URI or hyperlink to a resource that includes the required information. + c. No term or condition of this Public License will be waived and no + failure to comply consented to unless expressly agreed to by the + Licensor. - 3. If requested by the Licensor, You must remove any of the information required by Section 3(a)(1)(A) to the extent reasonably practicable. + d. Nothing in this Public License constitutes or may be interpreted + as a limitation upon, or waiver of, any privileges and immunities + that apply to the Licensor or You, including from the legal + processes of any jurisdiction or authority. -b. ___ShareAlike.___ -In addition to the conditions in Section 3(a), if You Share Adapted Material You produce, the following conditions also apply. +======================================================================= + +Creative Commons is not a party to its public +licenses. Notwithstanding, Creative Commons may elect to apply one of +its public licenses to material it publishes and in those instances +will be considered the “Licensor.” The text of the Creative Commons +public licenses is dedicated to the public domain under the CC0 Public +Domain Dedication. Except for the limited purpose of indicating that +material is shared under a Creative Commons public license or as +otherwise permitted by the Creative Commons policies published at +creativecommons.org/policies, Creative Commons does not authorize the +use of the trademark "Creative Commons" or any other trademark or logo +of Creative Commons without its prior written consent including, +without limitation, in connection with any unauthorized modifications +to any of its public licenses or any other arrangements, +understandings, or agreements concerning use of licensed material. For +the avoidance of doubt, this paragraph does not form part of the +public licenses. -1. The Adapter’s License You apply must be a Creative Commons license with the same License Elements, this version or later, or a BY-NC-SA Compatible License. - -2. You must include the text of, or the URI or hyperlink to, the Adapter's License You apply. You may satisfy this condition in any reasonable manner based on the medium, means, and context in which You Share Adapted Material. - -3. You may not offer or impose any additional or different terms or conditions on, or apply any Effective Technological Measures to, Adapted Material that restrict exercise of the rights granted under the Adapter's License You apply. - -### Section 4 – Sui Generis Database Rights. - -Where the Licensed Rights include Sui Generis Database Rights that apply to Your use of the Licensed Material: - -a. for the avoidance of doubt, Section 2(a)(1) grants You the right to extract, reuse, reproduce, and Share all or a substantial portion of the contents of the database for NonCommercial purposes only; - -b. if You include all or a substantial portion of the database contents in a database in which You have Sui Generis Database Rights, then the database in which You have Sui Generis Database Rights (but not its individual contents) is Adapted Material, including for purposes of Section 3(b); and - -c. You must comply with the conditions in Section 3(a) if You Share all or a substantial portion of the contents of the database. - -For the avoidance of doubt, this Section 4 supplements and does not replace Your obligations under this Public License where the Licensed Rights include other Copyright and Similar Rights. - -### Section 5 – Disclaimer of Warranties and Limitation of Liability. - -a. __Unless otherwise separately undertaken by the Licensor, to the extent possible, the Licensor offers the Licensed Material as-is and as-available, and makes no representations or warranties of any kind concerning the Licensed Material, whether express, implied, statutory, or other. This includes, without limitation, warranties of title, merchantability, fitness for a particular purpose, non-infringement, absence of latent or other defects, accuracy, or the presence or absence of errors, whether or not known or discoverable. Where disclaimers of warranties are not allowed in full or in part, this disclaimer may not apply to You.__ - -b. __To the extent possible, in no event will the Licensor be liable to You on any legal theory (including, without limitation, negligence) or otherwise for any direct, special, indirect, incidental, consequential, punitive, exemplary, or other losses, costs, expenses, or damages arising out of this Public License or use of the Licensed Material, even if the Licensor has been advised of the possibility of such losses, costs, expenses, or damages. Where a limitation of liability is not allowed in full or in part, this limitation may not apply to You.__ - -c. The disclaimer of warranties and limitation of liability provided above shall be interpreted in a manner that, to the extent possible, most closely approximates an absolute disclaimer and waiver of all liability. - -### Section 6 – Term and Termination. - -a. This Public License applies for the term of the Copyright and Similar Rights licensed here. However, if You fail to comply with this Public License, then Your rights under this Public License terminate automatically. - -b. Where Your right to use the Licensed Material has terminated under Section 6(a), it reinstates: - - 1. automatically as of the date the violation is cured, provided it is cured within 30 days of Your discovery of the violation; or - - 2. upon express reinstatement by the Licensor. - - For the avoidance of doubt, this Section 6(b) does not affect any right the Licensor may have to seek remedies for Your violations of this Public License. - -c. For the avoidance of doubt, the Licensor may also offer the Licensed Material under separate terms or conditions or stop distributing the Licensed Material at any time; however, doing so will not terminate this Public License. - -d. Sections 1, 5, 6, 7, and 8 survive termination of this Public License. - -### Section 7 – Other Terms and Conditions. - -a. The Licensor shall not be bound by any additional or different terms or conditions communicated by You unless expressly agreed. - -b. Any arrangements, understandings, or agreements regarding the Licensed Material not stated herein are separate from and independent of the terms and conditions of this Public License. - -### Section 8 – Interpretation. - -a. For the avoidance of doubt, this Public License does not, and shall not be interpreted to, reduce, limit, restrict, or impose conditions on any use of the Licensed Material that could lawfully be made without permission under this Public License. - -b. To the extent possible, if any provision of this Public License is deemed unenforceable, it shall be automatically reformed to the minimum extent necessary to make it enforceable. If the provision cannot be reformed, it shall be severed from this Public License without affecting the enforceability of the remaining terms and conditions. - -c. No term or condition of this Public License will be waived and no failure to comply consented to unless expressly agreed to by the Licensor. - -d. Nothing in this Public License constitutes or may be interpreted as a limitation upon, or waiver of, any privileges and immunities that apply to the Licensor or You, including from the legal processes of any jurisdiction or authority. - -> Creative Commons is not a party to its public licenses. Notwithstanding, Creative Commons may elect to apply one of its public licenses to material it publishes and in those instances will be considered the “Licensor.” Except for the limited purpose of indicating that material is shared under a Creative Commons public license or as otherwise permitted by the Creative Commons policies published at [creativecommons.org/policies](http://creativecommons.org/policies), Creative Commons does not authorize the use of the trademark “Creative Commons” or any other trademark or logo of Creative Commons without its prior written consent including, without limitation, in connection with any unauthorized modifications to any of its public licenses or any other arrangements, understandings, or agreements concerning use of licensed material. For the avoidance of doubt, this paragraph does not form part of the public licenses. -> -> Creative Commons may be contacted at creativecommons.org +Creative Commons may be contacted at creativecommons.org. diff --git a/README.md b/README.md index 395d94c..c399657 100644 --- a/README.md +++ b/README.md @@ -1,9 +1,9 @@ -# :book: 图解算法数据结构 - -[![stars](https://img.shields.io/github/stars/krahets/LeetCode-Book?style=social)](https://github.com/krahets/LeetCode-Book) [![forks](https://img.shields.io/github/forks/krahets/LeetCode-Book?style=social)](https://github.com/krahets/LeetCode-Book) [![followers](https://img.shields.io/github/followers/krahets?style=social)](https://github.com/krahets) +# :book: 图解算法数据结构 ![leetcode-tests](https://img.shields.io/badge/LeetCode%20tests-75%20accepted%2C%200%20failed-brightgreen) [![Python](https://img.shields.io/badge/Language-Python3-teal)](https://github.com/krahets/LeetCode-Book/tree/main/python) [![Java](https://img.shields.io/badge/Language-Java-orange)](https://github.com/krahets/LeetCode-Book/tree/main/java) [![C++](https://img.shields.io/badge/Language-C++-blue)](https://github.com/krahets/LeetCode-Book/tree/main/cpp) +[![stars](https://img.shields.io/github/stars/krahets/LeetCode-Book?style=social)](https://github.com/krahets/LeetCode-Book) [![forks](https://img.shields.io/github/forks/krahets/LeetCode-Book?style=social)](https://github.com/krahets/LeetCode-Book) + [](https://leetcode-cn.com/leetbook/detail/illustration-of-algorithm/) LeetBook《图解算法数据结构》面向算法初学者、互联网求职者设计，主要内容包括： @@ -183,4 +183,6 @@ LeetBook《图解算法数据结构》面向算法初学者、互联网求职者 若本书对您有所帮助，麻烦请您点个 Star :star: 啦，谢谢！ -Creative Commons License
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. +## License + +This work is licensed under CC-BY-SA-4.0 license. From ba00e1fc80220d07d3a300193fe7658a81f69423 Mon Sep 17 00:00:00 2001 From: Yudong Jin Date: Mon, 28 Nov 2022 03:54:10 +0800 Subject: [PATCH 02/23] Update README.md --- README.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/README.md b/README.md index c399657..4b44dd7 100644 --- a/README.md +++ b/README.md @@ -19,7 +19,7 @@ LeetBook《图解算法数据结构》面向算法初学者、互联网求职者 - **基础知识：** 时间复杂度、空间复杂度等算法知识。 - **数据结构：** 数组、栈、队列、字符串、链表、树、图、堆、哈希表。 -- **算法专题：** 分治算法、动态规划、搜索与回溯算法、查找算法、贪心算法、排序、位运算、双指针、模拟、数学。 +- **算法专题：** 分治算法、动态规划、搜索与回溯、查找算法、贪心算法、排序、位运算、双指针、模拟、数学。 ## :student: 适合人群 From b1015406fc10c1a3bc74e23851ea9237632c06d7 Mon Sep 17 00:00:00 2001 From: Yudong Jin Date: Sat, 3 Dec 2022 02:34:38 +0800 Subject: [PATCH 03/23] Update README.md --- README.md | 69 ++++++++++++++++++++++++++----------------------------- 1 file changed, 32 insertions(+), 37 deletions(-) diff --git a/README.md b/README.md index 4b44dd7..3602847 100644 --- a/README.md +++ b/README.md @@ -12,13 +12,13 @@ LeetBook《图解算法数据结构》面向算法初学者、互联网求职者 - 图文详解 75 道题目，覆盖主要算法知识点，非常适合作为算法学习的 **第一份题库** 。 - 题库活跃于各大互联网公司招聘中，可使笔面试准备事半功倍。 -- 致力于行文深入浅出、图文搭配，提供简洁的 **Python3, Java, C++** 解题代码。 -- 笔者整理了 **30 天刷题计划**、**题目分类**，让刷题有迹可循。 +- 行文深入浅出、图文搭配，提供简洁的 **Python, Java, C++** 解题代码。 +- 提供 **30 天刷题计划**、**题目分类**，让刷题有迹可循。 -### :blue_book: 数据结构与算法专栏 +### :blue_book: 基础知识与算法专题 -- **基础知识：** 时间复杂度、空间复杂度等算法知识。 -- **数据结构：** 数组、栈、队列、字符串、链表、树、图、堆、哈希表。 +- **基础知识：** 时间复杂度、空间复杂度的概念与推算方法。 +- **数据结构：** 数组、栈、队列、字符串、链表、树、图、堆、哈希表的基本用法。 - **算法专题：** 分治算法、动态规划、搜索与回溯、查找算法、贪心算法、排序、位运算、双指针、模拟、数学。 ## :student: 适合人群 @@ -29,46 +29,41 @@ LeetBook《图解算法数据结构》面向算法初学者、互联网求职者 --- -## :ledger: 算法专栏 +# 算法学习路线 -### 「[数据结构](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/50e446/)」 +## :book: 第一步：算法快速入门 -> 建议对数据结构不熟悉的同学，先看这篇熟悉用法。 +如果您是算法初学者，建议先读一本算法入门书，可大幅提升 :rocket: 刷题效率。 -- 常用数据结构的**分类**和**基本特点**。 -- 在算法解题中，数据结构的**常用操作**。 -- 在 Python3 , Java , C++ 语言中，各数据结构的**初始化与构建方法**。 +

+ + + +

-### 「[算法复杂度](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/r84gmi/)」 +

+ + 前往阅读 > + + hello-algo.com + + +

-> 复杂度是算法优劣性的有力评价指标，对于理解算法起着至关重要的作用。 +

+ + 前往仓库 > + + github.com/krahets/hello-algo + + +

-- 什么是时间复杂度、空间复杂度？ -- 时间复杂度和空间复杂度的**概念定义**、**符号表示**、**常见种类**、**时空权衡**。 -- 时间与空间复杂度的**示例题目与解析**。 +## :hourglass: 第二步：按计划刷题 -### 「[动态规划](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/m5zfc1/)」 +本《剑指 Offer》刷题计划已被力扣（LeetCode）官方采用，理念为从易到难、从基础类题目到综合类题目。 -> 动态规划是算法重难点，也是笔面试重要考点，需要重点理解与练习。 - -- 动态规划问题特点，动态规划和分治算法的联系与区别； -- 借助例题介绍**重叠子问题**和**最优子结构**分别是什么，以及动态规划是如何解决它们的； -- 动态规划的**解题框架**总结； -- 动态规划的**练习例题**，从易到难排序； - -### 「[排序算法](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/pxal47/)」 - -> 排序是最经典的算法问题之一，由浅入深的多种算法涵盖多个算法知识点（例如暴力搜索、分治算法、堆数据结构等）。 - -- 排序算法分类方法，包括稳定性 、就地性 、自适应性； -- 排序算法与二分查找、双指针算法之间的关系； -- 各主要排序算法的时间复杂度与空间复杂度； - ---- - -## :hourglass: 刷题计划 - -笔者整理了《剑指 Offer 》刷题计划，核心理念为从易到难、从基础类题目到综合类题目，供希望按照知识点类型顺序刷题的小伙伴们参考。行百里者半九十，坚持一个月刷完，一起加油！ +行百里者半九十，坚持一个月刷完，一起加油！ | 日程 | 题目 | | :--------: | :------------------------------------------------ | From 027869401f56c91c2938da051b2b6e2edf19ea5d Mon Sep 17 00:00:00 2001 From: krahets Date: Wed, 15 Feb 2023 04:15:10 +0800 Subject: [PATCH 04/23] Update README.md --- README.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) mode change 100644 => 100755 README.md diff --git a/README.md b/README.md old mode 100644 new mode 100755 index 3602847..a040211 --- a/README.md +++ b/README.md @@ -61,7 +61,7 @@ LeetBook《图解算法数据结构》面向算法初学者、互联网求职者 ## :hourglass: 第二步：按计划刷题 -本《剑指 Offer》刷题计划已被力扣（LeetCode）官方采用，理念为从易到难、从基础类题目到综合类题目。 +本《剑指 Offer》刷题计划已被[力扣（LeetCode）官方题单](https://leetcode.cn/study-plan/lcof/)采用，理念为从易到难、从基础类题目到综合类题目。 行百里者半九十，坚持一个月刷完，一起加油！ From f79c85e6047c25656953aaaefd5e5cfd69cda51e Mon Sep 17 00:00:00 2001 From: Yudong Jin Date: Wed, 14 Jun 2023 18:00:16 +0800 Subject: [PATCH 05/23] Update README.md --- README.md | 39 ++++++++------------------------------- 1 file changed, 8 insertions(+), 31 deletions(-) diff --git a/README.md b/README.md index a040211..2754ca9 100755 --- a/README.md +++ b/README.md @@ -1,46 +1,25 @@ -# :book: 图解算法数据结构 +[](https://leetcode-cn.com/leetbook/detail/illustration-of-algorithm/) ![leetcode-tests](https://img.shields.io/badge/LeetCode%20tests-75%20accepted%2C%200%20failed-brightgreen) [![Python](https://img.shields.io/badge/Language-Python3-teal)](https://github.com/krahets/LeetCode-Book/tree/main/python) [![Java](https://img.shields.io/badge/Language-Java-orange)](https://github.com/krahets/LeetCode-Book/tree/main/java) [![C++](https://img.shields.io/badge/Language-C++-blue)](https://github.com/krahets/LeetCode-Book/tree/main/cpp) -[![stars](https://img.shields.io/github/stars/krahets/LeetCode-Book?style=social)](https://github.com/krahets/LeetCode-Book) [![forks](https://img.shields.io/github/forks/krahets/LeetCode-Book?style=social)](https://github.com/krahets/LeetCode-Book) +[LeetBook《图解算法数据结构》](https://leetcode-cn.com/leetbook/detail/illustration-of-algorithm/)是一本面向算法初学者、互联网求职者编写的算法题目解析手册。 -[](https://leetcode-cn.com/leetbook/detail/illustration-of-algorithm/) - -LeetBook《图解算法数据结构》面向算法初学者、互联网求职者设计，主要内容包括： - -### :green_book: 剑指 Offer 图文题解 - -- 图文详解 75 道题目，覆盖主要算法知识点，非常适合作为算法学习的 **第一份题库** 。 -- 题库活跃于各大互联网公司招聘中，可使笔面试准备事半功倍。 +- 图文详解《剑指 Offer》的所有题目，题库活跃于各大互联网公司笔面试。 - 行文深入浅出、图文搭配，提供简洁的 **Python, Java, C++** 解题代码。 - 提供 **30 天刷题计划**、**题目分类**，让刷题有迹可循。 -### :blue_book: 基础知识与算法专题 - -- **基础知识：** 时间复杂度、空间复杂度的概念与推算方法。 -- **数据结构：** 数组、栈、队列、字符串、链表、树、图、堆、哈希表的基本用法。 -- **算法专题：** 分治算法、动态规划、搜索与回溯、查找算法、贪心算法、排序、位运算、双指针、模拟、数学。 - -## :student: 适合人群 - -- 互联网算法、软件岗位求职者。 -- 从零开始接触数据结构与算法的同学。 -- 具有一定编程基础，计划系统学习算法的同学。 +> 若本 LeetBook 对您有所帮助，请在页面右上角点个 Star :star: 支持一下，谢谢！ --- -# 算法学习路线 - -## :book: 第一步：算法快速入门 - -如果您是算法初学者，建议先读一本算法入门书，可大幅提升 :rocket: 刷题效率。 -

+如果您是算法初学者，推荐阅读《Hello 算法》，夯实基础，可大幅提升刷题效率。 +

前往阅读 > @@ -59,9 +38,9 @@ LeetBook《图解算法数据结构》面向算法初学者、互联网求职者

-## :hourglass: 第二步：按计划刷题 +--- -本《剑指 Offer》刷题计划已被[力扣（LeetCode）官方题单](https://leetcode.cn/study-plan/lcof/)采用，理念为从易到难、从基础类题目到综合类题目。 +本《剑指 Offer》刷题计划已被[力扣（LeetCode）官方题单](https://leetcode.cn/study-plan/lcof/)采用。 行百里者半九十，坚持一个月刷完，一起加油！ @@ -176,8 +155,6 @@ LeetBook《图解算法数据结构》面向算法初学者、互联网求职者 --- -若本书对您有所帮助，麻烦请您点个 Star :star: 啦，谢谢！ - ## License This work is licensed under CC-BY-SA-4.0 license. From d3f7e15edbc254428d7318a2bac89b60c0a4a55e Mon Sep 17 00:00:00 2001 From: Yudong Jin Date: Mon, 19 Jun 2023 17:06:54 +0800 Subject: [PATCH 06/23] Update README.md --- README.md | 54 ++++++++++++++++++++++++++++-------------------------- 1 file changed, 28 insertions(+), 26 deletions(-) diff --git a/README.md b/README.md index 2754ca9..2e27e4b 100755 --- a/README.md +++ b/README.md @@ -1,42 +1,44 @@ -[](https://leetcode-cn.com/leetbook/detail/illustration-of-algorithm/) +

+ + + +

-![leetcode-tests](https://img.shields.io/badge/LeetCode%20tests-75%20accepted%2C%200%20failed-brightgreen) [![Python](https://img.shields.io/badge/Language-Python3-teal)](https://github.com/krahets/LeetCode-Book/tree/main/python) [![Java](https://img.shields.io/badge/Language-Java-orange)](https://github.com/krahets/LeetCode-Book/tree/main/java) [![C++](https://img.shields.io/badge/Language-C++-blue)](https://github.com/krahets/LeetCode-Book/tree/main/cpp) +

+ + + + +

-[LeetBook《图解算法数据结构》](https://leetcode-cn.com/leetbook/detail/illustration-of-algorithm/)是一本面向算法初学者、互联网求职者编写的算法题目解析手册。 +[《图解算法数据结构》](https://leetcode-cn.com/leetbook/detail/illustration-of-algorithm/)是一本面向算法初学者和互联网求职者编写的 LeetBook 手册。 -- 图文详解《剑指 Offer》的所有题目，题库活跃于各大互联网公司笔面试。 -- 行文深入浅出、图文搭配，提供简洁的 **Python, Java, C++** 解题代码。 -- 提供 **30 天刷题计划**、**题目分类**，让刷题有迹可循。 +- 图文并茂地讲解《剑指 Offer》75 道题目，题库活跃于各大互联网公司笔面试。 +- 重点解析部分算法知识，包括复杂度分析、数据结构、排序算法和动态规划等。 +- 提供简洁的 Python, Java, C++ 解题代码和一个月刷题计划，让刷题有迹可循。 > 若本 LeetBook 对您有所帮助，请在页面右上角点个 Star :star: 支持一下，谢谢！ --- -

- - - +

+ +

-如果您是算法初学者，推荐阅读《Hello 算法》，夯实基础，可大幅提升刷题效率。 - -

- - 前往阅读 > - - hello-algo.com - - +

+ 推荐算法初学者阅读《Hello 算法》，夯实基础后可大幅提升刷题效率

-

- +

+ 前往仓库 > - - github.com/krahets/hello-algo - - -

+ +     + + 在线阅读 > + +

--- From cec2ddfbb965964ea430b5f7ac784bfc3bdeba15 Mon Sep 17 00:00:00 2001 From: Yudong Jin Date: Sat, 22 Jul 2023 04:34:52 +0800 Subject: [PATCH 07/23] Update README.md --- README.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/README.md b/README.md index 2e27e4b..8b48835 100755 --- a/README.md +++ b/README.md @@ -22,8 +22,8 @@ ---

- - + +

From 3b24a1eb93d44d4699e9ec5f755689f698bb076e Mon Sep 17 00:00:00 2001 From: krahets Date: Tue, 10 Oct 2023 19:48:36 +0800 Subject: [PATCH 08/23] Move the code to sword_for_offer directory. --- {cpp => sword_for_offer/codes/cpp}/include/ListNode.hpp | 0 {cpp => sword_for_offer/codes/cpp}/include/PrintUtil.hpp | 0 {cpp => sword_for_offer/codes/cpp}/include/TreeNode.hpp | 0 {cpp => sword_for_offer/codes/cpp}/include/include.hpp | 0 .../sfo_03_find_duplicate_numbers_in_an_array_s1.cpp | 0 .../sfo_03_find_duplicate_numbers_in_an_array_s2.cpp | 0 .../sfo_04_find_a_number_in_2d_matrix_s1.cpp | 0 .../cpp}/sfo_05_replace_spaces_s1/sfo_05_replace_spaces_s1.cpp | 0 .../sfo_06_print_a_linked_list_in_reverse_order_s1.cpp | 0 .../sfo_06_print_a_linked_list_in_reverse_order_s2.cpp | 0 .../sfo_07_reconstruct_binary_tree_s1.cpp | 0 .../sfo_09_implement_a_queue_using_two_stacks_s1.cpp | 0 .../sfo_10i_fibonacci_numbers_s1/sfo_10i_fibonacci_numbers_s1.cpp | 0 .../codes/cpp}/sfo_10ii_frog_jump_s1/sfo_10ii_frog_jump_s1.cpp | 0 .../sfo_11_find_minimum_in_rotated_sorted_array_s1.cpp | 0 .../sfo_11_find_minimum_in_rotated_sorted_array_s2.cpp | 0 .../codes/cpp}/sfo_12_word_search_s1/sfo_12_word_search_s1.cpp | 0 .../sfo_13_range_of_motion_of_a_robot_s1.cpp | 0 .../sfo_13_range_of_motion_of_a_robot_s2.cpp | 0 .../cpp}/sfo_14i_cut_the_rope_i_s1/sfo_14i_cut_the_rope_i_s1.cpp | 0 .../sfo_14ii_cut_the_rope_ii_s1/sfo_14ii_cut_the_rope_ii_s1.cpp | 0 .../sfo_15_number_of_1_bits_s1/sfo_15_number_of_1_bits_s1.cpp | 0 .../sfo_15_number_of_1_bits_s2/sfo_15_number_of_1_bits_s2.cpp | 0 .../sfo_16_powers_of_integers_s1/sfo_16_powers_of_integers_s1.cpp | 0 .../sfo_18_delete_a_node_from_a_linked_list_s1.cpp | 0 .../sfo_19_regular_expression_matching_s1.cpp | 0 .../sfo_19_regular_expression_matching_s2.cpp | 0 .../sfo_21_adjust_the_order_of_numbers_in_an_array_s1.cpp | 0 .../sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1.cpp | 0 .../sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2.cpp | 0 .../sfo_24_reverse_a_linked_list_s1.cpp | 0 .../sfo_24_reverse_a_linked_list_s2.cpp | 0 .../sfo_25_combine_two_sorted_linked_lists_s1.cpp | 0 .../sfo_26_substructure_of_a_binary_tree_s1.cpp | 0 .../sfo_27_mirror_of_a_binary_tree_s1.cpp | 0 .../sfo_27_mirror_of_a_binary_tree_s2.cpp | 0 .../sfo_28_symmetric_binary_tree_s1.cpp | 0 .../sfo_29_print_a_given_matrix_in_spiral_form_s1.cpp | 0 .../codes/cpp}/sfo_30_min_stack_s1/sfo_30_min_stack_s1.cpp | 0 .../sfo_31_validate_stack_sequences_s1.cpp | 0 .../sfo_32i_print_a_binary_tree_topbottom_i_s1.cpp | 0 .../sfo_32ii_print_a_binary_tree_topbottom_ii_s1.cpp | 0 .../sfo_32iii_print_a_binary_tree_topbottom_iii_s1.cpp | 0 .../sfo_32iii_print_a_binary_tree_topbottom_iii_s2.cpp | 0 .../sfo_33_postorder_traversal_of_a_binary_search_tree_s1.cpp | 0 .../sfo_33_postorder_traversal_of_a_binary_search_tree_s2.cpp | 0 .../sfo_34_all_xsum_paths_in_a_binary_tree_s1.cpp | 0 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.../codes/python}/sfo_59i_sliding_window_maximum_s1.py | 0 .../codes/python}/sfo_59i_sliding_window_maximum_s2.py | 0 {python => sword_for_offer/codes/python}/sfo_59ii_max_queue_s1.py | 0 .../codes/python}/sfo_60_probabilities_for_rolling_n_dices_s1.py | 0 .../codes/python}/sfo_61_straight_in_poker_s1.py | 0 .../codes/python}/sfo_61_straight_in_poker_s2.py | 0 .../codes/python}/sfo_62_josephus_problem_s1.py | 0 .../codes/python}/sfo_63_the_maximum_profit_of_stocks_s1.py | 0 .../codes/python}/sfo_64_solve_1_2___n_s1.py | 0 ...mplement_addition_operation_without_arithmetic_operators_s1.py | 0 .../codes/python}/sfo_66_a_product_array_puzzle_s1.py | 0 .../codes/python}/sfo_67_convert_string_to_int_s1.py | 0 .../codes/python}/sfo_67_convert_string_to_int_s2.py | 0 ..._68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1.py | 0 ..._68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2.py | 0 ..._68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3.py | 0 .../sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1.py | 0 .../sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2.py | 0 357 files changed, 0 insertions(+), 0 deletions(-) rename {cpp => sword_for_offer/codes/cpp}/include/ListNode.hpp (100%) rename {cpp => sword_for_offer/codes/cpp}/include/PrintUtil.hpp (100%) rename {cpp => sword_for_offer/codes/cpp}/include/TreeNode.hpp (100%) rename {cpp => sword_for_offer/codes/cpp}/include/include.hpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_03_find_duplicate_numbers_in_an_array_s1/sfo_03_find_duplicate_numbers_in_an_array_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_03_find_duplicate_numbers_in_an_array_s2/sfo_03_find_duplicate_numbers_in_an_array_s2.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_04_find_a_number_in_2d_matrix_s1/sfo_04_find_a_number_in_2d_matrix_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_05_replace_spaces_s1/sfo_05_replace_spaces_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_06_print_a_linked_list_in_reverse_order_s1/sfo_06_print_a_linked_list_in_reverse_order_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_06_print_a_linked_list_in_reverse_order_s2/sfo_06_print_a_linked_list_in_reverse_order_s2.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_07_reconstruct_binary_tree_s1/sfo_07_reconstruct_binary_tree_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_09_implement_a_queue_using_two_stacks_s1/sfo_09_implement_a_queue_using_two_stacks_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_10i_fibonacci_numbers_s1/sfo_10i_fibonacci_numbers_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_10ii_frog_jump_s1/sfo_10ii_frog_jump_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_11_find_minimum_in_rotated_sorted_array_s1/sfo_11_find_minimum_in_rotated_sorted_array_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_11_find_minimum_in_rotated_sorted_array_s2/sfo_11_find_minimum_in_rotated_sorted_array_s2.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_12_word_search_s1/sfo_12_word_search_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_13_range_of_motion_of_a_robot_s1/sfo_13_range_of_motion_of_a_robot_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_13_range_of_motion_of_a_robot_s2/sfo_13_range_of_motion_of_a_robot_s2.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_14i_cut_the_rope_i_s1/sfo_14i_cut_the_rope_i_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_14ii_cut_the_rope_ii_s1/sfo_14ii_cut_the_rope_ii_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_15_number_of_1_bits_s1/sfo_15_number_of_1_bits_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_15_number_of_1_bits_s2/sfo_15_number_of_1_bits_s2.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_16_powers_of_integers_s1/sfo_16_powers_of_integers_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_18_delete_a_node_from_a_linked_list_s1/sfo_18_delete_a_node_from_a_linked_list_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_19_regular_expression_matching_s1/sfo_19_regular_expression_matching_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_19_regular_expression_matching_s2/sfo_19_regular_expression_matching_s2.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_21_adjust_the_order_of_numbers_in_an_array_s1/sfo_21_adjust_the_order_of_numbers_in_an_array_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2.cpp (100%) rename {cpp => 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{cpp => sword_for_offer/codes/cpp}/sfo_30_min_stack_s1/sfo_30_min_stack_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_31_validate_stack_sequences_s1/sfo_31_validate_stack_sequences_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_32i_print_a_binary_tree_topbottom_i_s1/sfo_32i_print_a_binary_tree_topbottom_i_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_32ii_print_a_binary_tree_topbottom_ii_s1/sfo_32ii_print_a_binary_tree_topbottom_ii_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_32iii_print_a_binary_tree_topbottom_iii_s1/sfo_32iii_print_a_binary_tree_topbottom_iii_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_32iii_print_a_binary_tree_topbottom_iii_s2/sfo_32iii_print_a_binary_tree_topbottom_iii_s2.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_33_postorder_traversal_of_a_binary_search_tree_s1/sfo_33_postorder_traversal_of_a_binary_search_tree_s1.cpp (100%) rename {cpp => 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=> sword_for_offer/codes/cpp}/sfo_56ii_single_number_ii_s2/sfo_56ii_single_number_ii_s2.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_57_two_numbers_with_sum_s_s1/sfo_57_two_numbers_with_sum_s_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_57ii_consecutive_numbers_with_sum_s_s1/sfo_57ii_consecutive_numbers_with_sum_s_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_57ii_consecutive_numbers_with_sum_s_s2/sfo_57ii_consecutive_numbers_with_sum_s_s2.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_58ii_left_rotation_of_a_string_s1/sfo_58ii_left_rotation_of_a_string_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_58ii_left_rotation_of_a_string_s2/sfo_58ii_left_rotation_of_a_string_s2.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_58ii_left_rotation_of_a_string_s3/sfo_58ii_left_rotation_of_a_string_s3.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_59ii_max_queue_s1/sfo_59ii_max_queue_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_60_probabilities_for_rolling_n_dices_s1/sfo_60_probabilities_for_rolling_n_dices_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_61_straight_in_poker_s1/sfo_61_straight_in_poker_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_61_straight_in_poker_s2/sfo_61_straight_in_poker_s2.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_62_josephus_problem_s1/sfo_62_josephus_problem_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_63_the_maximum_profit_of_stocks_s1/sfo_63_the_maximum_profit_of_stocks_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_64_solve_1_2___n_s1/sfo_64_solve_1_2___n_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_65_implement_addition_operation_without_arithmetic_operators_s1/sfo_65_implement_addition_operation_without_arithmetic_operators_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_66_a_product_array_puzzle_s1/sfo_66_a_product_array_puzzle_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_67_convert_string_to_int_s1/sfo_67_convert_string_to_int_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1.cpp (100%) rename {cpp => sword_for_offer/codes/cpp}/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2.cpp (100%) rename {java => sword_for_offer/codes/java}/include/ListNode.java (100%) rename {java => sword_for_offer/codes/java}/include/PrintUtil.java (100%) rename {java => sword_for_offer/codes/java}/include/TreeNode.java (100%) rename {java => sword_for_offer/codes/java}/sfo_03_find_duplicate_numbers_in_an_array_s1/sfo_03_find_duplicate_numbers_in_an_array_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_03_find_duplicate_numbers_in_an_array_s2/sfo_03_find_duplicate_numbers_in_an_array_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_04_find_a_number_in_2d_matrix_s1/sfo_04_find_a_number_in_2d_matrix_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_05_replace_spaces_s1/sfo_05_replace_spaces_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_06_print_a_linked_list_in_reverse_order_s1/sfo_06_print_a_linked_list_in_reverse_order_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_06_print_a_linked_list_in_reverse_order_s2/sfo_06_print_a_linked_list_in_reverse_order_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_07_reconstruct_binary_tree_s1/sfo_07_reconstruct_binary_tree_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_09_implement_a_queue_using_two_stacks_s1/sfo_09_implement_a_queue_using_two_stacks_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_10i_fibonacci_numbers_s1/sfo_10i_fibonacci_numbers_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_10ii_frog_jump_s1/sfo_10ii_frog_jump_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_11_find_minimum_in_rotated_sorted_array_s1/sfo_11_find_minimum_in_rotated_sorted_array_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_11_find_minimum_in_rotated_sorted_array_s2/sfo_11_find_minimum_in_rotated_sorted_array_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_12_word_search_s1/sfo_12_word_search_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_13_range_of_motion_of_a_robot_s1/sfo_13_range_of_motion_of_a_robot_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_13_range_of_motion_of_a_robot_s2/sfo_13_range_of_motion_of_a_robot_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_14i_cut_the_rope_i_s1/sfo_14i_cut_the_rope_i_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_14ii_cut_the_rope_ii_s1/sfo_14ii_cut_the_rope_ii_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_15_number_of_1_bits_s1/sfo_15_number_of_1_bits_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_15_number_of_1_bits_s2/sfo_15_number_of_1_bits_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_16_powers_of_integers_s1/sfo_16_powers_of_integers_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_17_print_from_1_to_the_largest_n_digits_s1/sfo_17_print_from_1_to_the_largest_n_digits_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_17_print_from_1_to_the_largest_n_digits_s2/sfo_17_print_from_1_to_the_largest_n_digits_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_17_print_from_1_to_the_largest_n_digits_s3/sfo_17_print_from_1_to_the_largest_n_digits_s3.java (100%) rename {java => sword_for_offer/codes/java}/sfo_17_print_from_1_to_the_largest_n_digits_s4/sfo_17_print_from_1_to_the_largest_n_digits_s4.java (100%) rename {java => sword_for_offer/codes/java}/sfo_18_delete_a_node_from_a_linked_list_s1/sfo_18_delete_a_node_from_a_linked_list_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_19_regular_expression_matching_s1/sfo_19_regular_expression_matching_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_19_regular_expression_matching_s2/sfo_19_regular_expression_matching_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_20_a_string_representing_a_numeric_value_s1/sfo_20_a_string_representing_a_numeric_value_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_21_adjust_the_order_of_numbers_in_an_array_s1/sfo_21_adjust_the_order_of_numbers_in_an_array_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_24_reverse_a_linked_list_s1/sfo_24_reverse_a_linked_list_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_24_reverse_a_linked_list_s2/sfo_24_reverse_a_linked_list_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_25_combine_two_sorted_linked_lists_s1/sfo_25_combine_two_sorted_linked_lists_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_26_substructure_of_a_binary_tree_s1/sfo_26_substructure_of_a_binary_tree_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_27_mirror_of_a_binary_tree_s1/sfo_27_mirror_of_a_binary_tree_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_27_mirror_of_a_binary_tree_s2/sfo_27_mirror_of_a_binary_tree_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_28_symmetric_binary_tree_s1/sfo_28_symmetric_binary_tree_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_29_print_a_given_matrix_in_spiral_form_s1/sfo_29_print_a_given_matrix_in_spiral_form_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_30_min_stack_s1/sfo_30_min_stack_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_31_validate_stack_sequences_s1/sfo_31_validate_stack_sequences_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_32i_print_a_binary_tree_topbottom_i_s1/sfo_32i_print_a_binary_tree_topbottom_i_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_32ii_print_a_binary_tree_topbottom_ii_s1/sfo_32ii_print_a_binary_tree_topbottom_ii_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_32iii_print_a_binary_tree_topbottom_iii_s1/sfo_32iii_print_a_binary_tree_topbottom_iii_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_32iii_print_a_binary_tree_topbottom_iii_s2/sfo_32iii_print_a_binary_tree_topbottom_iii_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_32iii_print_a_binary_tree_topbottom_iii_s3/sfo_32iii_print_a_binary_tree_topbottom_iii_s3.java (100%) rename {java => sword_for_offer/codes/java}/sfo_33_postorder_traversal_of_a_binary_search_tree_s1/sfo_33_postorder_traversal_of_a_binary_search_tree_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_33_postorder_traversal_of_a_binary_search_tree_s2/sfo_33_postorder_traversal_of_a_binary_search_tree_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_34_all_xsum_paths_in_a_binary_tree_s1/sfo_34_all_xsum_paths_in_a_binary_tree_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_36_binary_search_tree_and_doubly_linked_list_s1/sfo_36_binary_search_tree_and_doubly_linked_list_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_37_serialize_and_deserialize_a_binary_tree_s1/sfo_37_serialize_and_deserialize_a_binary_tree_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_38_all_permutations_of_a_string_s1/sfo_38_all_permutations_of_a_string_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_39_the_majority_element_in_an_array_s1/sfo_39_the_majority_element_in_an_array_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_39_the_majority_element_in_an_array_s2/sfo_39_the_majority_element_in_an_array_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_40_the_smallest_k_numbers_s1/sfo_40_the_smallest_k_numbers_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_40_the_smallest_k_numbers_s2/sfo_40_the_smallest_k_numbers_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_41_find_median_from_data_stream_s1/sfo_41_find_median_from_data_stream_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_42_largest_sum_contiguous_subarray_s1/sfo_42_largest_sum_contiguous_subarray_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_44_nth_digit_s1/sfo_44_nth_digit_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_45_arrange_an_array_into_the_smallest_number_s1/sfo_45_arrange_an_array_into_the_smallest_number_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_45_arrange_an_array_into_the_smallest_number_s2/sfo_45_arrange_an_array_into_the_smallest_number_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_46_translate_numbers_into_strings_s1/sfo_46_translate_numbers_into_strings_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_46_translate_numbers_into_strings_s2/sfo_46_translate_numbers_into_strings_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_46_translate_numbers_into_strings_s3/sfo_46_translate_numbers_into_strings_s3.java (100%) rename {java => sword_for_offer/codes/java}/sfo_47_the_maximum_value_of_gifts_s1/sfo_47_the_maximum_value_of_gifts_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_47_the_maximum_value_of_gifts_s2/sfo_47_the_maximum_value_of_gifts_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_48_the_longest_substring_without_repeated_characters_s1/sfo_48_the_longest_substring_without_repeated_characters_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_48_the_longest_substring_without_repeated_characters_s2/sfo_48_the_longest_substring_without_repeated_characters_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_48_the_longest_substring_without_repeated_characters_s3/sfo_48_the_longest_substring_without_repeated_characters_s3.java (100%) rename {java => sword_for_offer/codes/java}/sfo_49_ugly_numbers_s1/sfo_49_ugly_numbers_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_51_reversed_pairs_in_an_array_s1/sfo_51_reversed_pairs_in_an_array_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_52_the_first_common_node_in_two_linked_lists_s1/sfo_52_the_first_common_node_in_two_linked_lists_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_53i_find_a_number_in_a_sorted_array_s1/sfo_53i_find_a_number_in_a_sorted_array_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_53i_find_a_number_in_a_sorted_array_s2/sfo_53i_find_a_number_in_a_sorted_array_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_53ii_the_missing_number_from_0_to_n1_s1/sfo_53ii_the_missing_number_from_0_to_n1_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_55i_depth_of_a_binary_tree_s1/sfo_55i_depth_of_a_binary_tree_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_55i_depth_of_a_binary_tree_s2/sfo_55i_depth_of_a_binary_tree_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_55ii_balanced_binary_tree_s1/sfo_55ii_balanced_binary_tree_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_55ii_balanced_binary_tree_s2/sfo_55ii_balanced_binary_tree_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_56i_single_number_i_s1/sfo_56i_single_number_i_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_56ii_single_number_ii_s1/sfo_56ii_single_number_ii_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_56ii_single_number_ii_s2/sfo_56ii_single_number_ii_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_57_two_numbers_with_sum_s_s1/sfo_57_two_numbers_with_sum_s_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_57ii_consecutive_numbers_with_sum_s_s1/sfo_57ii_consecutive_numbers_with_sum_s_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_57ii_consecutive_numbers_with_sum_s_s2/sfo_57ii_consecutive_numbers_with_sum_s_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_58i_reverse_order_of_words_s1/sfo_58i_reverse_order_of_words_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_58i_reverse_order_of_words_s2/sfo_58i_reverse_order_of_words_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_58ii_left_rotation_of_a_string_s1/sfo_58ii_left_rotation_of_a_string_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_58ii_left_rotation_of_a_string_s2/sfo_58ii_left_rotation_of_a_string_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_58ii_left_rotation_of_a_string_s3/sfo_58ii_left_rotation_of_a_string_s3.java (100%) rename {java => sword_for_offer/codes/java}/sfo_58ii_left_rotation_of_a_string_s4/sfo_58ii_left_rotation_of_a_string_s4.java (100%) rename {java => sword_for_offer/codes/java}/sfo_58ii_left_rotation_of_a_string_s5/sfo_58ii_left_rotation_of_a_string_s5.java (100%) rename {java => sword_for_offer/codes/java}/sfo_59i_sliding_window_maximum_s1/sfo_59i_sliding_window_maximum_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_59i_sliding_window_maximum_s2/sfo_59i_sliding_window_maximum_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_59ii_max_queue_s1/sfo_59ii_max_queue_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_60_probabilities_for_rolling_n_dices_s1/sfo_60_probabilities_for_rolling_n_dices_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_61_straight_in_poker_s1/sfo_61_straight_in_poker_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_61_straight_in_poker_s2/sfo_61_straight_in_poker_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_62_josephus_problem_s1/sfo_62_josephus_problem_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_63_the_maximum_profit_of_stocks_s1/sfo_63_the_maximum_profit_of_stocks_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_64_solve_1_2___n_s1/sfo_64_solve_1_2___n_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_64_solve_1_2___n_s2/sfo_64_solve_1_2___n_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_65_implement_addition_operation_without_arithmetic_operators_s1/sfo_65_implement_addition_operation_without_arithmetic_operators_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_66_a_product_array_puzzle_s1/sfo_66_a_product_array_puzzle_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_67_convert_string_to_int_s1/sfo_67_convert_string_to_int_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_67_convert_string_to_int_s2/sfo_67_convert_string_to_int_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2.java (100%) rename {java => sword_for_offer/codes/java}/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3.java (100%) rename {java => sword_for_offer/codes/java}/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1.java (100%) rename {java => sword_for_offer/codes/java}/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2.java (100%) rename {python => sword_for_offer/codes/python}/include/__init__.py (100%) rename {python => sword_for_offer/codes/python}/include/binary_tree.py (100%) rename {python => sword_for_offer/codes/python}/include/linked_list.py (100%) rename {python => sword_for_offer/codes/python}/include/print_util.py (100%) rename {python => sword_for_offer/codes/python}/sfo_03_find_duplicate_numbers_in_an_array_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_03_find_duplicate_numbers_in_an_array_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_04_find_a_number_in_2d_matrix_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_05_replace_spaces_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_06_print_a_linked_list_in_reverse_order_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_06_print_a_linked_list_in_reverse_order_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_07_reconstruct_binary_tree_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_09_implement_a_queue_using_two_stacks_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_10i_fibonacci_numbers_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_10i_fibonacci_numbers_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_10ii_frog_jump_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_10ii_frog_jump_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_11_find_minimum_in_rotated_sorted_array_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_11_find_minimum_in_rotated_sorted_array_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_12_word_search_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_13_range_of_motion_of_a_robot_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_13_range_of_motion_of_a_robot_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_14i_cut_the_rope_i_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_14ii_cut_the_rope_ii_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_14ii_cut_the_rope_ii_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_15_number_of_1_bits_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_15_number_of_1_bits_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_16_powers_of_integers_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_17_print_from_1_to_the_largest_n_digits_s1.py (100%) rename {python => 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sword_for_offer/codes/python}/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_24_reverse_a_linked_list_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_24_reverse_a_linked_list_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_24_reverse_a_linked_list_s3.py (100%) rename {python => sword_for_offer/codes/python}/sfo_25_combine_two_sorted_linked_lists_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_26_substructure_of_a_binary_tree_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_27_mirror_of_a_binary_tree_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_27_mirror_of_a_binary_tree_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_27_mirror_of_a_binary_tree_s3.py (100%) rename {python => sword_for_offer/codes/python}/sfo_28_symmetric_binary_tree_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_29_print_a_given_matrix_in_spiral_form_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_30_min_stack_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_31_validate_stack_sequences_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_32i_print_a_binary_tree_topbottom_i_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_32ii_print_a_binary_tree_topbottom_ii_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_32iii_print_a_binary_tree_topbottom_iii_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_32iii_print_a_binary_tree_topbottom_iii_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_32iii_print_a_binary_tree_topbottom_iii_s3.py (100%) rename {python => sword_for_offer/codes/python}/sfo_33_postorder_traversal_of_a_binary_search_tree_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_33_postorder_traversal_of_a_binary_search_tree_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_34_all_xsum_paths_in_a_binary_tree_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_36_binary_search_tree_and_doubly_linked_list_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_37_serialize_and_deserialize_a_binary_tree_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_38_all_permutations_of_a_string_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_39_the_majority_element_in_an_array_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_39_the_majority_element_in_an_array_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_40_the_smallest_k_numbers_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_40_the_smallest_k_numbers_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_41_find_median_from_data_stream_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_41_find_median_from_data_stream_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_42_largest_sum_contiguous_subarray_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_44_nth_digit_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_45_arrange_an_array_into_the_smallest_number_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_45_arrange_an_array_into_the_smallest_number_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_46_translate_numbers_into_strings_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_46_translate_numbers_into_strings_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_46_translate_numbers_into_strings_s3.py (100%) rename {python => sword_for_offer/codes/python}/sfo_46_translate_numbers_into_strings_s4.py (100%) rename {python => sword_for_offer/codes/python}/sfo_46_translate_numbers_into_strings_s5.py (100%) rename {python => sword_for_offer/codes/python}/sfo_47_the_maximum_value_of_gifts_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_47_the_maximum_value_of_gifts_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_48_the_longest_substring_without_repeated_characters_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_48_the_longest_substring_without_repeated_characters_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_48_the_longest_substring_without_repeated_characters_s3.py (100%) rename {python => sword_for_offer/codes/python}/sfo_49_ugly_numbers_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_50_find_the_first_nonrepeating_character_in_a_string_s3.py (100%) rename {python => sword_for_offer/codes/python}/sfo_51_reversed_pairs_in_an_array_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_52_the_first_common_node_in_two_linked_lists_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_53i_find_a_number_in_a_sorted_array_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_53i_find_a_number_in_a_sorted_array_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_53ii_the_missing_number_from_0_to_n1_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_55i_depth_of_a_binary_tree_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_55i_depth_of_a_binary_tree_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_55ii_balanced_binary_tree_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_55ii_balanced_binary_tree_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_56i_single_number_i_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_56ii_single_number_ii_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_56ii_single_number_ii_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_57_two_numbers_with_sum_s_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_57ii_consecutive_numbers_with_sum_s_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_57ii_consecutive_numbers_with_sum_s_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_58i_reverse_order_of_words_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_58i_reverse_order_of_words_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_58i_reverse_order_of_words_s3.py (100%) rename {python => sword_for_offer/codes/python}/sfo_58ii_left_rotation_of_a_string_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_58ii_left_rotation_of_a_string_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_58ii_left_rotation_of_a_string_s3.py (100%) rename {python => sword_for_offer/codes/python}/sfo_58ii_left_rotation_of_a_string_s4.py (100%) rename {python => sword_for_offer/codes/python}/sfo_58ii_left_rotation_of_a_string_s5.py (100%) rename {python => sword_for_offer/codes/python}/sfo_59i_sliding_window_maximum_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_59i_sliding_window_maximum_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_59ii_max_queue_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_60_probabilities_for_rolling_n_dices_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_61_straight_in_poker_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_61_straight_in_poker_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_62_josephus_problem_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_63_the_maximum_profit_of_stocks_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_64_solve_1_2___n_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_65_implement_addition_operation_without_arithmetic_operators_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_66_a_product_array_puzzle_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_67_convert_string_to_int_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_67_convert_string_to_int_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2.py (100%) rename {python => sword_for_offer/codes/python}/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3.py (100%) rename {python => sword_for_offer/codes/python}/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1.py (100%) rename {python => sword_for_offer/codes/python}/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2.py (100%) diff --git a/cpp/include/ListNode.hpp b/sword_for_offer/codes/cpp/include/ListNode.hpp similarity index 100% rename from cpp/include/ListNode.hpp rename to sword_for_offer/codes/cpp/include/ListNode.hpp diff --git a/cpp/include/PrintUtil.hpp b/sword_for_offer/codes/cpp/include/PrintUtil.hpp similarity index 100% rename from cpp/include/PrintUtil.hpp rename to sword_for_offer/codes/cpp/include/PrintUtil.hpp diff --git a/cpp/include/TreeNode.hpp b/sword_for_offer/codes/cpp/include/TreeNode.hpp similarity index 100% rename from cpp/include/TreeNode.hpp rename to sword_for_offer/codes/cpp/include/TreeNode.hpp diff --git a/cpp/include/include.hpp b/sword_for_offer/codes/cpp/include/include.hpp similarity index 100% rename from cpp/include/include.hpp rename to sword_for_offer/codes/cpp/include/include.hpp diff --git a/cpp/sfo_03_find_duplicate_numbers_in_an_array_s1/sfo_03_find_duplicate_numbers_in_an_array_s1.cpp b/sword_for_offer/codes/cpp/sfo_03_find_duplicate_numbers_in_an_array_s1/sfo_03_find_duplicate_numbers_in_an_array_s1.cpp similarity index 100% rename from cpp/sfo_03_find_duplicate_numbers_in_an_array_s1/sfo_03_find_duplicate_numbers_in_an_array_s1.cpp rename to sword_for_offer/codes/cpp/sfo_03_find_duplicate_numbers_in_an_array_s1/sfo_03_find_duplicate_numbers_in_an_array_s1.cpp diff --git a/cpp/sfo_03_find_duplicate_numbers_in_an_array_s2/sfo_03_find_duplicate_numbers_in_an_array_s2.cpp b/sword_for_offer/codes/cpp/sfo_03_find_duplicate_numbers_in_an_array_s2/sfo_03_find_duplicate_numbers_in_an_array_s2.cpp similarity index 100% rename from cpp/sfo_03_find_duplicate_numbers_in_an_array_s2/sfo_03_find_duplicate_numbers_in_an_array_s2.cpp rename to sword_for_offer/codes/cpp/sfo_03_find_duplicate_numbers_in_an_array_s2/sfo_03_find_duplicate_numbers_in_an_array_s2.cpp diff --git a/cpp/sfo_04_find_a_number_in_2d_matrix_s1/sfo_04_find_a_number_in_2d_matrix_s1.cpp b/sword_for_offer/codes/cpp/sfo_04_find_a_number_in_2d_matrix_s1/sfo_04_find_a_number_in_2d_matrix_s1.cpp similarity index 100% rename from cpp/sfo_04_find_a_number_in_2d_matrix_s1/sfo_04_find_a_number_in_2d_matrix_s1.cpp rename to sword_for_offer/codes/cpp/sfo_04_find_a_number_in_2d_matrix_s1/sfo_04_find_a_number_in_2d_matrix_s1.cpp diff --git a/cpp/sfo_05_replace_spaces_s1/sfo_05_replace_spaces_s1.cpp b/sword_for_offer/codes/cpp/sfo_05_replace_spaces_s1/sfo_05_replace_spaces_s1.cpp similarity index 100% rename from cpp/sfo_05_replace_spaces_s1/sfo_05_replace_spaces_s1.cpp rename to sword_for_offer/codes/cpp/sfo_05_replace_spaces_s1/sfo_05_replace_spaces_s1.cpp diff --git a/cpp/sfo_06_print_a_linked_list_in_reverse_order_s1/sfo_06_print_a_linked_list_in_reverse_order_s1.cpp b/sword_for_offer/codes/cpp/sfo_06_print_a_linked_list_in_reverse_order_s1/sfo_06_print_a_linked_list_in_reverse_order_s1.cpp similarity index 100% rename from cpp/sfo_06_print_a_linked_list_in_reverse_order_s1/sfo_06_print_a_linked_list_in_reverse_order_s1.cpp rename to sword_for_offer/codes/cpp/sfo_06_print_a_linked_list_in_reverse_order_s1/sfo_06_print_a_linked_list_in_reverse_order_s1.cpp diff --git a/cpp/sfo_06_print_a_linked_list_in_reverse_order_s2/sfo_06_print_a_linked_list_in_reverse_order_s2.cpp b/sword_for_offer/codes/cpp/sfo_06_print_a_linked_list_in_reverse_order_s2/sfo_06_print_a_linked_list_in_reverse_order_s2.cpp similarity index 100% rename from cpp/sfo_06_print_a_linked_list_in_reverse_order_s2/sfo_06_print_a_linked_list_in_reverse_order_s2.cpp rename to sword_for_offer/codes/cpp/sfo_06_print_a_linked_list_in_reverse_order_s2/sfo_06_print_a_linked_list_in_reverse_order_s2.cpp diff --git a/cpp/sfo_07_reconstruct_binary_tree_s1/sfo_07_reconstruct_binary_tree_s1.cpp b/sword_for_offer/codes/cpp/sfo_07_reconstruct_binary_tree_s1/sfo_07_reconstruct_binary_tree_s1.cpp similarity index 100% rename from cpp/sfo_07_reconstruct_binary_tree_s1/sfo_07_reconstruct_binary_tree_s1.cpp rename to sword_for_offer/codes/cpp/sfo_07_reconstruct_binary_tree_s1/sfo_07_reconstruct_binary_tree_s1.cpp diff --git a/cpp/sfo_09_implement_a_queue_using_two_stacks_s1/sfo_09_implement_a_queue_using_two_stacks_s1.cpp b/sword_for_offer/codes/cpp/sfo_09_implement_a_queue_using_two_stacks_s1/sfo_09_implement_a_queue_using_two_stacks_s1.cpp similarity index 100% rename from cpp/sfo_09_implement_a_queue_using_two_stacks_s1/sfo_09_implement_a_queue_using_two_stacks_s1.cpp rename to sword_for_offer/codes/cpp/sfo_09_implement_a_queue_using_two_stacks_s1/sfo_09_implement_a_queue_using_two_stacks_s1.cpp diff --git a/cpp/sfo_10i_fibonacci_numbers_s1/sfo_10i_fibonacci_numbers_s1.cpp b/sword_for_offer/codes/cpp/sfo_10i_fibonacci_numbers_s1/sfo_10i_fibonacci_numbers_s1.cpp similarity index 100% rename from cpp/sfo_10i_fibonacci_numbers_s1/sfo_10i_fibonacci_numbers_s1.cpp rename to sword_for_offer/codes/cpp/sfo_10i_fibonacci_numbers_s1/sfo_10i_fibonacci_numbers_s1.cpp diff --git a/cpp/sfo_10ii_frog_jump_s1/sfo_10ii_frog_jump_s1.cpp b/sword_for_offer/codes/cpp/sfo_10ii_frog_jump_s1/sfo_10ii_frog_jump_s1.cpp similarity index 100% rename from cpp/sfo_10ii_frog_jump_s1/sfo_10ii_frog_jump_s1.cpp rename to sword_for_offer/codes/cpp/sfo_10ii_frog_jump_s1/sfo_10ii_frog_jump_s1.cpp diff --git a/cpp/sfo_11_find_minimum_in_rotated_sorted_array_s1/sfo_11_find_minimum_in_rotated_sorted_array_s1.cpp b/sword_for_offer/codes/cpp/sfo_11_find_minimum_in_rotated_sorted_array_s1/sfo_11_find_minimum_in_rotated_sorted_array_s1.cpp similarity index 100% rename from cpp/sfo_11_find_minimum_in_rotated_sorted_array_s1/sfo_11_find_minimum_in_rotated_sorted_array_s1.cpp rename to sword_for_offer/codes/cpp/sfo_11_find_minimum_in_rotated_sorted_array_s1/sfo_11_find_minimum_in_rotated_sorted_array_s1.cpp diff --git a/cpp/sfo_11_find_minimum_in_rotated_sorted_array_s2/sfo_11_find_minimum_in_rotated_sorted_array_s2.cpp b/sword_for_offer/codes/cpp/sfo_11_find_minimum_in_rotated_sorted_array_s2/sfo_11_find_minimum_in_rotated_sorted_array_s2.cpp similarity index 100% rename from cpp/sfo_11_find_minimum_in_rotated_sorted_array_s2/sfo_11_find_minimum_in_rotated_sorted_array_s2.cpp rename to sword_for_offer/codes/cpp/sfo_11_find_minimum_in_rotated_sorted_array_s2/sfo_11_find_minimum_in_rotated_sorted_array_s2.cpp diff --git a/cpp/sfo_12_word_search_s1/sfo_12_word_search_s1.cpp b/sword_for_offer/codes/cpp/sfo_12_word_search_s1/sfo_12_word_search_s1.cpp similarity index 100% rename from cpp/sfo_12_word_search_s1/sfo_12_word_search_s1.cpp rename to sword_for_offer/codes/cpp/sfo_12_word_search_s1/sfo_12_word_search_s1.cpp diff --git a/cpp/sfo_13_range_of_motion_of_a_robot_s1/sfo_13_range_of_motion_of_a_robot_s1.cpp b/sword_for_offer/codes/cpp/sfo_13_range_of_motion_of_a_robot_s1/sfo_13_range_of_motion_of_a_robot_s1.cpp similarity index 100% rename from cpp/sfo_13_range_of_motion_of_a_robot_s1/sfo_13_range_of_motion_of_a_robot_s1.cpp rename to sword_for_offer/codes/cpp/sfo_13_range_of_motion_of_a_robot_s1/sfo_13_range_of_motion_of_a_robot_s1.cpp diff --git a/cpp/sfo_13_range_of_motion_of_a_robot_s2/sfo_13_range_of_motion_of_a_robot_s2.cpp b/sword_for_offer/codes/cpp/sfo_13_range_of_motion_of_a_robot_s2/sfo_13_range_of_motion_of_a_robot_s2.cpp similarity index 100% rename from cpp/sfo_13_range_of_motion_of_a_robot_s2/sfo_13_range_of_motion_of_a_robot_s2.cpp rename to sword_for_offer/codes/cpp/sfo_13_range_of_motion_of_a_robot_s2/sfo_13_range_of_motion_of_a_robot_s2.cpp diff --git a/cpp/sfo_14i_cut_the_rope_i_s1/sfo_14i_cut_the_rope_i_s1.cpp b/sword_for_offer/codes/cpp/sfo_14i_cut_the_rope_i_s1/sfo_14i_cut_the_rope_i_s1.cpp similarity index 100% rename from cpp/sfo_14i_cut_the_rope_i_s1/sfo_14i_cut_the_rope_i_s1.cpp rename to sword_for_offer/codes/cpp/sfo_14i_cut_the_rope_i_s1/sfo_14i_cut_the_rope_i_s1.cpp diff --git a/cpp/sfo_14ii_cut_the_rope_ii_s1/sfo_14ii_cut_the_rope_ii_s1.cpp b/sword_for_offer/codes/cpp/sfo_14ii_cut_the_rope_ii_s1/sfo_14ii_cut_the_rope_ii_s1.cpp similarity index 100% rename from cpp/sfo_14ii_cut_the_rope_ii_s1/sfo_14ii_cut_the_rope_ii_s1.cpp rename to sword_for_offer/codes/cpp/sfo_14ii_cut_the_rope_ii_s1/sfo_14ii_cut_the_rope_ii_s1.cpp diff --git a/cpp/sfo_15_number_of_1_bits_s1/sfo_15_number_of_1_bits_s1.cpp b/sword_for_offer/codes/cpp/sfo_15_number_of_1_bits_s1/sfo_15_number_of_1_bits_s1.cpp similarity index 100% rename from cpp/sfo_15_number_of_1_bits_s1/sfo_15_number_of_1_bits_s1.cpp rename to sword_for_offer/codes/cpp/sfo_15_number_of_1_bits_s1/sfo_15_number_of_1_bits_s1.cpp diff --git a/cpp/sfo_15_number_of_1_bits_s2/sfo_15_number_of_1_bits_s2.cpp b/sword_for_offer/codes/cpp/sfo_15_number_of_1_bits_s2/sfo_15_number_of_1_bits_s2.cpp similarity index 100% rename from cpp/sfo_15_number_of_1_bits_s2/sfo_15_number_of_1_bits_s2.cpp rename to sword_for_offer/codes/cpp/sfo_15_number_of_1_bits_s2/sfo_15_number_of_1_bits_s2.cpp diff --git a/cpp/sfo_16_powers_of_integers_s1/sfo_16_powers_of_integers_s1.cpp b/sword_for_offer/codes/cpp/sfo_16_powers_of_integers_s1/sfo_16_powers_of_integers_s1.cpp similarity index 100% rename from cpp/sfo_16_powers_of_integers_s1/sfo_16_powers_of_integers_s1.cpp rename to sword_for_offer/codes/cpp/sfo_16_powers_of_integers_s1/sfo_16_powers_of_integers_s1.cpp diff --git a/cpp/sfo_18_delete_a_node_from_a_linked_list_s1/sfo_18_delete_a_node_from_a_linked_list_s1.cpp b/sword_for_offer/codes/cpp/sfo_18_delete_a_node_from_a_linked_list_s1/sfo_18_delete_a_node_from_a_linked_list_s1.cpp similarity index 100% rename from cpp/sfo_18_delete_a_node_from_a_linked_list_s1/sfo_18_delete_a_node_from_a_linked_list_s1.cpp rename to sword_for_offer/codes/cpp/sfo_18_delete_a_node_from_a_linked_list_s1/sfo_18_delete_a_node_from_a_linked_list_s1.cpp diff --git a/cpp/sfo_19_regular_expression_matching_s1/sfo_19_regular_expression_matching_s1.cpp b/sword_for_offer/codes/cpp/sfo_19_regular_expression_matching_s1/sfo_19_regular_expression_matching_s1.cpp similarity index 100% rename from cpp/sfo_19_regular_expression_matching_s1/sfo_19_regular_expression_matching_s1.cpp rename to sword_for_offer/codes/cpp/sfo_19_regular_expression_matching_s1/sfo_19_regular_expression_matching_s1.cpp diff --git a/cpp/sfo_19_regular_expression_matching_s2/sfo_19_regular_expression_matching_s2.cpp b/sword_for_offer/codes/cpp/sfo_19_regular_expression_matching_s2/sfo_19_regular_expression_matching_s2.cpp similarity index 100% rename from cpp/sfo_19_regular_expression_matching_s2/sfo_19_regular_expression_matching_s2.cpp rename to sword_for_offer/codes/cpp/sfo_19_regular_expression_matching_s2/sfo_19_regular_expression_matching_s2.cpp diff --git a/cpp/sfo_21_adjust_the_order_of_numbers_in_an_array_s1/sfo_21_adjust_the_order_of_numbers_in_an_array_s1.cpp b/sword_for_offer/codes/cpp/sfo_21_adjust_the_order_of_numbers_in_an_array_s1/sfo_21_adjust_the_order_of_numbers_in_an_array_s1.cpp similarity index 100% rename from cpp/sfo_21_adjust_the_order_of_numbers_in_an_array_s1/sfo_21_adjust_the_order_of_numbers_in_an_array_s1.cpp rename to sword_for_offer/codes/cpp/sfo_21_adjust_the_order_of_numbers_in_an_array_s1/sfo_21_adjust_the_order_of_numbers_in_an_array_s1.cpp diff --git a/cpp/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1.cpp b/sword_for_offer/codes/cpp/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1.cpp similarity index 100% rename from cpp/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1.cpp rename to sword_for_offer/codes/cpp/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1.cpp diff --git a/cpp/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2.cpp b/sword_for_offer/codes/cpp/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2.cpp similarity index 100% rename from cpp/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2.cpp rename to sword_for_offer/codes/cpp/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2.cpp diff --git a/cpp/sfo_24_reverse_a_linked_list_s1/sfo_24_reverse_a_linked_list_s1.cpp b/sword_for_offer/codes/cpp/sfo_24_reverse_a_linked_list_s1/sfo_24_reverse_a_linked_list_s1.cpp similarity index 100% rename from cpp/sfo_24_reverse_a_linked_list_s1/sfo_24_reverse_a_linked_list_s1.cpp rename to sword_for_offer/codes/cpp/sfo_24_reverse_a_linked_list_s1/sfo_24_reverse_a_linked_list_s1.cpp diff --git a/cpp/sfo_24_reverse_a_linked_list_s2/sfo_24_reverse_a_linked_list_s2.cpp b/sword_for_offer/codes/cpp/sfo_24_reverse_a_linked_list_s2/sfo_24_reverse_a_linked_list_s2.cpp similarity index 100% rename from cpp/sfo_24_reverse_a_linked_list_s2/sfo_24_reverse_a_linked_list_s2.cpp rename to sword_for_offer/codes/cpp/sfo_24_reverse_a_linked_list_s2/sfo_24_reverse_a_linked_list_s2.cpp diff --git a/cpp/sfo_25_combine_two_sorted_linked_lists_s1/sfo_25_combine_two_sorted_linked_lists_s1.cpp b/sword_for_offer/codes/cpp/sfo_25_combine_two_sorted_linked_lists_s1/sfo_25_combine_two_sorted_linked_lists_s1.cpp similarity index 100% rename from cpp/sfo_25_combine_two_sorted_linked_lists_s1/sfo_25_combine_two_sorted_linked_lists_s1.cpp rename to sword_for_offer/codes/cpp/sfo_25_combine_two_sorted_linked_lists_s1/sfo_25_combine_two_sorted_linked_lists_s1.cpp diff --git a/cpp/sfo_26_substructure_of_a_binary_tree_s1/sfo_26_substructure_of_a_binary_tree_s1.cpp b/sword_for_offer/codes/cpp/sfo_26_substructure_of_a_binary_tree_s1/sfo_26_substructure_of_a_binary_tree_s1.cpp similarity index 100% rename from cpp/sfo_26_substructure_of_a_binary_tree_s1/sfo_26_substructure_of_a_binary_tree_s1.cpp rename to sword_for_offer/codes/cpp/sfo_26_substructure_of_a_binary_tree_s1/sfo_26_substructure_of_a_binary_tree_s1.cpp diff --git a/cpp/sfo_27_mirror_of_a_binary_tree_s1/sfo_27_mirror_of_a_binary_tree_s1.cpp b/sword_for_offer/codes/cpp/sfo_27_mirror_of_a_binary_tree_s1/sfo_27_mirror_of_a_binary_tree_s1.cpp similarity index 100% rename from cpp/sfo_27_mirror_of_a_binary_tree_s1/sfo_27_mirror_of_a_binary_tree_s1.cpp rename to sword_for_offer/codes/cpp/sfo_27_mirror_of_a_binary_tree_s1/sfo_27_mirror_of_a_binary_tree_s1.cpp diff --git a/cpp/sfo_27_mirror_of_a_binary_tree_s2/sfo_27_mirror_of_a_binary_tree_s2.cpp b/sword_for_offer/codes/cpp/sfo_27_mirror_of_a_binary_tree_s2/sfo_27_mirror_of_a_binary_tree_s2.cpp similarity index 100% rename from cpp/sfo_27_mirror_of_a_binary_tree_s2/sfo_27_mirror_of_a_binary_tree_s2.cpp rename to sword_for_offer/codes/cpp/sfo_27_mirror_of_a_binary_tree_s2/sfo_27_mirror_of_a_binary_tree_s2.cpp diff --git a/cpp/sfo_28_symmetric_binary_tree_s1/sfo_28_symmetric_binary_tree_s1.cpp b/sword_for_offer/codes/cpp/sfo_28_symmetric_binary_tree_s1/sfo_28_symmetric_binary_tree_s1.cpp similarity index 100% rename from cpp/sfo_28_symmetric_binary_tree_s1/sfo_28_symmetric_binary_tree_s1.cpp rename to sword_for_offer/codes/cpp/sfo_28_symmetric_binary_tree_s1/sfo_28_symmetric_binary_tree_s1.cpp diff --git a/cpp/sfo_29_print_a_given_matrix_in_spiral_form_s1/sfo_29_print_a_given_matrix_in_spiral_form_s1.cpp b/sword_for_offer/codes/cpp/sfo_29_print_a_given_matrix_in_spiral_form_s1/sfo_29_print_a_given_matrix_in_spiral_form_s1.cpp similarity index 100% rename from cpp/sfo_29_print_a_given_matrix_in_spiral_form_s1/sfo_29_print_a_given_matrix_in_spiral_form_s1.cpp rename to sword_for_offer/codes/cpp/sfo_29_print_a_given_matrix_in_spiral_form_s1/sfo_29_print_a_given_matrix_in_spiral_form_s1.cpp diff --git a/cpp/sfo_30_min_stack_s1/sfo_30_min_stack_s1.cpp b/sword_for_offer/codes/cpp/sfo_30_min_stack_s1/sfo_30_min_stack_s1.cpp similarity index 100% rename from cpp/sfo_30_min_stack_s1/sfo_30_min_stack_s1.cpp rename to sword_for_offer/codes/cpp/sfo_30_min_stack_s1/sfo_30_min_stack_s1.cpp diff --git a/cpp/sfo_31_validate_stack_sequences_s1/sfo_31_validate_stack_sequences_s1.cpp b/sword_for_offer/codes/cpp/sfo_31_validate_stack_sequences_s1/sfo_31_validate_stack_sequences_s1.cpp similarity index 100% rename from cpp/sfo_31_validate_stack_sequences_s1/sfo_31_validate_stack_sequences_s1.cpp rename to sword_for_offer/codes/cpp/sfo_31_validate_stack_sequences_s1/sfo_31_validate_stack_sequences_s1.cpp diff --git a/cpp/sfo_32i_print_a_binary_tree_topbottom_i_s1/sfo_32i_print_a_binary_tree_topbottom_i_s1.cpp b/sword_for_offer/codes/cpp/sfo_32i_print_a_binary_tree_topbottom_i_s1/sfo_32i_print_a_binary_tree_topbottom_i_s1.cpp similarity index 100% rename from cpp/sfo_32i_print_a_binary_tree_topbottom_i_s1/sfo_32i_print_a_binary_tree_topbottom_i_s1.cpp rename to sword_for_offer/codes/cpp/sfo_32i_print_a_binary_tree_topbottom_i_s1/sfo_32i_print_a_binary_tree_topbottom_i_s1.cpp diff --git a/cpp/sfo_32ii_print_a_binary_tree_topbottom_ii_s1/sfo_32ii_print_a_binary_tree_topbottom_ii_s1.cpp b/sword_for_offer/codes/cpp/sfo_32ii_print_a_binary_tree_topbottom_ii_s1/sfo_32ii_print_a_binary_tree_topbottom_ii_s1.cpp similarity index 100% rename from cpp/sfo_32ii_print_a_binary_tree_topbottom_ii_s1/sfo_32ii_print_a_binary_tree_topbottom_ii_s1.cpp rename to sword_for_offer/codes/cpp/sfo_32ii_print_a_binary_tree_topbottom_ii_s1/sfo_32ii_print_a_binary_tree_topbottom_ii_s1.cpp diff --git a/cpp/sfo_32iii_print_a_binary_tree_topbottom_iii_s1/sfo_32iii_print_a_binary_tree_topbottom_iii_s1.cpp b/sword_for_offer/codes/cpp/sfo_32iii_print_a_binary_tree_topbottom_iii_s1/sfo_32iii_print_a_binary_tree_topbottom_iii_s1.cpp similarity index 100% rename from cpp/sfo_32iii_print_a_binary_tree_topbottom_iii_s1/sfo_32iii_print_a_binary_tree_topbottom_iii_s1.cpp rename to sword_for_offer/codes/cpp/sfo_32iii_print_a_binary_tree_topbottom_iii_s1/sfo_32iii_print_a_binary_tree_topbottom_iii_s1.cpp diff --git a/cpp/sfo_32iii_print_a_binary_tree_topbottom_iii_s2/sfo_32iii_print_a_binary_tree_topbottom_iii_s2.cpp b/sword_for_offer/codes/cpp/sfo_32iii_print_a_binary_tree_topbottom_iii_s2/sfo_32iii_print_a_binary_tree_topbottom_iii_s2.cpp similarity index 100% rename from cpp/sfo_32iii_print_a_binary_tree_topbottom_iii_s2/sfo_32iii_print_a_binary_tree_topbottom_iii_s2.cpp rename to sword_for_offer/codes/cpp/sfo_32iii_print_a_binary_tree_topbottom_iii_s2/sfo_32iii_print_a_binary_tree_topbottom_iii_s2.cpp diff --git a/cpp/sfo_33_postorder_traversal_of_a_binary_search_tree_s1/sfo_33_postorder_traversal_of_a_binary_search_tree_s1.cpp b/sword_for_offer/codes/cpp/sfo_33_postorder_traversal_of_a_binary_search_tree_s1/sfo_33_postorder_traversal_of_a_binary_search_tree_s1.cpp similarity index 100% rename from cpp/sfo_33_postorder_traversal_of_a_binary_search_tree_s1/sfo_33_postorder_traversal_of_a_binary_search_tree_s1.cpp rename to sword_for_offer/codes/cpp/sfo_33_postorder_traversal_of_a_binary_search_tree_s1/sfo_33_postorder_traversal_of_a_binary_search_tree_s1.cpp diff --git a/cpp/sfo_33_postorder_traversal_of_a_binary_search_tree_s2/sfo_33_postorder_traversal_of_a_binary_search_tree_s2.cpp b/sword_for_offer/codes/cpp/sfo_33_postorder_traversal_of_a_binary_search_tree_s2/sfo_33_postorder_traversal_of_a_binary_search_tree_s2.cpp similarity index 100% rename from cpp/sfo_33_postorder_traversal_of_a_binary_search_tree_s2/sfo_33_postorder_traversal_of_a_binary_search_tree_s2.cpp rename to sword_for_offer/codes/cpp/sfo_33_postorder_traversal_of_a_binary_search_tree_s2/sfo_33_postorder_traversal_of_a_binary_search_tree_s2.cpp diff --git a/cpp/sfo_34_all_xsum_paths_in_a_binary_tree_s1/sfo_34_all_xsum_paths_in_a_binary_tree_s1.cpp b/sword_for_offer/codes/cpp/sfo_34_all_xsum_paths_in_a_binary_tree_s1/sfo_34_all_xsum_paths_in_a_binary_tree_s1.cpp similarity index 100% rename from cpp/sfo_34_all_xsum_paths_in_a_binary_tree_s1/sfo_34_all_xsum_paths_in_a_binary_tree_s1.cpp rename to sword_for_offer/codes/cpp/sfo_34_all_xsum_paths_in_a_binary_tree_s1/sfo_34_all_xsum_paths_in_a_binary_tree_s1.cpp diff --git a/cpp/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1.cpp b/sword_for_offer/codes/cpp/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1.cpp similarity index 100% rename from cpp/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1.cpp rename to sword_for_offer/codes/cpp/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1.cpp diff --git a/cpp/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2.cpp b/sword_for_offer/codes/cpp/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2.cpp similarity index 100% rename from cpp/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2.cpp rename to sword_for_offer/codes/cpp/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2.cpp diff --git a/cpp/sfo_36_binary_search_tree_and_doubly_linked_list_s1/sfo_36_binary_search_tree_and_doubly_linked_list_s1.cpp b/sword_for_offer/codes/cpp/sfo_36_binary_search_tree_and_doubly_linked_list_s1/sfo_36_binary_search_tree_and_doubly_linked_list_s1.cpp similarity index 100% rename from cpp/sfo_36_binary_search_tree_and_doubly_linked_list_s1/sfo_36_binary_search_tree_and_doubly_linked_list_s1.cpp rename to sword_for_offer/codes/cpp/sfo_36_binary_search_tree_and_doubly_linked_list_s1/sfo_36_binary_search_tree_and_doubly_linked_list_s1.cpp diff --git a/cpp/sfo_37_serialize_and_deserialize_a_binary_tree_s1/sfo_37_serialize_and_deserialize_a_binary_tree_s1.cpp b/sword_for_offer/codes/cpp/sfo_37_serialize_and_deserialize_a_binary_tree_s1/sfo_37_serialize_and_deserialize_a_binary_tree_s1.cpp similarity index 100% rename from cpp/sfo_37_serialize_and_deserialize_a_binary_tree_s1/sfo_37_serialize_and_deserialize_a_binary_tree_s1.cpp rename to sword_for_offer/codes/cpp/sfo_37_serialize_and_deserialize_a_binary_tree_s1/sfo_37_serialize_and_deserialize_a_binary_tree_s1.cpp diff --git a/cpp/sfo_38_all_permutations_of_a_string_s1/sfo_38_all_permutations_of_a_string_s1.cpp b/sword_for_offer/codes/cpp/sfo_38_all_permutations_of_a_string_s1/sfo_38_all_permutations_of_a_string_s1.cpp similarity index 100% rename from cpp/sfo_38_all_permutations_of_a_string_s1/sfo_38_all_permutations_of_a_string_s1.cpp rename to sword_for_offer/codes/cpp/sfo_38_all_permutations_of_a_string_s1/sfo_38_all_permutations_of_a_string_s1.cpp diff --git a/cpp/sfo_39_the_majority_element_in_an_array_s1/sfo_39_the_majority_element_in_an_array_s1.cpp b/sword_for_offer/codes/cpp/sfo_39_the_majority_element_in_an_array_s1/sfo_39_the_majority_element_in_an_array_s1.cpp similarity index 100% rename from cpp/sfo_39_the_majority_element_in_an_array_s1/sfo_39_the_majority_element_in_an_array_s1.cpp rename to sword_for_offer/codes/cpp/sfo_39_the_majority_element_in_an_array_s1/sfo_39_the_majority_element_in_an_array_s1.cpp diff --git a/cpp/sfo_39_the_majority_element_in_an_array_s2/sfo_39_the_majority_element_in_an_array_s2.cpp b/sword_for_offer/codes/cpp/sfo_39_the_majority_element_in_an_array_s2/sfo_39_the_majority_element_in_an_array_s2.cpp similarity index 100% rename from cpp/sfo_39_the_majority_element_in_an_array_s2/sfo_39_the_majority_element_in_an_array_s2.cpp rename to sword_for_offer/codes/cpp/sfo_39_the_majority_element_in_an_array_s2/sfo_39_the_majority_element_in_an_array_s2.cpp diff --git a/cpp/sfo_40_the_smallest_k_numbers_s1/sfo_40_the_smallest_k_numbers_s1.cpp b/sword_for_offer/codes/cpp/sfo_40_the_smallest_k_numbers_s1/sfo_40_the_smallest_k_numbers_s1.cpp similarity index 100% rename from cpp/sfo_40_the_smallest_k_numbers_s1/sfo_40_the_smallest_k_numbers_s1.cpp rename to sword_for_offer/codes/cpp/sfo_40_the_smallest_k_numbers_s1/sfo_40_the_smallest_k_numbers_s1.cpp diff --git a/cpp/sfo_40_the_smallest_k_numbers_s2/sfo_40_the_smallest_k_numbers_s2.cpp b/sword_for_offer/codes/cpp/sfo_40_the_smallest_k_numbers_s2/sfo_40_the_smallest_k_numbers_s2.cpp similarity index 100% rename from cpp/sfo_40_the_smallest_k_numbers_s2/sfo_40_the_smallest_k_numbers_s2.cpp rename to sword_for_offer/codes/cpp/sfo_40_the_smallest_k_numbers_s2/sfo_40_the_smallest_k_numbers_s2.cpp diff --git a/cpp/sfo_41_find_median_from_data_stream_s1/sfo_41_find_median_from_data_stream_s1.cpp b/sword_for_offer/codes/cpp/sfo_41_find_median_from_data_stream_s1/sfo_41_find_median_from_data_stream_s1.cpp similarity index 100% rename from cpp/sfo_41_find_median_from_data_stream_s1/sfo_41_find_median_from_data_stream_s1.cpp rename to sword_for_offer/codes/cpp/sfo_41_find_median_from_data_stream_s1/sfo_41_find_median_from_data_stream_s1.cpp diff --git a/cpp/sfo_42_largest_sum_contiguous_subarray_s1/sfo_42_largest_sum_contiguous_subarray_s1.cpp b/sword_for_offer/codes/cpp/sfo_42_largest_sum_contiguous_subarray_s1/sfo_42_largest_sum_contiguous_subarray_s1.cpp similarity index 100% rename from cpp/sfo_42_largest_sum_contiguous_subarray_s1/sfo_42_largest_sum_contiguous_subarray_s1.cpp rename to sword_for_offer/codes/cpp/sfo_42_largest_sum_contiguous_subarray_s1/sfo_42_largest_sum_contiguous_subarray_s1.cpp diff --git a/cpp/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1.cpp b/sword_for_offer/codes/cpp/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1.cpp similarity index 100% rename from cpp/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1.cpp rename to sword_for_offer/codes/cpp/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1.cpp diff --git a/cpp/sfo_44_nth_digit_s1/sfo_44_nth_digit_s1.cpp b/sword_for_offer/codes/cpp/sfo_44_nth_digit_s1/sfo_44_nth_digit_s1.cpp similarity index 100% rename from cpp/sfo_44_nth_digit_s1/sfo_44_nth_digit_s1.cpp rename to sword_for_offer/codes/cpp/sfo_44_nth_digit_s1/sfo_44_nth_digit_s1.cpp diff --git a/cpp/sfo_45_arrange_an_array_into_the_smallest_number_s1/sfo_45_arrange_an_array_into_the_smallest_number_s1.cpp b/sword_for_offer/codes/cpp/sfo_45_arrange_an_array_into_the_smallest_number_s1/sfo_45_arrange_an_array_into_the_smallest_number_s1.cpp similarity index 100% rename from cpp/sfo_45_arrange_an_array_into_the_smallest_number_s1/sfo_45_arrange_an_array_into_the_smallest_number_s1.cpp rename to sword_for_offer/codes/cpp/sfo_45_arrange_an_array_into_the_smallest_number_s1/sfo_45_arrange_an_array_into_the_smallest_number_s1.cpp diff --git a/cpp/sfo_45_arrange_an_array_into_the_smallest_number_s2/sfo_45_arrange_an_array_into_the_smallest_number_s2.cpp b/sword_for_offer/codes/cpp/sfo_45_arrange_an_array_into_the_smallest_number_s2/sfo_45_arrange_an_array_into_the_smallest_number_s2.cpp similarity index 100% rename from cpp/sfo_45_arrange_an_array_into_the_smallest_number_s2/sfo_45_arrange_an_array_into_the_smallest_number_s2.cpp rename to sword_for_offer/codes/cpp/sfo_45_arrange_an_array_into_the_smallest_number_s2/sfo_45_arrange_an_array_into_the_smallest_number_s2.cpp diff --git a/cpp/sfo_46_translate_numbers_into_strings_s1/sfo_46_translate_numbers_into_strings_s1.cpp b/sword_for_offer/codes/cpp/sfo_46_translate_numbers_into_strings_s1/sfo_46_translate_numbers_into_strings_s1.cpp similarity index 100% rename from cpp/sfo_46_translate_numbers_into_strings_s1/sfo_46_translate_numbers_into_strings_s1.cpp rename to sword_for_offer/codes/cpp/sfo_46_translate_numbers_into_strings_s1/sfo_46_translate_numbers_into_strings_s1.cpp diff --git a/cpp/sfo_46_translate_numbers_into_strings_s2/sfo_46_translate_numbers_into_strings_s2.cpp b/sword_for_offer/codes/cpp/sfo_46_translate_numbers_into_strings_s2/sfo_46_translate_numbers_into_strings_s2.cpp similarity index 100% rename from cpp/sfo_46_translate_numbers_into_strings_s2/sfo_46_translate_numbers_into_strings_s2.cpp rename to sword_for_offer/codes/cpp/sfo_46_translate_numbers_into_strings_s2/sfo_46_translate_numbers_into_strings_s2.cpp diff --git a/cpp/sfo_46_translate_numbers_into_strings_s3/sfo_46_translate_numbers_into_strings_s3.cpp b/sword_for_offer/codes/cpp/sfo_46_translate_numbers_into_strings_s3/sfo_46_translate_numbers_into_strings_s3.cpp similarity index 100% rename from cpp/sfo_46_translate_numbers_into_strings_s3/sfo_46_translate_numbers_into_strings_s3.cpp rename to sword_for_offer/codes/cpp/sfo_46_translate_numbers_into_strings_s3/sfo_46_translate_numbers_into_strings_s3.cpp diff --git a/cpp/sfo_47_the_maximum_value_of_gifts_s1/sfo_47_the_maximum_value_of_gifts_s1.cpp b/sword_for_offer/codes/cpp/sfo_47_the_maximum_value_of_gifts_s1/sfo_47_the_maximum_value_of_gifts_s1.cpp similarity index 100% rename from cpp/sfo_47_the_maximum_value_of_gifts_s1/sfo_47_the_maximum_value_of_gifts_s1.cpp rename to sword_for_offer/codes/cpp/sfo_47_the_maximum_value_of_gifts_s1/sfo_47_the_maximum_value_of_gifts_s1.cpp diff --git a/cpp/sfo_47_the_maximum_value_of_gifts_s2/sfo_47_the_maximum_value_of_gifts_s2.cpp b/sword_for_offer/codes/cpp/sfo_47_the_maximum_value_of_gifts_s2/sfo_47_the_maximum_value_of_gifts_s2.cpp similarity index 100% rename from cpp/sfo_47_the_maximum_value_of_gifts_s2/sfo_47_the_maximum_value_of_gifts_s2.cpp rename to sword_for_offer/codes/cpp/sfo_47_the_maximum_value_of_gifts_s2/sfo_47_the_maximum_value_of_gifts_s2.cpp diff --git a/cpp/sfo_48_the_longest_substring_without_repeated_characters_s1/sfo_48_the_longest_substring_without_repeated_characters_s1.cpp b/sword_for_offer/codes/cpp/sfo_48_the_longest_substring_without_repeated_characters_s1/sfo_48_the_longest_substring_without_repeated_characters_s1.cpp similarity index 100% rename from cpp/sfo_48_the_longest_substring_without_repeated_characters_s1/sfo_48_the_longest_substring_without_repeated_characters_s1.cpp rename to sword_for_offer/codes/cpp/sfo_48_the_longest_substring_without_repeated_characters_s1/sfo_48_the_longest_substring_without_repeated_characters_s1.cpp diff --git a/cpp/sfo_48_the_longest_substring_without_repeated_characters_s2/sfo_48_the_longest_substring_without_repeated_characters_s2.cpp b/sword_for_offer/codes/cpp/sfo_48_the_longest_substring_without_repeated_characters_s2/sfo_48_the_longest_substring_without_repeated_characters_s2.cpp similarity index 100% rename from cpp/sfo_48_the_longest_substring_without_repeated_characters_s2/sfo_48_the_longest_substring_without_repeated_characters_s2.cpp rename to sword_for_offer/codes/cpp/sfo_48_the_longest_substring_without_repeated_characters_s2/sfo_48_the_longest_substring_without_repeated_characters_s2.cpp diff --git a/cpp/sfo_48_the_longest_substring_without_repeated_characters_s3/sfo_48_the_longest_substring_without_repeated_characters_s3.cpp b/sword_for_offer/codes/cpp/sfo_48_the_longest_substring_without_repeated_characters_s3/sfo_48_the_longest_substring_without_repeated_characters_s3.cpp similarity index 100% rename from cpp/sfo_48_the_longest_substring_without_repeated_characters_s3/sfo_48_the_longest_substring_without_repeated_characters_s3.cpp rename to sword_for_offer/codes/cpp/sfo_48_the_longest_substring_without_repeated_characters_s3/sfo_48_the_longest_substring_without_repeated_characters_s3.cpp diff --git a/cpp/sfo_49_ugly_numbers_s1/sfo_49_ugly_numbers_s1.cpp b/sword_for_offer/codes/cpp/sfo_49_ugly_numbers_s1/sfo_49_ugly_numbers_s1.cpp similarity index 100% rename from cpp/sfo_49_ugly_numbers_s1/sfo_49_ugly_numbers_s1.cpp rename to sword_for_offer/codes/cpp/sfo_49_ugly_numbers_s1/sfo_49_ugly_numbers_s1.cpp diff --git a/cpp/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1.cpp b/sword_for_offer/codes/cpp/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1.cpp similarity index 100% rename from cpp/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1.cpp rename to sword_for_offer/codes/cpp/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1.cpp diff --git a/cpp/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2.cpp b/sword_for_offer/codes/cpp/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2.cpp similarity index 100% rename from cpp/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2.cpp rename to sword_for_offer/codes/cpp/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2.cpp diff --git a/cpp/sfo_51_reversed_pairs_in_an_array_s1/sfo_51_reversed_pairs_in_an_array_s1.cpp b/sword_for_offer/codes/cpp/sfo_51_reversed_pairs_in_an_array_s1/sfo_51_reversed_pairs_in_an_array_s1.cpp similarity index 100% rename from cpp/sfo_51_reversed_pairs_in_an_array_s1/sfo_51_reversed_pairs_in_an_array_s1.cpp rename to sword_for_offer/codes/cpp/sfo_51_reversed_pairs_in_an_array_s1/sfo_51_reversed_pairs_in_an_array_s1.cpp diff --git a/cpp/sfo_52_the_first_common_node_in_two_linked_lists_s1/sfo_52_the_first_common_node_in_two_linked_lists_s1.cpp b/sword_for_offer/codes/cpp/sfo_52_the_first_common_node_in_two_linked_lists_s1/sfo_52_the_first_common_node_in_two_linked_lists_s1.cpp similarity index 100% rename from cpp/sfo_52_the_first_common_node_in_two_linked_lists_s1/sfo_52_the_first_common_node_in_two_linked_lists_s1.cpp rename to sword_for_offer/codes/cpp/sfo_52_the_first_common_node_in_two_linked_lists_s1/sfo_52_the_first_common_node_in_two_linked_lists_s1.cpp diff --git a/cpp/sfo_53i_find_a_number_in_a_sorted_array_s1/sfo_53i_find_a_number_in_a_sorted_array_s1.cpp b/sword_for_offer/codes/cpp/sfo_53i_find_a_number_in_a_sorted_array_s1/sfo_53i_find_a_number_in_a_sorted_array_s1.cpp similarity index 100% rename from cpp/sfo_53i_find_a_number_in_a_sorted_array_s1/sfo_53i_find_a_number_in_a_sorted_array_s1.cpp rename to sword_for_offer/codes/cpp/sfo_53i_find_a_number_in_a_sorted_array_s1/sfo_53i_find_a_number_in_a_sorted_array_s1.cpp diff --git a/cpp/sfo_53i_find_a_number_in_a_sorted_array_s2/sfo_53i_find_a_number_in_a_sorted_array_s2.cpp b/sword_for_offer/codes/cpp/sfo_53i_find_a_number_in_a_sorted_array_s2/sfo_53i_find_a_number_in_a_sorted_array_s2.cpp similarity index 100% rename from cpp/sfo_53i_find_a_number_in_a_sorted_array_s2/sfo_53i_find_a_number_in_a_sorted_array_s2.cpp rename to sword_for_offer/codes/cpp/sfo_53i_find_a_number_in_a_sorted_array_s2/sfo_53i_find_a_number_in_a_sorted_array_s2.cpp diff --git a/cpp/sfo_53ii_the_missing_number_from_0_to_n1_s1/sfo_53ii_the_missing_number_from_0_to_n1_s1.cpp b/sword_for_offer/codes/cpp/sfo_53ii_the_missing_number_from_0_to_n1_s1/sfo_53ii_the_missing_number_from_0_to_n1_s1.cpp similarity index 100% rename from cpp/sfo_53ii_the_missing_number_from_0_to_n1_s1/sfo_53ii_the_missing_number_from_0_to_n1_s1.cpp rename to sword_for_offer/codes/cpp/sfo_53ii_the_missing_number_from_0_to_n1_s1/sfo_53ii_the_missing_number_from_0_to_n1_s1.cpp diff --git a/cpp/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1.cpp b/sword_for_offer/codes/cpp/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1.cpp similarity index 100% rename from cpp/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1.cpp rename to sword_for_offer/codes/cpp/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1.cpp diff --git a/cpp/sfo_55i_depth_of_a_binary_tree_s1/sfo_55i_depth_of_a_binary_tree_s1.cpp b/sword_for_offer/codes/cpp/sfo_55i_depth_of_a_binary_tree_s1/sfo_55i_depth_of_a_binary_tree_s1.cpp similarity index 100% rename from cpp/sfo_55i_depth_of_a_binary_tree_s1/sfo_55i_depth_of_a_binary_tree_s1.cpp rename to sword_for_offer/codes/cpp/sfo_55i_depth_of_a_binary_tree_s1/sfo_55i_depth_of_a_binary_tree_s1.cpp diff --git a/cpp/sfo_55i_depth_of_a_binary_tree_s2/sfo_55i_depth_of_a_binary_tree_s2.cpp b/sword_for_offer/codes/cpp/sfo_55i_depth_of_a_binary_tree_s2/sfo_55i_depth_of_a_binary_tree_s2.cpp similarity index 100% rename from cpp/sfo_55i_depth_of_a_binary_tree_s2/sfo_55i_depth_of_a_binary_tree_s2.cpp rename to sword_for_offer/codes/cpp/sfo_55i_depth_of_a_binary_tree_s2/sfo_55i_depth_of_a_binary_tree_s2.cpp diff --git a/cpp/sfo_55ii_balanced_binary_tree_s1/sfo_55ii_balanced_binary_tree_s1.cpp b/sword_for_offer/codes/cpp/sfo_55ii_balanced_binary_tree_s1/sfo_55ii_balanced_binary_tree_s1.cpp similarity index 100% rename from cpp/sfo_55ii_balanced_binary_tree_s1/sfo_55ii_balanced_binary_tree_s1.cpp rename to sword_for_offer/codes/cpp/sfo_55ii_balanced_binary_tree_s1/sfo_55ii_balanced_binary_tree_s1.cpp diff --git a/cpp/sfo_55ii_balanced_binary_tree_s2/sfo_55ii_balanced_binary_tree_s2.cpp b/sword_for_offer/codes/cpp/sfo_55ii_balanced_binary_tree_s2/sfo_55ii_balanced_binary_tree_s2.cpp similarity index 100% rename from cpp/sfo_55ii_balanced_binary_tree_s2/sfo_55ii_balanced_binary_tree_s2.cpp rename to sword_for_offer/codes/cpp/sfo_55ii_balanced_binary_tree_s2/sfo_55ii_balanced_binary_tree_s2.cpp diff --git a/cpp/sfo_56i_single_number_i_s1/sfo_56i_single_number_i_s1.cpp b/sword_for_offer/codes/cpp/sfo_56i_single_number_i_s1/sfo_56i_single_number_i_s1.cpp similarity index 100% rename from cpp/sfo_56i_single_number_i_s1/sfo_56i_single_number_i_s1.cpp rename to sword_for_offer/codes/cpp/sfo_56i_single_number_i_s1/sfo_56i_single_number_i_s1.cpp diff --git a/cpp/sfo_56ii_single_number_ii_s1/sfo_56ii_single_number_ii_s1.cpp b/sword_for_offer/codes/cpp/sfo_56ii_single_number_ii_s1/sfo_56ii_single_number_ii_s1.cpp similarity index 100% rename from cpp/sfo_56ii_single_number_ii_s1/sfo_56ii_single_number_ii_s1.cpp rename to sword_for_offer/codes/cpp/sfo_56ii_single_number_ii_s1/sfo_56ii_single_number_ii_s1.cpp diff --git a/cpp/sfo_56ii_single_number_ii_s2/sfo_56ii_single_number_ii_s2.cpp b/sword_for_offer/codes/cpp/sfo_56ii_single_number_ii_s2/sfo_56ii_single_number_ii_s2.cpp similarity index 100% rename from cpp/sfo_56ii_single_number_ii_s2/sfo_56ii_single_number_ii_s2.cpp rename to sword_for_offer/codes/cpp/sfo_56ii_single_number_ii_s2/sfo_56ii_single_number_ii_s2.cpp diff --git a/cpp/sfo_57_two_numbers_with_sum_s_s1/sfo_57_two_numbers_with_sum_s_s1.cpp b/sword_for_offer/codes/cpp/sfo_57_two_numbers_with_sum_s_s1/sfo_57_two_numbers_with_sum_s_s1.cpp similarity index 100% rename from cpp/sfo_57_two_numbers_with_sum_s_s1/sfo_57_two_numbers_with_sum_s_s1.cpp rename to sword_for_offer/codes/cpp/sfo_57_two_numbers_with_sum_s_s1/sfo_57_two_numbers_with_sum_s_s1.cpp diff --git a/cpp/sfo_57ii_consecutive_numbers_with_sum_s_s1/sfo_57ii_consecutive_numbers_with_sum_s_s1.cpp b/sword_for_offer/codes/cpp/sfo_57ii_consecutive_numbers_with_sum_s_s1/sfo_57ii_consecutive_numbers_with_sum_s_s1.cpp similarity index 100% rename from cpp/sfo_57ii_consecutive_numbers_with_sum_s_s1/sfo_57ii_consecutive_numbers_with_sum_s_s1.cpp rename to sword_for_offer/codes/cpp/sfo_57ii_consecutive_numbers_with_sum_s_s1/sfo_57ii_consecutive_numbers_with_sum_s_s1.cpp diff --git a/cpp/sfo_57ii_consecutive_numbers_with_sum_s_s2/sfo_57ii_consecutive_numbers_with_sum_s_s2.cpp b/sword_for_offer/codes/cpp/sfo_57ii_consecutive_numbers_with_sum_s_s2/sfo_57ii_consecutive_numbers_with_sum_s_s2.cpp similarity index 100% rename from cpp/sfo_57ii_consecutive_numbers_with_sum_s_s2/sfo_57ii_consecutive_numbers_with_sum_s_s2.cpp rename to sword_for_offer/codes/cpp/sfo_57ii_consecutive_numbers_with_sum_s_s2/sfo_57ii_consecutive_numbers_with_sum_s_s2.cpp diff --git a/cpp/sfo_58ii_left_rotation_of_a_string_s1/sfo_58ii_left_rotation_of_a_string_s1.cpp b/sword_for_offer/codes/cpp/sfo_58ii_left_rotation_of_a_string_s1/sfo_58ii_left_rotation_of_a_string_s1.cpp similarity index 100% rename from cpp/sfo_58ii_left_rotation_of_a_string_s1/sfo_58ii_left_rotation_of_a_string_s1.cpp rename to sword_for_offer/codes/cpp/sfo_58ii_left_rotation_of_a_string_s1/sfo_58ii_left_rotation_of_a_string_s1.cpp diff --git a/cpp/sfo_58ii_left_rotation_of_a_string_s2/sfo_58ii_left_rotation_of_a_string_s2.cpp b/sword_for_offer/codes/cpp/sfo_58ii_left_rotation_of_a_string_s2/sfo_58ii_left_rotation_of_a_string_s2.cpp similarity index 100% rename from cpp/sfo_58ii_left_rotation_of_a_string_s2/sfo_58ii_left_rotation_of_a_string_s2.cpp rename to sword_for_offer/codes/cpp/sfo_58ii_left_rotation_of_a_string_s2/sfo_58ii_left_rotation_of_a_string_s2.cpp diff --git a/cpp/sfo_58ii_left_rotation_of_a_string_s3/sfo_58ii_left_rotation_of_a_string_s3.cpp b/sword_for_offer/codes/cpp/sfo_58ii_left_rotation_of_a_string_s3/sfo_58ii_left_rotation_of_a_string_s3.cpp similarity index 100% rename from cpp/sfo_58ii_left_rotation_of_a_string_s3/sfo_58ii_left_rotation_of_a_string_s3.cpp rename to sword_for_offer/codes/cpp/sfo_58ii_left_rotation_of_a_string_s3/sfo_58ii_left_rotation_of_a_string_s3.cpp diff --git a/cpp/sfo_59ii_max_queue_s1/sfo_59ii_max_queue_s1.cpp b/sword_for_offer/codes/cpp/sfo_59ii_max_queue_s1/sfo_59ii_max_queue_s1.cpp similarity index 100% rename from cpp/sfo_59ii_max_queue_s1/sfo_59ii_max_queue_s1.cpp rename to sword_for_offer/codes/cpp/sfo_59ii_max_queue_s1/sfo_59ii_max_queue_s1.cpp diff --git a/cpp/sfo_60_probabilities_for_rolling_n_dices_s1/sfo_60_probabilities_for_rolling_n_dices_s1.cpp b/sword_for_offer/codes/cpp/sfo_60_probabilities_for_rolling_n_dices_s1/sfo_60_probabilities_for_rolling_n_dices_s1.cpp similarity index 100% rename from cpp/sfo_60_probabilities_for_rolling_n_dices_s1/sfo_60_probabilities_for_rolling_n_dices_s1.cpp rename to sword_for_offer/codes/cpp/sfo_60_probabilities_for_rolling_n_dices_s1/sfo_60_probabilities_for_rolling_n_dices_s1.cpp diff --git a/cpp/sfo_61_straight_in_poker_s1/sfo_61_straight_in_poker_s1.cpp b/sword_for_offer/codes/cpp/sfo_61_straight_in_poker_s1/sfo_61_straight_in_poker_s1.cpp similarity index 100% rename from cpp/sfo_61_straight_in_poker_s1/sfo_61_straight_in_poker_s1.cpp rename to sword_for_offer/codes/cpp/sfo_61_straight_in_poker_s1/sfo_61_straight_in_poker_s1.cpp diff --git a/cpp/sfo_61_straight_in_poker_s2/sfo_61_straight_in_poker_s2.cpp b/sword_for_offer/codes/cpp/sfo_61_straight_in_poker_s2/sfo_61_straight_in_poker_s2.cpp similarity index 100% rename from cpp/sfo_61_straight_in_poker_s2/sfo_61_straight_in_poker_s2.cpp rename to sword_for_offer/codes/cpp/sfo_61_straight_in_poker_s2/sfo_61_straight_in_poker_s2.cpp diff --git a/cpp/sfo_62_josephus_problem_s1/sfo_62_josephus_problem_s1.cpp b/sword_for_offer/codes/cpp/sfo_62_josephus_problem_s1/sfo_62_josephus_problem_s1.cpp similarity index 100% rename from cpp/sfo_62_josephus_problem_s1/sfo_62_josephus_problem_s1.cpp rename to sword_for_offer/codes/cpp/sfo_62_josephus_problem_s1/sfo_62_josephus_problem_s1.cpp diff --git a/cpp/sfo_63_the_maximum_profit_of_stocks_s1/sfo_63_the_maximum_profit_of_stocks_s1.cpp b/sword_for_offer/codes/cpp/sfo_63_the_maximum_profit_of_stocks_s1/sfo_63_the_maximum_profit_of_stocks_s1.cpp similarity index 100% rename from cpp/sfo_63_the_maximum_profit_of_stocks_s1/sfo_63_the_maximum_profit_of_stocks_s1.cpp rename to sword_for_offer/codes/cpp/sfo_63_the_maximum_profit_of_stocks_s1/sfo_63_the_maximum_profit_of_stocks_s1.cpp diff --git a/cpp/sfo_64_solve_1_2___n_s1/sfo_64_solve_1_2___n_s1.cpp b/sword_for_offer/codes/cpp/sfo_64_solve_1_2___n_s1/sfo_64_solve_1_2___n_s1.cpp similarity index 100% rename from cpp/sfo_64_solve_1_2___n_s1/sfo_64_solve_1_2___n_s1.cpp rename to sword_for_offer/codes/cpp/sfo_64_solve_1_2___n_s1/sfo_64_solve_1_2___n_s1.cpp diff --git a/cpp/sfo_65_implement_addition_operation_without_arithmetic_operators_s1/sfo_65_implement_addition_operation_without_arithmetic_operators_s1.cpp b/sword_for_offer/codes/cpp/sfo_65_implement_addition_operation_without_arithmetic_operators_s1/sfo_65_implement_addition_operation_without_arithmetic_operators_s1.cpp similarity index 100% rename from cpp/sfo_65_implement_addition_operation_without_arithmetic_operators_s1/sfo_65_implement_addition_operation_without_arithmetic_operators_s1.cpp rename to sword_for_offer/codes/cpp/sfo_65_implement_addition_operation_without_arithmetic_operators_s1/sfo_65_implement_addition_operation_without_arithmetic_operators_s1.cpp diff --git a/cpp/sfo_66_a_product_array_puzzle_s1/sfo_66_a_product_array_puzzle_s1.cpp b/sword_for_offer/codes/cpp/sfo_66_a_product_array_puzzle_s1/sfo_66_a_product_array_puzzle_s1.cpp similarity index 100% rename from cpp/sfo_66_a_product_array_puzzle_s1/sfo_66_a_product_array_puzzle_s1.cpp rename to sword_for_offer/codes/cpp/sfo_66_a_product_array_puzzle_s1/sfo_66_a_product_array_puzzle_s1.cpp diff --git a/cpp/sfo_67_convert_string_to_int_s1/sfo_67_convert_string_to_int_s1.cpp b/sword_for_offer/codes/cpp/sfo_67_convert_string_to_int_s1/sfo_67_convert_string_to_int_s1.cpp similarity index 100% rename from cpp/sfo_67_convert_string_to_int_s1/sfo_67_convert_string_to_int_s1.cpp rename to sword_for_offer/codes/cpp/sfo_67_convert_string_to_int_s1/sfo_67_convert_string_to_int_s1.cpp diff --git a/cpp/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1.cpp b/sword_for_offer/codes/cpp/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1.cpp similarity index 100% rename from cpp/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1.cpp rename to sword_for_offer/codes/cpp/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1.cpp diff --git a/cpp/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2.cpp b/sword_for_offer/codes/cpp/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2.cpp similarity index 100% rename from cpp/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2.cpp rename to sword_for_offer/codes/cpp/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2.cpp diff --git a/cpp/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3.cpp b/sword_for_offer/codes/cpp/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3.cpp similarity index 100% rename from cpp/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3.cpp rename to sword_for_offer/codes/cpp/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3.cpp diff --git a/cpp/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1.cpp b/sword_for_offer/codes/cpp/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1.cpp similarity index 100% rename from cpp/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1.cpp rename to sword_for_offer/codes/cpp/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1.cpp diff --git a/cpp/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2.cpp b/sword_for_offer/codes/cpp/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2.cpp similarity index 100% rename from cpp/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2.cpp rename to sword_for_offer/codes/cpp/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2.cpp diff --git a/java/include/ListNode.java b/sword_for_offer/codes/java/include/ListNode.java similarity index 100% rename from java/include/ListNode.java rename to sword_for_offer/codes/java/include/ListNode.java diff --git a/java/include/PrintUtil.java b/sword_for_offer/codes/java/include/PrintUtil.java similarity index 100% rename from java/include/PrintUtil.java rename to sword_for_offer/codes/java/include/PrintUtil.java diff --git a/java/include/TreeNode.java b/sword_for_offer/codes/java/include/TreeNode.java similarity index 100% rename from java/include/TreeNode.java rename to sword_for_offer/codes/java/include/TreeNode.java diff --git a/java/sfo_03_find_duplicate_numbers_in_an_array_s1/sfo_03_find_duplicate_numbers_in_an_array_s1.java b/sword_for_offer/codes/java/sfo_03_find_duplicate_numbers_in_an_array_s1/sfo_03_find_duplicate_numbers_in_an_array_s1.java similarity index 100% rename from java/sfo_03_find_duplicate_numbers_in_an_array_s1/sfo_03_find_duplicate_numbers_in_an_array_s1.java rename to sword_for_offer/codes/java/sfo_03_find_duplicate_numbers_in_an_array_s1/sfo_03_find_duplicate_numbers_in_an_array_s1.java diff --git a/java/sfo_03_find_duplicate_numbers_in_an_array_s2/sfo_03_find_duplicate_numbers_in_an_array_s2.java b/sword_for_offer/codes/java/sfo_03_find_duplicate_numbers_in_an_array_s2/sfo_03_find_duplicate_numbers_in_an_array_s2.java similarity index 100% rename from java/sfo_03_find_duplicate_numbers_in_an_array_s2/sfo_03_find_duplicate_numbers_in_an_array_s2.java rename to sword_for_offer/codes/java/sfo_03_find_duplicate_numbers_in_an_array_s2/sfo_03_find_duplicate_numbers_in_an_array_s2.java diff --git a/java/sfo_04_find_a_number_in_2d_matrix_s1/sfo_04_find_a_number_in_2d_matrix_s1.java b/sword_for_offer/codes/java/sfo_04_find_a_number_in_2d_matrix_s1/sfo_04_find_a_number_in_2d_matrix_s1.java similarity index 100% rename from java/sfo_04_find_a_number_in_2d_matrix_s1/sfo_04_find_a_number_in_2d_matrix_s1.java rename to sword_for_offer/codes/java/sfo_04_find_a_number_in_2d_matrix_s1/sfo_04_find_a_number_in_2d_matrix_s1.java diff --git a/java/sfo_05_replace_spaces_s1/sfo_05_replace_spaces_s1.java b/sword_for_offer/codes/java/sfo_05_replace_spaces_s1/sfo_05_replace_spaces_s1.java similarity index 100% rename from java/sfo_05_replace_spaces_s1/sfo_05_replace_spaces_s1.java rename to sword_for_offer/codes/java/sfo_05_replace_spaces_s1/sfo_05_replace_spaces_s1.java diff --git a/java/sfo_06_print_a_linked_list_in_reverse_order_s1/sfo_06_print_a_linked_list_in_reverse_order_s1.java b/sword_for_offer/codes/java/sfo_06_print_a_linked_list_in_reverse_order_s1/sfo_06_print_a_linked_list_in_reverse_order_s1.java similarity index 100% rename from java/sfo_06_print_a_linked_list_in_reverse_order_s1/sfo_06_print_a_linked_list_in_reverse_order_s1.java rename to sword_for_offer/codes/java/sfo_06_print_a_linked_list_in_reverse_order_s1/sfo_06_print_a_linked_list_in_reverse_order_s1.java diff --git a/java/sfo_06_print_a_linked_list_in_reverse_order_s2/sfo_06_print_a_linked_list_in_reverse_order_s2.java b/sword_for_offer/codes/java/sfo_06_print_a_linked_list_in_reverse_order_s2/sfo_06_print_a_linked_list_in_reverse_order_s2.java similarity index 100% rename from java/sfo_06_print_a_linked_list_in_reverse_order_s2/sfo_06_print_a_linked_list_in_reverse_order_s2.java rename to sword_for_offer/codes/java/sfo_06_print_a_linked_list_in_reverse_order_s2/sfo_06_print_a_linked_list_in_reverse_order_s2.java diff --git a/java/sfo_07_reconstruct_binary_tree_s1/sfo_07_reconstruct_binary_tree_s1.java b/sword_for_offer/codes/java/sfo_07_reconstruct_binary_tree_s1/sfo_07_reconstruct_binary_tree_s1.java similarity index 100% rename from java/sfo_07_reconstruct_binary_tree_s1/sfo_07_reconstruct_binary_tree_s1.java rename to sword_for_offer/codes/java/sfo_07_reconstruct_binary_tree_s1/sfo_07_reconstruct_binary_tree_s1.java diff --git a/java/sfo_09_implement_a_queue_using_two_stacks_s1/sfo_09_implement_a_queue_using_two_stacks_s1.java b/sword_for_offer/codes/java/sfo_09_implement_a_queue_using_two_stacks_s1/sfo_09_implement_a_queue_using_two_stacks_s1.java similarity index 100% rename from java/sfo_09_implement_a_queue_using_two_stacks_s1/sfo_09_implement_a_queue_using_two_stacks_s1.java rename to sword_for_offer/codes/java/sfo_09_implement_a_queue_using_two_stacks_s1/sfo_09_implement_a_queue_using_two_stacks_s1.java diff --git a/java/sfo_10i_fibonacci_numbers_s1/sfo_10i_fibonacci_numbers_s1.java b/sword_for_offer/codes/java/sfo_10i_fibonacci_numbers_s1/sfo_10i_fibonacci_numbers_s1.java similarity index 100% rename from java/sfo_10i_fibonacci_numbers_s1/sfo_10i_fibonacci_numbers_s1.java rename to sword_for_offer/codes/java/sfo_10i_fibonacci_numbers_s1/sfo_10i_fibonacci_numbers_s1.java diff --git a/java/sfo_10ii_frog_jump_s1/sfo_10ii_frog_jump_s1.java b/sword_for_offer/codes/java/sfo_10ii_frog_jump_s1/sfo_10ii_frog_jump_s1.java similarity index 100% rename from java/sfo_10ii_frog_jump_s1/sfo_10ii_frog_jump_s1.java rename to sword_for_offer/codes/java/sfo_10ii_frog_jump_s1/sfo_10ii_frog_jump_s1.java diff --git a/java/sfo_11_find_minimum_in_rotated_sorted_array_s1/sfo_11_find_minimum_in_rotated_sorted_array_s1.java b/sword_for_offer/codes/java/sfo_11_find_minimum_in_rotated_sorted_array_s1/sfo_11_find_minimum_in_rotated_sorted_array_s1.java similarity index 100% rename from java/sfo_11_find_minimum_in_rotated_sorted_array_s1/sfo_11_find_minimum_in_rotated_sorted_array_s1.java rename to sword_for_offer/codes/java/sfo_11_find_minimum_in_rotated_sorted_array_s1/sfo_11_find_minimum_in_rotated_sorted_array_s1.java diff --git a/java/sfo_11_find_minimum_in_rotated_sorted_array_s2/sfo_11_find_minimum_in_rotated_sorted_array_s2.java b/sword_for_offer/codes/java/sfo_11_find_minimum_in_rotated_sorted_array_s2/sfo_11_find_minimum_in_rotated_sorted_array_s2.java similarity index 100% rename from java/sfo_11_find_minimum_in_rotated_sorted_array_s2/sfo_11_find_minimum_in_rotated_sorted_array_s2.java rename to sword_for_offer/codes/java/sfo_11_find_minimum_in_rotated_sorted_array_s2/sfo_11_find_minimum_in_rotated_sorted_array_s2.java diff --git a/java/sfo_12_word_search_s1/sfo_12_word_search_s1.java b/sword_for_offer/codes/java/sfo_12_word_search_s1/sfo_12_word_search_s1.java similarity index 100% rename from java/sfo_12_word_search_s1/sfo_12_word_search_s1.java rename to sword_for_offer/codes/java/sfo_12_word_search_s1/sfo_12_word_search_s1.java diff --git a/java/sfo_13_range_of_motion_of_a_robot_s1/sfo_13_range_of_motion_of_a_robot_s1.java b/sword_for_offer/codes/java/sfo_13_range_of_motion_of_a_robot_s1/sfo_13_range_of_motion_of_a_robot_s1.java similarity index 100% rename from java/sfo_13_range_of_motion_of_a_robot_s1/sfo_13_range_of_motion_of_a_robot_s1.java rename to sword_for_offer/codes/java/sfo_13_range_of_motion_of_a_robot_s1/sfo_13_range_of_motion_of_a_robot_s1.java diff --git a/java/sfo_13_range_of_motion_of_a_robot_s2/sfo_13_range_of_motion_of_a_robot_s2.java b/sword_for_offer/codes/java/sfo_13_range_of_motion_of_a_robot_s2/sfo_13_range_of_motion_of_a_robot_s2.java similarity index 100% rename from java/sfo_13_range_of_motion_of_a_robot_s2/sfo_13_range_of_motion_of_a_robot_s2.java rename to sword_for_offer/codes/java/sfo_13_range_of_motion_of_a_robot_s2/sfo_13_range_of_motion_of_a_robot_s2.java diff --git a/java/sfo_14i_cut_the_rope_i_s1/sfo_14i_cut_the_rope_i_s1.java b/sword_for_offer/codes/java/sfo_14i_cut_the_rope_i_s1/sfo_14i_cut_the_rope_i_s1.java similarity index 100% rename from java/sfo_14i_cut_the_rope_i_s1/sfo_14i_cut_the_rope_i_s1.java rename to sword_for_offer/codes/java/sfo_14i_cut_the_rope_i_s1/sfo_14i_cut_the_rope_i_s1.java diff --git a/java/sfo_14ii_cut_the_rope_ii_s1/sfo_14ii_cut_the_rope_ii_s1.java b/sword_for_offer/codes/java/sfo_14ii_cut_the_rope_ii_s1/sfo_14ii_cut_the_rope_ii_s1.java similarity index 100% rename from java/sfo_14ii_cut_the_rope_ii_s1/sfo_14ii_cut_the_rope_ii_s1.java rename to sword_for_offer/codes/java/sfo_14ii_cut_the_rope_ii_s1/sfo_14ii_cut_the_rope_ii_s1.java diff --git a/java/sfo_15_number_of_1_bits_s1/sfo_15_number_of_1_bits_s1.java b/sword_for_offer/codes/java/sfo_15_number_of_1_bits_s1/sfo_15_number_of_1_bits_s1.java similarity index 100% rename from java/sfo_15_number_of_1_bits_s1/sfo_15_number_of_1_bits_s1.java rename to sword_for_offer/codes/java/sfo_15_number_of_1_bits_s1/sfo_15_number_of_1_bits_s1.java diff --git a/java/sfo_15_number_of_1_bits_s2/sfo_15_number_of_1_bits_s2.java b/sword_for_offer/codes/java/sfo_15_number_of_1_bits_s2/sfo_15_number_of_1_bits_s2.java similarity index 100% rename from java/sfo_15_number_of_1_bits_s2/sfo_15_number_of_1_bits_s2.java rename to sword_for_offer/codes/java/sfo_15_number_of_1_bits_s2/sfo_15_number_of_1_bits_s2.java diff --git a/java/sfo_16_powers_of_integers_s1/sfo_16_powers_of_integers_s1.java b/sword_for_offer/codes/java/sfo_16_powers_of_integers_s1/sfo_16_powers_of_integers_s1.java similarity index 100% rename from java/sfo_16_powers_of_integers_s1/sfo_16_powers_of_integers_s1.java rename to sword_for_offer/codes/java/sfo_16_powers_of_integers_s1/sfo_16_powers_of_integers_s1.java diff --git a/java/sfo_17_print_from_1_to_the_largest_n_digits_s1/sfo_17_print_from_1_to_the_largest_n_digits_s1.java b/sword_for_offer/codes/java/sfo_17_print_from_1_to_the_largest_n_digits_s1/sfo_17_print_from_1_to_the_largest_n_digits_s1.java similarity index 100% rename from java/sfo_17_print_from_1_to_the_largest_n_digits_s1/sfo_17_print_from_1_to_the_largest_n_digits_s1.java rename to sword_for_offer/codes/java/sfo_17_print_from_1_to_the_largest_n_digits_s1/sfo_17_print_from_1_to_the_largest_n_digits_s1.java diff --git a/java/sfo_17_print_from_1_to_the_largest_n_digits_s2/sfo_17_print_from_1_to_the_largest_n_digits_s2.java b/sword_for_offer/codes/java/sfo_17_print_from_1_to_the_largest_n_digits_s2/sfo_17_print_from_1_to_the_largest_n_digits_s2.java similarity index 100% rename from java/sfo_17_print_from_1_to_the_largest_n_digits_s2/sfo_17_print_from_1_to_the_largest_n_digits_s2.java rename to sword_for_offer/codes/java/sfo_17_print_from_1_to_the_largest_n_digits_s2/sfo_17_print_from_1_to_the_largest_n_digits_s2.java diff --git a/java/sfo_17_print_from_1_to_the_largest_n_digits_s3/sfo_17_print_from_1_to_the_largest_n_digits_s3.java b/sword_for_offer/codes/java/sfo_17_print_from_1_to_the_largest_n_digits_s3/sfo_17_print_from_1_to_the_largest_n_digits_s3.java similarity index 100% rename from java/sfo_17_print_from_1_to_the_largest_n_digits_s3/sfo_17_print_from_1_to_the_largest_n_digits_s3.java rename to sword_for_offer/codes/java/sfo_17_print_from_1_to_the_largest_n_digits_s3/sfo_17_print_from_1_to_the_largest_n_digits_s3.java diff --git a/java/sfo_17_print_from_1_to_the_largest_n_digits_s4/sfo_17_print_from_1_to_the_largest_n_digits_s4.java b/sword_for_offer/codes/java/sfo_17_print_from_1_to_the_largest_n_digits_s4/sfo_17_print_from_1_to_the_largest_n_digits_s4.java similarity index 100% rename from java/sfo_17_print_from_1_to_the_largest_n_digits_s4/sfo_17_print_from_1_to_the_largest_n_digits_s4.java rename to sword_for_offer/codes/java/sfo_17_print_from_1_to_the_largest_n_digits_s4/sfo_17_print_from_1_to_the_largest_n_digits_s4.java diff --git a/java/sfo_18_delete_a_node_from_a_linked_list_s1/sfo_18_delete_a_node_from_a_linked_list_s1.java b/sword_for_offer/codes/java/sfo_18_delete_a_node_from_a_linked_list_s1/sfo_18_delete_a_node_from_a_linked_list_s1.java similarity index 100% rename from java/sfo_18_delete_a_node_from_a_linked_list_s1/sfo_18_delete_a_node_from_a_linked_list_s1.java rename to sword_for_offer/codes/java/sfo_18_delete_a_node_from_a_linked_list_s1/sfo_18_delete_a_node_from_a_linked_list_s1.java diff --git a/java/sfo_19_regular_expression_matching_s1/sfo_19_regular_expression_matching_s1.java b/sword_for_offer/codes/java/sfo_19_regular_expression_matching_s1/sfo_19_regular_expression_matching_s1.java similarity index 100% rename from java/sfo_19_regular_expression_matching_s1/sfo_19_regular_expression_matching_s1.java rename to sword_for_offer/codes/java/sfo_19_regular_expression_matching_s1/sfo_19_regular_expression_matching_s1.java diff --git a/java/sfo_19_regular_expression_matching_s2/sfo_19_regular_expression_matching_s2.java b/sword_for_offer/codes/java/sfo_19_regular_expression_matching_s2/sfo_19_regular_expression_matching_s2.java similarity index 100% rename from java/sfo_19_regular_expression_matching_s2/sfo_19_regular_expression_matching_s2.java rename to sword_for_offer/codes/java/sfo_19_regular_expression_matching_s2/sfo_19_regular_expression_matching_s2.java diff --git a/java/sfo_20_a_string_representing_a_numeric_value_s1/sfo_20_a_string_representing_a_numeric_value_s1.java b/sword_for_offer/codes/java/sfo_20_a_string_representing_a_numeric_value_s1/sfo_20_a_string_representing_a_numeric_value_s1.java similarity index 100% rename from java/sfo_20_a_string_representing_a_numeric_value_s1/sfo_20_a_string_representing_a_numeric_value_s1.java rename to sword_for_offer/codes/java/sfo_20_a_string_representing_a_numeric_value_s1/sfo_20_a_string_representing_a_numeric_value_s1.java diff --git a/java/sfo_21_adjust_the_order_of_numbers_in_an_array_s1/sfo_21_adjust_the_order_of_numbers_in_an_array_s1.java b/sword_for_offer/codes/java/sfo_21_adjust_the_order_of_numbers_in_an_array_s1/sfo_21_adjust_the_order_of_numbers_in_an_array_s1.java similarity index 100% rename from java/sfo_21_adjust_the_order_of_numbers_in_an_array_s1/sfo_21_adjust_the_order_of_numbers_in_an_array_s1.java rename to sword_for_offer/codes/java/sfo_21_adjust_the_order_of_numbers_in_an_array_s1/sfo_21_adjust_the_order_of_numbers_in_an_array_s1.java diff --git a/java/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1.java b/sword_for_offer/codes/java/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1.java similarity index 100% rename from java/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1.java rename to sword_for_offer/codes/java/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1.java diff --git a/java/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2.java b/sword_for_offer/codes/java/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2.java similarity index 100% rename from java/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2.java rename to sword_for_offer/codes/java/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2.java diff --git a/java/sfo_24_reverse_a_linked_list_s1/sfo_24_reverse_a_linked_list_s1.java b/sword_for_offer/codes/java/sfo_24_reverse_a_linked_list_s1/sfo_24_reverse_a_linked_list_s1.java similarity index 100% rename from java/sfo_24_reverse_a_linked_list_s1/sfo_24_reverse_a_linked_list_s1.java rename to sword_for_offer/codes/java/sfo_24_reverse_a_linked_list_s1/sfo_24_reverse_a_linked_list_s1.java diff --git a/java/sfo_24_reverse_a_linked_list_s2/sfo_24_reverse_a_linked_list_s2.java b/sword_for_offer/codes/java/sfo_24_reverse_a_linked_list_s2/sfo_24_reverse_a_linked_list_s2.java similarity index 100% rename from java/sfo_24_reverse_a_linked_list_s2/sfo_24_reverse_a_linked_list_s2.java rename to sword_for_offer/codes/java/sfo_24_reverse_a_linked_list_s2/sfo_24_reverse_a_linked_list_s2.java diff --git a/java/sfo_25_combine_two_sorted_linked_lists_s1/sfo_25_combine_two_sorted_linked_lists_s1.java b/sword_for_offer/codes/java/sfo_25_combine_two_sorted_linked_lists_s1/sfo_25_combine_two_sorted_linked_lists_s1.java similarity index 100% rename from java/sfo_25_combine_two_sorted_linked_lists_s1/sfo_25_combine_two_sorted_linked_lists_s1.java rename to sword_for_offer/codes/java/sfo_25_combine_two_sorted_linked_lists_s1/sfo_25_combine_two_sorted_linked_lists_s1.java diff --git a/java/sfo_26_substructure_of_a_binary_tree_s1/sfo_26_substructure_of_a_binary_tree_s1.java b/sword_for_offer/codes/java/sfo_26_substructure_of_a_binary_tree_s1/sfo_26_substructure_of_a_binary_tree_s1.java similarity index 100% rename from java/sfo_26_substructure_of_a_binary_tree_s1/sfo_26_substructure_of_a_binary_tree_s1.java rename to sword_for_offer/codes/java/sfo_26_substructure_of_a_binary_tree_s1/sfo_26_substructure_of_a_binary_tree_s1.java diff --git a/java/sfo_27_mirror_of_a_binary_tree_s1/sfo_27_mirror_of_a_binary_tree_s1.java b/sword_for_offer/codes/java/sfo_27_mirror_of_a_binary_tree_s1/sfo_27_mirror_of_a_binary_tree_s1.java similarity index 100% rename from java/sfo_27_mirror_of_a_binary_tree_s1/sfo_27_mirror_of_a_binary_tree_s1.java rename to sword_for_offer/codes/java/sfo_27_mirror_of_a_binary_tree_s1/sfo_27_mirror_of_a_binary_tree_s1.java diff --git a/java/sfo_27_mirror_of_a_binary_tree_s2/sfo_27_mirror_of_a_binary_tree_s2.java b/sword_for_offer/codes/java/sfo_27_mirror_of_a_binary_tree_s2/sfo_27_mirror_of_a_binary_tree_s2.java similarity index 100% rename from java/sfo_27_mirror_of_a_binary_tree_s2/sfo_27_mirror_of_a_binary_tree_s2.java rename to sword_for_offer/codes/java/sfo_27_mirror_of_a_binary_tree_s2/sfo_27_mirror_of_a_binary_tree_s2.java diff --git a/java/sfo_28_symmetric_binary_tree_s1/sfo_28_symmetric_binary_tree_s1.java b/sword_for_offer/codes/java/sfo_28_symmetric_binary_tree_s1/sfo_28_symmetric_binary_tree_s1.java similarity index 100% rename from java/sfo_28_symmetric_binary_tree_s1/sfo_28_symmetric_binary_tree_s1.java rename to sword_for_offer/codes/java/sfo_28_symmetric_binary_tree_s1/sfo_28_symmetric_binary_tree_s1.java diff --git a/java/sfo_29_print_a_given_matrix_in_spiral_form_s1/sfo_29_print_a_given_matrix_in_spiral_form_s1.java b/sword_for_offer/codes/java/sfo_29_print_a_given_matrix_in_spiral_form_s1/sfo_29_print_a_given_matrix_in_spiral_form_s1.java similarity index 100% rename from java/sfo_29_print_a_given_matrix_in_spiral_form_s1/sfo_29_print_a_given_matrix_in_spiral_form_s1.java rename to sword_for_offer/codes/java/sfo_29_print_a_given_matrix_in_spiral_form_s1/sfo_29_print_a_given_matrix_in_spiral_form_s1.java diff --git a/java/sfo_30_min_stack_s1/sfo_30_min_stack_s1.java b/sword_for_offer/codes/java/sfo_30_min_stack_s1/sfo_30_min_stack_s1.java similarity index 100% rename from java/sfo_30_min_stack_s1/sfo_30_min_stack_s1.java rename to sword_for_offer/codes/java/sfo_30_min_stack_s1/sfo_30_min_stack_s1.java diff --git a/java/sfo_31_validate_stack_sequences_s1/sfo_31_validate_stack_sequences_s1.java b/sword_for_offer/codes/java/sfo_31_validate_stack_sequences_s1/sfo_31_validate_stack_sequences_s1.java similarity index 100% rename from java/sfo_31_validate_stack_sequences_s1/sfo_31_validate_stack_sequences_s1.java rename to sword_for_offer/codes/java/sfo_31_validate_stack_sequences_s1/sfo_31_validate_stack_sequences_s1.java diff --git a/java/sfo_32i_print_a_binary_tree_topbottom_i_s1/sfo_32i_print_a_binary_tree_topbottom_i_s1.java b/sword_for_offer/codes/java/sfo_32i_print_a_binary_tree_topbottom_i_s1/sfo_32i_print_a_binary_tree_topbottom_i_s1.java similarity index 100% rename from java/sfo_32i_print_a_binary_tree_topbottom_i_s1/sfo_32i_print_a_binary_tree_topbottom_i_s1.java rename to sword_for_offer/codes/java/sfo_32i_print_a_binary_tree_topbottom_i_s1/sfo_32i_print_a_binary_tree_topbottom_i_s1.java diff --git a/java/sfo_32ii_print_a_binary_tree_topbottom_ii_s1/sfo_32ii_print_a_binary_tree_topbottom_ii_s1.java b/sword_for_offer/codes/java/sfo_32ii_print_a_binary_tree_topbottom_ii_s1/sfo_32ii_print_a_binary_tree_topbottom_ii_s1.java similarity index 100% rename from java/sfo_32ii_print_a_binary_tree_topbottom_ii_s1/sfo_32ii_print_a_binary_tree_topbottom_ii_s1.java rename to sword_for_offer/codes/java/sfo_32ii_print_a_binary_tree_topbottom_ii_s1/sfo_32ii_print_a_binary_tree_topbottom_ii_s1.java diff --git a/java/sfo_32iii_print_a_binary_tree_topbottom_iii_s1/sfo_32iii_print_a_binary_tree_topbottom_iii_s1.java b/sword_for_offer/codes/java/sfo_32iii_print_a_binary_tree_topbottom_iii_s1/sfo_32iii_print_a_binary_tree_topbottom_iii_s1.java similarity index 100% rename from java/sfo_32iii_print_a_binary_tree_topbottom_iii_s1/sfo_32iii_print_a_binary_tree_topbottom_iii_s1.java rename to sword_for_offer/codes/java/sfo_32iii_print_a_binary_tree_topbottom_iii_s1/sfo_32iii_print_a_binary_tree_topbottom_iii_s1.java diff --git a/java/sfo_32iii_print_a_binary_tree_topbottom_iii_s2/sfo_32iii_print_a_binary_tree_topbottom_iii_s2.java b/sword_for_offer/codes/java/sfo_32iii_print_a_binary_tree_topbottom_iii_s2/sfo_32iii_print_a_binary_tree_topbottom_iii_s2.java similarity index 100% rename from java/sfo_32iii_print_a_binary_tree_topbottom_iii_s2/sfo_32iii_print_a_binary_tree_topbottom_iii_s2.java rename to sword_for_offer/codes/java/sfo_32iii_print_a_binary_tree_topbottom_iii_s2/sfo_32iii_print_a_binary_tree_topbottom_iii_s2.java diff --git a/java/sfo_32iii_print_a_binary_tree_topbottom_iii_s3/sfo_32iii_print_a_binary_tree_topbottom_iii_s3.java b/sword_for_offer/codes/java/sfo_32iii_print_a_binary_tree_topbottom_iii_s3/sfo_32iii_print_a_binary_tree_topbottom_iii_s3.java similarity index 100% rename from java/sfo_32iii_print_a_binary_tree_topbottom_iii_s3/sfo_32iii_print_a_binary_tree_topbottom_iii_s3.java rename to sword_for_offer/codes/java/sfo_32iii_print_a_binary_tree_topbottom_iii_s3/sfo_32iii_print_a_binary_tree_topbottom_iii_s3.java diff --git a/java/sfo_33_postorder_traversal_of_a_binary_search_tree_s1/sfo_33_postorder_traversal_of_a_binary_search_tree_s1.java b/sword_for_offer/codes/java/sfo_33_postorder_traversal_of_a_binary_search_tree_s1/sfo_33_postorder_traversal_of_a_binary_search_tree_s1.java similarity index 100% rename from java/sfo_33_postorder_traversal_of_a_binary_search_tree_s1/sfo_33_postorder_traversal_of_a_binary_search_tree_s1.java rename to sword_for_offer/codes/java/sfo_33_postorder_traversal_of_a_binary_search_tree_s1/sfo_33_postorder_traversal_of_a_binary_search_tree_s1.java diff --git a/java/sfo_33_postorder_traversal_of_a_binary_search_tree_s2/sfo_33_postorder_traversal_of_a_binary_search_tree_s2.java b/sword_for_offer/codes/java/sfo_33_postorder_traversal_of_a_binary_search_tree_s2/sfo_33_postorder_traversal_of_a_binary_search_tree_s2.java similarity index 100% rename from java/sfo_33_postorder_traversal_of_a_binary_search_tree_s2/sfo_33_postorder_traversal_of_a_binary_search_tree_s2.java rename to sword_for_offer/codes/java/sfo_33_postorder_traversal_of_a_binary_search_tree_s2/sfo_33_postorder_traversal_of_a_binary_search_tree_s2.java diff --git a/java/sfo_34_all_xsum_paths_in_a_binary_tree_s1/sfo_34_all_xsum_paths_in_a_binary_tree_s1.java b/sword_for_offer/codes/java/sfo_34_all_xsum_paths_in_a_binary_tree_s1/sfo_34_all_xsum_paths_in_a_binary_tree_s1.java similarity index 100% rename from java/sfo_34_all_xsum_paths_in_a_binary_tree_s1/sfo_34_all_xsum_paths_in_a_binary_tree_s1.java rename to sword_for_offer/codes/java/sfo_34_all_xsum_paths_in_a_binary_tree_s1/sfo_34_all_xsum_paths_in_a_binary_tree_s1.java diff --git a/java/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1.java b/sword_for_offer/codes/java/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1.java similarity index 100% rename from java/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1.java rename to sword_for_offer/codes/java/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1.java diff --git a/java/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2.java b/sword_for_offer/codes/java/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2.java similarity index 100% rename from java/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2.java rename to sword_for_offer/codes/java/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2.java diff --git a/java/sfo_36_binary_search_tree_and_doubly_linked_list_s1/sfo_36_binary_search_tree_and_doubly_linked_list_s1.java b/sword_for_offer/codes/java/sfo_36_binary_search_tree_and_doubly_linked_list_s1/sfo_36_binary_search_tree_and_doubly_linked_list_s1.java similarity index 100% rename from java/sfo_36_binary_search_tree_and_doubly_linked_list_s1/sfo_36_binary_search_tree_and_doubly_linked_list_s1.java rename to sword_for_offer/codes/java/sfo_36_binary_search_tree_and_doubly_linked_list_s1/sfo_36_binary_search_tree_and_doubly_linked_list_s1.java diff --git a/java/sfo_37_serialize_and_deserialize_a_binary_tree_s1/sfo_37_serialize_and_deserialize_a_binary_tree_s1.java b/sword_for_offer/codes/java/sfo_37_serialize_and_deserialize_a_binary_tree_s1/sfo_37_serialize_and_deserialize_a_binary_tree_s1.java similarity index 100% rename from java/sfo_37_serialize_and_deserialize_a_binary_tree_s1/sfo_37_serialize_and_deserialize_a_binary_tree_s1.java rename to sword_for_offer/codes/java/sfo_37_serialize_and_deserialize_a_binary_tree_s1/sfo_37_serialize_and_deserialize_a_binary_tree_s1.java diff --git a/java/sfo_38_all_permutations_of_a_string_s1/sfo_38_all_permutations_of_a_string_s1.java b/sword_for_offer/codes/java/sfo_38_all_permutations_of_a_string_s1/sfo_38_all_permutations_of_a_string_s1.java similarity index 100% rename from java/sfo_38_all_permutations_of_a_string_s1/sfo_38_all_permutations_of_a_string_s1.java rename to sword_for_offer/codes/java/sfo_38_all_permutations_of_a_string_s1/sfo_38_all_permutations_of_a_string_s1.java diff --git a/java/sfo_39_the_majority_element_in_an_array_s1/sfo_39_the_majority_element_in_an_array_s1.java b/sword_for_offer/codes/java/sfo_39_the_majority_element_in_an_array_s1/sfo_39_the_majority_element_in_an_array_s1.java similarity index 100% rename from java/sfo_39_the_majority_element_in_an_array_s1/sfo_39_the_majority_element_in_an_array_s1.java rename to sword_for_offer/codes/java/sfo_39_the_majority_element_in_an_array_s1/sfo_39_the_majority_element_in_an_array_s1.java diff --git a/java/sfo_39_the_majority_element_in_an_array_s2/sfo_39_the_majority_element_in_an_array_s2.java b/sword_for_offer/codes/java/sfo_39_the_majority_element_in_an_array_s2/sfo_39_the_majority_element_in_an_array_s2.java similarity index 100% rename from java/sfo_39_the_majority_element_in_an_array_s2/sfo_39_the_majority_element_in_an_array_s2.java rename to sword_for_offer/codes/java/sfo_39_the_majority_element_in_an_array_s2/sfo_39_the_majority_element_in_an_array_s2.java diff --git a/java/sfo_40_the_smallest_k_numbers_s1/sfo_40_the_smallest_k_numbers_s1.java b/sword_for_offer/codes/java/sfo_40_the_smallest_k_numbers_s1/sfo_40_the_smallest_k_numbers_s1.java similarity index 100% rename from java/sfo_40_the_smallest_k_numbers_s1/sfo_40_the_smallest_k_numbers_s1.java rename to sword_for_offer/codes/java/sfo_40_the_smallest_k_numbers_s1/sfo_40_the_smallest_k_numbers_s1.java diff --git a/java/sfo_40_the_smallest_k_numbers_s2/sfo_40_the_smallest_k_numbers_s2.java b/sword_for_offer/codes/java/sfo_40_the_smallest_k_numbers_s2/sfo_40_the_smallest_k_numbers_s2.java similarity index 100% rename from java/sfo_40_the_smallest_k_numbers_s2/sfo_40_the_smallest_k_numbers_s2.java rename to sword_for_offer/codes/java/sfo_40_the_smallest_k_numbers_s2/sfo_40_the_smallest_k_numbers_s2.java diff --git a/java/sfo_41_find_median_from_data_stream_s1/sfo_41_find_median_from_data_stream_s1.java b/sword_for_offer/codes/java/sfo_41_find_median_from_data_stream_s1/sfo_41_find_median_from_data_stream_s1.java similarity index 100% rename from java/sfo_41_find_median_from_data_stream_s1/sfo_41_find_median_from_data_stream_s1.java rename to sword_for_offer/codes/java/sfo_41_find_median_from_data_stream_s1/sfo_41_find_median_from_data_stream_s1.java diff --git a/java/sfo_42_largest_sum_contiguous_subarray_s1/sfo_42_largest_sum_contiguous_subarray_s1.java b/sword_for_offer/codes/java/sfo_42_largest_sum_contiguous_subarray_s1/sfo_42_largest_sum_contiguous_subarray_s1.java similarity index 100% rename from java/sfo_42_largest_sum_contiguous_subarray_s1/sfo_42_largest_sum_contiguous_subarray_s1.java rename to sword_for_offer/codes/java/sfo_42_largest_sum_contiguous_subarray_s1/sfo_42_largest_sum_contiguous_subarray_s1.java diff --git a/java/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1.java b/sword_for_offer/codes/java/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1.java similarity index 100% rename from java/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1.java rename to sword_for_offer/codes/java/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1.java diff --git a/java/sfo_44_nth_digit_s1/sfo_44_nth_digit_s1.java b/sword_for_offer/codes/java/sfo_44_nth_digit_s1/sfo_44_nth_digit_s1.java similarity index 100% rename from java/sfo_44_nth_digit_s1/sfo_44_nth_digit_s1.java rename to sword_for_offer/codes/java/sfo_44_nth_digit_s1/sfo_44_nth_digit_s1.java diff --git a/java/sfo_45_arrange_an_array_into_the_smallest_number_s1/sfo_45_arrange_an_array_into_the_smallest_number_s1.java b/sword_for_offer/codes/java/sfo_45_arrange_an_array_into_the_smallest_number_s1/sfo_45_arrange_an_array_into_the_smallest_number_s1.java similarity index 100% rename from java/sfo_45_arrange_an_array_into_the_smallest_number_s1/sfo_45_arrange_an_array_into_the_smallest_number_s1.java rename to sword_for_offer/codes/java/sfo_45_arrange_an_array_into_the_smallest_number_s1/sfo_45_arrange_an_array_into_the_smallest_number_s1.java diff --git a/java/sfo_45_arrange_an_array_into_the_smallest_number_s2/sfo_45_arrange_an_array_into_the_smallest_number_s2.java b/sword_for_offer/codes/java/sfo_45_arrange_an_array_into_the_smallest_number_s2/sfo_45_arrange_an_array_into_the_smallest_number_s2.java similarity index 100% rename from java/sfo_45_arrange_an_array_into_the_smallest_number_s2/sfo_45_arrange_an_array_into_the_smallest_number_s2.java rename to sword_for_offer/codes/java/sfo_45_arrange_an_array_into_the_smallest_number_s2/sfo_45_arrange_an_array_into_the_smallest_number_s2.java diff --git a/java/sfo_46_translate_numbers_into_strings_s1/sfo_46_translate_numbers_into_strings_s1.java b/sword_for_offer/codes/java/sfo_46_translate_numbers_into_strings_s1/sfo_46_translate_numbers_into_strings_s1.java similarity index 100% rename from java/sfo_46_translate_numbers_into_strings_s1/sfo_46_translate_numbers_into_strings_s1.java rename to sword_for_offer/codes/java/sfo_46_translate_numbers_into_strings_s1/sfo_46_translate_numbers_into_strings_s1.java diff --git a/java/sfo_46_translate_numbers_into_strings_s2/sfo_46_translate_numbers_into_strings_s2.java b/sword_for_offer/codes/java/sfo_46_translate_numbers_into_strings_s2/sfo_46_translate_numbers_into_strings_s2.java similarity index 100% rename from java/sfo_46_translate_numbers_into_strings_s2/sfo_46_translate_numbers_into_strings_s2.java rename to sword_for_offer/codes/java/sfo_46_translate_numbers_into_strings_s2/sfo_46_translate_numbers_into_strings_s2.java diff --git a/java/sfo_46_translate_numbers_into_strings_s3/sfo_46_translate_numbers_into_strings_s3.java b/sword_for_offer/codes/java/sfo_46_translate_numbers_into_strings_s3/sfo_46_translate_numbers_into_strings_s3.java similarity index 100% rename from java/sfo_46_translate_numbers_into_strings_s3/sfo_46_translate_numbers_into_strings_s3.java rename to sword_for_offer/codes/java/sfo_46_translate_numbers_into_strings_s3/sfo_46_translate_numbers_into_strings_s3.java diff --git a/java/sfo_47_the_maximum_value_of_gifts_s1/sfo_47_the_maximum_value_of_gifts_s1.java b/sword_for_offer/codes/java/sfo_47_the_maximum_value_of_gifts_s1/sfo_47_the_maximum_value_of_gifts_s1.java similarity index 100% rename from java/sfo_47_the_maximum_value_of_gifts_s1/sfo_47_the_maximum_value_of_gifts_s1.java rename to sword_for_offer/codes/java/sfo_47_the_maximum_value_of_gifts_s1/sfo_47_the_maximum_value_of_gifts_s1.java diff --git a/java/sfo_47_the_maximum_value_of_gifts_s2/sfo_47_the_maximum_value_of_gifts_s2.java b/sword_for_offer/codes/java/sfo_47_the_maximum_value_of_gifts_s2/sfo_47_the_maximum_value_of_gifts_s2.java similarity index 100% rename from java/sfo_47_the_maximum_value_of_gifts_s2/sfo_47_the_maximum_value_of_gifts_s2.java rename to sword_for_offer/codes/java/sfo_47_the_maximum_value_of_gifts_s2/sfo_47_the_maximum_value_of_gifts_s2.java diff --git a/java/sfo_48_the_longest_substring_without_repeated_characters_s1/sfo_48_the_longest_substring_without_repeated_characters_s1.java b/sword_for_offer/codes/java/sfo_48_the_longest_substring_without_repeated_characters_s1/sfo_48_the_longest_substring_without_repeated_characters_s1.java similarity index 100% rename from java/sfo_48_the_longest_substring_without_repeated_characters_s1/sfo_48_the_longest_substring_without_repeated_characters_s1.java rename to sword_for_offer/codes/java/sfo_48_the_longest_substring_without_repeated_characters_s1/sfo_48_the_longest_substring_without_repeated_characters_s1.java diff --git a/java/sfo_48_the_longest_substring_without_repeated_characters_s2/sfo_48_the_longest_substring_without_repeated_characters_s2.java b/sword_for_offer/codes/java/sfo_48_the_longest_substring_without_repeated_characters_s2/sfo_48_the_longest_substring_without_repeated_characters_s2.java similarity index 100% rename from java/sfo_48_the_longest_substring_without_repeated_characters_s2/sfo_48_the_longest_substring_without_repeated_characters_s2.java rename to sword_for_offer/codes/java/sfo_48_the_longest_substring_without_repeated_characters_s2/sfo_48_the_longest_substring_without_repeated_characters_s2.java diff --git a/java/sfo_48_the_longest_substring_without_repeated_characters_s3/sfo_48_the_longest_substring_without_repeated_characters_s3.java b/sword_for_offer/codes/java/sfo_48_the_longest_substring_without_repeated_characters_s3/sfo_48_the_longest_substring_without_repeated_characters_s3.java similarity index 100% rename from java/sfo_48_the_longest_substring_without_repeated_characters_s3/sfo_48_the_longest_substring_without_repeated_characters_s3.java rename to sword_for_offer/codes/java/sfo_48_the_longest_substring_without_repeated_characters_s3/sfo_48_the_longest_substring_without_repeated_characters_s3.java diff --git a/java/sfo_49_ugly_numbers_s1/sfo_49_ugly_numbers_s1.java b/sword_for_offer/codes/java/sfo_49_ugly_numbers_s1/sfo_49_ugly_numbers_s1.java similarity index 100% rename from java/sfo_49_ugly_numbers_s1/sfo_49_ugly_numbers_s1.java rename to sword_for_offer/codes/java/sfo_49_ugly_numbers_s1/sfo_49_ugly_numbers_s1.java diff --git a/java/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1.java b/sword_for_offer/codes/java/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1.java similarity index 100% rename from java/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1.java rename to sword_for_offer/codes/java/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1.java diff --git a/java/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2.java b/sword_for_offer/codes/java/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2.java similarity index 100% rename from java/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2.java rename to sword_for_offer/codes/java/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2.java diff --git a/java/sfo_51_reversed_pairs_in_an_array_s1/sfo_51_reversed_pairs_in_an_array_s1.java b/sword_for_offer/codes/java/sfo_51_reversed_pairs_in_an_array_s1/sfo_51_reversed_pairs_in_an_array_s1.java similarity index 100% rename from java/sfo_51_reversed_pairs_in_an_array_s1/sfo_51_reversed_pairs_in_an_array_s1.java rename to sword_for_offer/codes/java/sfo_51_reversed_pairs_in_an_array_s1/sfo_51_reversed_pairs_in_an_array_s1.java diff --git a/java/sfo_52_the_first_common_node_in_two_linked_lists_s1/sfo_52_the_first_common_node_in_two_linked_lists_s1.java b/sword_for_offer/codes/java/sfo_52_the_first_common_node_in_two_linked_lists_s1/sfo_52_the_first_common_node_in_two_linked_lists_s1.java similarity index 100% rename from java/sfo_52_the_first_common_node_in_two_linked_lists_s1/sfo_52_the_first_common_node_in_two_linked_lists_s1.java rename to sword_for_offer/codes/java/sfo_52_the_first_common_node_in_two_linked_lists_s1/sfo_52_the_first_common_node_in_two_linked_lists_s1.java diff --git a/java/sfo_53i_find_a_number_in_a_sorted_array_s1/sfo_53i_find_a_number_in_a_sorted_array_s1.java b/sword_for_offer/codes/java/sfo_53i_find_a_number_in_a_sorted_array_s1/sfo_53i_find_a_number_in_a_sorted_array_s1.java similarity index 100% rename from java/sfo_53i_find_a_number_in_a_sorted_array_s1/sfo_53i_find_a_number_in_a_sorted_array_s1.java rename to sword_for_offer/codes/java/sfo_53i_find_a_number_in_a_sorted_array_s1/sfo_53i_find_a_number_in_a_sorted_array_s1.java diff --git a/java/sfo_53i_find_a_number_in_a_sorted_array_s2/sfo_53i_find_a_number_in_a_sorted_array_s2.java b/sword_for_offer/codes/java/sfo_53i_find_a_number_in_a_sorted_array_s2/sfo_53i_find_a_number_in_a_sorted_array_s2.java similarity index 100% rename from java/sfo_53i_find_a_number_in_a_sorted_array_s2/sfo_53i_find_a_number_in_a_sorted_array_s2.java rename to sword_for_offer/codes/java/sfo_53i_find_a_number_in_a_sorted_array_s2/sfo_53i_find_a_number_in_a_sorted_array_s2.java diff --git a/java/sfo_53ii_the_missing_number_from_0_to_n1_s1/sfo_53ii_the_missing_number_from_0_to_n1_s1.java b/sword_for_offer/codes/java/sfo_53ii_the_missing_number_from_0_to_n1_s1/sfo_53ii_the_missing_number_from_0_to_n1_s1.java similarity index 100% rename from java/sfo_53ii_the_missing_number_from_0_to_n1_s1/sfo_53ii_the_missing_number_from_0_to_n1_s1.java rename to sword_for_offer/codes/java/sfo_53ii_the_missing_number_from_0_to_n1_s1/sfo_53ii_the_missing_number_from_0_to_n1_s1.java diff --git a/java/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1.java b/sword_for_offer/codes/java/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1.java similarity index 100% rename from java/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1.java rename to sword_for_offer/codes/java/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1.java diff --git a/java/sfo_55i_depth_of_a_binary_tree_s1/sfo_55i_depth_of_a_binary_tree_s1.java b/sword_for_offer/codes/java/sfo_55i_depth_of_a_binary_tree_s1/sfo_55i_depth_of_a_binary_tree_s1.java similarity index 100% rename from java/sfo_55i_depth_of_a_binary_tree_s1/sfo_55i_depth_of_a_binary_tree_s1.java rename to sword_for_offer/codes/java/sfo_55i_depth_of_a_binary_tree_s1/sfo_55i_depth_of_a_binary_tree_s1.java diff --git a/java/sfo_55i_depth_of_a_binary_tree_s2/sfo_55i_depth_of_a_binary_tree_s2.java b/sword_for_offer/codes/java/sfo_55i_depth_of_a_binary_tree_s2/sfo_55i_depth_of_a_binary_tree_s2.java similarity index 100% rename from java/sfo_55i_depth_of_a_binary_tree_s2/sfo_55i_depth_of_a_binary_tree_s2.java rename to sword_for_offer/codes/java/sfo_55i_depth_of_a_binary_tree_s2/sfo_55i_depth_of_a_binary_tree_s2.java diff --git a/java/sfo_55ii_balanced_binary_tree_s1/sfo_55ii_balanced_binary_tree_s1.java b/sword_for_offer/codes/java/sfo_55ii_balanced_binary_tree_s1/sfo_55ii_balanced_binary_tree_s1.java similarity index 100% rename from java/sfo_55ii_balanced_binary_tree_s1/sfo_55ii_balanced_binary_tree_s1.java rename to sword_for_offer/codes/java/sfo_55ii_balanced_binary_tree_s1/sfo_55ii_balanced_binary_tree_s1.java diff --git a/java/sfo_55ii_balanced_binary_tree_s2/sfo_55ii_balanced_binary_tree_s2.java b/sword_for_offer/codes/java/sfo_55ii_balanced_binary_tree_s2/sfo_55ii_balanced_binary_tree_s2.java similarity index 100% rename from java/sfo_55ii_balanced_binary_tree_s2/sfo_55ii_balanced_binary_tree_s2.java rename to sword_for_offer/codes/java/sfo_55ii_balanced_binary_tree_s2/sfo_55ii_balanced_binary_tree_s2.java diff --git a/java/sfo_56i_single_number_i_s1/sfo_56i_single_number_i_s1.java b/sword_for_offer/codes/java/sfo_56i_single_number_i_s1/sfo_56i_single_number_i_s1.java similarity index 100% rename from java/sfo_56i_single_number_i_s1/sfo_56i_single_number_i_s1.java rename to sword_for_offer/codes/java/sfo_56i_single_number_i_s1/sfo_56i_single_number_i_s1.java diff --git a/java/sfo_56ii_single_number_ii_s1/sfo_56ii_single_number_ii_s1.java b/sword_for_offer/codes/java/sfo_56ii_single_number_ii_s1/sfo_56ii_single_number_ii_s1.java similarity index 100% rename from java/sfo_56ii_single_number_ii_s1/sfo_56ii_single_number_ii_s1.java rename to sword_for_offer/codes/java/sfo_56ii_single_number_ii_s1/sfo_56ii_single_number_ii_s1.java diff --git a/java/sfo_56ii_single_number_ii_s2/sfo_56ii_single_number_ii_s2.java b/sword_for_offer/codes/java/sfo_56ii_single_number_ii_s2/sfo_56ii_single_number_ii_s2.java similarity index 100% rename from java/sfo_56ii_single_number_ii_s2/sfo_56ii_single_number_ii_s2.java rename to sword_for_offer/codes/java/sfo_56ii_single_number_ii_s2/sfo_56ii_single_number_ii_s2.java diff --git a/java/sfo_57_two_numbers_with_sum_s_s1/sfo_57_two_numbers_with_sum_s_s1.java b/sword_for_offer/codes/java/sfo_57_two_numbers_with_sum_s_s1/sfo_57_two_numbers_with_sum_s_s1.java similarity index 100% rename from java/sfo_57_two_numbers_with_sum_s_s1/sfo_57_two_numbers_with_sum_s_s1.java rename to sword_for_offer/codes/java/sfo_57_two_numbers_with_sum_s_s1/sfo_57_two_numbers_with_sum_s_s1.java diff --git a/java/sfo_57ii_consecutive_numbers_with_sum_s_s1/sfo_57ii_consecutive_numbers_with_sum_s_s1.java b/sword_for_offer/codes/java/sfo_57ii_consecutive_numbers_with_sum_s_s1/sfo_57ii_consecutive_numbers_with_sum_s_s1.java similarity index 100% rename from java/sfo_57ii_consecutive_numbers_with_sum_s_s1/sfo_57ii_consecutive_numbers_with_sum_s_s1.java rename to sword_for_offer/codes/java/sfo_57ii_consecutive_numbers_with_sum_s_s1/sfo_57ii_consecutive_numbers_with_sum_s_s1.java diff --git a/java/sfo_57ii_consecutive_numbers_with_sum_s_s2/sfo_57ii_consecutive_numbers_with_sum_s_s2.java b/sword_for_offer/codes/java/sfo_57ii_consecutive_numbers_with_sum_s_s2/sfo_57ii_consecutive_numbers_with_sum_s_s2.java similarity index 100% rename from java/sfo_57ii_consecutive_numbers_with_sum_s_s2/sfo_57ii_consecutive_numbers_with_sum_s_s2.java rename to sword_for_offer/codes/java/sfo_57ii_consecutive_numbers_with_sum_s_s2/sfo_57ii_consecutive_numbers_with_sum_s_s2.java diff --git a/java/sfo_58i_reverse_order_of_words_s1/sfo_58i_reverse_order_of_words_s1.java b/sword_for_offer/codes/java/sfo_58i_reverse_order_of_words_s1/sfo_58i_reverse_order_of_words_s1.java similarity index 100% rename from java/sfo_58i_reverse_order_of_words_s1/sfo_58i_reverse_order_of_words_s1.java rename to sword_for_offer/codes/java/sfo_58i_reverse_order_of_words_s1/sfo_58i_reverse_order_of_words_s1.java diff --git a/java/sfo_58i_reverse_order_of_words_s2/sfo_58i_reverse_order_of_words_s2.java b/sword_for_offer/codes/java/sfo_58i_reverse_order_of_words_s2/sfo_58i_reverse_order_of_words_s2.java similarity index 100% rename from java/sfo_58i_reverse_order_of_words_s2/sfo_58i_reverse_order_of_words_s2.java rename to sword_for_offer/codes/java/sfo_58i_reverse_order_of_words_s2/sfo_58i_reverse_order_of_words_s2.java diff --git a/java/sfo_58ii_left_rotation_of_a_string_s1/sfo_58ii_left_rotation_of_a_string_s1.java b/sword_for_offer/codes/java/sfo_58ii_left_rotation_of_a_string_s1/sfo_58ii_left_rotation_of_a_string_s1.java similarity index 100% rename from java/sfo_58ii_left_rotation_of_a_string_s1/sfo_58ii_left_rotation_of_a_string_s1.java rename to sword_for_offer/codes/java/sfo_58ii_left_rotation_of_a_string_s1/sfo_58ii_left_rotation_of_a_string_s1.java diff --git a/java/sfo_58ii_left_rotation_of_a_string_s2/sfo_58ii_left_rotation_of_a_string_s2.java b/sword_for_offer/codes/java/sfo_58ii_left_rotation_of_a_string_s2/sfo_58ii_left_rotation_of_a_string_s2.java similarity index 100% rename from java/sfo_58ii_left_rotation_of_a_string_s2/sfo_58ii_left_rotation_of_a_string_s2.java rename to sword_for_offer/codes/java/sfo_58ii_left_rotation_of_a_string_s2/sfo_58ii_left_rotation_of_a_string_s2.java diff --git a/java/sfo_58ii_left_rotation_of_a_string_s3/sfo_58ii_left_rotation_of_a_string_s3.java b/sword_for_offer/codes/java/sfo_58ii_left_rotation_of_a_string_s3/sfo_58ii_left_rotation_of_a_string_s3.java similarity index 100% rename from java/sfo_58ii_left_rotation_of_a_string_s3/sfo_58ii_left_rotation_of_a_string_s3.java rename to sword_for_offer/codes/java/sfo_58ii_left_rotation_of_a_string_s3/sfo_58ii_left_rotation_of_a_string_s3.java diff --git a/java/sfo_58ii_left_rotation_of_a_string_s4/sfo_58ii_left_rotation_of_a_string_s4.java b/sword_for_offer/codes/java/sfo_58ii_left_rotation_of_a_string_s4/sfo_58ii_left_rotation_of_a_string_s4.java similarity index 100% rename from java/sfo_58ii_left_rotation_of_a_string_s4/sfo_58ii_left_rotation_of_a_string_s4.java rename to sword_for_offer/codes/java/sfo_58ii_left_rotation_of_a_string_s4/sfo_58ii_left_rotation_of_a_string_s4.java diff --git a/java/sfo_58ii_left_rotation_of_a_string_s5/sfo_58ii_left_rotation_of_a_string_s5.java b/sword_for_offer/codes/java/sfo_58ii_left_rotation_of_a_string_s5/sfo_58ii_left_rotation_of_a_string_s5.java similarity index 100% rename from java/sfo_58ii_left_rotation_of_a_string_s5/sfo_58ii_left_rotation_of_a_string_s5.java rename to sword_for_offer/codes/java/sfo_58ii_left_rotation_of_a_string_s5/sfo_58ii_left_rotation_of_a_string_s5.java diff --git a/java/sfo_59i_sliding_window_maximum_s1/sfo_59i_sliding_window_maximum_s1.java b/sword_for_offer/codes/java/sfo_59i_sliding_window_maximum_s1/sfo_59i_sliding_window_maximum_s1.java similarity index 100% rename from java/sfo_59i_sliding_window_maximum_s1/sfo_59i_sliding_window_maximum_s1.java rename to sword_for_offer/codes/java/sfo_59i_sliding_window_maximum_s1/sfo_59i_sliding_window_maximum_s1.java diff --git a/java/sfo_59i_sliding_window_maximum_s2/sfo_59i_sliding_window_maximum_s2.java b/sword_for_offer/codes/java/sfo_59i_sliding_window_maximum_s2/sfo_59i_sliding_window_maximum_s2.java similarity index 100% rename from java/sfo_59i_sliding_window_maximum_s2/sfo_59i_sliding_window_maximum_s2.java rename to sword_for_offer/codes/java/sfo_59i_sliding_window_maximum_s2/sfo_59i_sliding_window_maximum_s2.java diff --git a/java/sfo_59ii_max_queue_s1/sfo_59ii_max_queue_s1.java b/sword_for_offer/codes/java/sfo_59ii_max_queue_s1/sfo_59ii_max_queue_s1.java similarity index 100% rename from java/sfo_59ii_max_queue_s1/sfo_59ii_max_queue_s1.java rename to sword_for_offer/codes/java/sfo_59ii_max_queue_s1/sfo_59ii_max_queue_s1.java diff --git a/java/sfo_60_probabilities_for_rolling_n_dices_s1/sfo_60_probabilities_for_rolling_n_dices_s1.java b/sword_for_offer/codes/java/sfo_60_probabilities_for_rolling_n_dices_s1/sfo_60_probabilities_for_rolling_n_dices_s1.java similarity index 100% rename from java/sfo_60_probabilities_for_rolling_n_dices_s1/sfo_60_probabilities_for_rolling_n_dices_s1.java rename to sword_for_offer/codes/java/sfo_60_probabilities_for_rolling_n_dices_s1/sfo_60_probabilities_for_rolling_n_dices_s1.java diff --git a/java/sfo_61_straight_in_poker_s1/sfo_61_straight_in_poker_s1.java b/sword_for_offer/codes/java/sfo_61_straight_in_poker_s1/sfo_61_straight_in_poker_s1.java similarity index 100% rename from java/sfo_61_straight_in_poker_s1/sfo_61_straight_in_poker_s1.java rename to sword_for_offer/codes/java/sfo_61_straight_in_poker_s1/sfo_61_straight_in_poker_s1.java diff --git a/java/sfo_61_straight_in_poker_s2/sfo_61_straight_in_poker_s2.java b/sword_for_offer/codes/java/sfo_61_straight_in_poker_s2/sfo_61_straight_in_poker_s2.java similarity index 100% rename from java/sfo_61_straight_in_poker_s2/sfo_61_straight_in_poker_s2.java rename to sword_for_offer/codes/java/sfo_61_straight_in_poker_s2/sfo_61_straight_in_poker_s2.java diff --git a/java/sfo_62_josephus_problem_s1/sfo_62_josephus_problem_s1.java b/sword_for_offer/codes/java/sfo_62_josephus_problem_s1/sfo_62_josephus_problem_s1.java similarity index 100% rename from java/sfo_62_josephus_problem_s1/sfo_62_josephus_problem_s1.java rename to sword_for_offer/codes/java/sfo_62_josephus_problem_s1/sfo_62_josephus_problem_s1.java diff --git a/java/sfo_63_the_maximum_profit_of_stocks_s1/sfo_63_the_maximum_profit_of_stocks_s1.java b/sword_for_offer/codes/java/sfo_63_the_maximum_profit_of_stocks_s1/sfo_63_the_maximum_profit_of_stocks_s1.java similarity index 100% rename from java/sfo_63_the_maximum_profit_of_stocks_s1/sfo_63_the_maximum_profit_of_stocks_s1.java rename to sword_for_offer/codes/java/sfo_63_the_maximum_profit_of_stocks_s1/sfo_63_the_maximum_profit_of_stocks_s1.java diff --git a/java/sfo_64_solve_1_2___n_s1/sfo_64_solve_1_2___n_s1.java b/sword_for_offer/codes/java/sfo_64_solve_1_2___n_s1/sfo_64_solve_1_2___n_s1.java similarity index 100% rename from java/sfo_64_solve_1_2___n_s1/sfo_64_solve_1_2___n_s1.java rename to sword_for_offer/codes/java/sfo_64_solve_1_2___n_s1/sfo_64_solve_1_2___n_s1.java diff --git a/java/sfo_64_solve_1_2___n_s2/sfo_64_solve_1_2___n_s2.java b/sword_for_offer/codes/java/sfo_64_solve_1_2___n_s2/sfo_64_solve_1_2___n_s2.java similarity index 100% rename from java/sfo_64_solve_1_2___n_s2/sfo_64_solve_1_2___n_s2.java rename to sword_for_offer/codes/java/sfo_64_solve_1_2___n_s2/sfo_64_solve_1_2___n_s2.java diff --git a/java/sfo_65_implement_addition_operation_without_arithmetic_operators_s1/sfo_65_implement_addition_operation_without_arithmetic_operators_s1.java b/sword_for_offer/codes/java/sfo_65_implement_addition_operation_without_arithmetic_operators_s1/sfo_65_implement_addition_operation_without_arithmetic_operators_s1.java similarity index 100% rename from java/sfo_65_implement_addition_operation_without_arithmetic_operators_s1/sfo_65_implement_addition_operation_without_arithmetic_operators_s1.java rename to sword_for_offer/codes/java/sfo_65_implement_addition_operation_without_arithmetic_operators_s1/sfo_65_implement_addition_operation_without_arithmetic_operators_s1.java diff --git a/java/sfo_66_a_product_array_puzzle_s1/sfo_66_a_product_array_puzzle_s1.java b/sword_for_offer/codes/java/sfo_66_a_product_array_puzzle_s1/sfo_66_a_product_array_puzzle_s1.java similarity index 100% rename from java/sfo_66_a_product_array_puzzle_s1/sfo_66_a_product_array_puzzle_s1.java rename to sword_for_offer/codes/java/sfo_66_a_product_array_puzzle_s1/sfo_66_a_product_array_puzzle_s1.java diff --git a/java/sfo_67_convert_string_to_int_s1/sfo_67_convert_string_to_int_s1.java b/sword_for_offer/codes/java/sfo_67_convert_string_to_int_s1/sfo_67_convert_string_to_int_s1.java similarity index 100% rename from java/sfo_67_convert_string_to_int_s1/sfo_67_convert_string_to_int_s1.java rename to sword_for_offer/codes/java/sfo_67_convert_string_to_int_s1/sfo_67_convert_string_to_int_s1.java diff --git a/java/sfo_67_convert_string_to_int_s2/sfo_67_convert_string_to_int_s2.java b/sword_for_offer/codes/java/sfo_67_convert_string_to_int_s2/sfo_67_convert_string_to_int_s2.java similarity index 100% rename from java/sfo_67_convert_string_to_int_s2/sfo_67_convert_string_to_int_s2.java rename to sword_for_offer/codes/java/sfo_67_convert_string_to_int_s2/sfo_67_convert_string_to_int_s2.java diff --git a/java/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1.java b/sword_for_offer/codes/java/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1.java similarity index 100% rename from java/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1.java rename to sword_for_offer/codes/java/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1.java diff --git a/java/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2.java b/sword_for_offer/codes/java/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2.java similarity index 100% rename from java/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2.java rename to sword_for_offer/codes/java/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2.java diff --git a/java/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3.java b/sword_for_offer/codes/java/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3.java similarity index 100% rename from java/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3.java rename to sword_for_offer/codes/java/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3.java diff --git 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a/python/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2.py b/sword_for_offer/codes/python/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2.py similarity index 100% rename from python/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2.py rename to sword_for_offer/codes/python/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2.py From 04658dd33363d6e7269aa4864136a0d4a1f39d84 Mon Sep 17 00:00:00 2001 From: krahets Date: Tue, 10 Oct 2023 19:49:12 +0800 Subject: [PATCH 09/23] Format the docs and codes. --- .../codes/cpp/include/ListNode.hpp | 23 +- .../codes/cpp/include/PrintUtil.hpp | 323 ++++++++-------- .../codes/cpp/include/TreeNode.hpp | 31 +- sword_for_offer/codes/cpp/include/include.hpp | 12 +- ..._find_duplicate_numbers_in_an_array_s2.cpp | 26 +- .../sfo_04_find_a_number_in_2d_matrix_s1.cpp | 35 +- .../sfo_05_replace_spaces_s1.cpp | 19 +- ...rint_a_linked_list_in_reverse_order_s1.cpp | 26 +- ...rint_a_linked_list_in_reverse_order_s2.cpp | 22 +- .../sfo_07_reconstruct_binary_tree_s1.cpp | 34 +- ..._implement_a_queue_using_two_stacks_s1.cpp | 22 +- .../sfo_10i_fibonacci_numbers_s1.cpp | 16 +- .../sfo_10ii_frog_jump_s1.cpp | 16 +- ...ind_minimum_in_rotated_sorted_array_s1.cpp | 27 +- ...ind_minimum_in_rotated_sorted_array_s2.cpp | 29 +- .../sfo_12_word_search_s1.cpp | 44 +-- .../sfo_13_range_of_motion_of_a_robot_s1.cpp | 22 +- .../sfo_13_range_of_motion_of_a_robot_s2.cpp | 25 +- .../sfo_14i_cut_the_rope_i_s1.cpp | 23 +- .../sfo_14ii_cut_the_rope_ii_s1.cpp | 28 +- .../sfo_15_number_of_1_bits_s1.cpp | 16 +- .../sfo_15_number_of_1_bits_s2.cpp | 16 +- .../sfo_16_powers_of_integers_s1.cpp | 24 +- ...18_delete_a_node_from_a_linked_list_s1.cpp | 28 +- .../sfo_19_regular_expression_matching_s1.cpp | 25 +- .../sfo_19_regular_expression_matching_s2.cpp | 37 +- ...st_the_order_of_numbers_in_an_array_s1.cpp | 27 +- ..._node_from_the_end_of_a_linked_list_s1.cpp | 24 +- ..._node_from_the_end_of_a_linked_list_s2.cpp | 27 +- .../sfo_24_reverse_a_linked_list_s1.cpp | 24 +- .../sfo_24_reverse_a_linked_list_s2.cpp | 32 +- ..._25_combine_two_sorted_linked_lists_s1.cpp | 33 +- ...fo_26_substructure_of_a_binary_tree_s1.cpp | 34 +- .../sfo_27_mirror_of_a_binary_tree_s1.cpp | 26 +- .../sfo_27_mirror_of_a_binary_tree_s2.cpp | 39 +- .../sfo_28_symmetric_binary_tree_s1.cpp | 30 +- ...print_a_given_matrix_in_spiral_form_s1.cpp | 52 +-- .../sfo_30_min_stack_s1.cpp | 19 +- .../sfo_31_validate_stack_sequences_s1.cpp | 26 +- ...32i_print_a_binary_tree_topbottom_i_s1.cpp | 33 +- ...ii_print_a_binary_tree_topbottom_ii_s1.cpp | 33 +- ...i_print_a_binary_tree_topbottom_iii_s1.cpp | 48 ++- ...i_print_a_binary_tree_topbottom_iii_s2.cpp | 38 +- ...r_traversal_of_a_binary_search_tree_s1.cpp | 32 +- ...r_traversal_of_a_binary_search_tree_s2.cpp | 25 +- ..._34_all_xsum_paths_in_a_binary_tree_s1.cpp | 30 +- ...d_list_with_next_and_random_pointer_s1.cpp | 46 +-- ...d_list_with_next_and_random_pointer_s2.cpp | 54 +-- ..._search_tree_and_doubly_linked_list_s1.cpp | 48 +-- ...alize_and_deserialize_a_binary_tree_s1.cpp | 43 ++- ...sfo_38_all_permutations_of_a_string_s1.cpp | 32 +- ...39_the_majority_element_in_an_array_s1.cpp | 23 +- ...39_the_majority_element_in_an_array_s2.cpp | 28 +- .../sfo_40_the_smallest_k_numbers_s1.cpp | 32 +- .../sfo_40_the_smallest_k_numbers_s2.cpp | 38 +- ...sfo_41_find_median_from_data_stream_s1.cpp | 19 +- ..._42_largest_sum_contiguous_subarray_s1.cpp | 28 +- ...number_of_1_in_integers_from_1_to_n_s1.cpp | 25 +- .../sfo_44_nth_digit_s1.cpp | 16 +- ...e_an_array_into_the_smallest_number_s1.cpp | 38 +- ...e_an_array_into_the_smallest_number_s2.cpp | 24 +- ...o_46_translate_numbers_into_strings_s1.cpp | 16 +- ...o_46_translate_numbers_into_strings_s2.cpp | 16 +- ...o_46_translate_numbers_into_strings_s3.cpp | 16 +- .../sfo_47_the_maximum_value_of_gifts_s1.cpp | 38 +- .../sfo_47_the_maximum_value_of_gifts_s2.cpp | 30 +- ...bstring_without_repeated_characters_s1.cpp | 26 +- ...bstring_without_repeated_characters_s2.cpp | 21 +- ...bstring_without_repeated_characters_s3.cpp | 22 +- .../sfo_49_ugly_numbers_s1.cpp | 25 +- ..._nonrepeating_character_in_a_string_s1.cpp | 21 +- ..._nonrepeating_character_in_a_string_s2.cpp | 23 +- .../sfo_51_reversed_pairs_in_an_array_s1.cpp | 26 +- ...rst_common_node_in_two_linked_lists_s1.cpp | 24 +- ...53i_find_a_number_in_a_sorted_array_s1.cpp | 40 +- ...53i_find_a_number_in_a_sorted_array_s2.cpp | 31 +- ...3ii_the_missing_number_from_0_to_n1_s1.cpp | 26 +- ...argest_node_of_a_binary_search_tree_s1.cpp | 32 +- .../sfo_55i_depth_of_a_binary_tree_s1.cpp | 21 +- .../sfo_55i_depth_of_a_binary_tree_s2.cpp | 35 +- .../sfo_55ii_balanced_binary_tree_s1.cpp | 32 +- .../sfo_55ii_balanced_binary_tree_s2.cpp | 29 +- .../sfo_56i_single_number_i_s1.cpp | 32 +- .../sfo_56ii_single_number_ii_s1.cpp | 20 +- .../sfo_56ii_single_number_ii_s2.cpp | 28 +- .../sfo_57_two_numbers_with_sum_s_s1.cpp | 29 +- ...57ii_consecutive_numbers_with_sum_s_s1.cpp | 22 +- ...57ii_consecutive_numbers_with_sum_s_s2.cpp | 22 +- .../sfo_58ii_left_rotation_of_a_string_s1.cpp | 14 +- .../sfo_58ii_left_rotation_of_a_string_s2.cpp | 22 +- .../sfo_58ii_left_rotation_of_a_string_s3.cpp | 14 +- .../sfo_59ii_max_queue_s1.cpp | 23 +- ...0_probabilities_for_rolling_n_dices_s1.cpp | 14 +- .../sfo_61_straight_in_poker_s1.cpp | 26 +- .../sfo_61_straight_in_poker_s2.cpp | 26 +- .../sfo_62_josephus_problem_s1.cpp | 14 +- ...sfo_63_the_maximum_profit_of_stocks_s1.cpp | 20 +- .../sfo_64_solve_1_2___n_s1.cpp | 14 +- ...ration_without_arithmetic_operators_s1.cpp | 17 +- .../sfo_66_a_product_array_puzzle_s1.cpp | 25 +- .../sfo_67_convert_string_to_int_s1.cpp | 35 +- ...on_ancestor_of_a_binary_search_tree_s1.cpp | 38 +- ...on_ancestor_of_a_binary_search_tree_s2.cpp | 40 +- ...on_ancestor_of_a_binary_search_tree_s3.cpp | 29 +- ...st_common_ancestor_of_a_binary_tree_s1.cpp | 34 +- ...st_common_ancestor_of_a_binary_tree_s2.cpp | 39 +- .../codes/java/include/ListNode.java | 4 +- .../codes/java/include/PrintUtil.java | 3 + .../codes/java/include/TreeNode.java | 33 +- ...find_duplicate_numbers_in_an_array_s1.java | 5 +- ...find_duplicate_numbers_in_an_array_s2.java | 7 +- .../sfo_04_find_a_number_in_2d_matrix_s1.java | 22 +- .../sfo_05_replace_spaces_s1.java | 9 +- ...int_a_linked_list_in_reverse_order_s1.java | 7 +- ...int_a_linked_list_in_reverse_order_s2.java | 6 +- .../sfo_07_reconstruct_binary_tree_s1.java | 15 +- ...implement_a_queue_using_two_stacks_s1.java | 11 +- .../sfo_10i_fibonacci_numbers_s1.java | 2 +- .../sfo_10ii_frog_jump_s1.java | 2 +- ...nd_minimum_in_rotated_sorted_array_s1.java | 9 +- ...nd_minimum_in_rotated_sorted_array_s2.java | 11 +- .../sfo_12_word_search_s1.java | 26 +- .../sfo_13_range_of_motion_of_a_robot_s1.java | 12 +- .../sfo_13_range_of_motion_of_a_robot_s2.java | 9 +- .../sfo_14i_cut_the_rope_i_s1.java | 11 +- .../sfo_14ii_cut_the_rope_ii_s1.java | 16 +- .../sfo_15_number_of_1_bits_s1.java | 2 +- .../sfo_15_number_of_1_bits_s2.java | 2 +- .../sfo_16_powers_of_integers_s1.java | 10 +- ...int_from_1_to_the_largest_n_digits_s1.java | 4 +- ...int_from_1_to_the_largest_n_digits_s2.java | 8 +- ...int_from_1_to_the_largest_n_digits_s3.java | 17 +- ...int_from_1_to_the_largest_n_digits_s4.java | 19 +- ...8_delete_a_node_from_a_linked_list_s1.java | 8 +- ...sfo_19_regular_expression_matching_s1.java | 12 +- ...sfo_19_regular_expression_matching_s2.java | 23 +- ...tring_representing_a_numeric_value_s1.java | 87 ++++- ...t_the_order_of_numbers_in_an_array_s1.java | 8 +- ...node_from_the_end_of_a_linked_list_s1.java | 4 +- ...node_from_the_end_of_a_linked_list_s2.java | 7 +- .../sfo_24_reverse_a_linked_list_s1.java | 8 +- .../sfo_24_reverse_a_linked_list_s2.java | 12 +- ...25_combine_two_sorted_linked_lists_s1.java | 7 +- ...o_26_substructure_of_a_binary_tree_s1.java | 7 +- .../sfo_27_mirror_of_a_binary_tree_s1.java | 6 +- .../sfo_27_mirror_of_a_binary_tree_s2.java | 20 +- .../sfo_28_symmetric_binary_tree_s1.java | 10 +- ...rint_a_given_matrix_in_spiral_form_s1.java | 37 +- .../sfo_30_min_stack_s1.java | 9 +- .../sfo_31_validate_stack_sequences_s1.java | 4 +- ...2i_print_a_binary_tree_topbottom_i_s1.java | 19 +- ...i_print_a_binary_tree_topbottom_ii_s1.java | 16 +- ..._print_a_binary_tree_topbottom_iii_s1.java | 22 +- ..._print_a_binary_tree_topbottom_iii_s2.java | 27 +- ..._print_a_binary_tree_topbottom_iii_s3.java | 19 +- ..._traversal_of_a_binary_search_tree_s1.java | 10 +- ..._traversal_of_a_binary_search_tree_s2.java | 7 +- ...34_all_xsum_paths_in_a_binary_tree_s1.java | 13 +- ..._list_with_next_and_random_pointer_s1.java | 7 +- ..._list_with_next_and_random_pointer_s2.java | 13 +- ...search_tree_and_doubly_linked_list_s1.java | 30 +- ...lize_and_deserialize_a_binary_tree_s1.java | 32 +- ...fo_38_all_permutations_of_a_string_s1.java | 18 +- ...9_the_majority_element_in_an_array_s1.java | 5 +- ...9_the_majority_element_in_an_array_s2.java | 10 +- .../sfo_40_the_smallest_k_numbers_s1.java | 11 +- .../sfo_40_the_smallest_k_numbers_s2.java | 17 +- ...fo_41_find_median_from_data_stream_s1.java | 5 +- ...42_largest_sum_contiguous_subarray_s1.java | 2 +- ...umber_of_1_in_integers_from_1_to_n_s1.java | 11 +- ..._an_array_into_the_smallest_number_s1.java | 16 +- ..._an_array_into_the_smallest_number_s2.java | 4 +- ..._46_translate_numbers_into_strings_s1.java | 2 +- ..._46_translate_numbers_into_strings_s2.java | 2 +- ..._46_translate_numbers_into_strings_s3.java | 2 +- .../sfo_47_the_maximum_value_of_gifts_s1.java | 24 +- .../sfo_47_the_maximum_value_of_gifts_s2.java | 16 +- ...string_without_repeated_characters_s1.java | 2 +- ...string_without_repeated_characters_s2.java | 5 +- ...string_without_repeated_characters_s3.java | 4 +- .../sfo_49_ugly_numbers_s1.java | 11 +- ...nonrepeating_character_in_a_string_s1.java | 7 +- ...nonrepeating_character_in_a_string_s2.java | 7 +- .../sfo_51_reversed_pairs_in_an_array_s1.java | 5 +- ...st_common_node_in_two_linked_lists_s1.java | 2 +- ...3i_find_a_number_in_a_sorted_array_s1.java | 22 +- ...3i_find_a_number_in_a_sorted_array_s2.java | 9 +- ...ii_the_missing_number_from_0_to_n1_s1.java | 8 +- ...rgest_node_of_a_binary_search_tree_s1.java | 11 +- .../sfo_55i_depth_of_a_binary_tree_s1.java | 3 +- .../sfo_55i_depth_of_a_binary_tree_s2.java | 19 +- .../sfo_55ii_balanced_binary_tree_s1.java | 9 +- .../sfo_55ii_balanced_binary_tree_s2.java | 6 +- .../sfo_56i_single_number_i_s1.java | 14 +- .../sfo_56ii_single_number_ii_s1.java | 2 +- .../sfo_56ii_single_number_ii_s2.java | 10 +- .../sfo_57_two_numbers_with_sum_s_s1.java | 11 +- ...7ii_consecutive_numbers_with_sum_s_s1.java | 8 +- ...7ii_consecutive_numbers_with_sum_s_s2.java | 8 +- .../sfo_58i_reverse_order_of_words_s1.java | 12 +- .../sfo_58i_reverse_order_of_words_s2.java | 9 +- ...sfo_58ii_left_rotation_of_a_string_s2.java | 4 +- 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\346\211\223\345\215\260\344\273\216 1 \345\210\260\346\234\200\345\244\247\347\232\204 n \344\275\215\346\225\260.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 18. \345\210\240\351\231\244\351\223\276\350\241\250\347\232\204\350\212\202\347\202\271.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 19. \346\255\243\345\210\231\350\241\250\350\276\276\345\274\217\345\214\271\351\205\215.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 20. \350\241\250\347\244\272\346\225\260\345\200\274\347\232\204\345\255\227\347\254\246\344\270\262.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 21. \350\260\203\346\225\264\346\225\260\347\273\204\351\241\272\345\272\217\344\275\277\345\245\207\346\225\260\344\275\215\344\272\216\345\201\266\346\225\260\345\211\215\351\235\242.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 22. \351\223\276\350\241\250\344\270\255\345\200\222\346\225\260\347\254\254 k \344\270\252\350\212\202\347\202\271.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 24. \345\217\215\350\275\254\351\223\276\350\241\250.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 25. \345\220\210\345\271\266\344\270\244\344\270\252\346\216\222\345\272\217\347\232\204\351\223\276\350\241\250.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 26. \346\240\221\347\232\204\345\255\220\347\273\223\346\236\204.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 27. \344\272\214\345\217\211\346\240\221\347\232\204\351\225\234\345\203\217.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 28. \345\257\271\347\247\260\347\232\204\344\272\214\345\217\211\346\240\221.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 29. \351\241\272\346\227\266\351\222\210\346\211\223\345\215\260\347\237\251\351\230\265.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 30. \345\214\205\345\220\253 min \345\207\275\346\225\260\347\232\204\346\240\210.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 31. \346\240\210\347\232\204\345\216\213\345\205\245\343\200\201\345\274\271\345\207\272\345\272\217\345\210\227.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 32 - I. \344\273\216\344\270\212\345\210\260\344\270\213\346\211\223\345\215\260\344\272\214\345\217\211\346\240\221.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 32 - II. \344\273\216\344\270\212\345\210\260\344\270\213\346\211\223\345\215\260\344\272\214\345\217\211\346\240\221 II.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 32 - III. \344\273\216\344\270\212\345\210\260\344\270\213\346\211\223\345\215\260\344\272\214\345\217\211\346\240\221 III.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 33. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\347\232\204\345\220\216\345\272\217\351\201\215\345\216\206\345\272\217\345\210\227.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 34. \344\272\214\345\217\211\346\240\221\344\270\255\345\222\214\344\270\272\346\237\220\344\270\200\345\200\274\347\232\204\350\267\257\345\276\204.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 35. \345\244\215\346\235\202\351\223\276\350\241\250\347\232\204\345\244\215\345\210\266.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 36. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\344\270\216\345\217\214\345\220\221\351\223\276\350\241\250.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 37. \345\272\217\345\210\227\345\214\226\344\272\214\345\217\211\346\240\221.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 38. \345\255\227\347\254\246\344\270\262\347\232\204\346\216\222\345\210\227.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 39. \346\225\260\347\273\204\344\270\255\345\207\272\347\216\260\346\254\241\346\225\260\350\266\205\350\277\207\344\270\200\345\215\212\347\232\204\346\225\260\345\255\227.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 40. \346\234\200\345\260\217\347\232\204 k \344\270\252\346\225\260.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 41. \346\225\260\346\215\256\346\265\201\344\270\255\347\232\204\344\270\255\344\275\215\346\225\260.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 42. \350\277\236\347\273\255\345\255\220\346\225\260\347\273\204\347\232\204\346\234\200\345\244\247\345\222\214.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 43. 1\357\275\236n \346\225\264\346\225\260\344\270\255 1 \345\207\272\347\216\260\347\232\204\346\254\241\346\225\260.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 44. \346\225\260\345\255\227\345\272\217\345\210\227\344\270\255\346\237\220\344\270\200\344\275\215\347\232\204\346\225\260\345\255\227.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 45. \346\212\212\346\225\260\347\273\204\346\216\222\346\210\220\346\234\200\345\260\217\347\232\204\346\225\260.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 46. \346\212\212\346\225\260\345\255\227\347\277\273\350\257\221\346\210\220\345\255\227\347\254\246\344\270\262.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 47. \347\244\274\347\211\251\347\232\204\346\234\200\345\244\247\344\273\267\345\200\274.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 48. \346\234\200\351\225\277\344\270\215\345\220\253\351\207\215\345\244\215\345\255\227\347\254\246\347\232\204\345\255\220\345\255\227\347\254\246\344\270\262.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 49. \344\270\221\346\225\260.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 50. \347\254\254\344\270\200\344\270\252\345\217\252\345\207\272\347\216\260\344\270\200\346\254\241\347\232\204\345\255\227\347\254\246.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 51. \346\225\260\347\273\204\344\270\255\347\232\204\351\200\206\345\272\217\345\257\271.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 52. \344\270\244\344\270\252\351\223\276\350\241\250\347\232\204\347\254\254\344\270\200\344\270\252\345\205\254\345\205\261\350\212\202\347\202\271.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 53 - I. \345\234\250\346\216\222\345\272\217\346\225\260\347\273\204\344\270\255\346\237\245\346\211\276\346\225\260\345\255\227 I.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 53 - II. 0\357\275\236n-1 \344\270\255\347\274\272\345\244\261\347\232\204\346\225\260\345\255\227.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 54. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\347\232\204\347\254\254 k \345\244\247\350\212\202\347\202\271.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 55 - I. \344\272\214\345\217\211\346\240\221\347\232\204\346\267\261\345\272\246.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 55 - II. \345\271\263\350\241\241\344\272\214\345\217\211\346\240\221.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 56 - I. \346\225\260\347\273\204\344\270\255\346\225\260\345\255\227\345\207\272\347\216\260\347\232\204\346\254\241\346\225\260.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 56 - II. \346\225\260\347\273\204\344\270\255\346\225\260\345\255\227\345\207\272\347\216\260\347\232\204\346\254\241\346\225\260 II.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 57 - II. \345\222\214\344\270\272 s \347\232\204\350\277\236\347\273\255\346\255\243\346\225\260\345\272\217\345\210\227.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 57. \345\222\214\344\270\272 s \347\232\204\344\270\244\344\270\252\346\225\260\345\255\227.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 58 - I. \347\277\273\350\275\254\345\215\225\350\257\215\351\241\272\345\272\217.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 58 - II. \345\267\246\346\227\213\350\275\254\345\255\227\347\254\246\344\270\262.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 59 - I. \346\273\221\345\212\250\347\252\227\345\217\243\347\232\204\346\234\200\345\244\247\345\200\274.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 59 - II. \351\230\237\345\210\227\347\232\204\346\234\200\345\244\247\345\200\274.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 60. n \344\270\252\351\252\260\345\255\220\347\232\204\347\202\271\346\225\260.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 61. \346\211\221\345\205\213\347\211\214\344\270\255\347\232\204\351\241\272\345\255\220.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 62. \345\234\206\345\234\210\344\270\255\346\234\200\345\220\216\345\211\251\344\270\213\347\232\204\346\225\260\345\255\227.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 63. \350\202\241\347\245\250\347\232\204\346\234\200\345\244\247\345\210\251\346\266\246.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 64. \346\261\202 1 + 2 + \342\200\246 + n.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 65. \344\270\215\347\224\250\345\212\240\345\207\217\344\271\230\351\231\244\345\201\232\345\212\240\346\263\225.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 66. \346\236\204\345\273\272\344\271\230\347\247\257\346\225\260\347\273\204.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 67. \346\212\212\345\255\227\347\254\246\344\270\262\350\275\254\346\215\242\346\210\220\346\225\264\346\225\260.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 68 - I. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\347\232\204\346\234\200\350\277\221\345\205\254\345\205\261\347\245\226\345\205\210.md" create mode 100644 "sword_for_offer/docs/\345\211\221\346\214\207 Offer 68 - II. \344\272\214\345\217\211\346\240\221\347\232\204\346\234\200\350\277\221\345\205\254\345\205\261\347\245\226\345\205\210.md" diff --git a/sword_for_offer/codes/cpp/include/ListNode.hpp b/sword_for_offer/codes/cpp/include/ListNode.hpp index 4191649..fdb66f4 100644 --- a/sword_for_offer/codes/cpp/include/ListNode.hpp +++ b/sword_for_offer/codes/cpp/include/ListNode.hpp @@ -11,21 +11,22 @@ using namespace std; /** * @brief Definition for a singly-linked list node - * + * */ struct ListNode { int val; ListNode *next; - ListNode(int x) : val(x), next(nullptr) {} + ListNode(int x) : val(x), next(nullptr) { + } }; /** * @brief Generate a linked list with a vector - * - * @param list - * @return ListNode* + * + * @param list + * @return ListNode* */ -ListNode* vectorToLinkedList(vector list) { +ListNode *vectorToLinkedList(vector list) { ListNode *dum = new ListNode(0); ListNode *head = dum; for (int val : list) { @@ -37,12 +38,12 @@ ListNode* vectorToLinkedList(vector list) { /** * @brief Get a list node with specific value from a linked list - * - * @param head - * @param val - * @return ListNode* + * + * @param head + * @param val + * @return ListNode* */ -ListNode* getListNode(ListNode *head, int val) { +ListNode *getListNode(ListNode *head, int val) { while (head != nullptr && head->val != val) { head = head->next; } diff --git a/sword_for_offer/codes/cpp/include/PrintUtil.hpp b/sword_for_offer/codes/cpp/include/PrintUtil.hpp index 34e8b86..4bc89fb 100644 --- a/sword_for_offer/codes/cpp/include/PrintUtil.hpp +++ b/sword_for_offer/codes/cpp/include/PrintUtil.hpp @@ -6,192 +6,185 @@ #pragma once -#include -#include -#include #include "ListNode.hpp" #include "TreeNode.hpp" +#include +#include +#include class PrintUtil { - public: - /** - * @brief Find an element in a vector - * - * @tparam T - * @param vec - * @param ele - * @return int - */ - template - static int vecFind(const vector& vec, T ele) { - int j = INT_MAX; - for (int i = 0; i < vec.size(); i++) { - if (vec[i] == ele) { - j = i; - } + public: + /** + * @brief Find an element in a vector + * + * @tparam T + * @param vec + * @param ele + * @return int + */ + template static int vecFind(const vector &vec, T ele) { + int j = INT_MAX; + for (int i = 0; i < vec.size(); i++) { + if (vec[i] == ele) { + j = i; } - return j; } + return j; + } - /** - * @brief Concatenate a vector with a delim - * - * @tparam T - * @param delim - * @param vec - * @return string - */ - template - static string strJoin(const string& delim, const T& vec) { - ostringstream s; - for (const auto& i : vec) { - if (&i != &vec[0]) { - s << delim; - } - s << i; + /** + * @brief Concatenate a vector with a delim + * + * @tparam T + * @param delim + * @param vec + * @return string + */ + template static string strJoin(const string &delim, const T &vec) { + ostringstream s; + for (const auto &i : vec) { + if (&i != &vec[0]) { + s << delim; } - return s.str(); + s << i; } + return s.str(); + } - /** - * @brief Repeat a string for n times - * - * @param str - * @param n - * @return string - */ - static string strRepeat(string str, int n) { - ostringstream os; - for(int i = 0; i < n; i++) - os << str; - return os.str(); - } + /** + * @brief Repeat a string for n times + * + * @param str + * @param n + * @return string + */ + static string strRepeat(string str, int n) { + ostringstream os; + for (int i = 0; i < n; i++) + os << str; + return os.str(); + } - /** - * @brief Get the Vector String object - * - * @tparam T - * @param list - * @return string - */ - template - static string getVectorString(vector &list) { - return "[" + strJoin(", ", list) + "]"; - } + /** + * @brief Get the Vector String object + * + * @tparam T + * @param list + * @return string + */ + template static string getVectorString(vector &list) { + return "[" + strJoin(", ", list) + "]"; + } + + /** + * @brief Print a vector + * + * @tparam T + * @param list + */ + template static void printVector(vector &list) { + cout << getVectorString(list) << '\n'; + } + + /** + * @brief Print a vector matrix + * + * @tparam T + * @param matrix + */ + template static void printVectorMatrix(vector> &matrix) { + cout << "[" << '\n'; + for (vector &list : matrix) + cout << " " + getVectorString(list) + "," << '\n'; + cout << "]" << '\n'; + } - /** - * @brief Print a vector - * - * @tparam T - * @param list - */ - template - static void printVector(vector &list) { - cout << getVectorString(list) << '\n'; + /** + * @brief Print a linked list + * + * @param head + */ + static void printLinkedList(ListNode *head) { + vector list; + while (head != nullptr) { + list.push_back(head->val); + head = head->next; } - /** - * @brief Print a vector matrix - * - * @tparam T - * @param matrix - */ - template - static void printVectorMatrix(vector> &matrix) { - cout << "[" << '\n'; - for (vector &list : matrix) - cout << " " + getVectorString(list) + "," << '\n'; - cout << "]" << '\n'; + cout << strJoin(" -> ", list) << '\n'; + } + + /** + * @brief This tree printer is borrowed from TECHIE DELIGHT + * https://www.techiedelight.com/c-program-print-binary-tree/ + */ + struct Trunk { + Trunk *prev; + string str; + Trunk(Trunk *prev, string str) { + this->prev = prev; + this->str = str; } + }; - /** - * @brief Print a linked list - * - * @param head - */ - static void printLinkedList(ListNode *head) { - vector list; - while (head != nullptr) { - list.push_back(head->val); - head = head->next; - } - - cout << strJoin(" -> ", list) << '\n'; + /** + * @brief Helper function to print branches of the binary tree + * + * @param p + */ + static void showTrunks(Trunk *p) { + if (p == nullptr) { + return; } - /** - * @brief This tree printer is borrowed from TECHIE DELIGHT - * https://www.techiedelight.com/c-program-print-binary-tree/ - */ - struct Trunk { - Trunk *prev; - string str; - Trunk(Trunk *prev, string str) { - this->prev = prev; - this->str = str; - } - }; - - /** - * @brief Helper function to print branches of the binary tree - * - * @param p - */ - static void showTrunks(Trunk *p) { - if (p == nullptr) { - return; - } - - showTrunks(p->prev); - cout << p->str; + showTrunks(p->prev); + cout << p->str; + } + + /** + * @brief The interface of the tree printer + * + * @param root + */ + static void printTree(TreeNode *root) { + printTree(root, nullptr, false); + } + + /** + * @brief Print a binary tree + * + * @param root + * @param prev + * @param isLeft + */ + static void printTree(TreeNode *root, Trunk *prev, bool isLeft) { + if (root == nullptr) { + return; } - /** - * @brief The interface of the tree printer - * - * @param root - */ - static void printTree(TreeNode *root) { - printTree(root, nullptr, false); + string prev_str = " "; + Trunk *trunk = new Trunk(prev, prev_str); + + printTree(root->right, trunk, true); + + if (!prev) { + trunk->str = "———"; + } else if (isLeft) { + trunk->str = "/———"; + prev_str = " |"; + } else { + trunk->str = "\\———"; + prev->str = prev_str; } - /** - * @brief Print a binary tree - * - * @param root - * @param prev - * @param isLeft - */ - static void printTree(TreeNode *root, Trunk *prev, bool isLeft) { - if (root == nullptr) { - return; - } - - string prev_str = " "; - Trunk *trunk = new Trunk(prev, prev_str); - - printTree(root->right, trunk, true); - - if (!prev) { - trunk->str = "———"; - } - else if (isLeft) { - trunk->str = "/———"; - prev_str = " |"; - } - else { - trunk->str = "\\———"; - prev->str = prev_str; - } - - showTrunks(trunk); - cout << " " << root->val << endl; - - if (prev) { - prev->str = prev_str; - } - trunk->str = " |"; - - printTree(root->left, trunk, false); + showTrunks(trunk); + cout << " " << root->val << endl; + + if (prev) { + prev->str = prev_str; } + trunk->str = " |"; + + printTree(root->left, trunk, false); + } }; diff --git a/sword_for_offer/codes/cpp/include/TreeNode.hpp b/sword_for_offer/codes/cpp/include/TreeNode.hpp index a8e9d04..cab592b 100644 --- a/sword_for_offer/codes/cpp/include/TreeNode.hpp +++ b/sword_for_offer/codes/cpp/include/TreeNode.hpp @@ -8,35 +8,36 @@ /** * @brief Definition for a binary tree node - * + * */ struct TreeNode { int val; TreeNode *left; TreeNode *right; - TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} + TreeNode(int x) : val(x), left(nullptr), right(nullptr) { + } }; /** * @brief Generate a binary tree with a vector - * - * @param list - * @return TreeNode* + * + * @param list + * @return TreeNode* */ -TreeNode* vectorToTree(vector list) { +TreeNode *vectorToTree(vector list) { TreeNode *root = new TreeNode(list[0]); - queue que; + queue que; que.emplace(root); int i = 1; - while(!que.empty()) { + while (!que.empty()) { TreeNode *node = que.front(); que.pop(); - if(list[i] != INT_MAX) { + if (list[i] != INT_MAX) { node->left = new TreeNode(list[i]); que.emplace(node->left); } i++; - if(list[i] != INT_MAX) { + if (list[i] != INT_MAX) { node->right = new TreeNode(list[i]); que.emplace(node->right); } @@ -47,12 +48,12 @@ TreeNode* vectorToTree(vector list) { /** * @brief Get a tree node with specific value in a binary tree - * - * @param root - * @param val - * @return TreeNode* + * + * @param root + * @param val + * @return TreeNode* */ -TreeNode* getTreeNode(TreeNode *root, int val) { +TreeNode *getTreeNode(TreeNode *root, int val) { if (root == nullptr) return nullptr; if (root->val == val) diff --git a/sword_for_offer/codes/cpp/include/include.hpp b/sword_for_offer/codes/cpp/include/include.hpp index 8e84583..5ca1e7b 100644 --- a/sword_for_offer/codes/cpp/include/include.hpp +++ b/sword_for_offer/codes/cpp/include/include.hpp @@ -6,18 +6,18 @@ #pragma once +#include #include -#include -#include -#include #include -#include +#include +#include +#include #include #include -#include +#include #include "ListNode.hpp" -#include "TreeNode.hpp" #include "PrintUtil.hpp" +#include "TreeNode.hpp" using namespace std; diff --git a/sword_for_offer/codes/cpp/sfo_03_find_duplicate_numbers_in_an_array_s2/sfo_03_find_duplicate_numbers_in_an_array_s2.cpp b/sword_for_offer/codes/cpp/sfo_03_find_duplicate_numbers_in_an_array_s2/sfo_03_find_duplicate_numbers_in_an_array_s2.cpp index 7975201..4f56d52 100644 --- a/sword_for_offer/codes/cpp/sfo_03_find_duplicate_numbers_in_an_array_s2/sfo_03_find_duplicate_numbers_in_an_array_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_03_find_duplicate_numbers_in_an_array_s2/sfo_03_find_duplicate_numbers_in_an_array_s2.cpp @@ -1,24 +1,24 @@ /* -* File: sfo_03_find_duplicate_numbers_in_an_array_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_03_find_duplicate_numbers_in_an_array_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - int findRepeatNumber(vector& nums) { + public: + int findRepeatNumber(vector &nums) { int i = 0; - while(i < nums.size()) { - if(nums[i] == i) { + while (i < nums.size()) { + if (nums[i] == i) { i++; continue; } - if(nums[nums[i]] == nums[i]) + if (nums[nums[i]] == nums[i]) return nums[i]; - swap(nums[i],nums[nums[i]]); + swap(nums[i], nums[nums[i]]); } return -1; } @@ -26,11 +26,11 @@ class Solution { int main() { // ======= Test Case ======= - vector nums = { 2, 3, 1, 0, 2, 5, 3 }; + vector nums = {2, 3, 1, 0, 2, 5, 3}; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->findRepeatNumber(nums); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_04_find_a_number_in_2d_matrix_s1/sfo_04_find_a_number_in_2d_matrix_s1.cpp b/sword_for_offer/codes/cpp/sfo_04_find_a_number_in_2d_matrix_s1/sfo_04_find_a_number_in_2d_matrix_s1.cpp index af7a76c..c329958 100644 --- a/sword_for_offer/codes/cpp/sfo_04_find_a_number_in_2d_matrix_s1/sfo_04_find_a_number_in_2d_matrix_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_04_find_a_number_in_2d_matrix_s1/sfo_04_find_a_number_in_2d_matrix_s1.cpp @@ -1,21 +1,23 @@ /* -* File: sfo_04_find_a_number_in_2d_matrix_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_04_find_a_number_in_2d_matrix_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - bool findNumberIn2DArray(vector>& matrix, int target) { + public: + bool findNumberIn2DArray(vector> &matrix, int target) { int i = matrix.size() - 1, j = 0; - while(i >= 0 && j < matrix[0].size()) - { - if(matrix[i][j] > target) i--; - else if(matrix[i][j] < target) j++; - else return true; + while (i >= 0 && j < matrix[0].size()) { + if (matrix[i][j] > target) + i--; + else if (matrix[i][j] < target) + j++; + else + return true; } return false; } @@ -24,17 +26,12 @@ class Solution { int main() { // ======= Test Case ======= vector> matrix = { - { 1, 4, 7, 11, 15 }, - { 2, 5, 8, 12, 19 }, - { 3, 6, 9, 16, 22 }, - { 10, 13, 14, 17, 24 }, - { 18, 21, 23, 26, 30 } -}; + {1, 4, 7, 11, 15}, {2, 5, 8, 12, 19}, {3, 6, 9, 16, 22}, {10, 13, 14, 17, 24}, {18, 21, 23, 26, 30}}; int target = 5; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); bool res = slt->findNumberIn2DArray(matrix, target); cout << boolalpha << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_05_replace_spaces_s1/sfo_05_replace_spaces_s1.cpp b/sword_for_offer/codes/cpp/sfo_05_replace_spaces_s1/sfo_05_replace_spaces_s1.cpp index 50cab0e..a467d7d 100644 --- a/sword_for_offer/codes/cpp/sfo_05_replace_spaces_s1/sfo_05_replace_spaces_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_05_replace_spaces_s1/sfo_05_replace_spaces_s1.cpp @@ -1,24 +1,25 @@ /* -* File: sfo_05_replace_spaces_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_05_replace_spaces_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: string replaceSpace(string s) { int count = 0, len = s.size(); // 统计空格数量 for (char c : s) { - if (c == ' ') count++; + if (c == ' ') + count++; } // 修改 s 长度 s.resize(len + 2 * count); // 倒序遍历修改 - for(int i = len - 1, j = s.size() - 1; i < j; i--, j--) { + for (int i = len - 1, j = s.size() - 1; i < j; i--, j--) { if (s[i] != ' ') s[j] = s[i]; else { @@ -36,9 +37,9 @@ int main() { // ======= Test Case ======= string s = "We are happy."; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); string res = slt->replaceSpace(s); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_06_print_a_linked_list_in_reverse_order_s1/sfo_06_print_a_linked_list_in_reverse_order_s1.cpp b/sword_for_offer/codes/cpp/sfo_06_print_a_linked_list_in_reverse_order_s1/sfo_06_print_a_linked_list_in_reverse_order_s1.cpp index 903fcd0..1edbf2c 100644 --- a/sword_for_offer/codes/cpp/sfo_06_print_a_linked_list_in_reverse_order_s1/sfo_06_print_a_linked_list_in_reverse_order_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_06_print_a_linked_list_in_reverse_order_s1/sfo_06_print_a_linked_list_in_reverse_order_s1.cpp @@ -1,22 +1,24 @@ /* -* File: sfo_06_print_a_linked_list_in_reverse_order_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_06_print_a_linked_list_in_reverse_order_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - vector reversePrint(ListNode* head) { + public: + vector reversePrint(ListNode *head) { recur(head); return res; } -private: + + private: vector res; - void recur(ListNode* head) { - if(head == nullptr) return; + void recur(ListNode *head) { + if (head == nullptr) + return; recur(head->next); res.push_back(head->val); } @@ -24,11 +26,11 @@ class Solution { int main() { // ======= Test Case ======= - ListNode* head = vectorToLinkedList(vector { 1, 3, 2 }); + ListNode *head = vectorToLinkedList(vector{1, 3, 2}); // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); vector res = slt->reversePrint(head); PrintUtil::printVector(res); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_06_print_a_linked_list_in_reverse_order_s2/sfo_06_print_a_linked_list_in_reverse_order_s2.cpp b/sword_for_offer/codes/cpp/sfo_06_print_a_linked_list_in_reverse_order_s2/sfo_06_print_a_linked_list_in_reverse_order_s2.cpp index 6e4cfce..ba69c3e 100644 --- a/sword_for_offer/codes/cpp/sfo_06_print_a_linked_list_in_reverse_order_s2/sfo_06_print_a_linked_list_in_reverse_order_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_06_print_a_linked_list_in_reverse_order_s2/sfo_06_print_a_linked_list_in_reverse_order_s2.cpp @@ -1,22 +1,22 @@ /* -* File: sfo_06_print_a_linked_list_in_reverse_order_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_06_print_a_linked_list_in_reverse_order_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - vector reversePrint(ListNode* head) { + public: + vector reversePrint(ListNode *head) { stack stk; - while(head != nullptr) { + while (head != nullptr) { stk.push(head->val); head = head->next; } vector res; - while(!stk.empty()) { + while (!stk.empty()) { res.push_back(stk.top()); stk.pop(); } @@ -26,11 +26,11 @@ class Solution { int main() { // ======= Test Case ======= - ListNode* head = vectorToLinkedList(vector { 1, 3, 2 }); + ListNode *head = vectorToLinkedList(vector{1, 3, 2}); // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); vector res = slt->reversePrint(head); PrintUtil::printVector(res); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_07_reconstruct_binary_tree_s1/sfo_07_reconstruct_binary_tree_s1.cpp b/sword_for_offer/codes/cpp/sfo_07_reconstruct_binary_tree_s1/sfo_07_reconstruct_binary_tree_s1.cpp index 4375d57..f795a67 100644 --- a/sword_for_offer/codes/cpp/sfo_07_reconstruct_binary_tree_s1/sfo_07_reconstruct_binary_tree_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_07_reconstruct_binary_tree_s1/sfo_07_reconstruct_binary_tree_s1.cpp @@ -1,26 +1,28 @@ /* -* File: sfo_07_reconstruct_binary_tree_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_07_reconstruct_binary_tree_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - TreeNode* buildTree(vector& preorder, vector& inorder) { + public: + TreeNode *buildTree(vector &preorder, vector &inorder) { this->preorder = preorder; - for(int i = 0; i < inorder.size(); i++) + for (int i = 0; i < inorder.size(); i++) dic[inorder[i]] = i; return recur(0, 0, inorder.size() - 1); } -private: + + private: vector preorder; unordered_map dic; - TreeNode* recur(int root, int left, int right) { - if(left > right) return nullptr; // 递归终止 - TreeNode* node = new TreeNode(preorder[root]); // 建立根节点 + TreeNode *recur(int root, int left, int right) { + if (left > right) + return nullptr; // 递归终止 + TreeNode *node = new TreeNode(preorder[root]); // 建立根节点 int i = dic[preorder[root]]; // 划分根节点、左子树、右子树 node->left = recur(root + 1, left, i - 1); // 开启左子树递归 node->right = recur(root + i - left + 1, i + 1, right); // 开启右子树递归 @@ -30,12 +32,12 @@ class Solution { int main() { // ======= Test Case ======= - vector preorder = { 3, 9, 20, 15, 7 }; - vector inorder = { 9, 3, 15, 20, 7 }; + vector preorder = {3, 9, 20, 15, 7}; + vector inorder = {9, 3, 15, 20, 7}; // ====== Driver Code ====== - Solution* slt = new Solution(); - TreeNode* res = slt->buildTree(preorder, inorder); + Solution *slt = new Solution(); + TreeNode *res = slt->buildTree(preorder, inorder); PrintUtil::printTree(res); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_09_implement_a_queue_using_two_stacks_s1/sfo_09_implement_a_queue_using_two_stacks_s1.cpp b/sword_for_offer/codes/cpp/sfo_09_implement_a_queue_using_two_stacks_s1/sfo_09_implement_a_queue_using_two_stacks_s1.cpp index 4c7da99..3b1f923 100644 --- a/sword_for_offer/codes/cpp/sfo_09_implement_a_queue_using_two_stacks_s1/sfo_09_implement_a_queue_using_two_stacks_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_09_implement_a_queue_using_two_stacks_s1/sfo_09_implement_a_queue_using_two_stacks_s1.cpp @@ -1,27 +1,29 @@ /* -* File: sfo_09_implement_a_queue_using_two_stacks_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_09_implement_a_queue_using_two_stacks_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class CQueue { -public: + public: stack A, B; - CQueue() {} + CQueue() { + } void appendTail(int value) { A.push(value); } int deleteHead() { - if(!B.empty()) { + if (!B.empty()) { int tmp = B.top(); B.pop(); return tmp; } - if(A.empty()) return -1; - while(!A.empty()) { + if (A.empty()) + return -1; + while (!A.empty()) { int tmp = A.top(); A.pop(); B.push(tmp); @@ -35,7 +37,7 @@ class CQueue { int main() { // ====== Driver Code ====== vector res; - CQueue* cQueue = new CQueue(); + CQueue *cQueue = new CQueue(); res.push_back(cQueue->deleteHead()); cQueue->appendTail(5); cQueue->appendTail(2); diff --git a/sword_for_offer/codes/cpp/sfo_10i_fibonacci_numbers_s1/sfo_10i_fibonacci_numbers_s1.cpp b/sword_for_offer/codes/cpp/sfo_10i_fibonacci_numbers_s1/sfo_10i_fibonacci_numbers_s1.cpp index c8bcb4e..33d67c5 100644 --- a/sword_for_offer/codes/cpp/sfo_10i_fibonacci_numbers_s1/sfo_10i_fibonacci_numbers_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_10i_fibonacci_numbers_s1/sfo_10i_fibonacci_numbers_s1.cpp @@ -1,17 +1,17 @@ /* -* File: sfo_10i_fibonacci_numbers_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_10i_fibonacci_numbers_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: int fib(int n) { int a = 0, b = 1, sum; - for(int i = 0; i < n; i++){ + for (int i = 0; i < n; i++) { sum = (a + b) % 1000000007; a = b; b = sum; @@ -24,9 +24,9 @@ int main() { // ======= Test Case ======= int n = 5; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->fib(n); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_10ii_frog_jump_s1/sfo_10ii_frog_jump_s1.cpp b/sword_for_offer/codes/cpp/sfo_10ii_frog_jump_s1/sfo_10ii_frog_jump_s1.cpp index edee18d..919174b 100644 --- a/sword_for_offer/codes/cpp/sfo_10ii_frog_jump_s1/sfo_10ii_frog_jump_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_10ii_frog_jump_s1/sfo_10ii_frog_jump_s1.cpp @@ -1,17 +1,17 @@ /* -* File: sfo_10ii_frog_jump_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_10ii_frog_jump_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: int numWays(int n) { int a = 1, b = 1, sum; - for(int i = 0; i < n; i++){ + for (int i = 0; i < n; i++) { sum = (a + b) % 1000000007; a = b; b = sum; @@ -24,9 +24,9 @@ int main() { // ======= Test Case ======= int n = 7; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->numWays(n); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_11_find_minimum_in_rotated_sorted_array_s1/sfo_11_find_minimum_in_rotated_sorted_array_s1.cpp b/sword_for_offer/codes/cpp/sfo_11_find_minimum_in_rotated_sorted_array_s1/sfo_11_find_minimum_in_rotated_sorted_array_s1.cpp index b23919d..87c6f6a 100644 --- a/sword_for_offer/codes/cpp/sfo_11_find_minimum_in_rotated_sorted_array_s1/sfo_11_find_minimum_in_rotated_sorted_array_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_11_find_minimum_in_rotated_sorted_array_s1/sfo_11_find_minimum_in_rotated_sorted_array_s1.cpp @@ -1,21 +1,24 @@ /* -* File: sfo_11_find_minimum_in_rotated_sorted_array_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_11_find_minimum_in_rotated_sorted_array_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - int minArray(vector& numbers) { + public: + int minArray(vector &numbers) { int i = 0, j = numbers.size() - 1; while (i < j) { int m = (i + j) / 2; - if (numbers[m] > numbers[j]) i = m + 1; - else if (numbers[m] < numbers[j]) j = m; - else j--; + if (numbers[m] > numbers[j]) + i = m + 1; + else if (numbers[m] < numbers[j]) + j = m; + else + j--; } return numbers[i]; } @@ -23,11 +26,11 @@ class Solution { int main() { // ======= Test Case ======= - vector numbers = { 3, 4, 5, 1, 2 }; + vector numbers = {3, 4, 5, 1, 2}; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->minArray(numbers); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_11_find_minimum_in_rotated_sorted_array_s2/sfo_11_find_minimum_in_rotated_sorted_array_s2.cpp b/sword_for_offer/codes/cpp/sfo_11_find_minimum_in_rotated_sorted_array_s2/sfo_11_find_minimum_in_rotated_sorted_array_s2.cpp index a910d92..88a2679 100644 --- a/sword_for_offer/codes/cpp/sfo_11_find_minimum_in_rotated_sorted_array_s2/sfo_11_find_minimum_in_rotated_sorted_array_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_11_find_minimum_in_rotated_sorted_array_s2/sfo_11_find_minimum_in_rotated_sorted_array_s2.cpp @@ -1,24 +1,27 @@ /* -* File: sfo_11_find_minimum_in_rotated_sorted_array_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_11_find_minimum_in_rotated_sorted_array_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - int minArray(vector& numbers) { + public: + int minArray(vector &numbers) { int i = 0, j = numbers.size() - 1; while (i < j) { int m = (i + j) / 2; - if (numbers[m] > numbers[j]) i = m + 1; - else if (numbers[m] < numbers[j]) j = m; + if (numbers[m] > numbers[j]) + i = m + 1; + else if (numbers[m] < numbers[j]) + j = m; else { int x = i; - for(int k = i + 1; k < j; k++) { - if(numbers[k] < numbers[x]) x = k; + for (int k = i + 1; k < j; k++) { + if (numbers[k] < numbers[x]) + x = k; } return numbers[x]; } @@ -29,11 +32,11 @@ class Solution { int main() { // ======= Test Case ======= - vector numbers = { 3, 4, 5, 1, 2 }; + vector numbers = {3, 4, 5, 1, 2}; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->minArray(numbers); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_12_word_search_s1/sfo_12_word_search_s1.cpp b/sword_for_offer/codes/cpp/sfo_12_word_search_s1/sfo_12_word_search_s1.cpp index 2018e0e..c299d5d 100644 --- a/sword_for_offer/codes/cpp/sfo_12_word_search_s1/sfo_12_word_search_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_12_word_search_s1/sfo_12_word_search_s1.cpp @@ -1,32 +1,36 @@ /* -* File: sfo_12_word_search_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_12_word_search_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - bool exist(vector>& board, string word) { + public: + bool exist(vector> &board, string word) { rows = board.size(); cols = board[0].size(); - for(int i = 0; i < rows; i++) { - for(int j = 0; j < cols; j++) { - if(dfs(board, word, i, j, 0)) return true; + for (int i = 0; i < rows; i++) { + for (int j = 0; j < cols; j++) { + if (dfs(board, word, i, j, 0)) + return true; } } return false; } -private: + + private: int rows, cols; - bool dfs(vector>& board, string word, int i, int j, int k) { - if(i >= rows || i < 0 || j >= cols || j < 0 || board[i][j] != word[k]) return false; - if(k == word.size() - 1) return true; + bool dfs(vector> &board, string word, int i, int j, int k) { + if (i >= rows || i < 0 || j >= cols || j < 0 || board[i][j] != word[k]) + return false; + if (k == word.size() - 1) + return true; board[i][j] = '\0'; - bool res = dfs(board, word, i + 1, j, k + 1) || dfs(board, word, i - 1, j, k + 1) || - dfs(board, word, i, j + 1, k + 1) || dfs(board, word, i , j - 1, k + 1); + bool res = dfs(board, word, i + 1, j, k + 1) || dfs(board, word, i - 1, j, k + 1) || + dfs(board, word, i, j + 1, k + 1) || dfs(board, word, i, j - 1, k + 1); board[i][j] = word[k]; return res; } @@ -34,16 +38,12 @@ class Solution { int main() { // ======= Test Case ======= - vector> board = { - { 'A', 'B', 'C', 'E' }, - { 'S', 'F', 'C', 'S' }, - { 'A', 'D', 'E', 'E' } -}; + vector> board = {{'A', 'B', 'C', 'E'}, {'S', 'F', 'C', 'S'}, {'A', 'D', 'E', 'E'}}; string word = "ABCCED"; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); bool res = slt->exist(board, word); cout << boolalpha << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_13_range_of_motion_of_a_robot_s1/sfo_13_range_of_motion_of_a_robot_s1.cpp b/sword_for_offer/codes/cpp/sfo_13_range_of_motion_of_a_robot_s1/sfo_13_range_of_motion_of_a_robot_s1.cpp index 85ff655..49e6353 100644 --- a/sword_for_offer/codes/cpp/sfo_13_range_of_motion_of_a_robot_s1/sfo_13_range_of_motion_of_a_robot_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_13_range_of_motion_of_a_robot_s1/sfo_13_range_of_motion_of_a_robot_s1.cpp @@ -1,24 +1,26 @@ /* -* File: sfo_13_range_of_motion_of_a_robot_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_13_range_of_motion_of_a_robot_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: int movingCount(int m, int n, int k) { vector> visited(m, vector(n, 0)); return dfs(0, 0, 0, 0, visited, m, n, k); } -private: + + private: int dfs(int i, int j, int si, int sj, vector> &visited, int m, int n, int k) { - if(i >= m || j >= n || k < si + sj || visited[i][j]) return 0; + if (i >= m || j >= n || k < si + sj || visited[i][j]) + return 0; visited[i][j] = true; return 1 + dfs(i + 1, j, (i + 1) % 10 != 0 ? si + 1 : si - 8, sj, visited, m, n, k) + - dfs(i, j + 1, si, (j + 1) % 10 != 0 ? sj + 1 : sj - 8, visited, m, n, k); + dfs(i, j + 1, si, (j + 1) % 10 != 0 ? sj + 1 : sj - 8, visited, m, n, k); } }; @@ -28,9 +30,9 @@ int main() { int n = 3; int k = 1; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->movingCount(m, n, k); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_13_range_of_motion_of_a_robot_s2/sfo_13_range_of_motion_of_a_robot_s2.cpp b/sword_for_offer/codes/cpp/sfo_13_range_of_motion_of_a_robot_s2/sfo_13_range_of_motion_of_a_robot_s2.cpp index 32fb20a..330cc26 100644 --- a/sword_for_offer/codes/cpp/sfo_13_range_of_motion_of_a_robot_s2/sfo_13_range_of_motion_of_a_robot_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_13_range_of_motion_of_a_robot_s2/sfo_13_range_of_motion_of_a_robot_s2.cpp @@ -1,28 +1,29 @@ /* -* File: sfo_13_range_of_motion_of_a_robot_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_13_range_of_motion_of_a_robot_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: int movingCount(int m, int n, int k) { vector> visited(m, vector(n, 0)); int res = 0; queue> que; - que.push({ 0, 0, 0, 0 }); - while(que.size() > 0) { + que.push({0, 0, 0, 0}); + while (que.size() > 0) { vector x = que.front(); que.pop(); int i = x[0], j = x[1], si = x[2], sj = x[3]; - if(i >= m || j >= n || k < si + sj || visited[i][j]) continue; + if (i >= m || j >= n || k < si + sj || visited[i][j]) + continue; visited[i][j] = true; res++; - que.push({ i + 1, j, (i + 1) % 10 != 0 ? si + 1 : si - 8, sj }); - que.push({ i, j + 1, si, (j + 1) % 10 != 0 ? sj + 1 : sj - 8 }); + que.push({i + 1, j, (i + 1) % 10 != 0 ? si + 1 : si - 8, sj}); + que.push({i, j + 1, si, (j + 1) % 10 != 0 ? sj + 1 : sj - 8}); } return res; } @@ -34,9 +35,9 @@ int main() { int n = 3; int k = 1; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->movingCount(m, n, k); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_14i_cut_the_rope_i_s1/sfo_14i_cut_the_rope_i_s1.cpp b/sword_for_offer/codes/cpp/sfo_14i_cut_the_rope_i_s1/sfo_14i_cut_the_rope_i_s1.cpp index 5c235ee..da0b46e 100644 --- a/sword_for_offer/codes/cpp/sfo_14i_cut_the_rope_i_s1/sfo_14i_cut_the_rope_i_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_14i_cut_the_rope_i_s1/sfo_14i_cut_the_rope_i_s1.cpp @@ -1,19 +1,22 @@ /* -* File: sfo_14i_cut_the_rope_i_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_14i_cut_the_rope_i_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: int cuttingRope(int n) { - if(n <= 3) return n - 1; + if (n <= 3) + return n - 1; int a = n / 3, b = n % 3; - if(b == 0) return pow(3, a); - if(b == 1) return pow(3, a - 1) * 4; + if (b == 0) + return pow(3, a); + if (b == 1) + return pow(3, a - 1) * 4; return pow(3, a) * 2; } }; @@ -22,9 +25,9 @@ int main() { // ======= Test Case ======= int n = 10; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->cuttingRope(n); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_14ii_cut_the_rope_ii_s1/sfo_14ii_cut_the_rope_ii_s1.cpp b/sword_for_offer/codes/cpp/sfo_14ii_cut_the_rope_ii_s1/sfo_14ii_cut_the_rope_ii_s1.cpp index b9cc1ec..2581ffe 100644 --- a/sword_for_offer/codes/cpp/sfo_14ii_cut_the_rope_ii_s1/sfo_14ii_cut_the_rope_ii_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_14ii_cut_the_rope_ii_s1/sfo_14ii_cut_the_rope_ii_s1.cpp @@ -1,24 +1,28 @@ /* -* File: sfo_14ii_cut_the_rope_ii_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_14ii_cut_the_rope_ii_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: int cuttingRope(int n) { - if(n <= 3) return n - 1; + if (n <= 3) + return n - 1; int b = n % 3, p = 1000000007; long rem = 1, x = 3; - for(int a = n / 3 - 1; a > 0; a /= 2) { - if(a % 2 == 1) rem = (rem * x) % p; + for (int a = n / 3 - 1; a > 0; a /= 2) { + if (a % 2 == 1) + rem = (rem * x) % p; x = (x * x) % p; } - if(b == 0) return (int)(rem * 3 % p); - if(b == 1) return (int)(rem * 4 % p); + if (b == 0) + return (int)(rem * 3 % p); + if (b == 1) + return (int)(rem * 4 % p); return (int)(rem * 6 % p); } }; @@ -27,9 +31,9 @@ int main() { // ======= Test Case ======= int n = 10; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->cuttingRope(n); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_15_number_of_1_bits_s1/sfo_15_number_of_1_bits_s1.cpp b/sword_for_offer/codes/cpp/sfo_15_number_of_1_bits_s1/sfo_15_number_of_1_bits_s1.cpp index dc8439c..7d7206d 100644 --- a/sword_for_offer/codes/cpp/sfo_15_number_of_1_bits_s1/sfo_15_number_of_1_bits_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_15_number_of_1_bits_s1/sfo_15_number_of_1_bits_s1.cpp @@ -1,17 +1,17 @@ /* -* File: sfo_15_number_of_1_bits_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_15_number_of_1_bits_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: int hammingWeight(uint32_t n) { unsigned int res = 0; // c++ 使用无符号数 - while(n != 0) { + while (n != 0) { res += n & 1; n >>= 1; } @@ -23,9 +23,9 @@ int main() { // ======= Test Case ======= uint32_t n = 11; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->hammingWeight(n); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_15_number_of_1_bits_s2/sfo_15_number_of_1_bits_s2.cpp b/sword_for_offer/codes/cpp/sfo_15_number_of_1_bits_s2/sfo_15_number_of_1_bits_s2.cpp index 8b7d8a9..8d4f2d1 100644 --- a/sword_for_offer/codes/cpp/sfo_15_number_of_1_bits_s2/sfo_15_number_of_1_bits_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_15_number_of_1_bits_s2/sfo_15_number_of_1_bits_s2.cpp @@ -1,17 +1,17 @@ /* -* File: sfo_15_number_of_1_bits_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_15_number_of_1_bits_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: int hammingWeight(uint32_t n) { int res = 0; - while(n != 0) { + while (n != 0) { res++; n &= n - 1; } @@ -23,9 +23,9 @@ int main() { // ======= Test Case ======= uint32_t n = 11; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->hammingWeight(n); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_16_powers_of_integers_s1/sfo_16_powers_of_integers_s1.cpp b/sword_for_offer/codes/cpp/sfo_16_powers_of_integers_s1/sfo_16_powers_of_integers_s1.cpp index c8e0edb..16f47c6 100644 --- a/sword_for_offer/codes/cpp/sfo_16_powers_of_integers_s1/sfo_16_powers_of_integers_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_16_powers_of_integers_s1/sfo_16_powers_of_integers_s1.cpp @@ -1,24 +1,26 @@ /* -* File: sfo_16_powers_of_integers_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_16_powers_of_integers_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: double myPow(double x, int n) { - if(x == 0.0f) return 0.0; + if (x == 0.0f) + return 0.0; long b = n; double res = 1.0; - if(b < 0) { + if (b < 0) { x = 1 / x; b = -b; } - while(b > 0) { - if((b & 1) == 1) res *= x; + while (b > 0) { + if ((b & 1) == 1) + res *= x; x *= x; b >>= 1; } @@ -31,9 +33,9 @@ int main() { double x = 2.0; int n = 10; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); double res = slt->myPow(x, n); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_18_delete_a_node_from_a_linked_list_s1/sfo_18_delete_a_node_from_a_linked_list_s1.cpp b/sword_for_offer/codes/cpp/sfo_18_delete_a_node_from_a_linked_list_s1/sfo_18_delete_a_node_from_a_linked_list_s1.cpp index 37dbc00..7591fce 100644 --- a/sword_for_offer/codes/cpp/sfo_18_delete_a_node_from_a_linked_list_s1/sfo_18_delete_a_node_from_a_linked_list_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_18_delete_a_node_from_a_linked_list_s1/sfo_18_delete_a_node_from_a_linked_list_s1.cpp @@ -1,34 +1,36 @@ /* -* File: sfo_18_delete_a_node_from_a_linked_list_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_18_delete_a_node_from_a_linked_list_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - ListNode* deleteNode(ListNode* head, int val) { - if(head->val == val) return head->next; + public: + ListNode *deleteNode(ListNode *head, int val) { + if (head->val == val) + return head->next; ListNode *pre = head, *cur = head->next; - while(cur != nullptr && cur->val != val) { + while (cur != nullptr && cur->val != val) { pre = cur; cur = cur->next; } - if(cur != nullptr) pre->next = cur->next; + if (cur != nullptr) + pre->next = cur->next; return head; } }; int main() { // ======= Test Case ======= - ListNode* head = vectorToLinkedList(vector { 4, 5, 1, 9 }); + ListNode *head = vectorToLinkedList(vector{4, 5, 1, 9}); int val = 5; // ====== Driver Code ====== - Solution* slt = new Solution(); - ListNode* res = slt->deleteNode(head, val); + Solution *slt = new Solution(); + ListNode *res = slt->deleteNode(head, val); PrintUtil::printLinkedList(res); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_19_regular_expression_matching_s1/sfo_19_regular_expression_matching_s1.cpp b/sword_for_offer/codes/cpp/sfo_19_regular_expression_matching_s1/sfo_19_regular_expression_matching_s1.cpp index f8c9aaa..31e8727 100644 --- a/sword_for_offer/codes/cpp/sfo_19_regular_expression_matching_s1/sfo_19_regular_expression_matching_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_19_regular_expression_matching_s1/sfo_19_regular_expression_matching_s1.cpp @@ -1,25 +1,24 @@ /* -* File: sfo_19_regular_expression_matching_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_19_regular_expression_matching_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: bool isMatch(string s, string p) { int m = s.size() + 1, n = p.size() + 1; vector> dp(m, vector(n, false)); dp[0][0] = true; - for(int j = 2; j < n; j += 2) + for (int j = 2; j < n; j += 2) dp[0][j] = dp[0][j - 2] && p[j - 1] == '*'; - for(int i = 1; i < m; i++) { - for(int j = 1; j < n; j++) { - dp[i][j] = p[j - 1] == '*' ? - dp[i][j - 2] || dp[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == '.'): - dp[i - 1][j - 1] && (p[j - 1] == '.' || s[i - 1] == p[j - 1]); + for (int i = 1; i < m; i++) { + for (int j = 1; j < n; j++) { + dp[i][j] = p[j - 1] == '*' ? dp[i][j - 2] || dp[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == '.') + : dp[i - 1][j - 1] && (p[j - 1] == '.' || s[i - 1] == p[j - 1]); } } return dp[m - 1][n - 1]; @@ -31,9 +30,9 @@ int main() { string s = "mississippi"; string p = "mis*is*p*."; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); bool res = slt->isMatch(s, p); cout << boolalpha << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_19_regular_expression_matching_s2/sfo_19_regular_expression_matching_s2.cpp b/sword_for_offer/codes/cpp/sfo_19_regular_expression_matching_s2/sfo_19_regular_expression_matching_s2.cpp index abad706..da3fea2 100644 --- a/sword_for_offer/codes/cpp/sfo_19_regular_expression_matching_s2/sfo_19_regular_expression_matching_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_19_regular_expression_matching_s2/sfo_19_regular_expression_matching_s2.cpp @@ -1,31 +1,36 @@ /* -* File: sfo_19_regular_expression_matching_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_19_regular_expression_matching_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: bool isMatch(string s, string p) { int m = s.size() + 1, n = p.size() + 1; vector> dp(m, vector(n, false)); dp[0][0] = true; // 初始化首行 - for(int j = 2; j < n; j += 2) + for (int j = 2; j < n; j += 2) dp[0][j] = dp[0][j - 2] && p[j - 1] == '*'; // 状态转移 - for(int i = 1; i < m; i++) { - for(int j = 1; j < n; j++) { - if(p[j - 1] == '*') { - if(dp[i][j - 2]) dp[i][j] = true; // 1. - else if(dp[i - 1][j] && s[i - 1] == p[j - 2]) dp[i][j] = true; // 2. - else if(dp[i - 1][j] && p[j - 2] == '.') dp[i][j] = true; // 3. + for (int i = 1; i < m; i++) { + for (int j = 1; j < n; j++) { + if (p[j - 1] == '*') { + if (dp[i][j - 2]) + dp[i][j] = true; // 1. + else if (dp[i - 1][j] && s[i - 1] == p[j - 2]) + dp[i][j] = true; // 2. + else if (dp[i - 1][j] && p[j - 2] == '.') + dp[i][j] = true; // 3. } else { - if(dp[i - 1][j - 1] && s[i - 1] == p[j - 1]) dp[i][j] = true; // 1. - else if(dp[i - 1][j - 1] && p[j - 1] == '.') dp[i][j] = true; // 2. + if (dp[i - 1][j - 1] && s[i - 1] == p[j - 1]) + dp[i][j] = true; // 1. + else if (dp[i - 1][j - 1] && p[j - 1] == '.') + dp[i][j] = true; // 2. } } } @@ -38,9 +43,9 @@ int main() { string s = "mississippi"; string p = "mis*is*p*."; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); bool res = slt->isMatch(s, p); cout << boolalpha << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_21_adjust_the_order_of_numbers_in_an_array_s1/sfo_21_adjust_the_order_of_numbers_in_an_array_s1.cpp b/sword_for_offer/codes/cpp/sfo_21_adjust_the_order_of_numbers_in_an_array_s1/sfo_21_adjust_the_order_of_numbers_in_an_array_s1.cpp index cc74780..433b940 100644 --- a/sword_for_offer/codes/cpp/sfo_21_adjust_the_order_of_numbers_in_an_array_s1/sfo_21_adjust_the_order_of_numbers_in_an_array_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_21_adjust_the_order_of_numbers_in_an_array_s1/sfo_21_adjust_the_order_of_numbers_in_an_array_s1.cpp @@ -1,20 +1,21 @@ /* -* File: sfo_21_adjust_the_order_of_numbers_in_an_array_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_21_adjust_the_order_of_numbers_in_an_array_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - vector exchange(vector& nums) { + public: + vector exchange(vector &nums) { int i = 0, j = nums.size() - 1; - while (i < j) - { - while(i < j && (nums[i] & 1) == 1) i++; - while(i < j && (nums[j] & 1) == 0) j--; + while (i < j) { + while (i < j && (nums[i] & 1) == 1) + i++; + while (i < j && (nums[j] & 1) == 0) + j--; swap(nums[i], nums[j]); } return nums; @@ -23,11 +24,11 @@ class Solution { int main() { // ======= Test Case ======= - vector nums = { 1, 2, 3, 4 }; + vector nums = {1, 2, 3, 4}; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); vector res = slt->exchange(nums); PrintUtil::printVector(res); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1.cpp b/sword_for_offer/codes/cpp/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1.cpp index 8e54ec8..8b383e2 100644 --- a/sword_for_offer/codes/cpp/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1.cpp @@ -1,19 +1,19 @@ /* -* File: sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - ListNode* getKthFromEnd(ListNode* head, int k) { + public: + ListNode *getKthFromEnd(ListNode *head, int k) { ListNode *former = head, *latter = head; - for(int i = 0; i < k; i++) + for (int i = 0; i < k; i++) former = former->next; - while(former != nullptr) { + while (former != nullptr) { former = former->next; latter = latter->next; } @@ -23,12 +23,12 @@ class Solution { int main() { // ======= Test Case ======= - ListNode* head = vectorToLinkedList(vector { 1, 2, 3, 4, 5 }); + ListNode *head = vectorToLinkedList(vector{1, 2, 3, 4, 5}); int k = 2; // ====== Driver Code ====== - Solution* slt = new Solution(); - ListNode* res = slt->getKthFromEnd(head, k); + Solution *slt = new Solution(); + ListNode *res = slt->getKthFromEnd(head, k); PrintUtil::printLinkedList(res); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2.cpp b/sword_for_offer/codes/cpp/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2.cpp index 1c80e0d..1556e95 100644 --- a/sword_for_offer/codes/cpp/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2.cpp @@ -1,21 +1,22 @@ /* -* File: sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - ListNode* getKthFromEnd(ListNode* head, int k) { + public: + ListNode *getKthFromEnd(ListNode *head, int k) { ListNode *former = head, *latter = head; - for(int i = 0; i < k; i++) { - if(former == nullptr) return nullptr; + for (int i = 0; i < k; i++) { + if (former == nullptr) + return nullptr; former = former->next; } - while(former != nullptr) { + while (former != nullptr) { former = former->next; latter = latter->next; } @@ -25,12 +26,12 @@ class Solution { int main() { // ======= Test Case ======= - ListNode* head = vectorToLinkedList(vector { 1, 2, 3, 4, 5 }); + ListNode *head = vectorToLinkedList(vector{1, 2, 3, 4, 5}); int k = 2; // ====== Driver Code ====== - Solution* slt = new Solution(); - ListNode* res = slt->getKthFromEnd(head, k); + Solution *slt = new Solution(); + ListNode *res = slt->getKthFromEnd(head, k); PrintUtil::printLinkedList(res); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_24_reverse_a_linked_list_s1/sfo_24_reverse_a_linked_list_s1.cpp b/sword_for_offer/codes/cpp/sfo_24_reverse_a_linked_list_s1/sfo_24_reverse_a_linked_list_s1.cpp index 24a0c66..f876eea 100644 --- a/sword_for_offer/codes/cpp/sfo_24_reverse_a_linked_list_s1/sfo_24_reverse_a_linked_list_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_24_reverse_a_linked_list_s1/sfo_24_reverse_a_linked_list_s1.cpp @@ -1,18 +1,18 @@ /* -* File: sfo_24_reverse_a_linked_list_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_24_reverse_a_linked_list_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - ListNode* reverseList(ListNode* head) { + public: + ListNode *reverseList(ListNode *head) { ListNode *cur = head, *pre = nullptr; - while(cur != nullptr) { - ListNode* tmp = cur->next; // 暂存后继节点 cur.next + while (cur != nullptr) { + ListNode *tmp = cur->next; // 暂存后继节点 cur.next cur->next = pre; // 修改 next 引用指向 pre = cur; // pre 暂存 cur cur = tmp; // cur 访问下一节点 @@ -23,11 +23,11 @@ class Solution { int main() { // ======= Test Case ======= - ListNode* head = vectorToLinkedList(vector { 1, 2, 3, 4, 5 }); + ListNode *head = vectorToLinkedList(vector{1, 2, 3, 4, 5}); // ====== Driver Code ====== - Solution* slt = new Solution(); - ListNode* res = slt->reverseList(head); + Solution *slt = new Solution(); + ListNode *res = slt->reverseList(head); PrintUtil::printLinkedList(res); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_24_reverse_a_linked_list_s2/sfo_24_reverse_a_linked_list_s2.cpp b/sword_for_offer/codes/cpp/sfo_24_reverse_a_linked_list_s2/sfo_24_reverse_a_linked_list_s2.cpp index 2ec993e..241d131 100644 --- a/sword_for_offer/codes/cpp/sfo_24_reverse_a_linked_list_s2/sfo_24_reverse_a_linked_list_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_24_reverse_a_linked_list_s2/sfo_24_reverse_a_linked_list_s2.cpp @@ -1,21 +1,23 @@ /* -* File: sfo_24_reverse_a_linked_list_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_24_reverse_a_linked_list_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - ListNode* reverseList(ListNode* head) { - return recur(head, nullptr); // 调用递归并返回 + public: + ListNode *reverseList(ListNode *head) { + return recur(head, nullptr); // 调用递归并返回 } -private: - ListNode* recur(ListNode* cur, ListNode* pre) { - if (cur == nullptr) return pre; // 终止条件 - ListNode* res = recur(cur->next, cur); // 递归后继节点 + + private: + ListNode *recur(ListNode *cur, ListNode *pre) { + if (cur == nullptr) + return pre; // 终止条件 + ListNode *res = recur(cur->next, cur); // 递归后继节点 cur->next = pre; // 修改节点引用指向 return res; // 返回反转链表的头节点 } @@ -23,11 +25,11 @@ class Solution { int main() { // ======= Test Case ======= - ListNode* head = vectorToLinkedList(vector { 1, 2, 3, 4, 5 }); + ListNode *head = vectorToLinkedList(vector{1, 2, 3, 4, 5}); // ====== Driver Code ====== - Solution* slt = new Solution(); - ListNode* res = slt->reverseList(head); + Solution *slt = new Solution(); + ListNode *res = slt->reverseList(head); PrintUtil::printLinkedList(res); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_25_combine_two_sorted_linked_lists_s1/sfo_25_combine_two_sorted_linked_lists_s1.cpp b/sword_for_offer/codes/cpp/sfo_25_combine_two_sorted_linked_lists_s1/sfo_25_combine_two_sorted_linked_lists_s1.cpp index 5ca2443..0ef5b1f 100644 --- a/sword_for_offer/codes/cpp/sfo_25_combine_two_sorted_linked_lists_s1/sfo_25_combine_two_sorted_linked_lists_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_25_combine_two_sorted_linked_lists_s1/sfo_25_combine_two_sorted_linked_lists_s1.cpp @@ -1,23 +1,22 @@ /* -* File: sfo_25_combine_two_sorted_linked_lists_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_25_combine_two_sorted_linked_lists_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { - ListNode* dum = new ListNode(0); - ListNode* cur = dum; - while(l1 != nullptr && l2 != nullptr) { - if(l1->val < l2->val) { + public: + ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) { + ListNode *dum = new ListNode(0); + ListNode *cur = dum; + while (l1 != nullptr && l2 != nullptr) { + if (l1->val < l2->val) { cur->next = l1; l1 = l1->next; - } - else { + } else { cur->next = l2; l2 = l2->next; } @@ -30,12 +29,12 @@ class Solution { int main() { // ======= Test Case ======= - ListNode* l1 = vectorToLinkedList(vector { 1, 2, 4 }); - ListNode* l2 = vectorToLinkedList(vector { 1, 3, 4 }); + ListNode *l1 = vectorToLinkedList(vector{1, 2, 4}); + ListNode *l2 = vectorToLinkedList(vector{1, 3, 4}); // ====== Driver Code ====== - Solution* slt = new Solution(); - ListNode* res = slt->mergeTwoLists(l1, l2); + Solution *slt = new Solution(); + ListNode *res = slt->mergeTwoLists(l1, l2); PrintUtil::printLinkedList(res); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_26_substructure_of_a_binary_tree_s1/sfo_26_substructure_of_a_binary_tree_s1.cpp b/sword_for_offer/codes/cpp/sfo_26_substructure_of_a_binary_tree_s1/sfo_26_substructure_of_a_binary_tree_s1.cpp index 9714f94..6bc1386 100644 --- a/sword_for_offer/codes/cpp/sfo_26_substructure_of_a_binary_tree_s1/sfo_26_substructure_of_a_binary_tree_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_26_substructure_of_a_binary_tree_s1/sfo_26_substructure_of_a_binary_tree_s1.cpp @@ -1,33 +1,37 @@ /* -* File: sfo_26_substructure_of_a_binary_tree_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_26_substructure_of_a_binary_tree_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - bool isSubStructure(TreeNode* A, TreeNode* B) { - return (A != nullptr && B != nullptr) && (recur(A, B) || isSubStructure(A->left, B) || isSubStructure(A->right, B)); + public: + bool isSubStructure(TreeNode *A, TreeNode *B) { + return (A != nullptr && B != nullptr) && + (recur(A, B) || isSubStructure(A->left, B) || isSubStructure(A->right, B)); } -private: - bool recur(TreeNode* A, TreeNode* B) { - if(B == nullptr) return true; - if(A == nullptr || A->val != B->val) return false; + + private: + bool recur(TreeNode *A, TreeNode *B) { + if (B == nullptr) + return true; + if (A == nullptr || A->val != B->val) + return false; return recur(A->left, B->left) && recur(A->right, B->right); } }; int main() { // ======= Test Case ======= - TreeNode* A = vectorToTree(vector { 3, 4, 5, 1, 2, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX }); - TreeNode* B = vectorToTree(vector { 4, 1, INT_MAX, INT_MAX, INT_MAX }); + TreeNode *A = vectorToTree(vector{3, 4, 5, 1, 2, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX}); + TreeNode *B = vectorToTree(vector{4, 1, INT_MAX, INT_MAX, INT_MAX}); // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); bool res = slt->isSubStructure(A, B); cout << boolalpha << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_27_mirror_of_a_binary_tree_s1/sfo_27_mirror_of_a_binary_tree_s1.cpp b/sword_for_offer/codes/cpp/sfo_27_mirror_of_a_binary_tree_s1/sfo_27_mirror_of_a_binary_tree_s1.cpp index 700f2cd..2dc872b 100644 --- a/sword_for_offer/codes/cpp/sfo_27_mirror_of_a_binary_tree_s1/sfo_27_mirror_of_a_binary_tree_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_27_mirror_of_a_binary_tree_s1/sfo_27_mirror_of_a_binary_tree_s1.cpp @@ -1,17 +1,18 @@ /* -* File: sfo_27_mirror_of_a_binary_tree_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_27_mirror_of_a_binary_tree_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - TreeNode* mirrorTree(TreeNode* root) { - if (root == nullptr) return nullptr; - TreeNode* tmp = root->left; + public: + TreeNode *mirrorTree(TreeNode *root) { + if (root == nullptr) + return nullptr; + TreeNode *tmp = root->left; root->left = mirrorTree(root->right); root->right = mirrorTree(tmp); return root; @@ -20,11 +21,12 @@ class Solution { int main() { // ======= Test Case ======= - TreeNode* root = vectorToTree(vector { 4, 2, 7, 1, 3, 6, 9, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX }); + TreeNode *root = vectorToTree( + vector{4, 2, 7, 1, 3, 6, 9, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX}); // ====== Driver Code ====== - Solution* slt = new Solution(); - TreeNode* res = slt->mirrorTree(root); + Solution *slt = new Solution(); + TreeNode *res = slt->mirrorTree(root); PrintUtil::printTree(res); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_27_mirror_of_a_binary_tree_s2/sfo_27_mirror_of_a_binary_tree_s2.cpp b/sword_for_offer/codes/cpp/sfo_27_mirror_of_a_binary_tree_s2/sfo_27_mirror_of_a_binary_tree_s2.cpp index a406e23..34b2df8 100644 --- a/sword_for_offer/codes/cpp/sfo_27_mirror_of_a_binary_tree_s2/sfo_27_mirror_of_a_binary_tree_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_27_mirror_of_a_binary_tree_s2/sfo_27_mirror_of_a_binary_tree_s2.cpp @@ -1,25 +1,27 @@ /* -* File: sfo_27_mirror_of_a_binary_tree_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_27_mirror_of_a_binary_tree_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - TreeNode* mirrorTree(TreeNode* root) { - if(root == nullptr) return nullptr; - stack stack; + public: + TreeNode *mirrorTree(TreeNode *root) { + if (root == nullptr) + return nullptr; + stack stack; stack.push(root); - while (!stack.empty()) - { - TreeNode* node = stack.top(); + while (!stack.empty()) { + TreeNode *node = stack.top(); stack.pop(); - if (node->left != nullptr) stack.push(node->left); - if (node->right != nullptr) stack.push(node->right); - TreeNode* tmp = node->left; + if (node->left != nullptr) + stack.push(node->left); + if (node->right != nullptr) + stack.push(node->right); + TreeNode *tmp = node->left; node->left = node->right; node->right = tmp; } @@ -29,11 +31,12 @@ class Solution { int main() { // ======= Test Case ======= - TreeNode* root = vectorToTree(vector { 4, 2, 7, 1, 3, 6, 9, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX }); + TreeNode *root = vectorToTree( + vector{4, 2, 7, 1, 3, 6, 9, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX}); // ====== Driver Code ====== - Solution* slt = new Solution(); - TreeNode* res = slt->mirrorTree(root); + Solution *slt = new Solution(); + TreeNode *res = slt->mirrorTree(root); PrintUtil::printTree(res); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_28_symmetric_binary_tree_s1/sfo_28_symmetric_binary_tree_s1.cpp b/sword_for_offer/codes/cpp/sfo_28_symmetric_binary_tree_s1/sfo_28_symmetric_binary_tree_s1.cpp index a893e2a..ba835b3 100644 --- a/sword_for_offer/codes/cpp/sfo_28_symmetric_binary_tree_s1/sfo_28_symmetric_binary_tree_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_28_symmetric_binary_tree_s1/sfo_28_symmetric_binary_tree_s1.cpp @@ -1,32 +1,36 @@ /* -* File: sfo_28_symmetric_binary_tree_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_28_symmetric_binary_tree_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - bool isSymmetric(TreeNode* root) { + public: + bool isSymmetric(TreeNode *root) { return root == nullptr || recur(root->left, root->right); } -private: - bool recur(TreeNode* L, TreeNode* R) { - if(L == nullptr && R == nullptr) return true; - if(L == nullptr || R == nullptr || L->val != R->val) return false; + + private: + bool recur(TreeNode *L, TreeNode *R) { + if (L == nullptr && R == nullptr) + return true; + if (L == nullptr || R == nullptr || L->val != R->val) + return false; return recur(L->left, R->right) && recur(L->right, R->left); } }; int main() { // ======= Test Case ======= - TreeNode* root = vectorToTree(vector { 1, 2, 2, 3, 4, 4, 3, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX }); + TreeNode *root = vectorToTree( + vector{1, 2, 2, 3, 4, 4, 3, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX}); // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); bool res = slt->isSymmetric(root); cout << boolalpha << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_29_print_a_given_matrix_in_spiral_form_s1/sfo_29_print_a_given_matrix_in_spiral_form_s1.cpp b/sword_for_offer/codes/cpp/sfo_29_print_a_given_matrix_in_spiral_form_s1/sfo_29_print_a_given_matrix_in_spiral_form_s1.cpp index d4d303b..aa507f4 100644 --- a/sword_for_offer/codes/cpp/sfo_29_print_a_given_matrix_in_spiral_form_s1/sfo_29_print_a_given_matrix_in_spiral_form_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_29_print_a_given_matrix_in_spiral_form_s1/sfo_29_print_a_given_matrix_in_spiral_form_s1.cpp @@ -1,28 +1,36 @@ /* -* File: sfo_29_print_a_given_matrix_in_spiral_form_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_29_print_a_given_matrix_in_spiral_form_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - vector spiralOrder(vector>& matrix) { - if (matrix.empty()) return {}; + public: + vector spiralOrder(vector> &matrix) { + if (matrix.empty()) + return {}; int l = 0, r = matrix[0].size() - 1, t = 0, b = matrix.size() - 1; vector res; - while(true) - { - for (int i = l; i <= r; i++) res.push_back(matrix[t][i]); // left to right - if (++t > b) break; - for (int i = t; i <= b; i++) res.push_back(matrix[i][r]); // top to bottom - if (l > --r) break; - for (int i = r; i >= l; i--) res.push_back(matrix[b][i]); // right to left - if (t > --b) break; - for (int i = b; i >= t; i--) res.push_back(matrix[i][l]); // bottom to top - if (++l > r) break; + while (true) { + for (int i = l; i <= r; i++) + res.push_back(matrix[t][i]); // left to right + if (++t > b) + break; + for (int i = t; i <= b; i++) + res.push_back(matrix[i][r]); // top to bottom + if (l > --r) + break; + for (int i = r; i >= l; i--) + res.push_back(matrix[b][i]); // right to left + if (t > --b) + break; + for (int i = b; i >= t; i--) + res.push_back(matrix[i][l]); // bottom to top + if (++l > r) + break; } return res; } @@ -30,15 +38,11 @@ class Solution { int main() { // ======= Test Case ======= - vector> matrix = { - { 1, 2, 3 }, - { 4, 5, 6 }, - { 7, 8, 9 } -}; + vector> matrix = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); vector res = slt->spiralOrder(matrix); PrintUtil::printVector(res); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_30_min_stack_s1/sfo_30_min_stack_s1.cpp b/sword_for_offer/codes/cpp/sfo_30_min_stack_s1/sfo_30_min_stack_s1.cpp index 05ae45d..1dfd9e9 100644 --- a/sword_for_offer/codes/cpp/sfo_30_min_stack_s1/sfo_30_min_stack_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_30_min_stack_s1/sfo_30_min_stack_s1.cpp @@ -1,23 +1,24 @@ /* -* File: sfo_30_min_stack_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_30_min_stack_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class MinStack { -public: + public: stack A, B; - MinStack() {} + MinStack() { + } void push(int x) { A.push(x); - if(B.empty() || B.top() >= x) + if (B.empty() || B.top() >= x) B.push(x); } void pop() { - if(A.top() == B.top()) + if (A.top() == B.top()) B.pop(); A.pop(); } @@ -32,7 +33,7 @@ class MinStack { int main() { // ====== Driver Code ====== vector res; - MinStack* minStack = new MinStack(); + MinStack *minStack = new MinStack(); minStack->push(-2); minStack->push(0); minStack->push(-3); diff --git a/sword_for_offer/codes/cpp/sfo_31_validate_stack_sequences_s1/sfo_31_validate_stack_sequences_s1.cpp b/sword_for_offer/codes/cpp/sfo_31_validate_stack_sequences_s1/sfo_31_validate_stack_sequences_s1.cpp index aaf210b..9d9868c 100644 --- a/sword_for_offer/codes/cpp/sfo_31_validate_stack_sequences_s1/sfo_31_validate_stack_sequences_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_31_validate_stack_sequences_s1/sfo_31_validate_stack_sequences_s1.cpp @@ -1,20 +1,20 @@ /* -* File: sfo_31_validate_stack_sequences_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_31_validate_stack_sequences_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - bool validateStackSequences(vector& pushed, vector& popped) { + public: + bool validateStackSequences(vector &pushed, vector &popped) { stack stk; int i = 0; - for(int num : pushed) { - stk.push(num); // num 入栈 - while(!stk.empty() && stk.top() == popped[i]) { // 循环判断与出栈 + for (int num : pushed) { + stk.push(num); // num 入栈 + while (!stk.empty() && stk.top() == popped[i]) { // 循环判断与出栈 stk.pop(); i++; } @@ -25,12 +25,12 @@ class Solution { int main() { // ======= Test Case ======= - vector pushed = { 1, 2, 3, 4, 5 }; - vector popped = { 4, 5, 3, 2, 1 }; + vector pushed = {1, 2, 3, 4, 5}; + vector popped = {4, 5, 3, 2, 1}; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); bool res = slt->validateStackSequences(pushed, popped); cout << boolalpha << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_32i_print_a_binary_tree_topbottom_i_s1/sfo_32i_print_a_binary_tree_topbottom_i_s1.cpp b/sword_for_offer/codes/cpp/sfo_32i_print_a_binary_tree_topbottom_i_s1/sfo_32i_print_a_binary_tree_topbottom_i_s1.cpp index e7ebcd1..41bc6ab 100644 --- a/sword_for_offer/codes/cpp/sfo_32i_print_a_binary_tree_topbottom_i_s1/sfo_32i_print_a_binary_tree_topbottom_i_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_32i_print_a_binary_tree_topbottom_i_s1/sfo_32i_print_a_binary_tree_topbottom_i_s1.cpp @@ -1,25 +1,28 @@ /* -* File: sfo_32i_print_a_binary_tree_topbottom_i_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_32i_print_a_binary_tree_topbottom_i_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - vector levelOrder(TreeNode* root) { + public: + vector levelOrder(TreeNode *root) { vector res; - if(!root) return res; - queue que; + if (!root) + return res; + queue que; que.push(root); - while(!que.empty()){ - TreeNode* node = que.front(); + while (!que.empty()) { + TreeNode *node = que.front(); que.pop(); res.push_back(node->val); - if(node->left) que.push(node->left); - if(node->right) que.push(node->right); + if (node->left) + que.push(node->left); + if (node->right) + que.push(node->right); } return res; } @@ -27,11 +30,11 @@ class Solution { int main() { // ======= Test Case ======= - TreeNode* root = vectorToTree(vector { 3, 9, 20, INT_MAX, INT_MAX, 15, 7, INT_MAX, INT_MAX, INT_MAX, INT_MAX }); + TreeNode *root = vectorToTree(vector{3, 9, 20, INT_MAX, INT_MAX, 15, 7, INT_MAX, INT_MAX, INT_MAX, INT_MAX}); // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); vector res = slt->levelOrder(root); PrintUtil::printVector(res); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_32ii_print_a_binary_tree_topbottom_ii_s1/sfo_32ii_print_a_binary_tree_topbottom_ii_s1.cpp b/sword_for_offer/codes/cpp/sfo_32ii_print_a_binary_tree_topbottom_ii_s1/sfo_32ii_print_a_binary_tree_topbottom_ii_s1.cpp index 98e77e6..12349f9 100644 --- a/sword_for_offer/codes/cpp/sfo_32ii_print_a_binary_tree_topbottom_ii_s1/sfo_32ii_print_a_binary_tree_topbottom_ii_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_32ii_print_a_binary_tree_topbottom_ii_s1/sfo_32ii_print_a_binary_tree_topbottom_ii_s1.cpp @@ -1,27 +1,30 @@ /* -* File: sfo_32ii_print_a_binary_tree_topbottom_ii_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_32ii_print_a_binary_tree_topbottom_ii_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - vector> levelOrder(TreeNode* root) { - queue que; + public: + vector> levelOrder(TreeNode *root) { + queue que; vector> res; int cnt = 0; - if(root != NULL) que.push(root); - while(!que.empty()) { + if (root != NULL) + que.push(root); + while (!que.empty()) { vector tmp; - for(int i = que.size(); i > 0; --i) { + for (int i = que.size(); i > 0; --i) { root = que.front(); que.pop(); tmp.push_back(root->val); - if(root->left != NULL) que.push(root->left); - if(root->right != NULL) que.push(root->right); + if (root->left != NULL) + que.push(root->left); + if (root->right != NULL) + que.push(root->right); } res.push_back(tmp); } @@ -31,11 +34,11 @@ class Solution { int main() { // ======= Test Case ======= - TreeNode* root = vectorToTree(vector { 3, 9, 20, INT_MAX, INT_MAX, 15, 7, INT_MAX, INT_MAX, INT_MAX, INT_MAX }); + TreeNode *root = vectorToTree(vector{3, 9, 20, INT_MAX, INT_MAX, 15, 7, INT_MAX, INT_MAX, INT_MAX, INT_MAX}); // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); vector> res = slt->levelOrder(root); PrintUtil::printVectorMatrix(res); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_32iii_print_a_binary_tree_topbottom_iii_s1/sfo_32iii_print_a_binary_tree_topbottom_iii_s1.cpp b/sword_for_offer/codes/cpp/sfo_32iii_print_a_binary_tree_topbottom_iii_s1/sfo_32iii_print_a_binary_tree_topbottom_iii_s1.cpp index 1d90ac8..382aa03 100644 --- a/sword_for_offer/codes/cpp/sfo_32iii_print_a_binary_tree_topbottom_iii_s1/sfo_32iii_print_a_binary_tree_topbottom_iii_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_32iii_print_a_binary_tree_topbottom_iii_s1/sfo_32iii_print_a_binary_tree_topbottom_iii_s1.cpp @@ -1,42 +1,48 @@ /* -* File: sfo_32iii_print_a_binary_tree_topbottom_iii_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_32iii_print_a_binary_tree_topbottom_iii_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - vector> levelOrder(TreeNode* root) { - deque deque; + public: + vector> levelOrder(TreeNode *root) { + deque deque; vector> res; - if(root != NULL) deque.push_back(root); - while(!deque.empty()) { + if (root != NULL) + deque.push_back(root); + while (!deque.empty()) { // 打印奇数层 vector tmp; - for(int i = deque.size(); i > 0; i--) { + for (int i = deque.size(); i > 0; i--) { // 从左向右打印 - TreeNode* node = deque.front(); + TreeNode *node = deque.front(); deque.pop_front(); tmp.push_back(node->val); // 先左后右加入下层节点 - if(node->left != NULL) deque.push_back(node->left); - if(node->right != NULL) deque.push_back(node->right); + if (node->left != NULL) + deque.push_back(node->left); + if (node->right != NULL) + deque.push_back(node->right); } res.push_back(tmp); - if(deque.empty()) break; // 若为空则提前跳出 + if (deque.empty()) + break; // 若为空则提前跳出 // 打印偶数层 tmp.clear(); - for(int i = deque.size(); i > 0; i--) { + for (int i = deque.size(); i > 0; i--) { // 从右向左打印 - TreeNode* node = deque.back(); + TreeNode *node = deque.back(); deque.pop_back(); tmp.push_back(node->val); // 先右后左加入下层节点 - if(node->right != NULL) deque.push_front(node->right); - if(node->left != NULL) deque.push_front(node->left); + if (node->right != NULL) + deque.push_front(node->right); + if (node->left != NULL) + deque.push_front(node->left); } res.push_back(tmp); } @@ -46,11 +52,11 @@ class Solution { int main() { // ======= Test Case ======= - TreeNode* root = vectorToTree(vector { 3, 9, 20, INT_MAX, INT_MAX, 15, 7, INT_MAX, INT_MAX, INT_MAX, INT_MAX }); + TreeNode *root = vectorToTree(vector{3, 9, 20, INT_MAX, INT_MAX, 15, 7, INT_MAX, INT_MAX, INT_MAX, INT_MAX}); // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); vector> res = slt->levelOrder(root); PrintUtil::printVectorMatrix(res); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_32iii_print_a_binary_tree_topbottom_iii_s2/sfo_32iii_print_a_binary_tree_topbottom_iii_s2.cpp b/sword_for_offer/codes/cpp/sfo_32iii_print_a_binary_tree_topbottom_iii_s2/sfo_32iii_print_a_binary_tree_topbottom_iii_s2.cpp index 660d0e9..db14771 100644 --- a/sword_for_offer/codes/cpp/sfo_32iii_print_a_binary_tree_topbottom_iii_s2/sfo_32iii_print_a_binary_tree_topbottom_iii_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_32iii_print_a_binary_tree_topbottom_iii_s2/sfo_32iii_print_a_binary_tree_topbottom_iii_s2.cpp @@ -1,28 +1,32 @@ /* -* File: sfo_32iii_print_a_binary_tree_topbottom_iii_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_32iii_print_a_binary_tree_topbottom_iii_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - vector> levelOrder(TreeNode* root) { - queue que; + public: + vector> levelOrder(TreeNode *root) { + queue que; vector> res; - if(root != NULL) que.push(root); - while(!que.empty()) { + if (root != NULL) + que.push(root); + while (!que.empty()) { vector tmp; - for(int i = que.size(); i > 0; i--) { - TreeNode* node = que.front(); + for (int i = que.size(); i > 0; i--) { + TreeNode *node = que.front(); que.pop(); tmp.push_back(node->val); - if(node->left != NULL) que.push(node->left); - if(node->right != NULL) que.push(node->right); + if (node->left != NULL) + que.push(node->left); + if (node->right != NULL) + que.push(node->right); } - if(res.size() % 2 == 1) reverse(tmp.begin(),tmp.end()); + if (res.size() % 2 == 1) + reverse(tmp.begin(), tmp.end()); res.push_back(tmp); } return res; @@ -31,11 +35,11 @@ class Solution { int main() { // ======= Test Case ======= - TreeNode* root = vectorToTree(vector { 3, 9, 20, INT_MAX, INT_MAX, 15, 7, INT_MAX, INT_MAX, INT_MAX, INT_MAX }); + TreeNode *root = vectorToTree(vector{3, 9, 20, INT_MAX, INT_MAX, 15, 7, INT_MAX, INT_MAX, INT_MAX, INT_MAX}); // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); vector> res = slt->levelOrder(root); PrintUtil::printVectorMatrix(res); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_33_postorder_traversal_of_a_binary_search_tree_s1/sfo_33_postorder_traversal_of_a_binary_search_tree_s1.cpp b/sword_for_offer/codes/cpp/sfo_33_postorder_traversal_of_a_binary_search_tree_s1/sfo_33_postorder_traversal_of_a_binary_search_tree_s1.cpp index 8f159e0..e708a45 100644 --- a/sword_for_offer/codes/cpp/sfo_33_postorder_traversal_of_a_binary_search_tree_s1/sfo_33_postorder_traversal_of_a_binary_search_tree_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_33_postorder_traversal_of_a_binary_search_tree_s1/sfo_33_postorder_traversal_of_a_binary_search_tree_s1.cpp @@ -1,35 +1,39 @@ /* -* File: sfo_33_postorder_traversal_of_a_binary_search_tree_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_33_postorder_traversal_of_a_binary_search_tree_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - bool verifyPostorder(vector& postorder) { + public: + bool verifyPostorder(vector &postorder) { return recur(postorder, 0, postorder.size() - 1); } -private: - bool recur(vector& postorder, int i, int j) { - if(i >= j) return true; + + private: + bool recur(vector &postorder, int i, int j) { + if (i >= j) + return true; int p = i; - while(postorder[p] < postorder[j]) p++; + while (postorder[p] < postorder[j]) + p++; int m = p; - while(postorder[p] > postorder[j]) p++; + while (postorder[p] > postorder[j]) + p++; return p == j && recur(postorder, i, m - 1) && recur(postorder, m, j - 1); } }; int main() { // ======= Test Case ======= - vector postorder = { 1, 6, 3, 2, 5 }; + vector postorder = {1, 6, 3, 2, 5}; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); bool res = slt->verifyPostorder(postorder); cout << boolalpha << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_33_postorder_traversal_of_a_binary_search_tree_s2/sfo_33_postorder_traversal_of_a_binary_search_tree_s2.cpp b/sword_for_offer/codes/cpp/sfo_33_postorder_traversal_of_a_binary_search_tree_s2/sfo_33_postorder_traversal_of_a_binary_search_tree_s2.cpp index 1e128b7..7636485 100644 --- a/sword_for_offer/codes/cpp/sfo_33_postorder_traversal_of_a_binary_search_tree_s2/sfo_33_postorder_traversal_of_a_binary_search_tree_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_33_postorder_traversal_of_a_binary_search_tree_s2/sfo_33_postorder_traversal_of_a_binary_search_tree_s2.cpp @@ -1,20 +1,21 @@ /* -* File: sfo_33_postorder_traversal_of_a_binary_search_tree_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_33_postorder_traversal_of_a_binary_search_tree_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - bool verifyPostorder(vector& postorder) { + public: + bool verifyPostorder(vector &postorder) { stack stk; int root = INT_MAX; - for(int i = postorder.size() - 1; i >= 0; i--) { - if(postorder[i] > root) return false; - while(!stk.empty() && stk.top() > postorder[i]) { + for (int i = postorder.size() - 1; i >= 0; i--) { + if (postorder[i] > root) + return false; + while (!stk.empty() && stk.top() > postorder[i]) { root = stk.top(); stk.pop(); } @@ -26,11 +27,11 @@ class Solution { int main() { // ======= Test Case ======= - vector postorder = { 1, 6, 3, 2, 5 }; + vector postorder = {1, 6, 3, 2, 5}; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); bool res = slt->verifyPostorder(postorder); cout << boolalpha << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_34_all_xsum_paths_in_a_binary_tree_s1/sfo_34_all_xsum_paths_in_a_binary_tree_s1.cpp b/sword_for_offer/codes/cpp/sfo_34_all_xsum_paths_in_a_binary_tree_s1/sfo_34_all_xsum_paths_in_a_binary_tree_s1.cpp index 301b13e..7632a34 100644 --- a/sword_for_offer/codes/cpp/sfo_34_all_xsum_paths_in_a_binary_tree_s1/sfo_34_all_xsum_paths_in_a_binary_tree_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_34_all_xsum_paths_in_a_binary_tree_s1/sfo_34_all_xsum_paths_in_a_binary_tree_s1.cpp @@ -1,26 +1,28 @@ /* -* File: sfo_34_all_xsum_paths_in_a_binary_tree_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_34_all_xsum_paths_in_a_binary_tree_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - vector> pathSum(TreeNode* root, int sum) { + public: + vector> pathSum(TreeNode *root, int sum) { recur(root, sum); return res; } -private: + + private: vector> res; vector path; - void recur(TreeNode* root, int tar) { - if(root == nullptr) return; + void recur(TreeNode *root, int tar) { + if (root == nullptr) + return; path.push_back(root->val); tar -= root->val; - if(tar == 0 && root->left == nullptr && root->right == nullptr) + if (tar == 0 && root->left == nullptr && root->right == nullptr) res.push_back(path); recur(root->left, tar); recur(root->right, tar); @@ -30,12 +32,14 @@ class Solution { int main() { // ======= Test Case ======= - TreeNode* root = vectorToTree(vector { 5, 4, 8, 11, INT_MAX, 13, 4, 7, 2, INT_MAX, INT_MAX, 5, 1, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX }); + TreeNode *root = + vectorToTree(vector{5, 4, 8, 11, INT_MAX, 13, 4, 7, 2, INT_MAX, INT_MAX, + 5, 1, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX}); int sum = 22; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); vector> res = slt->pathSum(root, sum); PrintUtil::printVectorMatrix(res); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1.cpp b/sword_for_offer/codes/cpp/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1.cpp index 41b199b..80cf965 100644 --- a/sword_for_offer/codes/cpp/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1.cpp @@ -1,8 +1,8 @@ /* -* File: sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" @@ -13,24 +13,26 @@ struct Node { int val; Node *next; Node *random; - Node(int x) : val(x), next(nullptr), random(nullptr) {} + Node(int x) : val(x), next(nullptr), random(nullptr) { + } }; // ===== Solution Code ===== class Solution { -public: - Node* copyRandomList(Node* head) { - if(head == nullptr) return nullptr; - Node* cur = head; - unordered_map map; + public: + Node *copyRandomList(Node *head) { + if (head == nullptr) + return nullptr; + Node *cur = head; + unordered_map map; // 3. 复制各节点，并建立 “原节点 -> 新节点” 的 Map 映射 - while(cur != nullptr) { + while (cur != nullptr) { map[cur] = new Node(cur->val); cur = cur->next; } cur = head; // 4. 构建新链表的 next 和 random 指向 - while(cur != nullptr) { + while (cur != nullptr) { map[cur]->next = map[cur->next]; map[cur]->random = map[cur->random]; cur = cur->next; @@ -42,10 +44,10 @@ class Solution { int main() { // ======= Test Case ======= - vector nodesVal { 7, 13, 11, 10, 1 }; - vector nodesRandom { INT_MAX, 0, 4, 2, 0 }; // Represent the 'null' Node with 'INT_MAX' + vector nodesVal{7, 13, 11, 10, 1}; + vector nodesRandom{INT_MAX, 0, 4, 2, 0}; // Represent the 'null' Node with 'INT_MAX' // Construct nodes - vector nodeList; + vector nodeList; for (int val : nodesVal) { nodeList.push_back(new Node(val)); } @@ -59,24 +61,24 @@ int main() { nodeList[i]->random = nodeList[nodesRandom[i]]; } // Get the head of the linked list - Node* head = nodeList[0]; + Node *head = nodeList[0]; // ====== Driver Code ====== - Solution* slt = new Solution(); - Node* res = slt->copyRandomList(head); + Solution *slt = new Solution(); + Node *res = slt->copyRandomList(head); // Print the copied linked list - vector nodeListNew; + vector nodeListNew; while (res != nullptr) { nodeListNew.push_back(res); res = res->next; } - vector> printArr(nodeListNew.size(), vector (2)); + vector> printArr(nodeListNew.size(), vector(2)); for (int i = 0; i < nodeListNew.size(); i++) { - Node* node = nodeListNew[i]; + Node *node = nodeListNew[i]; printArr[i][0] = node->val; printArr[i][1] = PrintUtil::vecFind(nodeListNew, node->random); } PrintUtil::printVectorMatrix(printArr); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2.cpp b/sword_for_offer/codes/cpp/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2.cpp index d56c2f0..b176760 100644 --- a/sword_for_offer/codes/cpp/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2.cpp @@ -1,8 +1,8 @@ /* -* File: sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" @@ -13,49 +13,51 @@ struct Node { int val; Node *next; Node *random; - Node(int x) : val(x), next(nullptr), random(nullptr) {} + Node(int x) : val(x), next(nullptr), random(nullptr) { + } }; // ===== Solution Code ===== class Solution { -public: - Node* copyRandomList(Node* head) { - if(head == nullptr) return nullptr; - Node* cur = head; + public: + Node *copyRandomList(Node *head) { + if (head == nullptr) + return nullptr; + Node *cur = head; // 1. 复制各节点，并构建拼接链表 - while(cur != nullptr) { - Node* tmp = new Node(cur->val); + while (cur != nullptr) { + Node *tmp = new Node(cur->val); tmp->next = cur->next; cur->next = tmp; cur = tmp->next; } // 2. 构建各新节点的 random 指向 cur = head; - while(cur != nullptr) { - if(cur->random != nullptr) + while (cur != nullptr) { + if (cur->random != nullptr) cur->next->random = cur->random->next; cur = cur->next->next; } // 3. 拆分两链表 cur = head->next; - Node* pre = head, *res = head->next; - while(cur->next != nullptr) { + Node *pre = head, *res = head->next; + while (cur->next != nullptr) { pre->next = pre->next->next; cur->next = cur->next->next; pre = pre->next; cur = cur->next; } pre->next = nullptr; // 单独处理原链表尾节点 - return res; // 返回新链表头节点 + return res; // 返回新链表头节点 } }; int main() { // ======= Test Case ======= - vector nodesVal { 7, 13, 11, 10, 1 }; - vector nodesRandom { INT_MAX, 0, 4, 2, 0 }; // Represent the 'null' Node with 'INT_MAX' + vector nodesVal{7, 13, 11, 10, 1}; + vector nodesRandom{INT_MAX, 0, 4, 2, 0}; // Represent the 'null' Node with 'INT_MAX' // Construct nodes - vector nodeList; + vector nodeList; for (int val : nodesVal) { nodeList.push_back(new Node(val)); } @@ -69,24 +71,24 @@ int main() { nodeList[i]->random = nodeList[nodesRandom[i]]; } // Get the head of the linked list - Node* head = nodeList[0]; + Node *head = nodeList[0]; // ====== Driver Code ====== - Solution* slt = new Solution(); - Node* res = slt->copyRandomList(head); + Solution *slt = new Solution(); + Node *res = slt->copyRandomList(head); // Print the copied linked list - vector nodeListNew; + vector nodeListNew; while (res != nullptr) { nodeListNew.push_back(res); res = res->next; } - vector> printArr(nodeListNew.size(), vector (2)); + vector> printArr(nodeListNew.size(), vector(2)); for (int i = 0; i < nodeListNew.size(); i++) { - Node* node = nodeListNew[i]; + Node *node = nodeListNew[i]; printArr[i][0] = node->val; printArr[i][1] = PrintUtil::vecFind(nodeListNew, node->random); } PrintUtil::printVectorMatrix(printArr); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_36_binary_search_tree_and_doubly_linked_list_s1/sfo_36_binary_search_tree_and_doubly_linked_list_s1.cpp b/sword_for_offer/codes/cpp/sfo_36_binary_search_tree_and_doubly_linked_list_s1/sfo_36_binary_search_tree_and_doubly_linked_list_s1.cpp index 43f354e..f8712e4 100644 --- a/sword_for_offer/codes/cpp/sfo_36_binary_search_tree_and_doubly_linked_list_s1/sfo_36_binary_search_tree_and_doubly_linked_list_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_36_binary_search_tree_and_doubly_linked_list_s1/sfo_36_binary_search_tree_and_doubly_linked_list_s1.cpp @@ -1,8 +1,8 @@ /* -* File: sfo_36_binary_search_tree_and_doubly_linked_list_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_36_binary_search_tree_and_doubly_linked_list_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" @@ -13,26 +13,32 @@ struct Node { int val; Node *left; Node *right; - Node(int x) : val(x), left(nullptr), right(nullptr) {} + Node(int x) : val(x), left(nullptr), right(nullptr) { + } }; // ===== Solution Code ===== class Solution { -public: - Node* treeToDoublyList(Node* root) { - if(root == nullptr) return nullptr; + public: + Node *treeToDoublyList(Node *root) { + if (root == nullptr) + return nullptr; dfs(root); head->left = pre; pre->right = head; return head; } -private: + + private: Node *pre, *head; - void dfs(Node* cur) { - if(cur == nullptr) return; + void dfs(Node *cur) { + if (cur == nullptr) + return; dfs(cur->left); - if(pre != nullptr) pre->right = cur; - else head = cur; + if (pre != nullptr) + pre->right = cur; + else + head = cur; cur->left = pre; pre = cur; dfs(cur->right); @@ -41,22 +47,16 @@ class Solution { int main() { // ======= Test Case ======= - vector nodeList = { - new Node(1), - new Node(2), - new Node(3), - new Node(4), - new Node(5) - }; + vector nodeList = {new Node(1), new Node(2), new Node(3), new Node(4), new Node(5)}; nodeList[3]->left = nodeList[1]; nodeList[3]->right = nodeList[4]; nodeList[1]->left = nodeList[0]; nodeList[1]->right = nodeList[2]; - Node* root = nodeList[3]; + Node *root = nodeList[3]; // ====== Driver Code ====== - Solution* slt = new Solution(); - Node* res = slt->treeToDoublyList(root); + Solution *slt = new Solution(); + Node *res = slt->treeToDoublyList(root); // Print the Doubly circular linked list vector nodesVal; for (int i = 0; i <= nodeList.size(); i++) { @@ -64,6 +64,6 @@ int main() { res = res->right; } cout << PrintUtil::strJoin(" <-> ", nodesVal) << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_37_serialize_and_deserialize_a_binary_tree_s1/sfo_37_serialize_and_deserialize_a_binary_tree_s1.cpp b/sword_for_offer/codes/cpp/sfo_37_serialize_and_deserialize_a_binary_tree_s1/sfo_37_serialize_and_deserialize_a_binary_tree_s1.cpp index d7677e9..d5d7578 100644 --- a/sword_for_offer/codes/cpp/sfo_37_serialize_and_deserialize_a_binary_tree_s1/sfo_37_serialize_and_deserialize_a_binary_tree_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_37_serialize_and_deserialize_a_binary_tree_s1/sfo_37_serialize_and_deserialize_a_binary_tree_s1.cpp @@ -1,28 +1,29 @@ /* -* File: sfo_37_serialize_and_deserialize_a_binary_tree_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_37_serialize_and_deserialize_a_binary_tree_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" class Codec { -public: + public: // Encodes a tree to a single string. - string serialize(TreeNode* root) { - if(root == nullptr) return "[]"; + string serialize(TreeNode *root) { + if (root == nullptr) + return "[]"; string res = "["; - queue que; + queue que; que.emplace(root); - while(!que.empty()) { - TreeNode* node = que.front(); + while (!que.empty()) { + TreeNode *node = que.front(); que.pop(); - if(node != nullptr) { + if (node != nullptr) { res += (to_string(node->val) + ","); que.emplace(node->left); que.emplace(node->right); - } - else res += "null,"; + } else + res += "null,"; } res.pop_back(); res += "]"; @@ -30,21 +31,21 @@ class Codec { } // Decodes your encoded data to tree. - TreeNode* deserialize(string data) { + TreeNode *deserialize(string data) { vector list = split(data.substr(1, data.length() - 2), ","); TreeNode *root = new TreeNode(std::stoi(list[0])); - queue que; + queue que; que.emplace(root); int i = 1; - while(!que.empty()) { + while (!que.empty()) { TreeNode *node = que.front(); que.pop(); - if(list[i] != "null") { + if (list[i] != "null") { node->left = new TreeNode(std::stoi(list[i])); que.emplace(node->left); } i++; - if(list[i] != "null") { + if (list[i] != "null") { node->right = new TreeNode(std::stoi(list[i])); que.emplace(node->right); } @@ -53,7 +54,7 @@ class Codec { return root; } -private: + private: // Split a str by a delim vector split(string str, string delim) { vector list; @@ -72,8 +73,8 @@ int main() { // ======= Test Case ======= string data = "[1,2,3,null,null,4,5,null,null,null,null]"; // ====== Driver Code ====== - Codec* codec = new Codec(); - TreeNode* root = codec->deserialize(data); + Codec *codec = new Codec(); + TreeNode *root = codec->deserialize(data); string res = codec->serialize(root); PrintUtil::printTree(root); cout << res << endl; diff --git a/sword_for_offer/codes/cpp/sfo_38_all_permutations_of_a_string_s1/sfo_38_all_permutations_of_a_string_s1.cpp b/sword_for_offer/codes/cpp/sfo_38_all_permutations_of_a_string_s1/sfo_38_all_permutations_of_a_string_s1.cpp index 6ab23af..2c3c7c7 100644 --- a/sword_for_offer/codes/cpp/sfo_38_all_permutations_of_a_string_s1/sfo_38_all_permutations_of_a_string_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_38_all_permutations_of_a_string_s1/sfo_38_all_permutations_of_a_string_s1.cpp @@ -1,32 +1,34 @@ /* -* File: sfo_38_all_permutations_of_a_string_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_38_all_permutations_of_a_string_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: vector permutation(string s) { dfs(s, 0); return res; } -private: + + private: vector res; void dfs(string s, int x) { - if(x == s.size() - 1) { - res.push_back(s); // 添加排列方案 + if (x == s.size() - 1) { + res.push_back(s); // 添加排列方案 return; } set st; - for(int i = x; i < s.size(); i++) { - if(st.find(s[i]) != st.end()) continue; // 重复，因此剪枝 + for (int i = x; i < s.size(); i++) { + if (st.find(s[i]) != st.end()) + continue; // 重复，因此剪枝 st.insert(s[i]); - swap(s[i], s[x]); // 交换，将 s[i] 固定在第 x 位 - dfs(s, x + 1); // 开启固定第 x + 1 位字符 - swap(s[i], s[x]); // 恢复交换 + swap(s[i], s[x]); // 交换，将 s[i] 固定在第 x 位 + dfs(s, x + 1); // 开启固定第 x + 1 位字符 + swap(s[i], s[x]); // 恢复交换 } } }; @@ -35,9 +37,9 @@ int main() { // ======= Test Case ======= string s = "abc"; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); vector res = slt->permutation(s); PrintUtil::printVector(res); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_39_the_majority_element_in_an_array_s1/sfo_39_the_majority_element_in_an_array_s1.cpp b/sword_for_offer/codes/cpp/sfo_39_the_majority_element_in_an_array_s1/sfo_39_the_majority_element_in_an_array_s1.cpp index 3a635e9..8d155e6 100644 --- a/sword_for_offer/codes/cpp/sfo_39_the_majority_element_in_an_array_s1/sfo_39_the_majority_element_in_an_array_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_39_the_majority_element_in_an_array_s1/sfo_39_the_majority_element_in_an_array_s1.cpp @@ -1,18 +1,19 @@ /* -* File: sfo_39_the_majority_element_in_an_array_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_39_the_majority_element_in_an_array_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - int majorityElement(vector& nums) { + public: + int majorityElement(vector &nums) { int x = 0, votes = 0; - for(int num : nums){ - if(votes == 0) x = num; + for (int num : nums) { + if (votes == 0) + x = num; votes += num == x ? 1 : -1; } return x; @@ -21,11 +22,11 @@ class Solution { int main() { // ======= Test Case ======= - vector nums = { 1, 2, 3, 2, 2, 2, 5, 4, 2 }; + vector nums = {1, 2, 3, 2, 2, 2, 5, 4, 2}; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->majorityElement(nums); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_39_the_majority_element_in_an_array_s2/sfo_39_the_majority_element_in_an_array_s2.cpp b/sword_for_offer/codes/cpp/sfo_39_the_majority_element_in_an_array_s2/sfo_39_the_majority_element_in_an_array_s2.cpp index f0f3c35..270aee3 100644 --- a/sword_for_offer/codes/cpp/sfo_39_the_majority_element_in_an_array_s2/sfo_39_the_majority_element_in_an_array_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_39_the_majority_element_in_an_array_s2/sfo_39_the_majority_element_in_an_array_s2.cpp @@ -1,34 +1,36 @@ /* -* File: sfo_39_the_majority_element_in_an_array_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_39_the_majority_element_in_an_array_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - int majorityElement(vector& nums) { + public: + int majorityElement(vector &nums) { int x = 0, votes = 0, count = 0; - for(int num : nums){ - if(votes == 0) x = num; + for (int num : nums) { + if (votes == 0) + x = num; votes += num == x ? 1 : -1; } // 验证 x 是否为众数 - for(int num : nums) - if(num == x) count++; + for (int num : nums) + if (num == x) + count++; return count > nums.size() / 2 ? x : 0; // 当无众数时返回 0 } }; int main() { // ======= Test Case ======= - vector nums = { 1, 2, 3, 2, 2, 2, 5, 4, 2 }; + vector nums = {1, 2, 3, 2, 2, 2, 5, 4, 2}; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->majorityElement(nums); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_40_the_smallest_k_numbers_s1/sfo_40_the_smallest_k_numbers_s1.cpp b/sword_for_offer/codes/cpp/sfo_40_the_smallest_k_numbers_s1/sfo_40_the_smallest_k_numbers_s1.cpp index b026047..07ff5a3 100644 --- a/sword_for_offer/codes/cpp/sfo_40_the_smallest_k_numbers_s1/sfo_40_the_smallest_k_numbers_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_40_the_smallest_k_numbers_s1/sfo_40_the_smallest_k_numbers_s1.cpp @@ -1,29 +1,33 @@ /* -* File: sfo_40_the_smallest_k_numbers_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_40_the_smallest_k_numbers_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - vector getLeastNumbers(vector& arr, int k) { + public: + vector getLeastNumbers(vector &arr, int k) { quickSort(arr, 0, arr.size() - 1); vector res; res.assign(arr.begin(), arr.begin() + k); return res; } -private: - void quickSort(vector& arr, int l, int r) { + + private: + void quickSort(vector &arr, int l, int r) { // 子数组长度为 1 时终止递归 - if (l >= r) return; + if (l >= r) + return; // 哨兵划分操作（以 arr[l] 作为基准数） int i = l, j = r; while (i < j) { - while (i < j && arr[j] >= arr[l]) j--; - while (i < j && arr[i] <= arr[l]) i++; + while (i < j && arr[j] >= arr[l]) + j--; + while (i < j && arr[i] <= arr[l]) + i++; swap(arr[i], arr[j]); } swap(arr[i], arr[l]); @@ -35,12 +39,12 @@ class Solution { int main() { // ======= Test Case ======= - vector arr = { 3, 2, 1 }; + vector arr = {3, 2, 1}; int k = 2; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); vector res = slt->getLeastNumbers(arr, k); PrintUtil::printVector(res); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_40_the_smallest_k_numbers_s2/sfo_40_the_smallest_k_numbers_s2.cpp b/sword_for_offer/codes/cpp/sfo_40_the_smallest_k_numbers_s2/sfo_40_the_smallest_k_numbers_s2.cpp index efb0ab6..9559c78 100644 --- a/sword_for_offer/codes/cpp/sfo_40_the_smallest_k_numbers_s2/sfo_40_the_smallest_k_numbers_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_40_the_smallest_k_numbers_s2/sfo_40_the_smallest_k_numbers_s2.cpp @@ -1,29 +1,35 @@ /* -* File: sfo_40_the_smallest_k_numbers_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_40_the_smallest_k_numbers_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - vector getLeastNumbers(vector& arr, int k) { - if (k >= arr.size()) return arr; + public: + vector getLeastNumbers(vector &arr, int k) { + if (k >= arr.size()) + return arr; return quickSort(arr, k, 0, arr.size() - 1); } -private: - vector quickSort(vector& arr, int k, int l, int r) { + + private: + vector quickSort(vector &arr, int k, int l, int r) { int i = l, j = r; while (i < j) { - while (i < j && arr[j] >= arr[l]) j--; - while (i < j && arr[i] <= arr[l]) i++; + while (i < j && arr[j] >= arr[l]) + j--; + while (i < j && arr[i] <= arr[l]) + i++; swap(arr[i], arr[j]); } swap(arr[i], arr[l]); - if (i > k) return quickSort(arr, k, l, i - 1); - if (i < k) return quickSort(arr, k, i + 1, r); + if (i > k) + return quickSort(arr, k, l, i - 1); + if (i < k) + return quickSort(arr, k, i + 1, r); vector res; res.assign(arr.begin(), arr.begin() + k); return res; @@ -32,12 +38,12 @@ class Solution { int main() { // ======= Test Case ======= - vector arr = { 3, 2, 1 }; + vector arr = {3, 2, 1}; int k = 2; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); vector res = slt->getLeastNumbers(arr, k); PrintUtil::printVector(res); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_41_find_median_from_data_stream_s1/sfo_41_find_median_from_data_stream_s1.cpp b/sword_for_offer/codes/cpp/sfo_41_find_median_from_data_stream_s1/sfo_41_find_median_from_data_stream_s1.cpp index 07a171d..31c204e 100644 --- a/sword_for_offer/codes/cpp/sfo_41_find_median_from_data_stream_s1/sfo_41_find_median_from_data_stream_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_41_find_median_from_data_stream_s1/sfo_41_find_median_from_data_stream_s1.cpp @@ -1,19 +1,20 @@ /* -* File: sfo_41_find_median_from_data_stream_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_41_find_median_from_data_stream_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class MedianFinder { -public: + public: priority_queue, greater> A; // 小顶堆，保存较大的一半 - priority_queue, less> B; // 大顶堆，保存较小的一半 - MedianFinder() { } + priority_queue, less> B; // 大顶堆，保存较小的一半 + MedianFinder() { + } void addNum(int num) { - if(A.size() != B.size()) { + if (A.size() != B.size()) { A.push(num); B.push(A.top()); A.pop(); @@ -31,7 +32,7 @@ class MedianFinder { int main() { // ====== Driver Code ====== vector res; - MedianFinder* medianFinder = new MedianFinder(); + MedianFinder *medianFinder = new MedianFinder(); medianFinder->addNum(1); medianFinder->addNum(2); res.push_back(medianFinder->findMedian()); diff --git a/sword_for_offer/codes/cpp/sfo_42_largest_sum_contiguous_subarray_s1/sfo_42_largest_sum_contiguous_subarray_s1.cpp b/sword_for_offer/codes/cpp/sfo_42_largest_sum_contiguous_subarray_s1/sfo_42_largest_sum_contiguous_subarray_s1.cpp index eb667c6..3a9d7cb 100644 --- a/sword_for_offer/codes/cpp/sfo_42_largest_sum_contiguous_subarray_s1/sfo_42_largest_sum_contiguous_subarray_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_42_largest_sum_contiguous_subarray_s1/sfo_42_largest_sum_contiguous_subarray_s1.cpp @@ -1,31 +1,33 @@ /* -* File: sfo_42_largest_sum_contiguous_subarray_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_42_largest_sum_contiguous_subarray_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - int maxSubArray(vector& nums) { + public: + int maxSubArray(vector &nums) { int res = nums[0]; - for(int i = 1; i < nums.size(); i++) { - if(nums[i - 1] > 0) nums[i] += nums[i - 1]; - if(nums[i] > res) res = nums[i]; + for (int i = 1; i < nums.size(); i++) { + if (nums[i - 1] > 0) + nums[i] += nums[i - 1]; + if (nums[i] > res) + res = nums[i]; } - return res; + return res; } }; int main() { // ======= Test Case ======= - vector nums = { -2, 1, -3, 4, -1, 2, 1, -5, 4 }; + vector nums = {-2, 1, -3, 4, -1, 2, 1, -5, 4}; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->maxSubArray(nums); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1.cpp b/sword_for_offer/codes/cpp/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1.cpp index f98701c..d4c4c40 100644 --- a/sword_for_offer/codes/cpp/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1.cpp @@ -1,21 +1,24 @@ /* -* File: sfo_43_total_number_of_1_in_integers_from_1_to_n_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_43_total_number_of_1_in_integers_from_1_to_n_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: int countDigitOne(int n) { long digit = 1; int high = n / 10, cur = n % 10, low = 0, res = 0; - while(high != 0 || cur != 0) { - if(cur == 0) res += high * digit; - else if(cur == 1) res += high * digit + low + 1; - else res += (high + 1) * digit; + while (high != 0 || cur != 0) { + if (cur == 0) + res += high * digit; + else if (cur == 1) + res += high * digit + low + 1; + else + res += (high + 1) * digit; low += cur * digit; cur = high % 10; high /= 10; @@ -29,9 +32,9 @@ int main() { // ======= Test Case ======= int n = 12; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->countDigitOne(n); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_44_nth_digit_s1/sfo_44_nth_digit_s1.cpp b/sword_for_offer/codes/cpp/sfo_44_nth_digit_s1/sfo_44_nth_digit_s1.cpp index 8da8891..b53783c 100644 --- a/sword_for_offer/codes/cpp/sfo_44_nth_digit_s1/sfo_44_nth_digit_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_44_nth_digit_s1/sfo_44_nth_digit_s1.cpp @@ -1,14 +1,14 @@ /* -* File: sfo_44_nth_digit_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_44_nth_digit_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: int findNthDigit(int n) { int digit = 1; long start = 1; @@ -19,7 +19,7 @@ class Solution { digit += 1; count = digit * start * 9; } - long num = start + (n - 1) / digit; // 2. + long num = start + (n - 1) / digit; // 2. return to_string(num)[(n - 1) % digit] - '0'; // 3. } }; @@ -28,9 +28,9 @@ int main() { // ======= Test Case ======= int n = 3; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->findNthDigit(n); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_45_arrange_an_array_into_the_smallest_number_s1/sfo_45_arrange_an_array_into_the_smallest_number_s1.cpp b/sword_for_offer/codes/cpp/sfo_45_arrange_an_array_into_the_smallest_number_s1/sfo_45_arrange_an_array_into_the_smallest_number_s1.cpp index 4b724a2..0e1383e 100644 --- a/sword_for_offer/codes/cpp/sfo_45_arrange_an_array_into_the_smallest_number_s1/sfo_45_arrange_an_array_into_the_smallest_number_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_45_arrange_an_array_into_the_smallest_number_s1/sfo_45_arrange_an_array_into_the_smallest_number_s1.cpp @@ -1,31 +1,35 @@ /* -* File: sfo_45_arrange_an_array_into_the_smallest_number_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_45_arrange_an_array_into_the_smallest_number_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - string minNumber(vector& nums) { + public: + string minNumber(vector &nums) { vector strs; - for(int i = 0; i < nums.size(); i++) + for (int i = 0; i < nums.size(); i++) strs.push_back(to_string(nums[i])); quickSort(strs, 0, strs.size() - 1); string res; - for(string s : strs) + for (string s : strs) res.append(s); return res; } -private: - void quickSort(vector& strs, int l, int r) { - if(l >= r) return; + + private: + void quickSort(vector &strs, int l, int r) { + if (l >= r) + return; int i = l, j = r; - while(i < j) { - while(strs[j] + strs[l] >= strs[l] + strs[j] && i < j) j--; - while(strs[i] + strs[l] <= strs[l] + strs[i] && i < j) i++; + while (i < j) { + while (strs[j] + strs[l] >= strs[l] + strs[j] && i < j) + j--; + while (strs[i] + strs[l] <= strs[l] + strs[i] && i < j) + i++; swap(strs[i], strs[j]); } swap(strs[i], strs[l]); @@ -36,11 +40,11 @@ class Solution { int main() { // ======= Test Case ======= - vector nums = { 3, 30, 34, 5, 9 }; + vector nums = {3, 30, 34, 5, 9}; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); string res = slt->minNumber(nums); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_45_arrange_an_array_into_the_smallest_number_s2/sfo_45_arrange_an_array_into_the_smallest_number_s2.cpp b/sword_for_offer/codes/cpp/sfo_45_arrange_an_array_into_the_smallest_number_s2/sfo_45_arrange_an_array_into_the_smallest_number_s2.cpp index e1723f2..98efd14 100644 --- a/sword_for_offer/codes/cpp/sfo_45_arrange_an_array_into_the_smallest_number_s2/sfo_45_arrange_an_array_into_the_smallest_number_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_45_arrange_an_array_into_the_smallest_number_s2/sfo_45_arrange_an_array_into_the_smallest_number_s2.cpp @@ -1,21 +1,21 @@ /* -* File: sfo_45_arrange_an_array_into_the_smallest_number_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_45_arrange_an_array_into_the_smallest_number_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - string minNumber(vector& nums) { + public: + string minNumber(vector &nums) { vector strs; string res; - for(int i = 0; i < nums.size(); i++) + for (int i = 0; i < nums.size(); i++) strs.push_back(to_string(nums[i])); - sort(strs.begin(), strs.end(), [](string& x, string& y){ return x + y < y + x; }); - for(int i = 0; i < strs.size(); i++) + sort(strs.begin(), strs.end(), [](string &x, string &y) { return x + y < y + x; }); + for (int i = 0; i < strs.size(); i++) res.append(strs[i]); return res; } @@ -23,11 +23,11 @@ class Solution { int main() { // ======= Test Case ======= - vector nums = { 3, 30, 34, 5, 9 }; + vector nums = {3, 30, 34, 5, 9}; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); string res = slt->minNumber(nums); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_46_translate_numbers_into_strings_s1/sfo_46_translate_numbers_into_strings_s1.cpp b/sword_for_offer/codes/cpp/sfo_46_translate_numbers_into_strings_s1/sfo_46_translate_numbers_into_strings_s1.cpp index d95b4d5..b995ba1 100644 --- a/sword_for_offer/codes/cpp/sfo_46_translate_numbers_into_strings_s1/sfo_46_translate_numbers_into_strings_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_46_translate_numbers_into_strings_s1/sfo_46_translate_numbers_into_strings_s1.cpp @@ -1,18 +1,18 @@ /* -* File: sfo_46_translate_numbers_into_strings_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_46_translate_numbers_into_strings_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: int translateNum(int num) { string s = to_string(num); int a = 1, b = 1, len = s.size(); - for(int i = 2; i <= len; i++) { + for (int i = 2; i <= len; i++) { string tmp = s.substr(i - 2, 2); int c = tmp.compare("10") >= 0 && tmp.compare("25") <= 0 ? a + b : a; b = a; @@ -26,9 +26,9 @@ int main() { // ======= Test Case ======= int num = 12258; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->translateNum(num); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_46_translate_numbers_into_strings_s2/sfo_46_translate_numbers_into_strings_s2.cpp b/sword_for_offer/codes/cpp/sfo_46_translate_numbers_into_strings_s2/sfo_46_translate_numbers_into_strings_s2.cpp index 5e4d147..941535c 100644 --- a/sword_for_offer/codes/cpp/sfo_46_translate_numbers_into_strings_s2/sfo_46_translate_numbers_into_strings_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_46_translate_numbers_into_strings_s2/sfo_46_translate_numbers_into_strings_s2.cpp @@ -1,18 +1,18 @@ /* -* File: sfo_46_translate_numbers_into_strings_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_46_translate_numbers_into_strings_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: int translateNum(int num) { string s = to_string(num); int a = 1, b = 1, len = s.size(); - for(int i = len - 2; i > -1; i--) { + for (int i = len - 2; i > -1; i--) { string tmp = s.substr(i, 2); int c = tmp.compare("10") >= 0 && tmp.compare("25") <= 0 ? a + b : a; b = a; @@ -26,9 +26,9 @@ int main() { // ======= Test Case ======= int num = 12258; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->translateNum(num); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_46_translate_numbers_into_strings_s3/sfo_46_translate_numbers_into_strings_s3.cpp b/sword_for_offer/codes/cpp/sfo_46_translate_numbers_into_strings_s3/sfo_46_translate_numbers_into_strings_s3.cpp index b118901..245fb69 100644 --- a/sword_for_offer/codes/cpp/sfo_46_translate_numbers_into_strings_s3/sfo_46_translate_numbers_into_strings_s3.cpp +++ b/sword_for_offer/codes/cpp/sfo_46_translate_numbers_into_strings_s3/sfo_46_translate_numbers_into_strings_s3.cpp @@ -1,17 +1,17 @@ /* -* File: sfo_46_translate_numbers_into_strings_s3.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_46_translate_numbers_into_strings_s3.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: int translateNum(int num) { int a = 1, b = 1, x, y = num % 10; - while(num > 9) { + while (num > 9) { num /= 10; x = num % 10; int tmp = 10 * x + y; @@ -28,9 +28,9 @@ int main() { // ======= Test Case ======= int num = 12258; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->translateNum(num); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_47_the_maximum_value_of_gifts_s1/sfo_47_the_maximum_value_of_gifts_s1.cpp b/sword_for_offer/codes/cpp/sfo_47_the_maximum_value_of_gifts_s1/sfo_47_the_maximum_value_of_gifts_s1.cpp index c4d9f4b..d7037f2 100644 --- a/sword_for_offer/codes/cpp/sfo_47_the_maximum_value_of_gifts_s1/sfo_47_the_maximum_value_of_gifts_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_47_the_maximum_value_of_gifts_s1/sfo_47_the_maximum_value_of_gifts_s1.cpp @@ -1,22 +1,26 @@ /* -* File: sfo_47_the_maximum_value_of_gifts_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_47_the_maximum_value_of_gifts_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - int maxValue(vector>& grid) { + public: + int maxValue(vector> &grid) { int m = grid.size(), n = grid[0].size(); - for(int i = 0; i < m; i++) { - for(int j = 0; j < n; j++) { - if(i == 0 && j == 0) continue; - if(i == 0) grid[i][j] += grid[i][j - 1] ; - else if(j == 0) grid[i][j] += grid[i - 1][j]; - else grid[i][j] += max(grid[i][j - 1], grid[i - 1][j]); + for (int i = 0; i < m; i++) { + for (int j = 0; j < n; j++) { + if (i == 0 && j == 0) + continue; + if (i == 0) + grid[i][j] += grid[i][j - 1]; + else if (j == 0) + grid[i][j] += grid[i - 1][j]; + else + grid[i][j] += max(grid[i][j - 1], grid[i - 1][j]); } } return grid[m - 1][n - 1]; @@ -25,15 +29,11 @@ class Solution { int main() { // ======= Test Case ======= - vector> grid = { - { 1, 3, 1 }, - { 1, 5, 1 }, - { 4, 2, 1 } -}; + vector> grid = {{1, 3, 1}, {1, 5, 1}, {4, 2, 1}}; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->maxValue(grid); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_47_the_maximum_value_of_gifts_s2/sfo_47_the_maximum_value_of_gifts_s2.cpp b/sword_for_offer/codes/cpp/sfo_47_the_maximum_value_of_gifts_s2/sfo_47_the_maximum_value_of_gifts_s2.cpp index 39b735b..570d594 100644 --- a/sword_for_offer/codes/cpp/sfo_47_the_maximum_value_of_gifts_s2/sfo_47_the_maximum_value_of_gifts_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_47_the_maximum_value_of_gifts_s2/sfo_47_the_maximum_value_of_gifts_s2.cpp @@ -1,22 +1,22 @@ /* -* File: sfo_47_the_maximum_value_of_gifts_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_47_the_maximum_value_of_gifts_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - int maxValue(vector>& grid) { + public: + int maxValue(vector> &grid) { int m = grid.size(), n = grid[0].size(); - for(int j = 1; j < n; j++) // 初始化第一行 + for (int j = 1; j < n; j++) // 初始化第一行 grid[0][j] += grid[0][j - 1]; - for(int i = 1; i < m; i++) // 初始化第一列 + for (int i = 1; i < m; i++) // 初始化第一列 grid[i][0] += grid[i - 1][0]; - for(int i = 1; i < m; i++) - for(int j = 1; j < n; j++) + for (int i = 1; i < m; i++) + for (int j = 1; j < n; j++) grid[i][j] += max(grid[i][j - 1], grid[i - 1][j]); return grid[m - 1][n - 1]; } @@ -24,15 +24,11 @@ class Solution { int main() { // ======= Test Case ======= - vector> grid = { - { 1, 3, 1 }, - { 1, 5, 1 }, - { 4, 2, 1 } -}; + vector> grid = {{1, 3, 1}, {1, 5, 1}, {4, 2, 1}}; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->maxValue(grid); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_48_the_longest_substring_without_repeated_characters_s1/sfo_48_the_longest_substring_without_repeated_characters_s1.cpp b/sword_for_offer/codes/cpp/sfo_48_the_longest_substring_without_repeated_characters_s1/sfo_48_the_longest_substring_without_repeated_characters_s1.cpp index 50cd7c1..cafa2ea 100644 --- a/sword_for_offer/codes/cpp/sfo_48_the_longest_substring_without_repeated_characters_s1/sfo_48_the_longest_substring_without_repeated_characters_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_48_the_longest_substring_without_repeated_characters_s1/sfo_48_the_longest_substring_without_repeated_characters_s1.cpp @@ -1,23 +1,25 @@ /* -* File: sfo_48_the_longest_substring_without_repeated_characters_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_48_the_longest_substring_without_repeated_characters_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: int lengthOfLongestSubstring(string s) { unordered_map dic; int res = 0, tmp = 0, len = s.size(), i; - for(int j = 0; j < len; j++) { - if(dic.find(s[j]) == dic.end()) i = - 1; - else i = dic.find(s[j])->second; // 获取索引 i - dic[s[j]] = j; // 更新哈希表 + for (int j = 0; j < len; j++) { + if (dic.find(s[j]) == dic.end()) + i = -1; + else + i = dic.find(s[j])->second; // 获取索引 i + dic[s[j]] = j; // 更新哈希表 tmp = tmp < j - i ? tmp + 1 : j - i; // dp[j - 1] -> dp[j] - res = max(res, tmp); // max(dp[j - 1], dp[j]) + res = max(res, tmp); // max(dp[j - 1], dp[j]) } return res; } @@ -27,9 +29,9 @@ int main() { // ======= Test Case ======= string s = "abcabcbb"; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->lengthOfLongestSubstring(s); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_48_the_longest_substring_without_repeated_characters_s2/sfo_48_the_longest_substring_without_repeated_characters_s2.cpp b/sword_for_offer/codes/cpp/sfo_48_the_longest_substring_without_repeated_characters_s2/sfo_48_the_longest_substring_without_repeated_characters_s2.cpp index 6360344..cbfc720 100644 --- a/sword_for_offer/codes/cpp/sfo_48_the_longest_substring_without_repeated_characters_s2/sfo_48_the_longest_substring_without_repeated_characters_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_48_the_longest_substring_without_repeated_characters_s2/sfo_48_the_longest_substring_without_repeated_characters_s2.cpp @@ -1,21 +1,22 @@ /* -* File: sfo_48_the_longest_substring_without_repeated_characters_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_48_the_longest_substring_without_repeated_characters_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: int lengthOfLongestSubstring(string s) { int res = 0, tmp = 0, len = s.size(); - for(int j = 0; j < len; j++) { + for (int j = 0; j < len; j++) { int i = j - 1; - while(i >= 0 && s[i] != s[j]) i--; // 线性查找 i + while (i >= 0 && s[i] != s[j]) + i--; // 线性查找 i tmp = tmp < j - i ? tmp + 1 : j - i; // dp[j - 1] -> dp[j] - res = max(res, tmp); // max(dp[j - 1], dp[j]) + res = max(res, tmp); // max(dp[j - 1], dp[j]) } return res; } @@ -25,9 +26,9 @@ int main() { // ======= Test Case ======= string s = "abcabcbb"; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->lengthOfLongestSubstring(s); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_48_the_longest_substring_without_repeated_characters_s3/sfo_48_the_longest_substring_without_repeated_characters_s3.cpp b/sword_for_offer/codes/cpp/sfo_48_the_longest_substring_without_repeated_characters_s3/sfo_48_the_longest_substring_without_repeated_characters_s3.cpp index 7fbd8e9..dd13ad4 100644 --- a/sword_for_offer/codes/cpp/sfo_48_the_longest_substring_without_repeated_characters_s3/sfo_48_the_longest_substring_without_repeated_characters_s3.cpp +++ b/sword_for_offer/codes/cpp/sfo_48_the_longest_substring_without_repeated_characters_s3/sfo_48_the_longest_substring_without_repeated_characters_s3.cpp @@ -1,22 +1,22 @@ /* -* File: sfo_48_the_longest_substring_without_repeated_characters_s3.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_48_the_longest_substring_without_repeated_characters_s3.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: int lengthOfLongestSubstring(string s) { unordered_map dic; int i = -1, res = 0, len = s.size(); - for(int j = 0; j < len; j++) { - if(dic.find(s[j]) != dic.end()) + for (int j = 0; j < len; j++) { + if (dic.find(s[j]) != dic.end()) i = max(i, dic.find(s[j])->second); // 更新左指针 - dic[s[j]] = j; // 哈希表记录 - res = max(res, j - i); // 更新结果 + dic[s[j]] = j; // 哈希表记录 + res = max(res, j - i); // 更新结果 } return res; } @@ -26,9 +26,9 @@ int main() { // ======= Test Case ======= string s = "abcabcbb"; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->lengthOfLongestSubstring(s); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_49_ugly_numbers_s1/sfo_49_ugly_numbers_s1.cpp b/sword_for_offer/codes/cpp/sfo_49_ugly_numbers_s1/sfo_49_ugly_numbers_s1.cpp index 4d3a8e7..74a1c45 100644 --- a/sword_for_offer/codes/cpp/sfo_49_ugly_numbers_s1/sfo_49_ugly_numbers_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_49_ugly_numbers_s1/sfo_49_ugly_numbers_s1.cpp @@ -1,24 +1,27 @@ /* -* File: sfo_49_ugly_numbers_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_49_ugly_numbers_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: int nthUglyNumber(int n) { int a = 0, b = 0, c = 0; int dp[n]; dp[0] = 1; - for(int i = 1; i < n; i++) { + for (int i = 1; i < n; i++) { int n2 = dp[a] * 2, n3 = dp[b] * 3, n5 = dp[c] * 5; dp[i] = min(min(n2, n3), n5); - if(dp[i] == n2) a++; - if(dp[i] == n3) b++; - if(dp[i] == n5) c++; + if (dp[i] == n2) + a++; + if (dp[i] == n3) + b++; + if (dp[i] == n5) + c++; } return dp[n - 1]; } @@ -28,9 +31,9 @@ int main() { // ======= Test Case ======= int n = 10; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->nthUglyNumber(n); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1.cpp b/sword_for_offer/codes/cpp/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1.cpp index a3bba29..29359c5 100644 --- a/sword_for_offer/codes/cpp/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1.cpp @@ -1,20 +1,21 @@ /* -* File: sfo_50_find_the_first_nonrepeating_character_in_a_string_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_50_find_the_first_nonrepeating_character_in_a_string_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: char firstUniqChar(string s) { unordered_map dic; - for(char c : s) + for (char c : s) dic[c] = dic.find(c) == dic.end(); - for(char c : s) - if(dic[c]) return c; + for (char c : s) + if (dic[c]) + return c; return ' '; } }; @@ -23,9 +24,9 @@ int main() { // ======= Test Case ======= string s = "abaccdeff"; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); char res = slt->firstUniqChar(s); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2.cpp b/sword_for_offer/codes/cpp/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2.cpp index 451971a..8f96034 100644 --- a/sword_for_offer/codes/cpp/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2.cpp @@ -1,24 +1,25 @@ /* -* File: sfo_50_find_the_first_nonrepeating_character_in_a_string_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_50_find_the_first_nonrepeating_character_in_a_string_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: char firstUniqChar(string s) { vector keys; unordered_map dic; - for(char c : s) { - if(dic.find(c) == dic.end()) + for (char c : s) { + if (dic.find(c) == dic.end()) keys.push_back(c); dic[c] = dic.find(c) == dic.end(); } - for(char c : keys) { - if(dic[c]) return c; + for (char c : keys) { + if (dic[c]) + return c; } return ' '; } @@ -28,9 +29,9 @@ int main() { // ======= Test Case ======= string s = "abaccdeff"; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); char res = slt->firstUniqChar(s); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_51_reversed_pairs_in_an_array_s1/sfo_51_reversed_pairs_in_an_array_s1.cpp b/sword_for_offer/codes/cpp/sfo_51_reversed_pairs_in_an_array_s1/sfo_51_reversed_pairs_in_an_array_s1.cpp index 5db1334..135e792 100644 --- a/sword_for_offer/codes/cpp/sfo_51_reversed_pairs_in_an_array_s1/sfo_51_reversed_pairs_in_an_array_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_51_reversed_pairs_in_an_array_s1/sfo_51_reversed_pairs_in_an_array_s1.cpp @@ -1,22 +1,24 @@ /* -* File: sfo_51_reversed_pairs_in_an_array_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_51_reversed_pairs_in_an_array_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - int reversePairs(vector& nums) { + public: + int reversePairs(vector &nums) { vector tmp(nums.size()); return mergeSort(0, nums.size() - 1, nums, tmp); } -private: - int mergeSort(int l, int r, vector& nums, vector& tmp) { + + private: + int mergeSort(int l, int r, vector &nums, vector &tmp) { // 终止条件 - if (l >= r) return 0; + if (l >= r) + return 0; // 递归划分 int m = (l + r) / 2; int res = mergeSort(l, m, nums, tmp) + mergeSort(m + 1, r, nums, tmp); @@ -40,11 +42,11 @@ class Solution { int main() { // ======= Test Case ======= - vector nums = { 7, 5, 6, 4 }; + vector nums = {7, 5, 6, 4}; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->reversePairs(nums); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_52_the_first_common_node_in_two_linked_lists_s1/sfo_52_the_first_common_node_in_two_linked_lists_s1.cpp b/sword_for_offer/codes/cpp/sfo_52_the_first_common_node_in_two_linked_lists_s1/sfo_52_the_first_common_node_in_two_linked_lists_s1.cpp index 7acf750..64417c9 100644 --- a/sword_for_offer/codes/cpp/sfo_52_the_first_common_node_in_two_linked_lists_s1/sfo_52_the_first_common_node_in_two_linked_lists_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_52_the_first_common_node_in_two_linked_lists_s1/sfo_52_the_first_common_node_in_two_linked_lists_s1.cpp @@ -1,15 +1,15 @@ /* -* File: sfo_52_the_first_common_node_in_two_linked_lists_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_52_the_first_common_node_in_two_linked_lists_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - ListNode* getIntersectionNode(ListNode *headA, ListNode *headB) { + public: + ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { ListNode *A = headA, *B = headB; while (A != B) { A = A != nullptr ? A->next : headB; @@ -25,14 +25,14 @@ int main() { // headA = 4 -> 1 -> 8 -> 4 -> 5 // ↑ // headB = 5 -> 0 -> 1 - ListNode* headA = vectorToLinkedList(vector { 4, 1, 8, 4, 5 }); - ListNode* headB = vectorToLinkedList(vector { 5, 0, 1, 8, 4, 5 }); - ListNode* intersectA = getListNode(headA, 8); - ListNode* intersectB = getListNode(headB, 1); + ListNode *headA = vectorToLinkedList(vector{4, 1, 8, 4, 5}); + ListNode *headB = vectorToLinkedList(vector{5, 0, 1, 8, 4, 5}); + ListNode *intersectA = getListNode(headA, 8); + ListNode *intersectB = getListNode(headB, 1); intersectB->next = intersectA; // Concatenate the two lists // ====== Driver Code ====== - Solution* slt = new Solution(); - ListNode* res = slt->getIntersectionNode(headA, headB); + Solution *slt = new Solution(); + ListNode *res = slt->getIntersectionNode(headA, headB); cout << res->val << endl; return 0; diff --git a/sword_for_offer/codes/cpp/sfo_53i_find_a_number_in_a_sorted_array_s1/sfo_53i_find_a_number_in_a_sorted_array_s1.cpp b/sword_for_offer/codes/cpp/sfo_53i_find_a_number_in_a_sorted_array_s1/sfo_53i_find_a_number_in_a_sorted_array_s1.cpp index eaa9ee2..7146021 100644 --- a/sword_for_offer/codes/cpp/sfo_53i_find_a_number_in_a_sorted_array_s1/sfo_53i_find_a_number_in_a_sorted_array_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_53i_find_a_number_in_a_sorted_array_s1/sfo_53i_find_a_number_in_a_sorted_array_s1.cpp @@ -1,31 +1,37 @@ /* -* File: sfo_53i_find_a_number_in_a_sorted_array_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_53i_find_a_number_in_a_sorted_array_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - int search(vector& nums, int target) { + public: + int search(vector &nums, int target) { // 搜索右边界 right int i = 0, j = nums.size() - 1; - while(i <= j) { + while (i <= j) { int m = (i + j) / 2; - if(nums[m] <= target) i = m + 1; - else j = m - 1; + if (nums[m] <= target) + i = m + 1; + else + j = m - 1; } int right = i; // 若数组中无 target ，则提前返回 - if(j >= 0 && nums[j] != target) return 0; + if (j >= 0 && nums[j] != target) + return 0; // 搜索左边界 right - i = 0; j = nums.size() - 1; - while(i <= j) { + i = 0; + j = nums.size() - 1; + while (i <= j) { int m = (i + j) / 2; - if(nums[m] < target) i = m + 1; - else j = m - 1; + if (nums[m] < target) + i = m + 1; + else + j = m - 1; } int left = j; return right - left - 1; @@ -34,12 +40,12 @@ class Solution { int main() { // ======= Test Case ======= - vector nums = { 5, 7, 7, 8, 8, 10 }; + vector nums = {5, 7, 7, 8, 8, 10}; int target = 8; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->search(nums, target); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_53i_find_a_number_in_a_sorted_array_s2/sfo_53i_find_a_number_in_a_sorted_array_s2.cpp b/sword_for_offer/codes/cpp/sfo_53i_find_a_number_in_a_sorted_array_s2/sfo_53i_find_a_number_in_a_sorted_array_s2.cpp index a7b8921..c52dc97 100644 --- a/sword_for_offer/codes/cpp/sfo_53i_find_a_number_in_a_sorted_array_s2/sfo_53i_find_a_number_in_a_sorted_array_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_53i_find_a_number_in_a_sorted_array_s2/sfo_53i_find_a_number_in_a_sorted_array_s2.cpp @@ -1,24 +1,27 @@ /* -* File: sfo_53i_find_a_number_in_a_sorted_array_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_53i_find_a_number_in_a_sorted_array_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - int search(vector& nums, int target) { + public: + int search(vector &nums, int target) { return helper(nums, target) - helper(nums, target - 1); } -private: - int helper(vector& nums, int tar) { + + private: + int helper(vector &nums, int tar) { int i = 0, j = nums.size() - 1; - while(i <= j) { + while (i <= j) { int m = (i + j) / 2; - if(nums[m] <= tar) i = m + 1; - else j = m - 1; + if (nums[m] <= tar) + i = m + 1; + else + j = m - 1; } return i; } @@ -26,12 +29,12 @@ class Solution { int main() { // ======= Test Case ======= - vector nums = { 5, 7, 7, 8, 8, 10 }; + vector nums = {5, 7, 7, 8, 8, 10}; int target = 8; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->search(nums, target); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_53ii_the_missing_number_from_0_to_n1_s1/sfo_53ii_the_missing_number_from_0_to_n1_s1.cpp b/sword_for_offer/codes/cpp/sfo_53ii_the_missing_number_from_0_to_n1_s1/sfo_53ii_the_missing_number_from_0_to_n1_s1.cpp index 59182cf..850a161 100644 --- a/sword_for_offer/codes/cpp/sfo_53ii_the_missing_number_from_0_to_n1_s1/sfo_53ii_the_missing_number_from_0_to_n1_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_53ii_the_missing_number_from_0_to_n1_s1/sfo_53ii_the_missing_number_from_0_to_n1_s1.cpp @@ -1,20 +1,22 @@ /* -* File: sfo_53ii_the_missing_number_from_0_to_n1_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_53ii_the_missing_number_from_0_to_n1_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - int missingNumber(vector& nums) { + public: + int missingNumber(vector &nums) { int i = 0, j = nums.size() - 1; - while(i <= j) { + while (i <= j) { int m = (i + j) / 2; - if(nums[m] == m) i = m + 1; - else j = m - 1; + if (nums[m] == m) + i = m + 1; + else + j = m - 1; } return i; } @@ -22,11 +24,11 @@ class Solution { int main() { // ======= Test Case ======= - vector nums = { 0, 1, 3 }; + vector nums = {0, 1, 3}; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->missingNumber(nums); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1.cpp b/sword_for_offer/codes/cpp/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1.cpp index 1f7a32d..5a9b7fc 100644 --- a/sword_for_offer/codes/cpp/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1.cpp @@ -1,38 +1,42 @@ /* -* File: sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - int kthLargest(TreeNode* root, int k) { + public: + int kthLargest(TreeNode *root, int k) { this->k = k; dfs(root); return res; } -private: + + private: int res, k; - void dfs(TreeNode* root) { - if(root == nullptr) return; + void dfs(TreeNode *root) { + if (root == nullptr) + return; dfs(root->right); - if(k == 0) return; - if(--k == 0) res = root->val; + if (k == 0) + return; + if (--k == 0) + res = root->val; dfs(root->left); } }; int main() { // ======= Test Case ======= - TreeNode* root = vectorToTree(vector { 3, 1, 4, INT_MAX, 2, INT_MAX, INT_MAX, INT_MAX, INT_MAX }); + TreeNode *root = vectorToTree(vector{3, 1, 4, INT_MAX, 2, INT_MAX, INT_MAX, INT_MAX, INT_MAX}); int k = 1; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->kthLargest(root, k); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_55i_depth_of_a_binary_tree_s1/sfo_55i_depth_of_a_binary_tree_s1.cpp b/sword_for_offer/codes/cpp/sfo_55i_depth_of_a_binary_tree_s1/sfo_55i_depth_of_a_binary_tree_s1.cpp index 282b207..6960fb7 100644 --- a/sword_for_offer/codes/cpp/sfo_55i_depth_of_a_binary_tree_s1/sfo_55i_depth_of_a_binary_tree_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_55i_depth_of_a_binary_tree_s1/sfo_55i_depth_of_a_binary_tree_s1.cpp @@ -1,27 +1,28 @@ /* -* File: sfo_55i_depth_of_a_binary_tree_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_55i_depth_of_a_binary_tree_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - int maxDepth(TreeNode* root) { - if(root == nullptr) return 0; + public: + int maxDepth(TreeNode *root) { + if (root == nullptr) + return 0; return max(maxDepth(root->left), maxDepth(root->right)) + 1; } }; int main() { // ======= Test Case ======= - TreeNode* root = vectorToTree(vector { 3, 9, 20, INT_MAX, INT_MAX, 15, 7, INT_MAX, INT_MAX, INT_MAX, INT_MAX }); + TreeNode *root = vectorToTree(vector{3, 9, 20, INT_MAX, INT_MAX, 15, 7, INT_MAX, INT_MAX, INT_MAX, INT_MAX}); // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->maxDepth(root); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_55i_depth_of_a_binary_tree_s2/sfo_55i_depth_of_a_binary_tree_s2.cpp b/sword_for_offer/codes/cpp/sfo_55i_depth_of_a_binary_tree_s2/sfo_55i_depth_of_a_binary_tree_s2.cpp index 7c79f40..6ffcea6 100644 --- a/sword_for_offer/codes/cpp/sfo_55i_depth_of_a_binary_tree_s2/sfo_55i_depth_of_a_binary_tree_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_55i_depth_of_a_binary_tree_s2/sfo_55i_depth_of_a_binary_tree_s2.cpp @@ -1,24 +1,27 @@ /* -* File: sfo_55i_depth_of_a_binary_tree_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_55i_depth_of_a_binary_tree_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - int maxDepth(TreeNode* root) { - if(root == nullptr) return 0; - vector que; + public: + int maxDepth(TreeNode *root) { + if (root == nullptr) + return 0; + vector que; que.push_back(root); int res = 0; - while(!que.empty()) { - vector tmp; - for(TreeNode* node : que) { - if(node->left != nullptr) tmp.push_back(node->left); - if(node->right != nullptr) tmp.push_back(node->right); + while (!que.empty()) { + vector tmp; + for (TreeNode *node : que) { + if (node->left != nullptr) + tmp.push_back(node->left); + if (node->right != nullptr) + tmp.push_back(node->right); } que = tmp; res++; @@ -29,11 +32,11 @@ class Solution { int main() { // ======= Test Case ======= - TreeNode* root = vectorToTree(vector { 3, 9, 20, INT_MAX, INT_MAX, 15, 7, INT_MAX, INT_MAX, INT_MAX, INT_MAX }); + TreeNode *root = vectorToTree(vector{3, 9, 20, INT_MAX, INT_MAX, 15, 7, INT_MAX, INT_MAX, INT_MAX, INT_MAX}); // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->maxDepth(root); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_55ii_balanced_binary_tree_s1/sfo_55ii_balanced_binary_tree_s1.cpp b/sword_for_offer/codes/cpp/sfo_55ii_balanced_binary_tree_s1/sfo_55ii_balanced_binary_tree_s1.cpp index 1eecb32..ba31b0c 100644 --- a/sword_for_offer/codes/cpp/sfo_55ii_balanced_binary_tree_s1/sfo_55ii_balanced_binary_tree_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_55ii_balanced_binary_tree_s1/sfo_55ii_balanced_binary_tree_s1.cpp @@ -1,35 +1,39 @@ /* -* File: sfo_55ii_balanced_binary_tree_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_55ii_balanced_binary_tree_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - bool isBalanced(TreeNode* root) { + public: + bool isBalanced(TreeNode *root) { return recur(root) != -1; } -private: - int recur(TreeNode* root) { - if (root == nullptr) return 0; + + private: + int recur(TreeNode *root) { + if (root == nullptr) + return 0; int left = recur(root->left); - if(left == -1) return -1; + if (left == -1) + return -1; int right = recur(root->right); - if(right == -1) return -1; + if (right == -1) + return -1; return abs(left - right) < 2 ? max(left, right) + 1 : -1; } }; int main() { // ======= Test Case ======= - TreeNode* root = vectorToTree(vector { 3, 9, 20, INT_MAX, INT_MAX, 15, 7, INT_MAX, INT_MAX, INT_MAX, INT_MAX }); + TreeNode *root = vectorToTree(vector{3, 9, 20, INT_MAX, INT_MAX, 15, 7, INT_MAX, INT_MAX, INT_MAX, INT_MAX}); // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); bool res = slt->isBalanced(root); cout << boolalpha << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_55ii_balanced_binary_tree_s2/sfo_55ii_balanced_binary_tree_s2.cpp b/sword_for_offer/codes/cpp/sfo_55ii_balanced_binary_tree_s2/sfo_55ii_balanced_binary_tree_s2.cpp index 065893c..557e461 100644 --- a/sword_for_offer/codes/cpp/sfo_55ii_balanced_binary_tree_s2/sfo_55ii_balanced_binary_tree_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_55ii_balanced_binary_tree_s2/sfo_55ii_balanced_binary_tree_s2.cpp @@ -1,32 +1,35 @@ /* -* File: sfo_55ii_balanced_binary_tree_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_55ii_balanced_binary_tree_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - bool isBalanced(TreeNode* root) { - if (root == nullptr) return true; + public: + bool isBalanced(TreeNode *root) { + if (root == nullptr) + return true; return abs(depth(root->left) - depth(root->right)) <= 1 && isBalanced(root->left) && isBalanced(root->right); } -private: - int depth(TreeNode* root) { - if (root == nullptr) return 0; + + private: + int depth(TreeNode *root) { + if (root == nullptr) + return 0; return max(depth(root->left), depth(root->right)) + 1; } }; int main() { // ======= Test Case ======= - TreeNode* root = vectorToTree(vector { 3, 9, 20, INT_MAX, INT_MAX, 15, 7, INT_MAX, INT_MAX, INT_MAX, INT_MAX }); + TreeNode *root = vectorToTree(vector{3, 9, 20, INT_MAX, INT_MAX, 15, 7, INT_MAX, INT_MAX, INT_MAX, INT_MAX}); // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); bool res = slt->isBalanced(root); cout << boolalpha << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_56i_single_number_i_s1/sfo_56i_single_number_i_s1.cpp b/sword_for_offer/codes/cpp/sfo_56i_single_number_i_s1/sfo_56i_single_number_i_s1.cpp index 8ea515e..005506d 100644 --- a/sword_for_offer/codes/cpp/sfo_56i_single_number_i_s1/sfo_56i_single_number_i_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_56i_single_number_i_s1/sfo_56i_single_number_i_s1.cpp @@ -1,35 +1,37 @@ /* -* File: sfo_56i_single_number_i_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_56i_single_number_i_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - vector singleNumbers(vector& nums) { + public: + vector singleNumbers(vector &nums) { int x = 0, y = 0, n = 0, m = 1; - for(int num : nums) // 1. 遍历异或 + for (int num : nums) // 1. 遍历异或 n ^= num; - while((n & m) == 0) // 2. 循环左移，计算 m + while ((n & m) == 0) // 2. 循环左移，计算 m m <<= 1; - for(int num : nums) { // 3. 遍历 nums 分组 - if(num & m) x ^= num; // 4. 当 num & m != 0 - else y ^= num; // 4. 当 num & m == 0 + for (int num : nums) { // 3. 遍历 nums 分组 + if (num & m) + x ^= num; // 4. 当 num & m != 0 + else + y ^= num; // 4. 当 num & m == 0 } - return vector {x, y}; // 5. 返回出现一次的数字 + return vector{x, y}; // 5. 返回出现一次的数字 } }; int main() { // ======= Test Case ======= - vector nums = { 4, 1, 4, 6 }; + vector nums = {4, 1, 4, 6}; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); vector res = slt->singleNumbers(nums); PrintUtil::printVector(res); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_56ii_single_number_ii_s1/sfo_56ii_single_number_ii_s1.cpp b/sword_for_offer/codes/cpp/sfo_56ii_single_number_ii_s1/sfo_56ii_single_number_ii_s1.cpp index 1157195..975d1a8 100644 --- a/sword_for_offer/codes/cpp/sfo_56ii_single_number_ii_s1/sfo_56ii_single_number_ii_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_56ii_single_number_ii_s1/sfo_56ii_single_number_ii_s1.cpp @@ -1,17 +1,17 @@ /* -* File: sfo_56ii_single_number_ii_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_56ii_single_number_ii_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - int singleNumber(vector& nums) { + public: + int singleNumber(vector &nums) { int ones = 0, twos = 0; - for(int num : nums){ + for (int num : nums) { ones = ones ^ num & ~twos; twos = twos ^ num & ~ones; } @@ -21,11 +21,11 @@ class Solution { int main() { // ======= Test Case ======= - vector nums = { 3, 4, 3, 3 }; + vector nums = {3, 4, 3, 3}; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->singleNumber(nums); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_56ii_single_number_ii_s2/sfo_56ii_single_number_ii_s2.cpp b/sword_for_offer/codes/cpp/sfo_56ii_single_number_ii_s2/sfo_56ii_single_number_ii_s2.cpp index 8af4b37..b20fcd4 100644 --- a/sword_for_offer/codes/cpp/sfo_56ii_single_number_ii_s2/sfo_56ii_single_number_ii_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_56ii_single_number_ii_s2/sfo_56ii_single_number_ii_s2.cpp @@ -1,26 +1,26 @@ /* -* File: sfo_56ii_single_number_ii_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_56ii_single_number_ii_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - int singleNumber(vector& nums) { - int counts[32] = {0}; // C++ 初始化数组需要写明初始值 0 - for(int num : nums) { - for(int i = 0; i < 32; i++) { + public: + int singleNumber(vector &nums) { + int counts[32] = {0}; // C++ 初始化数组需要写明初始值 0 + for (int num : nums) { + for (int i = 0; i < 32; i++) { counts[i] += num & 1; // 更新第 i 位 1 的个数之和 num >>= 1; // 第 i 位 --> 第 i 位 } } int res = 0, m = 3; - for(int i = 31; i >= 0; i--) { + for (int i = 31; i >= 0; i--) { res <<= 1; - res |= counts[i] % m; // 恢复第 i 位 + res |= counts[i] % m; // 恢复第 i 位 } return res; } @@ -28,11 +28,11 @@ class Solution { int main() { // ======= Test Case ======= - vector nums = { 3, 4, 3, 3 }; + vector nums = {3, 4, 3, 3}; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->singleNumber(nums); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_57_two_numbers_with_sum_s_s1/sfo_57_two_numbers_with_sum_s_s1.cpp b/sword_for_offer/codes/cpp/sfo_57_two_numbers_with_sum_s_s1/sfo_57_two_numbers_with_sum_s_s1.cpp index cea8b07..f86acae 100644 --- a/sword_for_offer/codes/cpp/sfo_57_two_numbers_with_sum_s_s1/sfo_57_two_numbers_with_sum_s_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_57_two_numbers_with_sum_s_s1/sfo_57_two_numbers_with_sum_s_s1.cpp @@ -1,21 +1,24 @@ /* -* File: sfo_57_two_numbers_with_sum_s_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_57_two_numbers_with_sum_s_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - vector twoSum(vector& nums, int target) { + public: + vector twoSum(vector &nums, int target) { int i = 0, j = nums.size() - 1; - while(i < j) { + while (i < j) { int s = nums[i] + nums[j]; - if(s < target) i++; - else if(s > target) j--; - else return { nums[i], nums[j] }; + if (s < target) + i++; + else if (s > target) + j--; + else + return {nums[i], nums[j]}; } return {}; } @@ -23,12 +26,12 @@ class Solution { int main() { // ======= Test Case ======= - vector nums = { 2, 7, 11, 15 }; + vector nums = {2, 7, 11, 15}; int target = 9; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); vector res = slt->twoSum(nums, target); PrintUtil::printVector(res); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_57ii_consecutive_numbers_with_sum_s_s1/sfo_57ii_consecutive_numbers_with_sum_s_s1.cpp b/sword_for_offer/codes/cpp/sfo_57ii_consecutive_numbers_with_sum_s_s1/sfo_57ii_consecutive_numbers_with_sum_s_s1.cpp index 1fdc139..5341d36 100644 --- a/sword_for_offer/codes/cpp/sfo_57ii_consecutive_numbers_with_sum_s_s1/sfo_57ii_consecutive_numbers_with_sum_s_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_57ii_consecutive_numbers_with_sum_s_s1/sfo_57ii_consecutive_numbers_with_sum_s_s1.cpp @@ -1,23 +1,23 @@ /* -* File: sfo_57ii_consecutive_numbers_with_sum_s_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_57ii_consecutive_numbers_with_sum_s_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: vector> findContinuousSequence(int target) { int i = 1; double j = 2.0; vector> res; - while(i < j) { - j = (-1 + sqrt(1 + 4 * (2 * target + (long) i * i - i))) / 2; - if(i < j && j == (int)j) { + while (i < j) { + j = (-1 + sqrt(1 + 4 * (2 * target + (long)i * i - i))) / 2; + if (i < j && j == (int)j) { vector ans; - for(int k = i; k <= (int)j; k++) + for (int k = i; k <= (int)j; k++) ans.push_back(k); res.push_back(ans); } @@ -31,9 +31,9 @@ int main() { // ======= Test Case ======= int target = 9; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); vector> res = slt->findContinuousSequence(target); PrintUtil::printVectorMatrix(res); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_57ii_consecutive_numbers_with_sum_s_s2/sfo_57ii_consecutive_numbers_with_sum_s_s2.cpp b/sword_for_offer/codes/cpp/sfo_57ii_consecutive_numbers_with_sum_s_s2/sfo_57ii_consecutive_numbers_with_sum_s_s2.cpp index bd522a9..32b5dc3 100644 --- a/sword_for_offer/codes/cpp/sfo_57ii_consecutive_numbers_with_sum_s_s2/sfo_57ii_consecutive_numbers_with_sum_s_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_57ii_consecutive_numbers_with_sum_s_s2/sfo_57ii_consecutive_numbers_with_sum_s_s2.cpp @@ -1,25 +1,25 @@ /* -* File: sfo_57ii_consecutive_numbers_with_sum_s_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_57ii_consecutive_numbers_with_sum_s_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: vector> findContinuousSequence(int target) { int i = 1, j = 2, s = 3; vector> res; - while(i < j) { - if(s == target) { + while (i < j) { + if (s == target) { vector ans; - for(int k = i; k <= j; k++) + for (int k = i; k <= j; k++) ans.push_back(k); res.push_back(ans); } - if(s >= target) { + if (s >= target) { s -= i; i++; } else { @@ -35,9 +35,9 @@ int main() { // ======= Test Case ======= int target = 9; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); vector> res = slt->findContinuousSequence(target); PrintUtil::printVectorMatrix(res); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_58ii_left_rotation_of_a_string_s1/sfo_58ii_left_rotation_of_a_string_s1.cpp b/sword_for_offer/codes/cpp/sfo_58ii_left_rotation_of_a_string_s1/sfo_58ii_left_rotation_of_a_string_s1.cpp index 339e8df..383d5d8 100644 --- a/sword_for_offer/codes/cpp/sfo_58ii_left_rotation_of_a_string_s1/sfo_58ii_left_rotation_of_a_string_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_58ii_left_rotation_of_a_string_s1/sfo_58ii_left_rotation_of_a_string_s1.cpp @@ -1,14 +1,14 @@ /* -* File: sfo_58ii_left_rotation_of_a_string_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_58ii_left_rotation_of_a_string_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: string reverseLeftWords(string s, int n) { return s.substr(n, s.size()) + s.substr(0, n); } @@ -19,9 +19,9 @@ int main() { string s = "abcdefg"; int n = 2; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); string res = slt->reverseLeftWords(s, n); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_58ii_left_rotation_of_a_string_s2/sfo_58ii_left_rotation_of_a_string_s2.cpp b/sword_for_offer/codes/cpp/sfo_58ii_left_rotation_of_a_string_s2/sfo_58ii_left_rotation_of_a_string_s2.cpp index 7113c28..a63ee08 100644 --- a/sword_for_offer/codes/cpp/sfo_58ii_left_rotation_of_a_string_s2/sfo_58ii_left_rotation_of_a_string_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_58ii_left_rotation_of_a_string_s2/sfo_58ii_left_rotation_of_a_string_s2.cpp @@ -1,23 +1,25 @@ /* -* File: sfo_58ii_left_rotation_of_a_string_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_58ii_left_rotation_of_a_string_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: string reverseLeftWords(string s, int n) { reverseString(s, 0, n - 1); reverseString(s, n, s.size() - 1); reverseString(s, 0, s.size() - 1); return s; } -private: - void reverseString(string& s, int i, int j) { - while(i < j) swap(s[i++], s[j--]); + + private: + void reverseString(string &s, int i, int j) { + while (i < j) + swap(s[i++], s[j--]); } }; @@ -26,9 +28,9 @@ int main() { string s = "abcdefg"; int n = 2; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); string res = slt->reverseLeftWords(s, n); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_58ii_left_rotation_of_a_string_s3/sfo_58ii_left_rotation_of_a_string_s3.cpp b/sword_for_offer/codes/cpp/sfo_58ii_left_rotation_of_a_string_s3/sfo_58ii_left_rotation_of_a_string_s3.cpp index 6a7dac3..358cd3d 100644 --- a/sword_for_offer/codes/cpp/sfo_58ii_left_rotation_of_a_string_s3/sfo_58ii_left_rotation_of_a_string_s3.cpp +++ b/sword_for_offer/codes/cpp/sfo_58ii_left_rotation_of_a_string_s3/sfo_58ii_left_rotation_of_a_string_s3.cpp @@ -1,14 +1,14 @@ /* -* File: sfo_58ii_left_rotation_of_a_string_s3.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_58ii_left_rotation_of_a_string_s3.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: string reverseLeftWords(string s, int n) { reverse(s.begin(), s.begin() + n); reverse(s.begin() + n, s.end()); @@ -22,9 +22,9 @@ int main() { string s = "abcdefg"; int n = 2; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); string res = slt->reverseLeftWords(s, n); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_59ii_max_queue_s1/sfo_59ii_max_queue_s1.cpp b/sword_for_offer/codes/cpp/sfo_59ii_max_queue_s1/sfo_59ii_max_queue_s1.cpp index 93a1388..2966b4c 100644 --- a/sword_for_offer/codes/cpp/sfo_59ii_max_queue_s1/sfo_59ii_max_queue_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_59ii_max_queue_s1/sfo_59ii_max_queue_s1.cpp @@ -1,8 +1,8 @@ /* -* File: sfo_59ii_max_queue_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_59ii_max_queue_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" @@ -10,21 +10,24 @@ class MaxQueue { queue que; deque deq; -public: - MaxQueue() { } + + public: + MaxQueue() { + } int max_value() { return deq.empty() ? -1 : deq.front(); } void push_back(int value) { que.push(value); - while(!deq.empty() && deq.back() < value) + while (!deq.empty() && deq.back() < value) deq.pop_back(); deq.push_back(value); } int pop_front() { - if(que.empty()) return -1; + if (que.empty()) + return -1; int val = que.front(); - if(val == deq.front()) + if (val == deq.front()) deq.pop_front(); que.pop(); return val; @@ -34,7 +37,7 @@ class MaxQueue { int main() { // ====== Driver Code ====== vector res; - MaxQueue* maxQueue = new MaxQueue(); + MaxQueue *maxQueue = new MaxQueue(); maxQueue->push_back(1); maxQueue->push_back(2); res.push_back(maxQueue->max_value()); diff --git a/sword_for_offer/codes/cpp/sfo_60_probabilities_for_rolling_n_dices_s1/sfo_60_probabilities_for_rolling_n_dices_s1.cpp b/sword_for_offer/codes/cpp/sfo_60_probabilities_for_rolling_n_dices_s1/sfo_60_probabilities_for_rolling_n_dices_s1.cpp index 13108c6..336bf83 100644 --- a/sword_for_offer/codes/cpp/sfo_60_probabilities_for_rolling_n_dices_s1/sfo_60_probabilities_for_rolling_n_dices_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_60_probabilities_for_rolling_n_dices_s1/sfo_60_probabilities_for_rolling_n_dices_s1.cpp @@ -1,14 +1,14 @@ /* -* File: sfo_60_probabilities_for_rolling_n_dices_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_60_probabilities_for_rolling_n_dices_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: vector dicesProbability(int n) { vector dp(6, 1.0 / 6.0); for (int i = 2; i <= n; i++) { @@ -28,9 +28,9 @@ int main() { // ======= Test Case ======= int n = 2; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); vector res = slt->dicesProbability(n); PrintUtil::printVector(res); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_61_straight_in_poker_s1/sfo_61_straight_in_poker_s1.cpp b/sword_for_offer/codes/cpp/sfo_61_straight_in_poker_s1/sfo_61_straight_in_poker_s1.cpp index 6bc4823..5413f2f 100644 --- a/sword_for_offer/codes/cpp/sfo_61_straight_in_poker_s1/sfo_61_straight_in_poker_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_61_straight_in_poker_s1/sfo_61_straight_in_poker_s1.cpp @@ -1,22 +1,24 @@ /* -* File: sfo_61_straight_in_poker_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_61_straight_in_poker_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - bool isStraight(vector& nums) { + public: + bool isStraight(vector &nums) { unordered_set repeat; int ma = 0, mi = 14; - for(int num : nums) { - if(num == 0) continue; // 跳过大小王 + for (int num : nums) { + if (num == 0) + continue; // 跳过大小王 ma = max(ma, num); // 最大牌 mi = min(mi, num); // 最小牌 - if(repeat.find(num) != repeat.end()) return false; // 若有重复，提前返回 false + if (repeat.find(num) != repeat.end()) + return false; // 若有重复，提前返回 false repeat.insert(num); // 添加此牌至 Set } return ma - mi < 5; // 最大牌 - 最小牌 < 5 则可构成顺子 @@ -25,11 +27,11 @@ class Solution { int main() { // ======= Test Case ======= - vector nums = { 0, 0, 1, 2, 5 }; + vector nums = {0, 0, 1, 2, 5}; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); bool res = slt->isStraight(nums); cout << boolalpha << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_61_straight_in_poker_s2/sfo_61_straight_in_poker_s2.cpp b/sword_for_offer/codes/cpp/sfo_61_straight_in_poker_s2/sfo_61_straight_in_poker_s2.cpp index 1e862ca..9c95a91 100644 --- a/sword_for_offer/codes/cpp/sfo_61_straight_in_poker_s2/sfo_61_straight_in_poker_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_61_straight_in_poker_s2/sfo_61_straight_in_poker_s2.cpp @@ -1,20 +1,22 @@ /* -* File: sfo_61_straight_in_poker_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_61_straight_in_poker_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - bool isStraight(vector& nums) { + public: + bool isStraight(vector &nums) { int joker = 0; sort(nums.begin(), nums.end()); // 数组排序 - for(int i = 0; i < 4; i++) { - if(nums[i] == 0) joker++; // 统计大小王数量 - else if(nums[i] == nums[i + 1]) return false; // 若有重复，提前返回 false + for (int i = 0; i < 4; i++) { + if (nums[i] == 0) + joker++; // 统计大小王数量 + else if (nums[i] == nums[i + 1]) + return false; // 若有重复，提前返回 false } return nums[4] - nums[joker] < 5; // 最大牌 - 最小牌 < 5 则可构成顺子 } @@ -22,11 +24,11 @@ class Solution { int main() { // ======= Test Case ======= - vector nums = { 0, 0, 1, 2, 5 }; + vector nums = {0, 0, 1, 2, 5}; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); bool res = slt->isStraight(nums); cout << boolalpha << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_62_josephus_problem_s1/sfo_62_josephus_problem_s1.cpp b/sword_for_offer/codes/cpp/sfo_62_josephus_problem_s1/sfo_62_josephus_problem_s1.cpp index b18e82b..82dd97f 100644 --- a/sword_for_offer/codes/cpp/sfo_62_josephus_problem_s1/sfo_62_josephus_problem_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_62_josephus_problem_s1/sfo_62_josephus_problem_s1.cpp @@ -1,14 +1,14 @@ /* -* File: sfo_62_josephus_problem_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_62_josephus_problem_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: int lastRemaining(int n, int m) { int x = 0; for (int i = 2; i <= n; i++) { @@ -23,9 +23,9 @@ int main() { int n = 5; int m = 3; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->lastRemaining(n, m); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_63_the_maximum_profit_of_stocks_s1/sfo_63_the_maximum_profit_of_stocks_s1.cpp b/sword_for_offer/codes/cpp/sfo_63_the_maximum_profit_of_stocks_s1/sfo_63_the_maximum_profit_of_stocks_s1.cpp index f791bb8..f1735a4 100644 --- a/sword_for_offer/codes/cpp/sfo_63_the_maximum_profit_of_stocks_s1/sfo_63_the_maximum_profit_of_stocks_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_63_the_maximum_profit_of_stocks_s1/sfo_63_the_maximum_profit_of_stocks_s1.cpp @@ -1,17 +1,17 @@ /* -* File: sfo_63_the_maximum_profit_of_stocks_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_63_the_maximum_profit_of_stocks_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - int maxProfit(vector& prices) { + public: + int maxProfit(vector &prices) { int cost = INT_MAX, profit = 0; - for(int price : prices) { + for (int price : prices) { cost = min(cost, price); profit = max(profit, price - cost); } @@ -21,11 +21,11 @@ class Solution { int main() { // ======= Test Case ======= - vector prices = { 7, 1, 5, 3, 6, 4 }; + vector prices = {7, 1, 5, 3, 6, 4}; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->maxProfit(prices); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_64_solve_1_2___n_s1/sfo_64_solve_1_2___n_s1.cpp b/sword_for_offer/codes/cpp/sfo_64_solve_1_2___n_s1/sfo_64_solve_1_2___n_s1.cpp index 4b7cd73..3f50253 100644 --- a/sword_for_offer/codes/cpp/sfo_64_solve_1_2___n_s1/sfo_64_solve_1_2___n_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_64_solve_1_2___n_s1/sfo_64_solve_1_2___n_s1.cpp @@ -1,14 +1,14 @@ /* -* File: sfo_64_solve_1_2___n_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_64_solve_1_2___n_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: int sumNums(int n) { n > 1 && (n += sumNums(n - 1)); return n; @@ -19,9 +19,9 @@ int main() { // ======= Test Case ======= int n = 3; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->sumNums(n); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_65_implement_addition_operation_without_arithmetic_operators_s1/sfo_65_implement_addition_operation_without_arithmetic_operators_s1.cpp b/sword_for_offer/codes/cpp/sfo_65_implement_addition_operation_without_arithmetic_operators_s1/sfo_65_implement_addition_operation_without_arithmetic_operators_s1.cpp index cb64ce9..b67eb53 100644 --- a/sword_for_offer/codes/cpp/sfo_65_implement_addition_operation_without_arithmetic_operators_s1/sfo_65_implement_addition_operation_without_arithmetic_operators_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_65_implement_addition_operation_without_arithmetic_operators_s1/sfo_65_implement_addition_operation_without_arithmetic_operators_s1.cpp @@ -1,17 +1,16 @@ /* -* File: sfo_65_implement_addition_operation_without_arithmetic_operators_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_65_implement_addition_operation_without_arithmetic_operators_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: int add(int a, int b) { - while(b != 0) - { + while (b != 0) { int c = (unsigned int)(a & b) << 1; a ^= b; b = c; @@ -25,9 +24,9 @@ int main() { int a = 1; int b = 1; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->add(a, b); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_66_a_product_array_puzzle_s1/sfo_66_a_product_array_puzzle_s1.cpp b/sword_for_offer/codes/cpp/sfo_66_a_product_array_puzzle_s1/sfo_66_a_product_array_puzzle_s1.cpp index 3db51e5..58ad324 100644 --- a/sword_for_offer/codes/cpp/sfo_66_a_product_array_puzzle_s1/sfo_66_a_product_array_puzzle_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_66_a_product_array_puzzle_s1/sfo_66_a_product_array_puzzle_s1.cpp @@ -1,24 +1,25 @@ /* -* File: sfo_66_a_product_array_puzzle_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_66_a_product_array_puzzle_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - vector constructArr(vector& a) { + public: + vector constructArr(vector &a) { int len = a.size(); - if(len == 0) return {}; + if (len == 0) + return {}; vector b(len, 1); b[0] = 1; int tmp = 1; - for(int i = 1; i < len; i++) { + for (int i = 1; i < len; i++) { b[i] = b[i - 1] * a[i - 1]; } - for(int i = len - 2; i >= 0; i--) { + for (int i = len - 2; i >= 0; i--) { tmp *= a[i + 1]; b[i] *= tmp; } @@ -28,11 +29,11 @@ class Solution { int main() { // ======= Test Case ======= - vector a = { 1, 2, 3, 4, 5 }; + vector a = {1, 2, 3, 4, 5}; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); vector res = slt->constructArr(a); PrintUtil::printVector(res); - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_67_convert_string_to_int_s1/sfo_67_convert_string_to_int_s1.cpp b/sword_for_offer/codes/cpp/sfo_67_convert_string_to_int_s1/sfo_67_convert_string_to_int_s1.cpp index 60ab737..6780249 100644 --- a/sword_for_offer/codes/cpp/sfo_67_convert_string_to_int_s1/sfo_67_convert_string_to_int_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_67_convert_string_to_int_s1/sfo_67_convert_string_to_int_s1.cpp @@ -1,25 +1,30 @@ /* -* File: sfo_67_convert_string_to_int_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_67_convert_string_to_int_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: + public: int strToInt(string str) { int res = 0, bndry = INT_MAX / 10; int i = 0, sign = 1, length = str.size(); - if(length == 0) return 0; - while(str[i] == ' ') - if(++i == length) return 0; - if(str[i] == '-') sign = -1; - if(str[i] == '-' || str[i] == '+') i++; - for(int j = i; j < length; j++) { - if(str[j] < '0' || str[j] > '9') break; - if(res > bndry || res == bndry && str[j] > '7') + if (length == 0) + return 0; + while (str[i] == ' ') + if (++i == length) + return 0; + if (str[i] == '-') + sign = -1; + if (str[i] == '-' || str[i] == '+') + i++; + for (int j = i; j < length; j++) { + if (str[j] < '0' || str[j] > '9') + break; + if (res > bndry || res == bndry && str[j] > '7') return sign == 1 ? INT_MAX : INT_MIN; res = res * 10 + (str[j] - '0'); } @@ -31,9 +36,9 @@ int main() { // ======= Test Case ======= string str = "42"; // ====== Driver Code ====== - Solution* slt = new Solution(); + Solution *slt = new Solution(); int res = slt->strToInt(str); cout << res << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1.cpp b/sword_for_offer/codes/cpp/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1.cpp index 55b9a7c..0089ec4 100644 --- a/sword_for_offer/codes/cpp/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1.cpp @@ -1,21 +1,22 @@ /* -* File: sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { - while(root != nullptr) { - if(root->val < p->val && root->val < q->val) // p,q 都在 root 的右子树中 - root = root->right; // 遍历至右子节点 - else if(root->val > p->val && root->val > q->val) // p,q 都在 root 的左子树中 - root = root->left; // 遍历至左子节点 - else break; + public: + TreeNode *lowestCommonAncestor(TreeNode *root, TreeNode *p, TreeNode *q) { + while (root != nullptr) { + if (root->val < p->val && root->val < q->val) // p,q 都在 root 的右子树中 + root = root->right; // 遍历至右子节点 + else if (root->val > p->val && root->val > q->val) // p,q 都在 root 的左子树中 + root = root->left; // 遍历至左子节点 + else + break; } return root; } @@ -23,13 +24,14 @@ class Solution { int main() { // ======= Test Case ======= - TreeNode* root = vectorToTree(vector { 6, 2, 8, 0, 4, 7, 9, INT_MAX, INT_MAX, 3, 5, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX }); - TreeNode* p = new TreeNode(2); - TreeNode* q = new TreeNode(8); + TreeNode *root = vectorToTree(vector{6, 2, 8, 0, 4, 7, 9, INT_MAX, INT_MAX, 3, 5, INT_MAX, INT_MAX, INT_MAX, + INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX}); + TreeNode *p = new TreeNode(2); + TreeNode *q = new TreeNode(8); // ====== Driver Code ====== - Solution* slt = new Solution(); - TreeNode* res = slt->lowestCommonAncestor(root, p, q); + Solution *slt = new Solution(); + TreeNode *res = slt->lowestCommonAncestor(root, p, q); cout << res->val << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2.cpp b/sword_for_offer/codes/cpp/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2.cpp index 5c03b89..4586339 100644 --- a/sword_for_offer/codes/cpp/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2.cpp @@ -1,23 +1,24 @@ /* -* File: sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { - if(p->val > q->val) + public: + TreeNode *lowestCommonAncestor(TreeNode *root, TreeNode *p, TreeNode *q) { + if (p->val > q->val) swap(p, q); - while(root != nullptr) { - if(root->val < p->val) // p,q 都在 root 的右子树中 - root = root->right; // 遍历至右子节点 - else if(root->val > q->val) // p,q 都在 root 的左子树中 - root = root->left; // 遍历至左子节点 - else break; + while (root != nullptr) { + if (root->val < p->val) // p,q 都在 root 的右子树中 + root = root->right; // 遍历至右子节点 + else if (root->val > q->val) // p,q 都在 root 的左子树中 + root = root->left; // 遍历至左子节点 + else + break; } return root; } @@ -25,13 +26,14 @@ class Solution { int main() { // ======= Test Case ======= - TreeNode* root = vectorToTree(vector { 6, 2, 8, 0, 4, 7, 9, INT_MAX, INT_MAX, 3, 5, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX }); - TreeNode* p = new TreeNode(2); - TreeNode* q = new TreeNode(8); + TreeNode *root = vectorToTree(vector{6, 2, 8, 0, 4, 7, 9, INT_MAX, INT_MAX, 3, 5, INT_MAX, INT_MAX, INT_MAX, + INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX}); + TreeNode *p = new TreeNode(2); + TreeNode *q = new TreeNode(8); // ====== Driver Code ====== - Solution* slt = new Solution(); - TreeNode* res = slt->lowestCommonAncestor(root, p, q); + Solution *slt = new Solution(); + TreeNode *res = slt->lowestCommonAncestor(root, p, q); cout << res->val << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3.cpp b/sword_for_offer/codes/cpp/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3.cpp index 4568d73..7350545 100644 --- a/sword_for_offer/codes/cpp/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3.cpp +++ b/sword_for_offer/codes/cpp/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3.cpp @@ -1,18 +1,18 @@ /* -* File: sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { - if(root->val < p->val && root->val < q->val) + public: + TreeNode *lowestCommonAncestor(TreeNode *root, TreeNode *p, TreeNode *q) { + if (root->val < p->val && root->val < q->val) return lowestCommonAncestor(root->right, p, q); - if(root->val > p->val && root->val > q->val) + if (root->val > p->val && root->val > q->val) return lowestCommonAncestor(root->left, p, q); return root; } @@ -20,13 +20,14 @@ class Solution { int main() { // ======= Test Case ======= - TreeNode* root = vectorToTree(vector { 6, 2, 8, 0, 4, 7, 9, INT_MAX, INT_MAX, 3, 5, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX }); - TreeNode* p = new TreeNode(2); - TreeNode* q = new TreeNode(8); + TreeNode *root = vectorToTree(vector{6, 2, 8, 0, 4, 7, 9, INT_MAX, INT_MAX, 3, 5, INT_MAX, INT_MAX, INT_MAX, + INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX}); + TreeNode *p = new TreeNode(2); + TreeNode *q = new TreeNode(8); // ====== Driver Code ====== - Solution* slt = new Solution(); - TreeNode* res = slt->lowestCommonAncestor(root, p, q); + Solution *slt = new Solution(); + TreeNode *res = slt->lowestCommonAncestor(root, p, q); cout << res->val << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1.cpp b/sword_for_offer/codes/cpp/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1.cpp index 7adc63a..ebc8e6c 100644 --- a/sword_for_offer/codes/cpp/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1.cpp +++ b/sword_for_offer/codes/cpp/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1.cpp @@ -1,33 +1,37 @@ /* -* File: sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { - if(root == nullptr || root == p || root == q) return root; + public: + TreeNode *lowestCommonAncestor(TreeNode *root, TreeNode *p, TreeNode *q) { + if (root == nullptr || root == p || root == q) + return root; TreeNode *left = lowestCommonAncestor(root->left, p, q); TreeNode *right = lowestCommonAncestor(root->right, p, q); - if(left == nullptr) return right; - if(right == nullptr) return left; + if (left == nullptr) + return right; + if (right == nullptr) + return left; return root; } }; int main() { // ======= Test Case ======= - TreeNode* root = vectorToTree(vector { 3, 5, 1, 6, 2, 0, 8, INT_MAX, INT_MAX, 7, 4, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX }); - TreeNode* p = getTreeNode(root, 5); - TreeNode* q = getTreeNode(root, 1); + TreeNode *root = vectorToTree(vector{3, 5, 1, 6, 2, 0, 8, INT_MAX, INT_MAX, 7, 4, INT_MAX, INT_MAX, INT_MAX, + INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX}); + TreeNode *p = getTreeNode(root, 5); + TreeNode *q = getTreeNode(root, 1); // ====== Driver Code ====== - Solution* slt = new Solution(); - TreeNode* res = slt->lowestCommonAncestor(root, p, q); + Solution *slt = new Solution(); + TreeNode *res = slt->lowestCommonAncestor(root, p, q); cout << res->val << endl; - + return 0; } diff --git a/sword_for_offer/codes/cpp/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2.cpp b/sword_for_offer/codes/cpp/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2.cpp index 6e57b95..b6c23c6 100644 --- a/sword_for_offer/codes/cpp/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2.cpp +++ b/sword_for_offer/codes/cpp/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2.cpp @@ -1,34 +1,39 @@ /* -* File: sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2.cpp -* Created Time: 2021-12-09 -* Author: Krahets (krahets@163.com) -*/ + * File: sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2.cpp + * Created Time: 2021-12-09 + * Author: Krahets (krahets@163.com) + */ #include "../include/include.hpp" // ===== Solution Code ===== class Solution { -public: - TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { - if(root == nullptr || root == p || root == q) return root; + public: + TreeNode *lowestCommonAncestor(TreeNode *root, TreeNode *p, TreeNode *q) { + if (root == nullptr || root == p || root == q) + return root; TreeNode *left = lowestCommonAncestor(root->left, p, q); TreeNode *right = lowestCommonAncestor(root->right, p, q); - if(left == nullptr && right == nullptr) return nullptr; // 1. - if(left == nullptr) return right; // 3. - if(right == nullptr) return left; // 4. - return root; // 2. if(left != null and right != null) + if (left == nullptr && right == nullptr) + return nullptr; // 1. + if (left == nullptr) + return right; // 3. + if (right == nullptr) + return left; // 4. + return root; // 2. if(left != null and right != null) } }; int main() { // ======= Test Case ======= - TreeNode* root = vectorToTree(vector { 3, 5, 1, 6, 2, 0, 8, INT_MAX, INT_MAX, 7, 4, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX }); - TreeNode* p = getTreeNode(root, 5); - TreeNode* q = getTreeNode(root, 1); + TreeNode *root = vectorToTree(vector{3, 5, 1, 6, 2, 0, 8, INT_MAX, INT_MAX, 7, 4, INT_MAX, INT_MAX, INT_MAX, + INT_MAX, INT_MAX, INT_MAX, INT_MAX, INT_MAX}); + TreeNode *p = getTreeNode(root, 5); + TreeNode *q = getTreeNode(root, 1); // ====== Driver Code ====== - Solution* slt = new Solution(); - TreeNode* res = slt->lowestCommonAncestor(root, p, q); + Solution *slt = new Solution(); + TreeNode *res = slt->lowestCommonAncestor(root, p, q); cout << res->val << endl; - + return 0; } diff --git a/sword_for_offer/codes/java/include/ListNode.java b/sword_for_offer/codes/java/include/ListNode.java index d3e7e84..4a0a2c6 100644 --- a/sword_for_offer/codes/java/include/ListNode.java +++ b/sword_for_offer/codes/java/include/ListNode.java @@ -12,9 +12,10 @@ public class ListNode { public ListNode(int x) { val = x; } - + /** * Generate a linked list with an array + * * @param arr * @return */ @@ -30,6 +31,7 @@ public static ListNode arrToLinkedList(int[] arr) { /** * Get a list node with specific value from a linked list + * * @param head * @param val * @return diff --git a/sword_for_offer/codes/java/include/PrintUtil.java b/sword_for_offer/codes/java/include/PrintUtil.java index d5b78e2..4438a29 100644 --- a/sword_for_offer/codes/java/include/PrintUtil.java +++ b/sword_for_offer/codes/java/include/PrintUtil.java @@ -15,6 +15,7 @@ class Trunk { public class PrintUtil { /** * Print a linked list + * * @param head */ public static void printLinkedList(ListNode head) { @@ -30,6 +31,7 @@ public static void printLinkedList(ListNode head) { * The interface of the tree printer * This tree printer is borrowed from TECHIE DELIGHT * https://www.techiedelight.com/c-program-print-binary-tree/ + * * @param root */ public static void printTree(TreeNode root) { @@ -38,6 +40,7 @@ public static void printTree(TreeNode root) { /** * Print a binary tree + * * @param root * @param prev * @param isLeft diff --git a/sword_for_offer/codes/java/include/TreeNode.java b/sword_for_offer/codes/java/include/TreeNode.java index 5ed7f33..fd46a58 100644 --- a/sword_for_offer/codes/java/include/TreeNode.java +++ b/sword_for_offer/codes/java/include/TreeNode.java @@ -16,21 +16,26 @@ public TreeNode(int x) { /** * Generate a binary tree with an array + * * @param arr * @return */ public static TreeNode arrToTree(Integer[] arr) { TreeNode root = new TreeNode(arr[0]); - Queue queue = new LinkedList() {{ add(root); }}; + Queue queue = new LinkedList() { + { + add(root); + } + }; int i = 1; - while(!queue.isEmpty()) { + while (!queue.isEmpty()) { TreeNode node = queue.poll(); - if(arr[i] != null) { + if (arr[i] != null) { node.left = new TreeNode(arr[i]); queue.add(node.left); } i++; - if(arr[i] != null) { + if (arr[i] != null) { node.right = new TreeNode(arr[i]); queue.add(node.right); } @@ -41,29 +46,35 @@ public static TreeNode arrToTree(Integer[] arr) { /** * Serialize a binary tree to a list + * * @param root * @return */ public static List treeToList(TreeNode root) { List list = new ArrayList<>(); - if(root == null) return list; - Queue queue = new LinkedList() {{ add(root); }}; - while(!queue.isEmpty()) { + if (root == null) + return list; + Queue queue = new LinkedList() { + { + add(root); + } + }; + while (!queue.isEmpty()) { TreeNode node = queue.poll(); - if(node != null) { + if (node != null) { list.add(node.val); queue.add(node.left); queue.add(node.right); - } - else { + } else { list.add(null); } } return list; } - + /** * Get a tree node with specific value in a binary tree + * * @param root * @param val * @return diff --git a/sword_for_offer/codes/java/sfo_03_find_duplicate_numbers_in_an_array_s1/sfo_03_find_duplicate_numbers_in_an_array_s1.java b/sword_for_offer/codes/java/sfo_03_find_duplicate_numbers_in_an_array_s1/sfo_03_find_duplicate_numbers_in_an_array_s1.java index 8c38ea5..34ac00f 100644 --- a/sword_for_offer/codes/java/sfo_03_find_duplicate_numbers_in_an_array_s1/sfo_03_find_duplicate_numbers_in_an_array_s1.java +++ b/sword_for_offer/codes/java/sfo_03_find_duplicate_numbers_in_an_array_s1/sfo_03_find_duplicate_numbers_in_an_array_s1.java @@ -13,8 +13,9 @@ class Solution { public int findRepeatNumber(int[] nums) { Set dic = new HashSet<>(); - for(int num : nums) { - if(dic.contains(num)) return num; + for (int num : nums) { + if (dic.contains(num)) + return num; dic.add(num); } return -1; diff --git a/sword_for_offer/codes/java/sfo_03_find_duplicate_numbers_in_an_array_s2/sfo_03_find_duplicate_numbers_in_an_array_s2.java b/sword_for_offer/codes/java/sfo_03_find_duplicate_numbers_in_an_array_s2/sfo_03_find_duplicate_numbers_in_an_array_s2.java index 52c6319..4ee6fc2 100644 --- a/sword_for_offer/codes/java/sfo_03_find_duplicate_numbers_in_an_array_s2/sfo_03_find_duplicate_numbers_in_an_array_s2.java +++ b/sword_for_offer/codes/java/sfo_03_find_duplicate_numbers_in_an_array_s2/sfo_03_find_duplicate_numbers_in_an_array_s2.java @@ -13,12 +13,13 @@ class Solution { public int findRepeatNumber(int[] nums) { int i = 0; - while(i < nums.length) { - if(nums[i] == i) { + while (i < nums.length) { + if (nums[i] == i) { i++; continue; } - if(nums[nums[i]] == nums[i]) return nums[i]; + if (nums[nums[i]] == nums[i]) + return nums[i]; int tmp = nums[i]; nums[i] = nums[tmp]; nums[tmp] = tmp; diff --git a/sword_for_offer/codes/java/sfo_04_find_a_number_in_2d_matrix_s1/sfo_04_find_a_number_in_2d_matrix_s1.java b/sword_for_offer/codes/java/sfo_04_find_a_number_in_2d_matrix_s1/sfo_04_find_a_number_in_2d_matrix_s1.java index 8f7b072..ea6e60b 100644 --- a/sword_for_offer/codes/java/sfo_04_find_a_number_in_2d_matrix_s1/sfo_04_find_a_number_in_2d_matrix_s1.java +++ b/sword_for_offer/codes/java/sfo_04_find_a_number_in_2d_matrix_s1/sfo_04_find_a_number_in_2d_matrix_s1.java @@ -13,11 +13,13 @@ class Solution { public boolean findNumberIn2DArray(int[][] matrix, int target) { int i = matrix.length - 1, j = 0; - while(i >= 0 && j < matrix[0].length) - { - if(matrix[i][j] > target) i--; - else if(matrix[i][j] < target) j++; - else return true; + while (i >= 0 && j < matrix[0].length) { + if (matrix[i][j] > target) + i--; + else if (matrix[i][j] < target) + j++; + else + return true; } return false; } @@ -27,11 +29,11 @@ public class sfo_04_find_a_number_in_2d_matrix_s1 { public static void main(String[] args) { // ======= Test Case ======= int[][] matrix = { - { 1, 4, 7, 11, 15 }, - { 2, 5, 8, 12, 19 }, - { 3, 6, 9, 16, 22 }, - { 10, 13, 14, 17, 24 }, - { 18, 21, 23, 26, 30 } + { 1, 4, 7, 11, 15 }, + { 2, 5, 8, 12, 19 }, + { 3, 6, 9, 16, 22 }, + { 10, 13, 14, 17, 24 }, + { 18, 21, 23, 26, 30 } }; int target = 5; // ====== Driver Code ====== diff --git a/sword_for_offer/codes/java/sfo_05_replace_spaces_s1/sfo_05_replace_spaces_s1.java b/sword_for_offer/codes/java/sfo_05_replace_spaces_s1/sfo_05_replace_spaces_s1.java index bb79bd1..9d18cdb 100644 --- a/sword_for_offer/codes/java/sfo_05_replace_spaces_s1/sfo_05_replace_spaces_s1.java +++ b/sword_for_offer/codes/java/sfo_05_replace_spaces_s1/sfo_05_replace_spaces_s1.java @@ -13,10 +13,11 @@ class Solution { public String replaceSpace(String s) { StringBuilder res = new StringBuilder(); - for(Character c : s.toCharArray()) - { - if(c == ' ') res.append("%20"); - else res.append(c); + for (Character c : s.toCharArray()) { + if (c == ' ') + res.append("%20"); + else + res.append(c); } return res.toString(); } diff --git a/sword_for_offer/codes/java/sfo_06_print_a_linked_list_in_reverse_order_s1/sfo_06_print_a_linked_list_in_reverse_order_s1.java b/sword_for_offer/codes/java/sfo_06_print_a_linked_list_in_reverse_order_s1/sfo_06_print_a_linked_list_in_reverse_order_s1.java index 1435329..c474285 100644 --- a/sword_for_offer/codes/java/sfo_06_print_a_linked_list_in_reverse_order_s1/sfo_06_print_a_linked_list_in_reverse_order_s1.java +++ b/sword_for_offer/codes/java/sfo_06_print_a_linked_list_in_reverse_order_s1/sfo_06_print_a_linked_list_in_reverse_order_s1.java @@ -12,15 +12,18 @@ // ===== Solution Code ===== class Solution { ArrayList tmp = new ArrayList(); + public int[] reversePrint(ListNode head) { recur(head); int[] res = new int[tmp.size()]; - for(int i = 0; i < res.length; i++) + for (int i = 0; i < res.length; i++) res[i] = tmp.get(i); return res; } + void recur(ListNode head) { - if(head == null) return; + if (head == null) + return; recur(head.next); tmp.add(head.val); } diff --git a/sword_for_offer/codes/java/sfo_06_print_a_linked_list_in_reverse_order_s2/sfo_06_print_a_linked_list_in_reverse_order_s2.java b/sword_for_offer/codes/java/sfo_06_print_a_linked_list_in_reverse_order_s2/sfo_06_print_a_linked_list_in_reverse_order_s2.java index fc3da6d..1466ff9 100644 --- a/sword_for_offer/codes/java/sfo_06_print_a_linked_list_in_reverse_order_s2/sfo_06_print_a_linked_list_in_reverse_order_s2.java +++ b/sword_for_offer/codes/java/sfo_06_print_a_linked_list_in_reverse_order_s2/sfo_06_print_a_linked_list_in_reverse_order_s2.java @@ -13,14 +13,14 @@ class Solution { public int[] reversePrint(ListNode head) { LinkedList stack = new LinkedList(); - while(head != null) { + while (head != null) { stack.addLast(head.val); head = head.next; } int[] res = new int[stack.size()]; - for(int i = 0; i < res.length; i++) + for (int i = 0; i < res.length; i++) res[i] = stack.removeLast(); - return res; + return res; } } diff --git a/sword_for_offer/codes/java/sfo_07_reconstruct_binary_tree_s1/sfo_07_reconstruct_binary_tree_s1.java b/sword_for_offer/codes/java/sfo_07_reconstruct_binary_tree_s1/sfo_07_reconstruct_binary_tree_s1.java index 442faf1..f13c1aa 100644 --- a/sword_for_offer/codes/java/sfo_07_reconstruct_binary_tree_s1/sfo_07_reconstruct_binary_tree_s1.java +++ b/sword_for_offer/codes/java/sfo_07_reconstruct_binary_tree_s1/sfo_07_reconstruct_binary_tree_s1.java @@ -13,19 +13,22 @@ class Solution { int[] preorder; HashMap dic = new HashMap<>(); + public TreeNode buildTree(int[] preorder, int[] inorder) { this.preorder = preorder; - for(int i = 0; i < inorder.length; i++) + for (int i = 0; i < inorder.length; i++) dic.put(inorder[i], i); return recur(0, 0, inorder.length - 1); } + TreeNode recur(int root, int left, int right) { - if(left > right) return null; // 递归终止 - TreeNode node = new TreeNode(preorder[root]); // 建立根节点 - int i = dic.get(preorder[root]); // 划分根节点、左子树、右子树 - node.left = recur(root + 1, left, i - 1); // 开启左子树递归 + if (left > right) + return null; // 递归终止 + TreeNode node = new TreeNode(preorder[root]); // 建立根节点 + int i = dic.get(preorder[root]); // 划分根节点、左子树、右子树 + node.left = recur(root + 1, left, i - 1); // 开启左子树递归 node.right = recur(root + i - left + 1, i + 1, right); // 开启右子树递归 - return node; // 回溯返回根节点 + return node; // 回溯返回根节点 } } diff --git a/sword_for_offer/codes/java/sfo_09_implement_a_queue_using_two_stacks_s1/sfo_09_implement_a_queue_using_two_stacks_s1.java b/sword_for_offer/codes/java/sfo_09_implement_a_queue_using_two_stacks_s1/sfo_09_implement_a_queue_using_two_stacks_s1.java index d8911d2..77e1fb0 100644 --- a/sword_for_offer/codes/java/sfo_09_implement_a_queue_using_two_stacks_s1/sfo_09_implement_a_queue_using_two_stacks_s1.java +++ b/sword_for_offer/codes/java/sfo_09_implement_a_queue_using_two_stacks_s1/sfo_09_implement_a_queue_using_two_stacks_s1.java @@ -12,17 +12,22 @@ // ===== Solution Code ===== class CQueue { LinkedList A, B; + public CQueue() { A = new LinkedList(); B = new LinkedList(); } + public void appendTail(int value) { A.addLast(value); } + public int deleteHead() { - if(!B.isEmpty()) return B.removeLast(); - if(A.isEmpty()) return -1; - while(!A.isEmpty()) + if (!B.isEmpty()) + return B.removeLast(); + if (A.isEmpty()) + return -1; + while (!A.isEmpty()) B.addLast(A.removeLast()); return B.removeLast(); } diff --git a/sword_for_offer/codes/java/sfo_10i_fibonacci_numbers_s1/sfo_10i_fibonacci_numbers_s1.java b/sword_for_offer/codes/java/sfo_10i_fibonacci_numbers_s1/sfo_10i_fibonacci_numbers_s1.java index 6442f9f..1d0ca49 100644 --- a/sword_for_offer/codes/java/sfo_10i_fibonacci_numbers_s1/sfo_10i_fibonacci_numbers_s1.java +++ b/sword_for_offer/codes/java/sfo_10i_fibonacci_numbers_s1/sfo_10i_fibonacci_numbers_s1.java @@ -13,7 +13,7 @@ class Solution { public int fib(int n) { int a = 0, b = 1, sum; - for(int i = 0; i < n; i++){ + for (int i = 0; i < n; i++) { sum = (a + b) % 1000000007; a = b; b = sum; diff --git a/sword_for_offer/codes/java/sfo_10ii_frog_jump_s1/sfo_10ii_frog_jump_s1.java b/sword_for_offer/codes/java/sfo_10ii_frog_jump_s1/sfo_10ii_frog_jump_s1.java index 85b3b9e..5e0740a 100644 --- a/sword_for_offer/codes/java/sfo_10ii_frog_jump_s1/sfo_10ii_frog_jump_s1.java +++ b/sword_for_offer/codes/java/sfo_10ii_frog_jump_s1/sfo_10ii_frog_jump_s1.java @@ -13,7 +13,7 @@ class Solution { public int numWays(int n) { int a = 1, b = 1, sum; - for(int i = 0; i < n; i++){ + for (int i = 0; i < n; i++) { sum = (a + b) % 1000000007; a = b; b = sum; diff --git a/sword_for_offer/codes/java/sfo_11_find_minimum_in_rotated_sorted_array_s1/sfo_11_find_minimum_in_rotated_sorted_array_s1.java b/sword_for_offer/codes/java/sfo_11_find_minimum_in_rotated_sorted_array_s1/sfo_11_find_minimum_in_rotated_sorted_array_s1.java index 400c7df..c58f893 100644 --- a/sword_for_offer/codes/java/sfo_11_find_minimum_in_rotated_sorted_array_s1/sfo_11_find_minimum_in_rotated_sorted_array_s1.java +++ b/sword_for_offer/codes/java/sfo_11_find_minimum_in_rotated_sorted_array_s1/sfo_11_find_minimum_in_rotated_sorted_array_s1.java @@ -15,9 +15,12 @@ public int minArray(int[] numbers) { int i = 0, j = numbers.length - 1; while (i < j) { int m = (i + j) / 2; - if (numbers[m] > numbers[j]) i = m + 1; - else if (numbers[m] < numbers[j]) j = m; - else j--; + if (numbers[m] > numbers[j]) + i = m + 1; + else if (numbers[m] < numbers[j]) + j = m; + else + j--; } return numbers[i]; } diff --git a/sword_for_offer/codes/java/sfo_11_find_minimum_in_rotated_sorted_array_s2/sfo_11_find_minimum_in_rotated_sorted_array_s2.java b/sword_for_offer/codes/java/sfo_11_find_minimum_in_rotated_sorted_array_s2/sfo_11_find_minimum_in_rotated_sorted_array_s2.java index b0b30ba..e17fe92 100644 --- a/sword_for_offer/codes/java/sfo_11_find_minimum_in_rotated_sorted_array_s2/sfo_11_find_minimum_in_rotated_sorted_array_s2.java +++ b/sword_for_offer/codes/java/sfo_11_find_minimum_in_rotated_sorted_array_s2/sfo_11_find_minimum_in_rotated_sorted_array_s2.java @@ -15,12 +15,15 @@ public int minArray(int[] numbers) { int i = 0, j = numbers.length - 1; while (i < j) { int m = (i + j) / 2; - if (numbers[m] > numbers[j]) i = m + 1; - else if (numbers[m] < numbers[j]) j = m; + if (numbers[m] > numbers[j]) + i = m + 1; + else if (numbers[m] < numbers[j]) + j = m; else { int x = i; - for(int k = i + 1; k < j; k++) { - if(numbers[k] < numbers[x]) x = k; + for (int k = i + 1; k < j; k++) { + if (numbers[k] < numbers[x]) + x = k; } return numbers[x]; } diff --git a/sword_for_offer/codes/java/sfo_12_word_search_s1/sfo_12_word_search_s1.java b/sword_for_offer/codes/java/sfo_12_word_search_s1/sfo_12_word_search_s1.java index 09abfff..c4db16e 100644 --- a/sword_for_offer/codes/java/sfo_12_word_search_s1/sfo_12_word_search_s1.java +++ b/sword_for_offer/codes/java/sfo_12_word_search_s1/sfo_12_word_search_s1.java @@ -13,19 +13,23 @@ class Solution { public boolean exist(char[][] board, String word) { char[] words = word.toCharArray(); - for(int i = 0; i < board.length; i++) { - for(int j = 0; j < board[0].length; j++) { - if(dfs(board, words, i, j, 0)) return true; + for (int i = 0; i < board.length; i++) { + for (int j = 0; j < board[0].length; j++) { + if (dfs(board, words, i, j, 0)) + return true; } } return false; } + boolean dfs(char[][] board, char[] word, int i, int j, int k) { - if(i >= board.length || i < 0 || j >= board[0].length || j < 0 || board[i][j] != word[k]) return false; - if(k == word.length - 1) return true; + if (i >= board.length || i < 0 || j >= board[0].length || j < 0 || board[i][j] != word[k]) + return false; + if (k == word.length - 1) + return true; board[i][j] = '\0'; - boolean res = dfs(board, word, i + 1, j, k + 1) || dfs(board, word, i - 1, j, k + 1) || - dfs(board, word, i, j + 1, k + 1) || dfs(board, word, i , j - 1, k + 1); + boolean res = dfs(board, word, i + 1, j, k + 1) || dfs(board, word, i - 1, j, k + 1) || + dfs(board, word, i, j + 1, k + 1) || dfs(board, word, i, j - 1, k + 1); board[i][j] = word[k]; return res; } @@ -35,10 +39,10 @@ public class sfo_12_word_search_s1 { public static void main(String[] args) { // ======= Test Case ======= char[][] board = { - { 'A', 'B', 'C', 'E' }, - { 'S', 'F', 'C', 'S' }, - { 'A', 'D', 'E', 'E' } -}; + { 'A', 'B', 'C', 'E' }, + { 'S', 'F', 'C', 'S' }, + { 'A', 'D', 'E', 'E' } + }; String word = "ABCCED"; // ====== Driver Code ====== Solution slt = new Solution(); diff --git a/sword_for_offer/codes/java/sfo_13_range_of_motion_of_a_robot_s1/sfo_13_range_of_motion_of_a_robot_s1.java b/sword_for_offer/codes/java/sfo_13_range_of_motion_of_a_robot_s1/sfo_13_range_of_motion_of_a_robot_s1.java index 40490c3..0e29cd1 100644 --- a/sword_for_offer/codes/java/sfo_13_range_of_motion_of_a_robot_s1/sfo_13_range_of_motion_of_a_robot_s1.java +++ b/sword_for_offer/codes/java/sfo_13_range_of_motion_of_a_robot_s1/sfo_13_range_of_motion_of_a_robot_s1.java @@ -13,15 +13,21 @@ class Solution { int m, n, k; boolean[][] visited; + public int movingCount(int m, int n, int k) { - this.m = m; this.n = n; this.k = k; + this.m = m; + this.n = n; + this.k = k; this.visited = new boolean[m][n]; return dfs(0, 0, 0, 0); } + public int dfs(int i, int j, int si, int sj) { - if(i >= m || j >= n || k < si + sj || visited[i][j]) return 0; + if (i >= m || j >= n || k < si + sj || visited[i][j]) + return 0; visited[i][j] = true; - return 1 + dfs(i + 1, j, (i + 1) % 10 != 0 ? si + 1 : si - 8, sj) + dfs(i, j + 1, si, (j + 1) % 10 != 0 ? sj + 1 : sj - 8); + return 1 + dfs(i + 1, j, (i + 1) % 10 != 0 ? si + 1 : si - 8, sj) + + dfs(i, j + 1, si, (j + 1) % 10 != 0 ? sj + 1 : sj - 8); } } diff --git a/sword_for_offer/codes/java/sfo_13_range_of_motion_of_a_robot_s2/sfo_13_range_of_motion_of_a_robot_s2.java b/sword_for_offer/codes/java/sfo_13_range_of_motion_of_a_robot_s2/sfo_13_range_of_motion_of_a_robot_s2.java index 946032e..92e2bf9 100644 --- a/sword_for_offer/codes/java/sfo_13_range_of_motion_of_a_robot_s2/sfo_13_range_of_motion_of_a_robot_s2.java +++ b/sword_for_offer/codes/java/sfo_13_range_of_motion_of_a_robot_s2/sfo_13_range_of_motion_of_a_robot_s2.java @@ -14,14 +14,15 @@ class Solution { public int movingCount(int m, int n, int k) { boolean[][] visited = new boolean[m][n]; int res = 0; - Queue queue= new LinkedList(); + Queue queue = new LinkedList(); queue.add(new int[] { 0, 0, 0, 0 }); - while(queue.size() > 0) { + while (queue.size() > 0) { int[] x = queue.poll(); int i = x[0], j = x[1], si = x[2], sj = x[3]; - if(i >= m || j >= n || k < si + sj || visited[i][j]) continue; + if (i >= m || j >= n || k < si + sj || visited[i][j]) + continue; visited[i][j] = true; - res ++; + res++; queue.add(new int[] { i + 1, j, (i + 1) % 10 != 0 ? si + 1 : si - 8, sj }); queue.add(new int[] { i, j + 1, si, (j + 1) % 10 != 0 ? sj + 1 : sj - 8 }); } diff --git a/sword_for_offer/codes/java/sfo_14i_cut_the_rope_i_s1/sfo_14i_cut_the_rope_i_s1.java b/sword_for_offer/codes/java/sfo_14i_cut_the_rope_i_s1/sfo_14i_cut_the_rope_i_s1.java index 0c4548b..fcbfe05 100644 --- a/sword_for_offer/codes/java/sfo_14i_cut_the_rope_i_s1/sfo_14i_cut_the_rope_i_s1.java +++ b/sword_for_offer/codes/java/sfo_14i_cut_the_rope_i_s1/sfo_14i_cut_the_rope_i_s1.java @@ -12,11 +12,14 @@ // ===== Solution Code ===== class Solution { public int cuttingRope(int n) { - if(n <= 3) return n - 1; + if (n <= 3) + return n - 1; int a = n / 3, b = n % 3; - if(b == 0) return (int)Math.pow(3, a); - if(b == 1) return (int)Math.pow(3, a - 1) * 4; - return (int)Math.pow(3, a) * 2; + if (b == 0) + return (int) Math.pow(3, a); + if (b == 1) + return (int) Math.pow(3, a - 1) * 4; + return (int) Math.pow(3, a) * 2; } } diff --git a/sword_for_offer/codes/java/sfo_14ii_cut_the_rope_ii_s1/sfo_14ii_cut_the_rope_ii_s1.java b/sword_for_offer/codes/java/sfo_14ii_cut_the_rope_ii_s1/sfo_14ii_cut_the_rope_ii_s1.java index 63b21c5..edd54ef 100644 --- a/sword_for_offer/codes/java/sfo_14ii_cut_the_rope_ii_s1/sfo_14ii_cut_the_rope_ii_s1.java +++ b/sword_for_offer/codes/java/sfo_14ii_cut_the_rope_ii_s1/sfo_14ii_cut_the_rope_ii_s1.java @@ -12,16 +12,20 @@ // ===== Solution Code ===== class Solution { public int cuttingRope(int n) { - if(n <= 3) return n - 1; + if (n <= 3) + return n - 1; int b = n % 3, p = 1000000007; long rem = 1, x = 3; - for(int a = n / 3 - 1; a > 0; a /= 2) { - if(a % 2 == 1) rem = (rem * x) % p; + for (int a = n / 3 - 1; a > 0; a /= 2) { + if (a % 2 == 1) + rem = (rem * x) % p; x = (x * x) % p; } - if(b == 0) return (int)(rem * 3 % p); - if(b == 1) return (int)(rem * 4 % p); - return (int)(rem * 6 % p); + if (b == 0) + return (int) (rem * 3 % p); + if (b == 1) + return (int) (rem * 4 % p); + return (int) (rem * 6 % p); } } diff --git a/sword_for_offer/codes/java/sfo_15_number_of_1_bits_s1/sfo_15_number_of_1_bits_s1.java b/sword_for_offer/codes/java/sfo_15_number_of_1_bits_s1/sfo_15_number_of_1_bits_s1.java index 719a28c..cb8cd37 100644 --- a/sword_for_offer/codes/java/sfo_15_number_of_1_bits_s1/sfo_15_number_of_1_bits_s1.java +++ b/sword_for_offer/codes/java/sfo_15_number_of_1_bits_s1/sfo_15_number_of_1_bits_s1.java @@ -13,7 +13,7 @@ class Solution { public int hammingWeight(int n) { int res = 0; - while(n != 0) { + while (n != 0) { res += n & 1; n >>>= 1; } diff --git a/sword_for_offer/codes/java/sfo_15_number_of_1_bits_s2/sfo_15_number_of_1_bits_s2.java b/sword_for_offer/codes/java/sfo_15_number_of_1_bits_s2/sfo_15_number_of_1_bits_s2.java index b36f811..b901527 100644 --- a/sword_for_offer/codes/java/sfo_15_number_of_1_bits_s2/sfo_15_number_of_1_bits_s2.java +++ b/sword_for_offer/codes/java/sfo_15_number_of_1_bits_s2/sfo_15_number_of_1_bits_s2.java @@ -13,7 +13,7 @@ class Solution { public int hammingWeight(int n) { int res = 0; - while(n != 0) { + while (n != 0) { res++; n &= n - 1; } diff --git a/sword_for_offer/codes/java/sfo_16_powers_of_integers_s1/sfo_16_powers_of_integers_s1.java b/sword_for_offer/codes/java/sfo_16_powers_of_integers_s1/sfo_16_powers_of_integers_s1.java index ec8a6b0..2f618d5 100644 --- a/sword_for_offer/codes/java/sfo_16_powers_of_integers_s1/sfo_16_powers_of_integers_s1.java +++ b/sword_for_offer/codes/java/sfo_16_powers_of_integers_s1/sfo_16_powers_of_integers_s1.java @@ -12,15 +12,17 @@ // ===== Solution Code ===== class Solution { public double myPow(double x, int n) { - if(x == 0.0f) return 0.0d; + if (x == 0.0f) + return 0.0d; long b = n; double res = 1.0; - if(b < 0) { + if (b < 0) { x = 1 / x; b = -b; } - while(b > 0) { - if((b & 1) == 1) res *= x; + while (b > 0) { + if ((b & 1) == 1) + res *= x; x *= x; b >>= 1; } diff --git a/sword_for_offer/codes/java/sfo_17_print_from_1_to_the_largest_n_digits_s1/sfo_17_print_from_1_to_the_largest_n_digits_s1.java b/sword_for_offer/codes/java/sfo_17_print_from_1_to_the_largest_n_digits_s1/sfo_17_print_from_1_to_the_largest_n_digits_s1.java index 7f97c4d..8691b7a 100644 --- a/sword_for_offer/codes/java/sfo_17_print_from_1_to_the_largest_n_digits_s1/sfo_17_print_from_1_to_the_largest_n_digits_s1.java +++ b/sword_for_offer/codes/java/sfo_17_print_from_1_to_the_largest_n_digits_s1/sfo_17_print_from_1_to_the_largest_n_digits_s1.java @@ -12,9 +12,9 @@ // ===== Solution Code ===== class Solution { public int[] printNumbers(int n) { - int end = (int)Math.pow(10, n) - 1; + int end = (int) Math.pow(10, n) - 1; int[] res = new int[end]; - for(int i = 0; i < end; i++) + for (int i = 0; i < end; i++) res[i] = i + 1; return res; } diff --git a/sword_for_offer/codes/java/sfo_17_print_from_1_to_the_largest_n_digits_s2/sfo_17_print_from_1_to_the_largest_n_digits_s2.java b/sword_for_offer/codes/java/sfo_17_print_from_1_to_the_largest_n_digits_s2/sfo_17_print_from_1_to_the_largest_n_digits_s2.java index cefa882..6e3d3b9 100644 --- a/sword_for_offer/codes/java/sfo_17_print_from_1_to_the_largest_n_digits_s2/sfo_17_print_from_1_to_the_largest_n_digits_s2.java +++ b/sword_for_offer/codes/java/sfo_17_print_from_1_to_the_largest_n_digits_s2/sfo_17_print_from_1_to_the_largest_n_digits_s2.java @@ -13,7 +13,8 @@ class Solution { StringBuilder res; int count = 0, n; - char[] num, loop = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9'}; + char[] num, loop = { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9' }; + public String printNumbers(int n) { this.n = n; res = new StringBuilder(); // 数字字符串集 @@ -22,12 +23,13 @@ public String printNumbers(int n) { res.deleteCharAt(res.length() - 1); // 删除最后多余的逗号 return res.toString(); // 转化为字符串并返回 } + void dfs(int x) { - if(x == n) { // 终止条件：已固定完所有位 + if (x == n) { // 终止条件：已固定完所有位 res.append(String.valueOf(num) + ","); // 拼接 num 并添加至 res 尾部，使用逗号隔开 return; } - for(char i : loop) { // 遍历 ‘0‘ - ’9‘ + for (char i : loop) { // 遍历 ‘0‘ - ’9‘ num[x] = i; // 固定第 x 位为 i dfs(x + 1); // 开启固定第 x + 1 位 } diff --git a/sword_for_offer/codes/java/sfo_17_print_from_1_to_the_largest_n_digits_s3/sfo_17_print_from_1_to_the_largest_n_digits_s3.java b/sword_for_offer/codes/java/sfo_17_print_from_1_to_the_largest_n_digits_s3/sfo_17_print_from_1_to_the_largest_n_digits_s3.java index d3d2366..c282c84 100644 --- a/sword_for_offer/codes/java/sfo_17_print_from_1_to_the_largest_n_digits_s3/sfo_17_print_from_1_to_the_largest_n_digits_s3.java +++ b/sword_for_offer/codes/java/sfo_17_print_from_1_to_the_largest_n_digits_s3/sfo_17_print_from_1_to_the_largest_n_digits_s3.java @@ -13,7 +13,8 @@ class Solution { StringBuilder res; int nine = 0, count = 0, start, n; - char[] num, loop = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9'}; + char[] num, loop = { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9' }; + public String printNumbers(int n) { this.n = n; res = new StringBuilder(); @@ -23,15 +24,19 @@ public String printNumbers(int n) { res.deleteCharAt(res.length() - 1); return res.toString(); } + void dfs(int x) { - if(x == n) { + if (x == n) { String s = String.valueOf(num).substring(start); - if(!s.equals("0")) res.append(s + ","); - if(n - start == nine) start--; + if (!s.equals("0")) + res.append(s + ","); + if (n - start == nine) + start--; return; } - for(char i : loop) { - if(i == '9') nine++; + for (char i : loop) { + if (i == '9') + nine++; num[x] = i; dfs(x + 1); } diff --git a/sword_for_offer/codes/java/sfo_17_print_from_1_to_the_largest_n_digits_s4/sfo_17_print_from_1_to_the_largest_n_digits_s4.java b/sword_for_offer/codes/java/sfo_17_print_from_1_to_the_largest_n_digits_s4/sfo_17_print_from_1_to_the_largest_n_digits_s4.java index 36b8d4a..6c2beef 100644 --- a/sword_for_offer/codes/java/sfo_17_print_from_1_to_the_largest_n_digits_s4/sfo_17_print_from_1_to_the_largest_n_digits_s4.java +++ b/sword_for_offer/codes/java/sfo_17_print_from_1_to_the_largest_n_digits_s4/sfo_17_print_from_1_to_the_largest_n_digits_s4.java @@ -13,24 +13,29 @@ class Solution { int[] res; int nine = 0, count = 0, start, n; - char[] num, loop = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9'}; + char[] num, loop = { '0', '1', '2', '3', '4', '5', '6', '7', '8', '9' }; + public int[] printNumbers(int n) { this.n = n; - res = new int[(int)Math.pow(10, n) - 1]; + res = new int[(int) Math.pow(10, n) - 1]; num = new char[n]; start = n - 1; dfs(0); return res; } + void dfs(int x) { - if(x == n) { + if (x == n) { String s = String.valueOf(num).substring(start); - if(!s.equals("0")) res[count++] = Integer.parseInt(s); - if(n - start == nine) start--; + if (!s.equals("0")) + res[count++] = Integer.parseInt(s); + if (n - start == nine) + start--; return; } - for(char i : loop) { - if(i == '9') nine++; + for (char i : loop) { + if (i == '9') + nine++; num[x] = i; dfs(x + 1); } diff --git a/sword_for_offer/codes/java/sfo_18_delete_a_node_from_a_linked_list_s1/sfo_18_delete_a_node_from_a_linked_list_s1.java b/sword_for_offer/codes/java/sfo_18_delete_a_node_from_a_linked_list_s1/sfo_18_delete_a_node_from_a_linked_list_s1.java index 4477b7f..40c1bb6 100644 --- a/sword_for_offer/codes/java/sfo_18_delete_a_node_from_a_linked_list_s1/sfo_18_delete_a_node_from_a_linked_list_s1.java +++ b/sword_for_offer/codes/java/sfo_18_delete_a_node_from_a_linked_list_s1/sfo_18_delete_a_node_from_a_linked_list_s1.java @@ -12,13 +12,15 @@ // ===== Solution Code ===== class Solution { public ListNode deleteNode(ListNode head, int val) { - if(head.val == val) return head.next; + if (head.val == val) + return head.next; ListNode pre = head, cur = head.next; - while(cur != null && cur.val != val) { + while (cur != null && cur.val != val) { pre = cur; cur = cur.next; } - if(cur != null) pre.next = cur.next; + if (cur != null) + pre.next = cur.next; return head; } } diff --git a/sword_for_offer/codes/java/sfo_19_regular_expression_matching_s1/sfo_19_regular_expression_matching_s1.java b/sword_for_offer/codes/java/sfo_19_regular_expression_matching_s1/sfo_19_regular_expression_matching_s1.java index c1bd612..a4520ff 100644 --- a/sword_for_offer/codes/java/sfo_19_regular_expression_matching_s1/sfo_19_regular_expression_matching_s1.java +++ b/sword_for_offer/codes/java/sfo_19_regular_expression_matching_s1/sfo_19_regular_expression_matching_s1.java @@ -15,13 +15,13 @@ public boolean isMatch(String s, String p) { int m = s.length() + 1, n = p.length() + 1; boolean[][] dp = new boolean[m][n]; dp[0][0] = true; - for(int j = 2; j < n; j += 2) + for (int j = 2; j < n; j += 2) dp[0][j] = dp[0][j - 2] && p.charAt(j - 1) == '*'; - for(int i = 1; i < m; i++) { - for(int j = 1; j < n; j++) { - dp[i][j] = p.charAt(j - 1) == '*' ? - dp[i][j - 2] || dp[i - 1][j] && (s.charAt(i - 1) == p.charAt(j - 2) || p.charAt(j - 2) == '.') : - dp[i - 1][j - 1] && (p.charAt(j - 1) == '.' || s.charAt(i - 1) == p.charAt(j - 1)); + for (int i = 1; i < m; i++) { + for (int j = 1; j < n; j++) { + dp[i][j] = p.charAt(j - 1) == '*' + ? dp[i][j - 2] || dp[i - 1][j] && (s.charAt(i - 1) == p.charAt(j - 2) || p.charAt(j - 2) == '.') + : dp[i - 1][j - 1] && (p.charAt(j - 1) == '.' || s.charAt(i - 1) == p.charAt(j - 1)); } } return dp[m - 1][n - 1]; diff --git a/sword_for_offer/codes/java/sfo_19_regular_expression_matching_s2/sfo_19_regular_expression_matching_s2.java b/sword_for_offer/codes/java/sfo_19_regular_expression_matching_s2/sfo_19_regular_expression_matching_s2.java index 371ca50..1d9ad15 100644 --- a/sword_for_offer/codes/java/sfo_19_regular_expression_matching_s2/sfo_19_regular_expression_matching_s2.java +++ b/sword_for_offer/codes/java/sfo_19_regular_expression_matching_s2/sfo_19_regular_expression_matching_s2.java @@ -16,18 +16,23 @@ public boolean isMatch(String s, String p) { boolean[][] dp = new boolean[m][n]; dp[0][0] = true; // 初始化首行 - for(int j = 2; j < n; j += 2) + for (int j = 2; j < n; j += 2) dp[0][j] = dp[0][j - 2] && p.charAt(j - 1) == '*'; // 状态转移 - for(int i = 1; i < m; i++) { - for(int j = 1; j < n; j++) { - if(p.charAt(j - 1) == '*') { - if(dp[i][j - 2]) dp[i][j] = true; // 1. - else if(dp[i - 1][j] && s.charAt(i - 1) == p.charAt(j - 2)) dp[i][j] = true; // 2. - else if(dp[i - 1][j] && p.charAt(j - 2) == '.') dp[i][j] = true; // 3. + for (int i = 1; i < m; i++) { + for (int j = 1; j < n; j++) { + if (p.charAt(j - 1) == '*') { + if (dp[i][j - 2]) + dp[i][j] = true; // 1. + else if (dp[i - 1][j] && s.charAt(i - 1) == p.charAt(j - 2)) + dp[i][j] = true; // 2. + else if (dp[i - 1][j] && p.charAt(j - 2) == '.') + dp[i][j] = true; // 3. } else { - if(dp[i - 1][j - 1] && s.charAt(i - 1) == p.charAt(j - 1)) dp[i][j] = true; // 1. - else if(dp[i - 1][j - 1] && p.charAt(j - 1) == '.') dp[i][j] = true; // 2. + if (dp[i - 1][j - 1] && s.charAt(i - 1) == p.charAt(j - 1)) + dp[i][j] = true; // 1. + else if (dp[i - 1][j - 1] && p.charAt(j - 1) == '.') + dp[i][j] = true; // 2. } } } diff --git a/sword_for_offer/codes/java/sfo_20_a_string_representing_a_numeric_value_s1/sfo_20_a_string_representing_a_numeric_value_s1.java b/sword_for_offer/codes/java/sfo_20_a_string_representing_a_numeric_value_s1/sfo_20_a_string_representing_a_numeric_value_s1.java index 63fda1a..dadd888 100644 --- a/sword_for_offer/codes/java/sfo_20_a_string_representing_a_numeric_value_s1/sfo_20_a_string_representing_a_numeric_value_s1.java +++ b/sword_for_offer/codes/java/sfo_20_a_string_representing_a_numeric_value_s1/sfo_20_a_string_representing_a_numeric_value_s1.java @@ -13,26 +13,79 @@ class Solution { public boolean isNumber(String s) { HashMap[] states = { - new HashMap() {{ put(' ', 0); put('s', 1); put('d', 2); put('.', 4); }}, // 0. - new HashMap() {{ put('d', 2); put('.', 4); }}, // 1. - new HashMap() {{ put('d', 2); put('.', 3); put('e', 5); put(' ', 8); }}, // 2. - new HashMap() {{ put('d', 3); put('e', 5); put(' ', 8); }}, // 3. - new HashMap() {{ put('d', 3); }}, // 4. - new HashMap() {{ put('s', 6); put('d', 7); }}, // 5. - new HashMap() {{ put('d', 7); }}, // 6. - new HashMap() {{ put('d', 7); put(' ', 8); }}, // 7. - new HashMap() {{ put(' ', 8); }} // 8. + new HashMap() { + { + put(' ', 0); + put('s', 1); + put('d', 2); + put('.', 4); + } + }, // 0. + new HashMap() { + { + put('d', 2); + put('.', 4); + } + }, // 1. + new HashMap() { + { + put('d', 2); + put('.', 3); + put('e', 5); + put(' ', 8); + } + }, // 2. + new HashMap() { + { + put('d', 3); + put('e', 5); + put(' ', 8); + } + }, // 3. + new HashMap() { + { + put('d', 3); + } + }, // 4. + new HashMap() { + { + put('s', 6); + put('d', 7); + } + }, // 5. + new HashMap() { + { + put('d', 7); + } + }, // 6. + new HashMap() { + { + put('d', 7); + put(' ', 8); + } + }, // 7. + new HashMap() { + { + put(' ', 8); + } + } // 8. }; int p = 0; char t; - for(char c : s.toCharArray()) { - if(c >= '0' && c <= '9') t = 'd'; - else if(c == '+' || c == '-') t = 's'; - else if(c == 'e' || c == 'E') t = 'e'; - else if(c == '.' || c == ' ') t = c; - else t = '?'; - if(!states[p].containsKey(t)) return false; - p = (int)states[p].get(t); + for (char c : s.toCharArray()) { + if (c >= '0' && c <= '9') + t = 'd'; + else if (c == '+' || c == '-') + t = 's'; + else if (c == 'e' || c == 'E') + t = 'e'; + else if (c == '.' || c == ' ') + t = c; + else + t = '?'; + if (!states[p].containsKey(t)) + return false; + p = (int) states[p].get(t); } return p == 2 || p == 3 || p == 7 || p == 8; } diff --git a/sword_for_offer/codes/java/sfo_21_adjust_the_order_of_numbers_in_an_array_s1/sfo_21_adjust_the_order_of_numbers_in_an_array_s1.java b/sword_for_offer/codes/java/sfo_21_adjust_the_order_of_numbers_in_an_array_s1/sfo_21_adjust_the_order_of_numbers_in_an_array_s1.java index c7c66bd..e94f5d5 100644 --- a/sword_for_offer/codes/java/sfo_21_adjust_the_order_of_numbers_in_an_array_s1/sfo_21_adjust_the_order_of_numbers_in_an_array_s1.java +++ b/sword_for_offer/codes/java/sfo_21_adjust_the_order_of_numbers_in_an_array_s1/sfo_21_adjust_the_order_of_numbers_in_an_array_s1.java @@ -13,9 +13,11 @@ class Solution { public int[] exchange(int[] nums) { int i = 0, j = nums.length - 1, tmp; - while(i < j) { - while(i < j && (nums[i] & 1) == 1) i++; - while(i < j && (nums[j] & 1) == 0) j--; + while (i < j) { + while (i < j && (nums[i] & 1) == 1) + i++; + while (i < j && (nums[j] & 1) == 0) + j--; tmp = nums[i]; nums[i] = nums[j]; nums[j] = tmp; diff --git a/sword_for_offer/codes/java/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1.java b/sword_for_offer/codes/java/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1.java index 5c6dbd6..a420208 100644 --- a/sword_for_offer/codes/java/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1.java +++ b/sword_for_offer/codes/java/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1.java @@ -13,9 +13,9 @@ class Solution { public ListNode getKthFromEnd(ListNode head, int k) { ListNode former = head, latter = head; - for(int i = 0; i < k; i++) + for (int i = 0; i < k; i++) former = former.next; - while(former != null) { + while (former != null) { former = former.next; latter = latter.next; } diff --git a/sword_for_offer/codes/java/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2.java b/sword_for_offer/codes/java/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2.java index c393d55..aa1f8db 100644 --- a/sword_for_offer/codes/java/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2.java +++ b/sword_for_offer/codes/java/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2.java @@ -13,11 +13,12 @@ class Solution { public ListNode getKthFromEnd(ListNode head, int k) { ListNode former = head, latter = head; - for(int i = 0; i < k; i++) { - if(former == null) return null; + for (int i = 0; i < k; i++) { + if (former == null) + return null; former = former.next; } - while(former != null) { + while (former != null) { former = former.next; latter = latter.next; } diff --git a/sword_for_offer/codes/java/sfo_24_reverse_a_linked_list_s1/sfo_24_reverse_a_linked_list_s1.java b/sword_for_offer/codes/java/sfo_24_reverse_a_linked_list_s1/sfo_24_reverse_a_linked_list_s1.java index ab30f38..1c72a86 100644 --- a/sword_for_offer/codes/java/sfo_24_reverse_a_linked_list_s1/sfo_24_reverse_a_linked_list_s1.java +++ b/sword_for_offer/codes/java/sfo_24_reverse_a_linked_list_s1/sfo_24_reverse_a_linked_list_s1.java @@ -13,11 +13,11 @@ class Solution { public ListNode reverseList(ListNode head) { ListNode cur = head, pre = null; - while(cur != null) { + while (cur != null) { ListNode tmp = cur.next; // 暂存后继节点 cur.next - cur.next = pre; // 修改 next 引用指向 - pre = cur; // pre 暂存 cur - cur = tmp; // cur 访问下一节点 + cur.next = pre; // 修改 next 引用指向 + pre = cur; // pre 暂存 cur + cur = tmp; // cur 访问下一节点 } return pre; } diff --git a/sword_for_offer/codes/java/sfo_24_reverse_a_linked_list_s2/sfo_24_reverse_a_linked_list_s2.java b/sword_for_offer/codes/java/sfo_24_reverse_a_linked_list_s2/sfo_24_reverse_a_linked_list_s2.java index a2cdb4a..d83df5c 100644 --- a/sword_for_offer/codes/java/sfo_24_reverse_a_linked_list_s2/sfo_24_reverse_a_linked_list_s2.java +++ b/sword_for_offer/codes/java/sfo_24_reverse_a_linked_list_s2/sfo_24_reverse_a_linked_list_s2.java @@ -12,13 +12,15 @@ // ===== Solution Code ===== class Solution { public ListNode reverseList(ListNode head) { - return recur(head, null); // 调用递归并返回 + return recur(head, null); // 调用递归并返回 } + private ListNode recur(ListNode cur, ListNode pre) { - if (cur == null) return pre; // 终止条件 - ListNode res = recur(cur.next, cur); // 递归后继节点 - cur.next = pre; // 修改节点引用指向 - return res; // 返回反转链表的头节点 + if (cur == null) + return pre; // 终止条件 + ListNode res = recur(cur.next, cur); // 递归后继节点 + cur.next = pre; // 修改节点引用指向 + return res; // 返回反转链表的头节点 } } diff --git a/sword_for_offer/codes/java/sfo_25_combine_two_sorted_linked_lists_s1/sfo_25_combine_two_sorted_linked_lists_s1.java b/sword_for_offer/codes/java/sfo_25_combine_two_sorted_linked_lists_s1/sfo_25_combine_two_sorted_linked_lists_s1.java index bc15f8a..f3e58b1 100644 --- a/sword_for_offer/codes/java/sfo_25_combine_two_sorted_linked_lists_s1/sfo_25_combine_two_sorted_linked_lists_s1.java +++ b/sword_for_offer/codes/java/sfo_25_combine_two_sorted_linked_lists_s1/sfo_25_combine_two_sorted_linked_lists_s1.java @@ -13,12 +13,11 @@ class Solution { public ListNode mergeTwoLists(ListNode l1, ListNode l2) { ListNode dum = new ListNode(0), cur = dum; - while(l1 != null && l2 != null) { - if(l1.val < l2.val) { + while (l1 != null && l2 != null) { + if (l1.val < l2.val) { cur.next = l1; l1 = l1.next; - } - else { + } else { cur.next = l2; l2 = l2.next; } diff --git a/sword_for_offer/codes/java/sfo_26_substructure_of_a_binary_tree_s1/sfo_26_substructure_of_a_binary_tree_s1.java b/sword_for_offer/codes/java/sfo_26_substructure_of_a_binary_tree_s1/sfo_26_substructure_of_a_binary_tree_s1.java index 56be37d..93ac9a5 100644 --- a/sword_for_offer/codes/java/sfo_26_substructure_of_a_binary_tree_s1/sfo_26_substructure_of_a_binary_tree_s1.java +++ b/sword_for_offer/codes/java/sfo_26_substructure_of_a_binary_tree_s1/sfo_26_substructure_of_a_binary_tree_s1.java @@ -14,9 +14,12 @@ class Solution { public boolean isSubStructure(TreeNode A, TreeNode B) { return (A != null && B != null) && (recur(A, B) || isSubStructure(A.left, B) || isSubStructure(A.right, B)); } + boolean recur(TreeNode A, TreeNode B) { - if(B == null) return true; - if(A == null || A.val != B.val) return false; + if (B == null) + return true; + if (A == null || A.val != B.val) + return false; return recur(A.left, B.left) && recur(A.right, B.right); } } diff --git a/sword_for_offer/codes/java/sfo_27_mirror_of_a_binary_tree_s1/sfo_27_mirror_of_a_binary_tree_s1.java b/sword_for_offer/codes/java/sfo_27_mirror_of_a_binary_tree_s1/sfo_27_mirror_of_a_binary_tree_s1.java index a186bcd..a6a916f 100644 --- a/sword_for_offer/codes/java/sfo_27_mirror_of_a_binary_tree_s1/sfo_27_mirror_of_a_binary_tree_s1.java +++ b/sword_for_offer/codes/java/sfo_27_mirror_of_a_binary_tree_s1/sfo_27_mirror_of_a_binary_tree_s1.java @@ -12,7 +12,8 @@ // ===== Solution Code ===== class Solution { public TreeNode mirrorTree(TreeNode root) { - if(root == null) return null; + if (root == null) + return null; TreeNode tmp = root.left; root.left = mirrorTree(root.right); root.right = mirrorTree(tmp); @@ -23,7 +24,8 @@ public TreeNode mirrorTree(TreeNode root) { public class sfo_27_mirror_of_a_binary_tree_s1 { public static void main(String[] args) { // ======= Test Case ======= - TreeNode root = TreeNode.arrToTree(new Integer[] { 4, 2, 7, 1, 3, 6, 9, null, null, null, null, null, null, null, null }); + TreeNode root = TreeNode + .arrToTree(new Integer[] { 4, 2, 7, 1, 3, 6, 9, null, null, null, null, null, null, null, null }); // ====== Driver Code ====== Solution slt = new Solution(); TreeNode res = slt.mirrorTree(root); diff --git a/sword_for_offer/codes/java/sfo_27_mirror_of_a_binary_tree_s2/sfo_27_mirror_of_a_binary_tree_s2.java b/sword_for_offer/codes/java/sfo_27_mirror_of_a_binary_tree_s2/sfo_27_mirror_of_a_binary_tree_s2.java index 52fc74a..cd43fc7 100644 --- a/sword_for_offer/codes/java/sfo_27_mirror_of_a_binary_tree_s2/sfo_27_mirror_of_a_binary_tree_s2.java +++ b/sword_for_offer/codes/java/sfo_27_mirror_of_a_binary_tree_s2/sfo_27_mirror_of_a_binary_tree_s2.java @@ -12,12 +12,19 @@ // ===== Solution Code ===== class Solution { public TreeNode mirrorTree(TreeNode root) { - if(root == null) return null; - Stack stack = new Stack() {{ add(root); }}; - while(!stack.isEmpty()) { + if (root == null) + return null; + Stack stack = new Stack() { + { + add(root); + } + }; + while (!stack.isEmpty()) { TreeNode node = stack.pop(); - if(node.left != null) stack.add(node.left); - if(node.right != null) stack.add(node.right); + if (node.left != null) + stack.add(node.left); + if (node.right != null) + stack.add(node.right); TreeNode tmp = node.left; node.left = node.right; node.right = tmp; @@ -29,7 +36,8 @@ public TreeNode mirrorTree(TreeNode root) { public class sfo_27_mirror_of_a_binary_tree_s2 { public static void main(String[] args) { // ======= Test Case ======= - TreeNode root = TreeNode.arrToTree(new Integer[] { 4, 2, 7, 1, 3, 6, 9, null, null, null, null, null, null, null, null }); + TreeNode root = TreeNode + .arrToTree(new Integer[] { 4, 2, 7, 1, 3, 6, 9, null, null, null, null, null, null, null, null }); // ====== Driver Code ====== Solution slt = new Solution(); TreeNode res = slt.mirrorTree(root); diff --git a/sword_for_offer/codes/java/sfo_28_symmetric_binary_tree_s1/sfo_28_symmetric_binary_tree_s1.java b/sword_for_offer/codes/java/sfo_28_symmetric_binary_tree_s1/sfo_28_symmetric_binary_tree_s1.java index 6db82ce..f1313cc 100644 --- a/sword_for_offer/codes/java/sfo_28_symmetric_binary_tree_s1/sfo_28_symmetric_binary_tree_s1.java +++ b/sword_for_offer/codes/java/sfo_28_symmetric_binary_tree_s1/sfo_28_symmetric_binary_tree_s1.java @@ -14,9 +14,12 @@ class Solution { public boolean isSymmetric(TreeNode root) { return root == null || recur(root.left, root.right); } + boolean recur(TreeNode L, TreeNode R) { - if(L == null && R == null) return true; - if(L == null || R == null || L.val != R.val) return false; + if (L == null && R == null) + return true; + if (L == null || R == null || L.val != R.val) + return false; return recur(L.left, R.right) && recur(L.right, R.left); } } @@ -24,7 +27,8 @@ boolean recur(TreeNode L, TreeNode R) { public class sfo_28_symmetric_binary_tree_s1 { public static void main(String[] args) { // ======= Test Case ======= - TreeNode root = TreeNode.arrToTree(new Integer[] { 1, 2, 2, 3, 4, 4, 3, null, null, null, null, null, null, null, null }); + TreeNode root = TreeNode + .arrToTree(new Integer[] { 1, 2, 2, 3, 4, 4, 3, null, null, null, null, null, null, null, null }); // ====== Driver Code ====== Solution slt = new Solution(); boolean res = slt.isSymmetric(root); diff --git a/sword_for_offer/codes/java/sfo_29_print_a_given_matrix_in_spiral_form_s1/sfo_29_print_a_given_matrix_in_spiral_form_s1.java b/sword_for_offer/codes/java/sfo_29_print_a_given_matrix_in_spiral_form_s1/sfo_29_print_a_given_matrix_in_spiral_form_s1.java index fe59b44..61977e7 100644 --- a/sword_for_offer/codes/java/sfo_29_print_a_given_matrix_in_spiral_form_s1/sfo_29_print_a_given_matrix_in_spiral_form_s1.java +++ b/sword_for_offer/codes/java/sfo_29_print_a_given_matrix_in_spiral_form_s1/sfo_29_print_a_given_matrix_in_spiral_form_s1.java @@ -12,18 +12,27 @@ // ===== Solution Code ===== class Solution { public int[] spiralOrder(int[][] matrix) { - if(matrix.length == 0) return new int[0]; + if (matrix.length == 0) + return new int[0]; int l = 0, r = matrix[0].length - 1, t = 0, b = matrix.length - 1, x = 0; int[] res = new int[(r + 1) * (b + 1)]; - while(true) { - for(int i = l; i <= r; i++) res[x++] = matrix[t][i]; // left to right - if(++t > b) break; - for(int i = t; i <= b; i++) res[x++] = matrix[i][r]; // top to bottom - if(l > --r) break; - for(int i = r; i >= l; i--) res[x++] = matrix[b][i]; // right to left - if(t > --b) break; - for(int i = b; i >= t; i--) res[x++] = matrix[i][l]; // bottom to top - if(++l > r) break; + while (true) { + for (int i = l; i <= r; i++) + res[x++] = matrix[t][i]; // left to right + if (++t > b) + break; + for (int i = t; i <= b; i++) + res[x++] = matrix[i][r]; // top to bottom + if (l > --r) + break; + for (int i = r; i >= l; i--) + res[x++] = matrix[b][i]; // right to left + if (t > --b) + break; + for (int i = b; i >= t; i--) + res[x++] = matrix[i][l]; // bottom to top + if (++l > r) + break; } return res; } @@ -33,10 +42,10 @@ public class sfo_29_print_a_given_matrix_in_spiral_form_s1 { public static void main(String[] args) { // ======= Test Case ======= int[][] matrix = { - { 1, 2, 3 }, - { 4, 5, 6 }, - { 7, 8, 9 } -}; + { 1, 2, 3 }, + { 4, 5, 6 }, + { 7, 8, 9 } + }; // ====== Driver Code ====== Solution slt = new Solution(); int[] res = slt.spiralOrder(matrix); diff --git a/sword_for_offer/codes/java/sfo_30_min_stack_s1/sfo_30_min_stack_s1.java b/sword_for_offer/codes/java/sfo_30_min_stack_s1/sfo_30_min_stack_s1.java index e3a30b9..20312a4 100644 --- a/sword_for_offer/codes/java/sfo_30_min_stack_s1/sfo_30_min_stack_s1.java +++ b/sword_for_offer/codes/java/sfo_30_min_stack_s1/sfo_30_min_stack_s1.java @@ -12,22 +12,27 @@ // ===== Solution Code ===== class MinStack { Stack A, B; + public MinStack() { A = new Stack<>(); B = new Stack<>(); } + public void push(int x) { A.add(x); - if(B.empty() || B.peek() >= x) + if (B.empty() || B.peek() >= x) B.add(x); } + public void pop() { - if(A.pop().equals(B.peek())) + if (A.pop().equals(B.peek())) B.pop(); } + public int top() { return A.peek(); } + public int min() { return B.peek(); } diff --git a/sword_for_offer/codes/java/sfo_31_validate_stack_sequences_s1/sfo_31_validate_stack_sequences_s1.java b/sword_for_offer/codes/java/sfo_31_validate_stack_sequences_s1/sfo_31_validate_stack_sequences_s1.java index 8d46c27..02bab2c 100644 --- a/sword_for_offer/codes/java/sfo_31_validate_stack_sequences_s1/sfo_31_validate_stack_sequences_s1.java +++ b/sword_for_offer/codes/java/sfo_31_validate_stack_sequences_s1/sfo_31_validate_stack_sequences_s1.java @@ -14,9 +14,9 @@ class Solution { public boolean validateStackSequences(int[] pushed, int[] popped) { Stack stack = new Stack<>(); int i = 0; - for(int num : pushed) { + for (int num : pushed) { stack.push(num); // num 入栈 - while(!stack.isEmpty() && stack.peek() == popped[i]) { // 循环判断与出栈 + while (!stack.isEmpty() && stack.peek() == popped[i]) { // 循环判断与出栈 stack.pop(); i++; } diff --git a/sword_for_offer/codes/java/sfo_32i_print_a_binary_tree_topbottom_i_s1/sfo_32i_print_a_binary_tree_topbottom_i_s1.java b/sword_for_offer/codes/java/sfo_32i_print_a_binary_tree_topbottom_i_s1/sfo_32i_print_a_binary_tree_topbottom_i_s1.java index 71ae2e6..7b906d3 100644 --- a/sword_for_offer/codes/java/sfo_32i_print_a_binary_tree_topbottom_i_s1/sfo_32i_print_a_binary_tree_topbottom_i_s1.java +++ b/sword_for_offer/codes/java/sfo_32i_print_a_binary_tree_topbottom_i_s1/sfo_32i_print_a_binary_tree_topbottom_i_s1.java @@ -12,17 +12,24 @@ // ===== Solution Code ===== class Solution { public int[] levelOrder(TreeNode root) { - if(root == null) return new int[0]; - Queue queue = new LinkedList(){{ add(root); }}; + if (root == null) + return new int[0]; + Queue queue = new LinkedList() { + { + add(root); + } + }; ArrayList ans = new ArrayList<>(); - while(!queue.isEmpty()) { + while (!queue.isEmpty()) { TreeNode node = queue.poll(); ans.add(node.val); - if(node.left != null) queue.add(node.left); - if(node.right != null) queue.add(node.right); + if (node.left != null) + queue.add(node.left); + if (node.right != null) + queue.add(node.right); } int[] res = new int[ans.size()]; - for(int i = 0; i < ans.size(); i++) + for (int i = 0; i < ans.size(); i++) res[i] = ans.get(i); return res; } diff --git a/sword_for_offer/codes/java/sfo_32ii_print_a_binary_tree_topbottom_ii_s1/sfo_32ii_print_a_binary_tree_topbottom_ii_s1.java b/sword_for_offer/codes/java/sfo_32ii_print_a_binary_tree_topbottom_ii_s1/sfo_32ii_print_a_binary_tree_topbottom_ii_s1.java index a1105ce..4904a84 100644 --- a/sword_for_offer/codes/java/sfo_32ii_print_a_binary_tree_topbottom_ii_s1/sfo_32ii_print_a_binary_tree_topbottom_ii_s1.java +++ b/sword_for_offer/codes/java/sfo_32ii_print_a_binary_tree_topbottom_ii_s1/sfo_32ii_print_a_binary_tree_topbottom_ii_s1.java @@ -14,14 +14,17 @@ class Solution { public List> levelOrder(TreeNode root) { Queue queue = new LinkedList<>(); List> res = new ArrayList<>(); - if(root != null) queue.add(root); - while(!queue.isEmpty()) { + if (root != null) + queue.add(root); + while (!queue.isEmpty()) { List tmp = new ArrayList<>(); - for(int i = queue.size(); i > 0; i--) { + for (int i = queue.size(); i > 0; i--) { TreeNode node = queue.poll(); tmp.add(node.val); - if(node.left != null) queue.add(node.left); - if(node.right != null) queue.add(node.right); + if (node.left != null) + queue.add(node.left); + if (node.right != null) + queue.add(node.right); } res.add(tmp); } @@ -36,6 +39,7 @@ public static void main(String[] args) { // ====== Driver Code ====== Solution slt = new Solution(); List> res = slt.levelOrder(root); - System.out.println(Arrays.deepToString(res.toArray()));; + System.out.println(Arrays.deepToString(res.toArray())); + ; } } diff --git a/sword_for_offer/codes/java/sfo_32iii_print_a_binary_tree_topbottom_iii_s1/sfo_32iii_print_a_binary_tree_topbottom_iii_s1.java b/sword_for_offer/codes/java/sfo_32iii_print_a_binary_tree_topbottom_iii_s1/sfo_32iii_print_a_binary_tree_topbottom_iii_s1.java index 1ff145c..f0df02b 100644 --- a/sword_for_offer/codes/java/sfo_32iii_print_a_binary_tree_topbottom_iii_s1/sfo_32iii_print_a_binary_tree_topbottom_iii_s1.java +++ b/sword_for_offer/codes/java/sfo_32iii_print_a_binary_tree_topbottom_iii_s1/sfo_32iii_print_a_binary_tree_topbottom_iii_s1.java @@ -14,15 +14,20 @@ class Solution { public List> levelOrder(TreeNode root) { Queue queue = new LinkedList<>(); List> res = new ArrayList<>(); - if(root != null) queue.add(root); - while(!queue.isEmpty()) { + if (root != null) + queue.add(root); + while (!queue.isEmpty()) { LinkedList tmp = new LinkedList<>(); - for(int i = queue.size(); i > 0; i--) { + for (int i = queue.size(); i > 0; i--) { TreeNode node = queue.poll(); - if(res.size() % 2 == 0) tmp.addLast(node.val); - else tmp.addFirst(node.val); - if(node.left != null) queue.add(node.left); - if(node.right != null) queue.add(node.right); + if (res.size() % 2 == 0) + tmp.addLast(node.val); + else + tmp.addFirst(node.val); + if (node.left != null) + queue.add(node.left); + if (node.right != null) + queue.add(node.right); } res.add(tmp); } @@ -37,6 +42,7 @@ public static void main(String[] args) { // ====== Driver Code ====== Solution slt = new Solution(); List> res = slt.levelOrder(root); - System.out.println(Arrays.deepToString(res.toArray()));; + System.out.println(Arrays.deepToString(res.toArray())); + ; } } diff --git a/sword_for_offer/codes/java/sfo_32iii_print_a_binary_tree_topbottom_iii_s2/sfo_32iii_print_a_binary_tree_topbottom_iii_s2.java b/sword_for_offer/codes/java/sfo_32iii_print_a_binary_tree_topbottom_iii_s2/sfo_32iii_print_a_binary_tree_topbottom_iii_s2.java index 28eb7a6..169e8da 100644 --- a/sword_for_offer/codes/java/sfo_32iii_print_a_binary_tree_topbottom_iii_s2/sfo_32iii_print_a_binary_tree_topbottom_iii_s2.java +++ b/sword_for_offer/codes/java/sfo_32iii_print_a_binary_tree_topbottom_iii_s2/sfo_32iii_print_a_binary_tree_topbottom_iii_s2.java @@ -14,29 +14,35 @@ class Solution { public List> levelOrder(TreeNode root) { Deque deque = new LinkedList<>(); List> res = new ArrayList<>(); - if(root != null) deque.add(root); - while(!deque.isEmpty()) { + if (root != null) + deque.add(root); + while (!deque.isEmpty()) { // 打印奇数层 List tmp = new ArrayList<>(); - for(int i = deque.size(); i > 0; i--) { + for (int i = deque.size(); i > 0; i--) { // 从左向右打印 TreeNode node = deque.removeFirst(); tmp.add(node.val); // 先左后右加入下层节点 - if(node.left != null) deque.addLast(node.left); - if(node.right != null) deque.addLast(node.right); + if (node.left != null) + deque.addLast(node.left); + if (node.right != null) + deque.addLast(node.right); } res.add(tmp); - if(deque.isEmpty()) break; // 若为空则提前跳出 + if (deque.isEmpty()) + break; // 若为空则提前跳出 // 打印偶数层 tmp = new ArrayList<>(); - for(int i = deque.size(); i > 0; i--) { + for (int i = deque.size(); i > 0; i--) { // 从右向左打印 TreeNode node = deque.removeLast(); tmp.add(node.val); // 先右后左加入下层节点 - if(node.right != null) deque.addFirst(node.right); - if(node.left != null) deque.addFirst(node.left); + if (node.right != null) + deque.addFirst(node.right); + if (node.left != null) + deque.addFirst(node.left); } res.add(tmp); } @@ -51,6 +57,7 @@ public static void main(String[] args) { // ====== Driver Code ====== Solution slt = new Solution(); List> res = slt.levelOrder(root); - System.out.println(Arrays.deepToString(res.toArray()));; + System.out.println(Arrays.deepToString(res.toArray())); + ; } } diff --git a/sword_for_offer/codes/java/sfo_32iii_print_a_binary_tree_topbottom_iii_s3/sfo_32iii_print_a_binary_tree_topbottom_iii_s3.java b/sword_for_offer/codes/java/sfo_32iii_print_a_binary_tree_topbottom_iii_s3/sfo_32iii_print_a_binary_tree_topbottom_iii_s3.java index 6c10d90..b45de4d 100644 --- a/sword_for_offer/codes/java/sfo_32iii_print_a_binary_tree_topbottom_iii_s3/sfo_32iii_print_a_binary_tree_topbottom_iii_s3.java +++ b/sword_for_offer/codes/java/sfo_32iii_print_a_binary_tree_topbottom_iii_s3/sfo_32iii_print_a_binary_tree_topbottom_iii_s3.java @@ -14,16 +14,20 @@ class Solution { public List> levelOrder(TreeNode root) { Queue queue = new LinkedList<>(); List> res = new ArrayList<>(); - if(root != null) queue.add(root); - while(!queue.isEmpty()) { + if (root != null) + queue.add(root); + while (!queue.isEmpty()) { List tmp = new ArrayList<>(); - for(int i = queue.size(); i > 0; i--) { + for (int i = queue.size(); i > 0; i--) { TreeNode node = queue.poll(); tmp.add(node.val); - if(node.left != null) queue.add(node.left); - if(node.right != null) queue.add(node.right); + if (node.left != null) + queue.add(node.left); + if (node.right != null) + queue.add(node.right); } - if(res.size() % 2 == 1) Collections.reverse(tmp); + if (res.size() % 2 == 1) + Collections.reverse(tmp); res.add(tmp); } return res; @@ -37,6 +41,7 @@ public static void main(String[] args) { // ====== Driver Code ====== Solution slt = new Solution(); List> res = slt.levelOrder(root); - System.out.println(Arrays.deepToString(res.toArray()));; + System.out.println(Arrays.deepToString(res.toArray())); + ; } } diff --git a/sword_for_offer/codes/java/sfo_33_postorder_traversal_of_a_binary_search_tree_s1/sfo_33_postorder_traversal_of_a_binary_search_tree_s1.java b/sword_for_offer/codes/java/sfo_33_postorder_traversal_of_a_binary_search_tree_s1/sfo_33_postorder_traversal_of_a_binary_search_tree_s1.java index 847ba83..733d2cf 100644 --- a/sword_for_offer/codes/java/sfo_33_postorder_traversal_of_a_binary_search_tree_s1/sfo_33_postorder_traversal_of_a_binary_search_tree_s1.java +++ b/sword_for_offer/codes/java/sfo_33_postorder_traversal_of_a_binary_search_tree_s1/sfo_33_postorder_traversal_of_a_binary_search_tree_s1.java @@ -14,12 +14,16 @@ class Solution { public boolean verifyPostorder(int[] postorder) { return recur(postorder, 0, postorder.length - 1); } + boolean recur(int[] postorder, int i, int j) { - if(i >= j) return true; + if (i >= j) + return true; int p = i; - while(postorder[p] < postorder[j]) p++; + while (postorder[p] < postorder[j]) + p++; int m = p; - while(postorder[p] > postorder[j]) p++; + while (postorder[p] > postorder[j]) + p++; return p == j && recur(postorder, i, m - 1) && recur(postorder, m, j - 1); } } diff --git a/sword_for_offer/codes/java/sfo_33_postorder_traversal_of_a_binary_search_tree_s2/sfo_33_postorder_traversal_of_a_binary_search_tree_s2.java b/sword_for_offer/codes/java/sfo_33_postorder_traversal_of_a_binary_search_tree_s2/sfo_33_postorder_traversal_of_a_binary_search_tree_s2.java index bee7743..e1ab3d6 100644 --- a/sword_for_offer/codes/java/sfo_33_postorder_traversal_of_a_binary_search_tree_s2/sfo_33_postorder_traversal_of_a_binary_search_tree_s2.java +++ b/sword_for_offer/codes/java/sfo_33_postorder_traversal_of_a_binary_search_tree_s2/sfo_33_postorder_traversal_of_a_binary_search_tree_s2.java @@ -14,9 +14,10 @@ class Solution { public boolean verifyPostorder(int[] postorder) { Stack stack = new Stack<>(); int root = Integer.MAX_VALUE; - for(int i = postorder.length - 1; i >= 0; i--) { - if(postorder[i] > root) return false; - while(!stack.isEmpty() && stack.peek() > postorder[i]) + for (int i = postorder.length - 1; i >= 0; i--) { + if (postorder[i] > root) + return false; + while (!stack.isEmpty() && stack.peek() > postorder[i]) root = stack.pop(); stack.add(postorder[i]); } diff --git a/sword_for_offer/codes/java/sfo_34_all_xsum_paths_in_a_binary_tree_s1/sfo_34_all_xsum_paths_in_a_binary_tree_s1.java b/sword_for_offer/codes/java/sfo_34_all_xsum_paths_in_a_binary_tree_s1/sfo_34_all_xsum_paths_in_a_binary_tree_s1.java index 63230ce..9a04a28 100644 --- a/sword_for_offer/codes/java/sfo_34_all_xsum_paths_in_a_binary_tree_s1/sfo_34_all_xsum_paths_in_a_binary_tree_s1.java +++ b/sword_for_offer/codes/java/sfo_34_all_xsum_paths_in_a_binary_tree_s1/sfo_34_all_xsum_paths_in_a_binary_tree_s1.java @@ -13,15 +13,18 @@ class Solution { LinkedList> res = new LinkedList<>(); LinkedList path = new LinkedList<>(); + public List> pathSum(TreeNode root, int sum) { recur(root, sum); return res; } + void recur(TreeNode root, int tar) { - if(root == null) return; + if (root == null) + return; path.add(root.val); tar -= root.val; - if(tar == 0 && root.left == null && root.right == null) + if (tar == 0 && root.left == null && root.right == null) res.add(new LinkedList(path)); recur(root.left, tar); recur(root.right, tar); @@ -32,11 +35,13 @@ void recur(TreeNode root, int tar) { public class sfo_34_all_xsum_paths_in_a_binary_tree_s1 { public static void main(String[] args) { // ======= Test Case ======= - TreeNode root = TreeNode.arrToTree(new Integer[] { 5, 4, 8, 11, null, 13, 4, 7, 2, null, null, 5, 1, null, null, null, null, null, null, null, null }); + TreeNode root = TreeNode.arrToTree(new Integer[] { 5, 4, 8, 11, null, 13, 4, 7, 2, null, null, 5, 1, null, null, + null, null, null, null, null, null }); int sum = 22; // ====== Driver Code ====== Solution slt = new Solution(); List> res = slt.pathSum(root, sum); - System.out.println(Arrays.deepToString(res.toArray()));; + System.out.println(Arrays.deepToString(res.toArray())); + ; } } diff --git a/sword_for_offer/codes/java/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1.java b/sword_for_offer/codes/java/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1.java index f440416..185fea9 100644 --- a/sword_for_offer/codes/java/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1.java +++ b/sword_for_offer/codes/java/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1.java @@ -25,17 +25,18 @@ public Node(int val) { // ===== Solution Code ===== class Solution { public Node copyRandomList(Node head) { - if(head == null) return null; + if (head == null) + return null; Node cur = head; Map map = new HashMap<>(); // 3. 复制各节点，并建立 “原节点 -> 新节点” 的 Map 映射 - while(cur != null) { + while (cur != null) { map.put(cur, new Node(cur.val)); cur = cur.next; } cur = head; // 4. 构建新链表的 next 和 random 指向 - while(cur != null) { + while (cur != null) { map.get(cur).next = map.get(cur.next); map.get(cur).random = map.get(cur.random); cur = cur.next; diff --git a/sword_for_offer/codes/java/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2.java b/sword_for_offer/codes/java/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2.java index 528ab20..331b413 100644 --- a/sword_for_offer/codes/java/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2.java +++ b/sword_for_offer/codes/java/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2.java @@ -25,10 +25,11 @@ public Node(int val) { // ===== Solution Code ===== class Solution { public Node copyRandomList(Node head) { - if(head == null) return null; + if (head == null) + return null; Node cur = head; // 1. 复制各节点，并构建拼接链表 - while(cur != null) { + while (cur != null) { Node tmp = new Node(cur.val); tmp.next = cur.next; cur.next = tmp; @@ -36,22 +37,22 @@ public Node copyRandomList(Node head) { } // 2. 构建各新节点的 random 指向 cur = head; - while(cur != null) { - if(cur.random != null) + while (cur != null) { + if (cur.random != null) cur.next.random = cur.random.next; cur = cur.next.next; } // 3. 拆分两链表 cur = head.next; Node pre = head, res = head.next; - while(cur.next != null) { + while (cur.next != null) { pre.next = pre.next.next; cur.next = cur.next.next; pre = pre.next; cur = cur.next; } pre.next = null; // 单独处理原链表尾节点 - return res; // 返回新链表头节点 + return res; // 返回新链表头节点 } } diff --git a/sword_for_offer/codes/java/sfo_36_binary_search_tree_and_doubly_linked_list_s1/sfo_36_binary_search_tree_and_doubly_linked_list_s1.java b/sword_for_offer/codes/java/sfo_36_binary_search_tree_and_doubly_linked_list_s1/sfo_36_binary_search_tree_and_doubly_linked_list_s1.java index 22480e0..0a8f879 100644 --- a/sword_for_offer/codes/java/sfo_36_binary_search_tree_and_doubly_linked_list_s1/sfo_36_binary_search_tree_and_doubly_linked_list_s1.java +++ b/sword_for_offer/codes/java/sfo_36_binary_search_tree_and_doubly_linked_list_s1/sfo_36_binary_search_tree_and_doubly_linked_list_s1.java @@ -25,18 +25,24 @@ public Node(int val) { // ===== Solution Code ===== class Solution { Node pre, head; + public Node treeToDoublyList(Node root) { - if(root == null) return null; + if (root == null) + return null; dfs(root); head.left = pre; pre.right = head; return head; } + void dfs(Node cur) { - if(cur == null) return; + if (cur == null) + return; dfs(cur.left); - if(pre != null) pre.right = cur; - else head = cur; + if (pre != null) + pre.right = cur; + else + head = cur; cur.left = pre; pre = cur; dfs(cur.right); @@ -46,13 +52,15 @@ void dfs(Node cur) { public class sfo_36_binary_search_tree_and_doubly_linked_list_s1 { public static void main(String[] args) { // ======= Test Case ======= - List nodeList = new ArrayList() {{ - add(new Node(1)); - add(new Node(2)); - add(new Node(3)); - add(new Node(4)); - add(new Node(5)); - }}; + List nodeList = new ArrayList() { + { + add(new Node(1)); + add(new Node(2)); + add(new Node(3)); + add(new Node(4)); + add(new Node(5)); + } + }; nodeList.get(3).left = nodeList.get(1); nodeList.get(3).right = nodeList.get(4); nodeList.get(1).left = nodeList.get(0); diff --git a/sword_for_offer/codes/java/sfo_37_serialize_and_deserialize_a_binary_tree_s1/sfo_37_serialize_and_deserialize_a_binary_tree_s1.java b/sword_for_offer/codes/java/sfo_37_serialize_and_deserialize_a_binary_tree_s1/sfo_37_serialize_and_deserialize_a_binary_tree_s1.java index b52230a..acf7696 100644 --- a/sword_for_offer/codes/java/sfo_37_serialize_and_deserialize_a_binary_tree_s1/sfo_37_serialize_and_deserialize_a_binary_tree_s1.java +++ b/sword_for_offer/codes/java/sfo_37_serialize_and_deserialize_a_binary_tree_s1/sfo_37_serialize_and_deserialize_a_binary_tree_s1.java @@ -12,17 +12,22 @@ // ===== Solution Code ===== class Codec { public String serialize(TreeNode root) { - if(root == null) return "[]"; + if (root == null) + return "[]"; StringBuilder res = new StringBuilder("["); - Queue queue = new LinkedList() {{ add(root); }}; - while(!queue.isEmpty()) { + Queue queue = new LinkedList() { + { + add(root); + } + }; + while (!queue.isEmpty()) { TreeNode node = queue.poll(); - if(node != null) { + if (node != null) { res.append(node.val + ","); queue.add(node.left); queue.add(node.right); - } - else res.append("null,"); + } else + res.append("null,"); } res.deleteCharAt(res.length() - 1); res.append("]"); @@ -30,19 +35,24 @@ public String serialize(TreeNode root) { } public TreeNode deserialize(String data) { - if(data.equals("[]")) return null; + if (data.equals("[]")) + return null; String[] vals = data.substring(1, data.length() - 1).split(","); TreeNode root = new TreeNode(Integer.parseInt(vals[0])); - Queue queue = new LinkedList() {{ add(root); }}; + Queue queue = new LinkedList() { + { + add(root); + } + }; int i = 1; - while(!queue.isEmpty()) { + while (!queue.isEmpty()) { TreeNode node = queue.poll(); - if(!vals[i].equals("null")) { + if (!vals[i].equals("null")) { node.left = new TreeNode(Integer.parseInt(vals[i])); queue.add(node.left); } i++; - if(!vals[i].equals("null")) { + if (!vals[i].equals("null")) { node.right = new TreeNode(Integer.parseInt(vals[i])); queue.add(node.right); } diff --git a/sword_for_offer/codes/java/sfo_38_all_permutations_of_a_string_s1/sfo_38_all_permutations_of_a_string_s1.java b/sword_for_offer/codes/java/sfo_38_all_permutations_of_a_string_s1/sfo_38_all_permutations_of_a_string_s1.java index 4d457f4..32a41e8 100644 --- a/sword_for_offer/codes/java/sfo_38_all_permutations_of_a_string_s1/sfo_38_all_permutations_of_a_string_s1.java +++ b/sword_for_offer/codes/java/sfo_38_all_permutations_of_a_string_s1/sfo_38_all_permutations_of_a_string_s1.java @@ -13,25 +13,29 @@ class Solution { List res = new LinkedList<>(); char[] c; + public String[] permutation(String s) { c = s.toCharArray(); dfs(0); return res.toArray(new String[res.size()]); } + void dfs(int x) { - if(x == c.length - 1) { - res.add(String.valueOf(c)); // 添加排列方案 + if (x == c.length - 1) { + res.add(String.valueOf(c)); // 添加排列方案 return; } HashSet set = new HashSet<>(); - for(int i = x; i < c.length; i++) { - if(set.contains(c[i])) continue; // 重复，因此剪枝 + for (int i = x; i < c.length; i++) { + if (set.contains(c[i])) + continue; // 重复，因此剪枝 set.add(c[i]); - swap(i, x); // 交换，将 c[i] 固定在第 x 位 - dfs(x + 1); // 开启固定第 x + 1 位字符 - swap(i, x); // 恢复交换 + swap(i, x); // 交换，将 c[i] 固定在第 x 位 + dfs(x + 1); // 开启固定第 x + 1 位字符 + swap(i, x); // 恢复交换 } } + void swap(int a, int b) { char tmp = c[a]; c[a] = c[b]; diff --git a/sword_for_offer/codes/java/sfo_39_the_majority_element_in_an_array_s1/sfo_39_the_majority_element_in_an_array_s1.java b/sword_for_offer/codes/java/sfo_39_the_majority_element_in_an_array_s1/sfo_39_the_majority_element_in_an_array_s1.java index 986a01c..2842079 100644 --- a/sword_for_offer/codes/java/sfo_39_the_majority_element_in_an_array_s1/sfo_39_the_majority_element_in_an_array_s1.java +++ b/sword_for_offer/codes/java/sfo_39_the_majority_element_in_an_array_s1/sfo_39_the_majority_element_in_an_array_s1.java @@ -13,8 +13,9 @@ class Solution { public int majorityElement(int[] nums) { int x = 0, votes = 0; - for(int num : nums){ - if(votes == 0) x = num; + for (int num : nums) { + if (votes == 0) + x = num; votes += num == x ? 1 : -1; } return x; diff --git a/sword_for_offer/codes/java/sfo_39_the_majority_element_in_an_array_s2/sfo_39_the_majority_element_in_an_array_s2.java b/sword_for_offer/codes/java/sfo_39_the_majority_element_in_an_array_s2/sfo_39_the_majority_element_in_an_array_s2.java index 522d884..7bc4a6d 100644 --- a/sword_for_offer/codes/java/sfo_39_the_majority_element_in_an_array_s2/sfo_39_the_majority_element_in_an_array_s2.java +++ b/sword_for_offer/codes/java/sfo_39_the_majority_element_in_an_array_s2/sfo_39_the_majority_element_in_an_array_s2.java @@ -13,13 +13,15 @@ class Solution { public int majorityElement(int[] nums) { int x = 0, votes = 0, count = 0; - for(int num : nums){ - if(votes == 0) x = num; + for (int num : nums) { + if (votes == 0) + x = num; votes += num == x ? 1 : -1; } // 验证 x 是否为众数 - for(int num : nums) - if(num == x) count++; + for (int num : nums) + if (num == x) + count++; return count > nums.length / 2 ? x : 0; // 当无众数时返回 0 } } diff --git a/sword_for_offer/codes/java/sfo_40_the_smallest_k_numbers_s1/sfo_40_the_smallest_k_numbers_s1.java b/sword_for_offer/codes/java/sfo_40_the_smallest_k_numbers_s1/sfo_40_the_smallest_k_numbers_s1.java index 52f8a87..705ea3b 100644 --- a/sword_for_offer/codes/java/sfo_40_the_smallest_k_numbers_s1/sfo_40_the_smallest_k_numbers_s1.java +++ b/sword_for_offer/codes/java/sfo_40_the_smallest_k_numbers_s1/sfo_40_the_smallest_k_numbers_s1.java @@ -15,14 +15,18 @@ public int[] getLeastNumbers(int[] arr, int k) { quickSort(arr, 0, arr.length - 1); return Arrays.copyOf(arr, k); } + private void quickSort(int[] arr, int l, int r) { // 子数组长度为 1 时终止递归 - if (l >= r) return; + if (l >= r) + return; // 哨兵划分操作（以 arr[l] 作为基准数） int i = l, j = r; while (i < j) { - while (i < j && arr[j] >= arr[l]) j--; - while (i < j && arr[i] <= arr[l]) i++; + while (i < j && arr[j] >= arr[l]) + j--; + while (i < j && arr[i] <= arr[l]) + i++; swap(arr, i, j); } swap(arr, i, l); @@ -30,6 +34,7 @@ private void quickSort(int[] arr, int l, int r) { quickSort(arr, l, i - 1); quickSort(arr, i + 1, r); } + private void swap(int[] arr, int i, int j) { int tmp = arr[i]; arr[i] = arr[j]; diff --git a/sword_for_offer/codes/java/sfo_40_the_smallest_k_numbers_s2/sfo_40_the_smallest_k_numbers_s2.java b/sword_for_offer/codes/java/sfo_40_the_smallest_k_numbers_s2/sfo_40_the_smallest_k_numbers_s2.java index 71ef40b..bd5c35c 100644 --- a/sword_for_offer/codes/java/sfo_40_the_smallest_k_numbers_s2/sfo_40_the_smallest_k_numbers_s2.java +++ b/sword_for_offer/codes/java/sfo_40_the_smallest_k_numbers_s2/sfo_40_the_smallest_k_numbers_s2.java @@ -12,21 +12,28 @@ // ===== Solution Code ===== class Solution { public int[] getLeastNumbers(int[] arr, int k) { - if (k >= arr.length) return arr; + if (k >= arr.length) + return arr; return quickSort(arr, k, 0, arr.length - 1); } + private int[] quickSort(int[] arr, int k, int l, int r) { int i = l, j = r; while (i < j) { - while (i < j && arr[j] >= arr[l]) j--; - while (i < j && arr[i] <= arr[l]) i++; + while (i < j && arr[j] >= arr[l]) + j--; + while (i < j && arr[i] <= arr[l]) + i++; swap(arr, i, j); } swap(arr, i, l); - if (i > k) return quickSort(arr, k, l, i - 1); - if (i < k) return quickSort(arr, k, i + 1, r); + if (i > k) + return quickSort(arr, k, l, i - 1); + if (i < k) + return quickSort(arr, k, i + 1, r); return Arrays.copyOf(arr, k); } + private void swap(int[] arr, int i, int j) { int tmp = arr[i]; arr[i] = arr[j]; diff --git a/sword_for_offer/codes/java/sfo_41_find_median_from_data_stream_s1/sfo_41_find_median_from_data_stream_s1.java b/sword_for_offer/codes/java/sfo_41_find_median_from_data_stream_s1/sfo_41_find_median_from_data_stream_s1.java index c91c6f0..d3dde6a 100644 --- a/sword_for_offer/codes/java/sfo_41_find_median_from_data_stream_s1/sfo_41_find_median_from_data_stream_s1.java +++ b/sword_for_offer/codes/java/sfo_41_find_median_from_data_stream_s1/sfo_41_find_median_from_data_stream_s1.java @@ -12,12 +12,14 @@ // ===== Solution Code ===== class MedianFinder { Queue A, B; + public MedianFinder() { A = new PriorityQueue<>(); // 小顶堆，保存较大的一半 B = new PriorityQueue<>((x, y) -> (y - x)); // 大顶堆，保存较小的一半 } + public void addNum(int num) { - if(A.size() != B.size()) { + if (A.size() != B.size()) { A.add(num); B.add(A.poll()); } else { @@ -25,6 +27,7 @@ public void addNum(int num) { A.add(B.poll()); } } + public double findMedian() { return A.size() != B.size() ? A.peek() : (A.peek() + B.peek()) / 2.0; } diff --git a/sword_for_offer/codes/java/sfo_42_largest_sum_contiguous_subarray_s1/sfo_42_largest_sum_contiguous_subarray_s1.java b/sword_for_offer/codes/java/sfo_42_largest_sum_contiguous_subarray_s1/sfo_42_largest_sum_contiguous_subarray_s1.java index e6a1862..abb663e 100644 --- a/sword_for_offer/codes/java/sfo_42_largest_sum_contiguous_subarray_s1/sfo_42_largest_sum_contiguous_subarray_s1.java +++ b/sword_for_offer/codes/java/sfo_42_largest_sum_contiguous_subarray_s1/sfo_42_largest_sum_contiguous_subarray_s1.java @@ -13,7 +13,7 @@ class Solution { public int maxSubArray(int[] nums) { int res = nums[0]; - for(int i = 1; i < nums.length; i++) { + for (int i = 1; i < nums.length; i++) { nums[i] += Math.max(nums[i - 1], 0); res = Math.max(res, nums[i]); } diff --git a/sword_for_offer/codes/java/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1.java b/sword_for_offer/codes/java/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1.java index b0b864d..9d6367d 100644 --- a/sword_for_offer/codes/java/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1.java +++ b/sword_for_offer/codes/java/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1.java @@ -14,10 +14,13 @@ class Solution { public int countDigitOne(int n) { int digit = 1, res = 0; int high = n / 10, cur = n % 10, low = 0; - while(high != 0 || cur != 0) { - if(cur == 0) res += high * digit; - else if(cur == 1) res += high * digit + low + 1; - else res += (high + 1) * digit; + while (high != 0 || cur != 0) { + if (cur == 0) + res += high * digit; + else if (cur == 1) + res += high * digit + low + 1; + else + res += (high + 1) * digit; low += cur * digit; cur = high % 10; high /= 10; diff --git a/sword_for_offer/codes/java/sfo_45_arrange_an_array_into_the_smallest_number_s1/sfo_45_arrange_an_array_into_the_smallest_number_s1.java b/sword_for_offer/codes/java/sfo_45_arrange_an_array_into_the_smallest_number_s1/sfo_45_arrange_an_array_into_the_smallest_number_s1.java index 8fd3e9d..582cea5 100644 --- a/sword_for_offer/codes/java/sfo_45_arrange_an_array_into_the_smallest_number_s1/sfo_45_arrange_an_array_into_the_smallest_number_s1.java +++ b/sword_for_offer/codes/java/sfo_45_arrange_an_array_into_the_smallest_number_s1/sfo_45_arrange_an_array_into_the_smallest_number_s1.java @@ -13,21 +13,25 @@ class Solution { public String minNumber(int[] nums) { String[] strs = new String[nums.length]; - for(int i = 0; i < nums.length; i++) + for (int i = 0; i < nums.length; i++) strs[i] = String.valueOf(nums[i]); quickSort(strs, 0, strs.length - 1); StringBuilder res = new StringBuilder(); - for(String s : strs) + for (String s : strs) res.append(s); return res.toString(); } + void quickSort(String[] strs, int l, int r) { - if(l >= r) return; + if (l >= r) + return; int i = l, j = r; String tmp = strs[i]; - while(i < j) { - while((strs[j] + strs[l]).compareTo(strs[l] + strs[j]) >= 0 && i < j) j--; - while((strs[i] + strs[l]).compareTo(strs[l] + strs[i]) <= 0 && i < j) i++; + while (i < j) { + while ((strs[j] + strs[l]).compareTo(strs[l] + strs[j]) >= 0 && i < j) + j--; + while ((strs[i] + strs[l]).compareTo(strs[l] + strs[i]) <= 0 && i < j) + i++; tmp = strs[i]; strs[i] = strs[j]; strs[j] = tmp; diff --git a/sword_for_offer/codes/java/sfo_45_arrange_an_array_into_the_smallest_number_s2/sfo_45_arrange_an_array_into_the_smallest_number_s2.java b/sword_for_offer/codes/java/sfo_45_arrange_an_array_into_the_smallest_number_s2/sfo_45_arrange_an_array_into_the_smallest_number_s2.java index 4539e62..5bf4cf4 100644 --- a/sword_for_offer/codes/java/sfo_45_arrange_an_array_into_the_smallest_number_s2/sfo_45_arrange_an_array_into_the_smallest_number_s2.java +++ b/sword_for_offer/codes/java/sfo_45_arrange_an_array_into_the_smallest_number_s2/sfo_45_arrange_an_array_into_the_smallest_number_s2.java @@ -13,11 +13,11 @@ class Solution { public String minNumber(int[] nums) { String[] strs = new String[nums.length]; - for(int i = 0; i < nums.length; i++) + for (int i = 0; i < nums.length; i++) strs[i] = String.valueOf(nums[i]); Arrays.sort(strs, (x, y) -> (x + y).compareTo(y + x)); StringBuilder res = new StringBuilder(); - for(String s : strs) + for (String s : strs) res.append(s); return res.toString(); } diff --git a/sword_for_offer/codes/java/sfo_46_translate_numbers_into_strings_s1/sfo_46_translate_numbers_into_strings_s1.java b/sword_for_offer/codes/java/sfo_46_translate_numbers_into_strings_s1/sfo_46_translate_numbers_into_strings_s1.java index 6040fef..a73fba6 100644 --- a/sword_for_offer/codes/java/sfo_46_translate_numbers_into_strings_s1/sfo_46_translate_numbers_into_strings_s1.java +++ b/sword_for_offer/codes/java/sfo_46_translate_numbers_into_strings_s1/sfo_46_translate_numbers_into_strings_s1.java @@ -14,7 +14,7 @@ class Solution { public int translateNum(int num) { String s = String.valueOf(num); int a = 1, b = 1; - for(int i = 2; i <= s.length(); i++) { + for (int i = 2; i <= s.length(); i++) { String tmp = s.substring(i - 2, i); int c = tmp.compareTo("10") >= 0 && tmp.compareTo("25") <= 0 ? a + b : a; b = a; diff --git a/sword_for_offer/codes/java/sfo_46_translate_numbers_into_strings_s2/sfo_46_translate_numbers_into_strings_s2.java b/sword_for_offer/codes/java/sfo_46_translate_numbers_into_strings_s2/sfo_46_translate_numbers_into_strings_s2.java index 4250977..0ef277b 100644 --- a/sword_for_offer/codes/java/sfo_46_translate_numbers_into_strings_s2/sfo_46_translate_numbers_into_strings_s2.java +++ b/sword_for_offer/codes/java/sfo_46_translate_numbers_into_strings_s2/sfo_46_translate_numbers_into_strings_s2.java @@ -14,7 +14,7 @@ class Solution { public int translateNum(int num) { String s = String.valueOf(num); int a = 1, b = 1; - for(int i = s.length() - 2; i > -1; i--) { + for (int i = s.length() - 2; i > -1; i--) { String tmp = s.substring(i, i + 2); int c = tmp.compareTo("10") >= 0 && tmp.compareTo("25") <= 0 ? a + b : a; b = a; diff --git a/sword_for_offer/codes/java/sfo_46_translate_numbers_into_strings_s3/sfo_46_translate_numbers_into_strings_s3.java b/sword_for_offer/codes/java/sfo_46_translate_numbers_into_strings_s3/sfo_46_translate_numbers_into_strings_s3.java index 4ea35ed..10f7f7a 100644 --- a/sword_for_offer/codes/java/sfo_46_translate_numbers_into_strings_s3/sfo_46_translate_numbers_into_strings_s3.java +++ b/sword_for_offer/codes/java/sfo_46_translate_numbers_into_strings_s3/sfo_46_translate_numbers_into_strings_s3.java @@ -13,7 +13,7 @@ class Solution { public int translateNum(int num) { int a = 1, b = 1, x, y = num % 10; - while(num > 9) { + while (num > 9) { num /= 10; x = num % 10; int tmp = 10 * x + y; diff --git a/sword_for_offer/codes/java/sfo_47_the_maximum_value_of_gifts_s1/sfo_47_the_maximum_value_of_gifts_s1.java b/sword_for_offer/codes/java/sfo_47_the_maximum_value_of_gifts_s1/sfo_47_the_maximum_value_of_gifts_s1.java index d7de671..6170d16 100644 --- a/sword_for_offer/codes/java/sfo_47_the_maximum_value_of_gifts_s1/sfo_47_the_maximum_value_of_gifts_s1.java +++ b/sword_for_offer/codes/java/sfo_47_the_maximum_value_of_gifts_s1/sfo_47_the_maximum_value_of_gifts_s1.java @@ -13,12 +13,16 @@ class Solution { public int maxValue(int[][] grid) { int m = grid.length, n = grid[0].length; - for(int i = 0; i < m; i++) { - for(int j = 0; j < n; j++) { - if(i == 0 && j == 0) continue; - if(i == 0) grid[i][j] += grid[i][j - 1] ; - else if(j == 0) grid[i][j] += grid[i - 1][j]; - else grid[i][j] += Math.max(grid[i][j - 1], grid[i - 1][j]); + for (int i = 0; i < m; i++) { + for (int j = 0; j < n; j++) { + if (i == 0 && j == 0) + continue; + if (i == 0) + grid[i][j] += grid[i][j - 1]; + else if (j == 0) + grid[i][j] += grid[i - 1][j]; + else + grid[i][j] += Math.max(grid[i][j - 1], grid[i - 1][j]); } } return grid[m - 1][n - 1]; @@ -29,10 +33,10 @@ public class sfo_47_the_maximum_value_of_gifts_s1 { public static void main(String[] args) { // ======= Test Case ======= int[][] grid = { - { 1, 3, 1 }, - { 1, 5, 1 }, - { 4, 2, 1 } -}; + { 1, 3, 1 }, + { 1, 5, 1 }, + { 4, 2, 1 } + }; // ====== Driver Code ====== Solution slt = new Solution(); int res = slt.maxValue(grid); diff --git a/sword_for_offer/codes/java/sfo_47_the_maximum_value_of_gifts_s2/sfo_47_the_maximum_value_of_gifts_s2.java b/sword_for_offer/codes/java/sfo_47_the_maximum_value_of_gifts_s2/sfo_47_the_maximum_value_of_gifts_s2.java index 5971383..5b66ab7 100644 --- a/sword_for_offer/codes/java/sfo_47_the_maximum_value_of_gifts_s2/sfo_47_the_maximum_value_of_gifts_s2.java +++ b/sword_for_offer/codes/java/sfo_47_the_maximum_value_of_gifts_s2/sfo_47_the_maximum_value_of_gifts_s2.java @@ -13,12 +13,12 @@ class Solution { public int maxValue(int[][] grid) { int m = grid.length, n = grid[0].length; - for(int j = 1; j < n; j++) // 初始化第一行 + for (int j = 1; j < n; j++) // 初始化第一行 grid[0][j] += grid[0][j - 1]; - for(int i = 1; i < m; i++) // 初始化第一列 + for (int i = 1; i < m; i++) // 初始化第一列 grid[i][0] += grid[i - 1][0]; - for(int i = 1; i < m; i++) - for(int j = 1; j < n; j++) + for (int i = 1; i < m; i++) + for (int j = 1; j < n; j++) grid[i][j] += Math.max(grid[i][j - 1], grid[i - 1][j]); return grid[m - 1][n - 1]; } @@ -28,10 +28,10 @@ public class sfo_47_the_maximum_value_of_gifts_s2 { public static void main(String[] args) { // ======= Test Case ======= int[][] grid = { - { 1, 3, 1 }, - { 1, 5, 1 }, - { 4, 2, 1 } -}; + { 1, 3, 1 }, + { 1, 5, 1 }, + { 4, 2, 1 } + }; // ====== Driver Code ====== Solution slt = new Solution(); int res = slt.maxValue(grid); diff --git a/sword_for_offer/codes/java/sfo_48_the_longest_substring_without_repeated_characters_s1/sfo_48_the_longest_substring_without_repeated_characters_s1.java b/sword_for_offer/codes/java/sfo_48_the_longest_substring_without_repeated_characters_s1/sfo_48_the_longest_substring_without_repeated_characters_s1.java index 41460d8..1454a49 100644 --- a/sword_for_offer/codes/java/sfo_48_the_longest_substring_without_repeated_characters_s1/sfo_48_the_longest_substring_without_repeated_characters_s1.java +++ b/sword_for_offer/codes/java/sfo_48_the_longest_substring_without_repeated_characters_s1/sfo_48_the_longest_substring_without_repeated_characters_s1.java @@ -14,7 +14,7 @@ class Solution { public int lengthOfLongestSubstring(String s) { Map dic = new HashMap<>(); int res = 0, tmp = 0, len = s.length(); - for(int j = 0; j < len; j++) { + for (int j = 0; j < len; j++) { int i = dic.getOrDefault(s.charAt(j), -1); // 获取索引 i dic.put(s.charAt(j), j); // 更新哈希表 tmp = tmp < j - i ? tmp + 1 : j - i; // dp[j - 1] -> dp[j] diff --git a/sword_for_offer/codes/java/sfo_48_the_longest_substring_without_repeated_characters_s2/sfo_48_the_longest_substring_without_repeated_characters_s2.java b/sword_for_offer/codes/java/sfo_48_the_longest_substring_without_repeated_characters_s2/sfo_48_the_longest_substring_without_repeated_characters_s2.java index 0141a66..c2e5f4e 100644 --- a/sword_for_offer/codes/java/sfo_48_the_longest_substring_without_repeated_characters_s2/sfo_48_the_longest_substring_without_repeated_characters_s2.java +++ b/sword_for_offer/codes/java/sfo_48_the_longest_substring_without_repeated_characters_s2/sfo_48_the_longest_substring_without_repeated_characters_s2.java @@ -13,9 +13,10 @@ class Solution { public int lengthOfLongestSubstring(String s) { int res = 0, tmp = 0, len = s.length(); - for(int j = 0; j < len; j++) { + for (int j = 0; j < len; j++) { int i = j - 1; - while(i >= 0 && s.charAt(i) != s.charAt(j)) i--; // 线性查找 i + while (i >= 0 && s.charAt(i) != s.charAt(j)) + i--; // 线性查找 i tmp = tmp < j - i ? tmp + 1 : j - i; // dp[j - 1] -> dp[j] res = Math.max(res, tmp); // max(dp[j - 1], dp[j]) } diff --git a/sword_for_offer/codes/java/sfo_48_the_longest_substring_without_repeated_characters_s3/sfo_48_the_longest_substring_without_repeated_characters_s3.java b/sword_for_offer/codes/java/sfo_48_the_longest_substring_without_repeated_characters_s3/sfo_48_the_longest_substring_without_repeated_characters_s3.java index 0399fad..819054b 100644 --- a/sword_for_offer/codes/java/sfo_48_the_longest_substring_without_repeated_characters_s3/sfo_48_the_longest_substring_without_repeated_characters_s3.java +++ b/sword_for_offer/codes/java/sfo_48_the_longest_substring_without_repeated_characters_s3/sfo_48_the_longest_substring_without_repeated_characters_s3.java @@ -14,8 +14,8 @@ class Solution { public int lengthOfLongestSubstring(String s) { Map dic = new HashMap<>(); int i = -1, res = 0, len = s.length(); - for(int j = 0; j < len; j++) { - if(dic.containsKey(s.charAt(j))) + for (int j = 0; j < len; j++) { + if (dic.containsKey(s.charAt(j))) i = Math.max(i, dic.get(s.charAt(j))); // 更新左指针 i dic.put(s.charAt(j), j); // 哈希表记录 res = Math.max(res, j - i); // 更新结果 diff --git a/sword_for_offer/codes/java/sfo_49_ugly_numbers_s1/sfo_49_ugly_numbers_s1.java b/sword_for_offer/codes/java/sfo_49_ugly_numbers_s1/sfo_49_ugly_numbers_s1.java index 39ed64d..59f7bbf 100644 --- a/sword_for_offer/codes/java/sfo_49_ugly_numbers_s1/sfo_49_ugly_numbers_s1.java +++ b/sword_for_offer/codes/java/sfo_49_ugly_numbers_s1/sfo_49_ugly_numbers_s1.java @@ -15,12 +15,15 @@ public int nthUglyNumber(int n) { int a = 0, b = 0, c = 0; int[] dp = new int[n]; dp[0] = 1; - for(int i = 1; i < n; i++) { + for (int i = 1; i < n; i++) { int n2 = dp[a] * 2, n3 = dp[b] * 3, n5 = dp[c] * 5; dp[i] = Math.min(Math.min(n2, n3), n5); - if(dp[i] == n2) a++; - if(dp[i] == n3) b++; - if(dp[i] == n5) c++; + if (dp[i] == n2) + a++; + if (dp[i] == n3) + b++; + if (dp[i] == n5) + c++; } return dp[n - 1]; } diff --git a/sword_for_offer/codes/java/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1.java b/sword_for_offer/codes/java/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1.java index a498979..bf11373 100644 --- a/sword_for_offer/codes/java/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1.java +++ b/sword_for_offer/codes/java/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1.java @@ -14,10 +14,11 @@ class Solution { public char firstUniqChar(String s) { HashMap dic = new HashMap<>(); char[] sc = s.toCharArray(); - for(char c : sc) + for (char c : sc) dic.put(c, !dic.containsKey(c)); - for(char c : sc) - if(dic.get(c)) return c; + for (char c : sc) + if (dic.get(c)) + return c; return ' '; } } diff --git a/sword_for_offer/codes/java/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2.java b/sword_for_offer/codes/java/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2.java index 7966921..891fcb3 100644 --- a/sword_for_offer/codes/java/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2.java +++ b/sword_for_offer/codes/java/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2.java @@ -14,10 +14,11 @@ class Solution { public char firstUniqChar(String s) { Map dic = new LinkedHashMap<>(); char[] sc = s.toCharArray(); - for(char c : sc) + for (char c : sc) dic.put(c, !dic.containsKey(c)); - for(Map.Entry d : dic.entrySet()){ - if(d.getValue()) return d.getKey(); + for (Map.Entry d : dic.entrySet()) { + if (d.getValue()) + return d.getKey(); } return ' '; } diff --git a/sword_for_offer/codes/java/sfo_51_reversed_pairs_in_an_array_s1/sfo_51_reversed_pairs_in_an_array_s1.java b/sword_for_offer/codes/java/sfo_51_reversed_pairs_in_an_array_s1/sfo_51_reversed_pairs_in_an_array_s1.java index 55d4e93..c2ad3d1 100644 --- a/sword_for_offer/codes/java/sfo_51_reversed_pairs_in_an_array_s1/sfo_51_reversed_pairs_in_an_array_s1.java +++ b/sword_for_offer/codes/java/sfo_51_reversed_pairs_in_an_array_s1/sfo_51_reversed_pairs_in_an_array_s1.java @@ -12,14 +12,17 @@ // ===== Solution Code ===== class Solution { int[] nums, tmp; + public int reversePairs(int[] nums) { this.nums = nums; tmp = new int[nums.length]; return mergeSort(0, nums.length - 1); } + private int mergeSort(int l, int r) { // 终止条件 - if (l >= r) return 0; + if (l >= r) + return 0; // 递归划分 int m = (l + r) / 2; int res = mergeSort(l, m) + mergeSort(m + 1, r); diff --git a/sword_for_offer/codes/java/sfo_52_the_first_common_node_in_two_linked_lists_s1/sfo_52_the_first_common_node_in_two_linked_lists_s1.java b/sword_for_offer/codes/java/sfo_52_the_first_common_node_in_two_linked_lists_s1/sfo_52_the_first_common_node_in_two_linked_lists_s1.java index b04feed..9d45def 100644 --- a/sword_for_offer/codes/java/sfo_52_the_first_common_node_in_two_linked_lists_s1/sfo_52_the_first_common_node_in_two_linked_lists_s1.java +++ b/sword_for_offer/codes/java/sfo_52_the_first_common_node_in_two_linked_lists_s1/sfo_52_the_first_common_node_in_two_linked_lists_s1.java @@ -26,7 +26,7 @@ public static void main(String[] args) { // ======= Test Case ======= // Build two linked lists with intersection of 8 // headA = 4 -> 1 -> 8 -> 4 -> 5 - // ↑ + // ↑ // headB = 5 -> 0 -> 1 ListNode headA = ListNode.arrToLinkedList(new int[] { 4, 1, 8, 4, 5 }); ListNode headB = ListNode.arrToLinkedList(new int[] { 5, 0, 1, 8, 4, 5 }); diff --git a/sword_for_offer/codes/java/sfo_53i_find_a_number_in_a_sorted_array_s1/sfo_53i_find_a_number_in_a_sorted_array_s1.java b/sword_for_offer/codes/java/sfo_53i_find_a_number_in_a_sorted_array_s1/sfo_53i_find_a_number_in_a_sorted_array_s1.java index 50d407d..36ca481 100644 --- a/sword_for_offer/codes/java/sfo_53i_find_a_number_in_a_sorted_array_s1/sfo_53i_find_a_number_in_a_sorted_array_s1.java +++ b/sword_for_offer/codes/java/sfo_53i_find_a_number_in_a_sorted_array_s1/sfo_53i_find_a_number_in_a_sorted_array_s1.java @@ -14,20 +14,26 @@ class Solution { public int search(int[] nums, int target) { // 搜索右边界 right int i = 0, j = nums.length - 1; - while(i <= j) { + while (i <= j) { int m = (i + j) / 2; - if(nums[m] <= target) i = m + 1; - else j = m - 1; + if (nums[m] <= target) + i = m + 1; + else + j = m - 1; } int right = i; // 若数组中无 target ，则提前返回 - if(j >= 0 && nums[j] != target) return 0; + if (j >= 0 && nums[j] != target) + return 0; // 搜索左边界 right - i = 0; j = nums.length - 1; - while(i <= j) { + i = 0; + j = nums.length - 1; + while (i <= j) { int m = (i + j) / 2; - if(nums[m] < target) i = m + 1; - else j = m - 1; + if (nums[m] < target) + i = m + 1; + else + j = m - 1; } int left = j; return right - left - 1; diff --git a/sword_for_offer/codes/java/sfo_53i_find_a_number_in_a_sorted_array_s2/sfo_53i_find_a_number_in_a_sorted_array_s2.java b/sword_for_offer/codes/java/sfo_53i_find_a_number_in_a_sorted_array_s2/sfo_53i_find_a_number_in_a_sorted_array_s2.java index 9613b15..cf336db 100644 --- a/sword_for_offer/codes/java/sfo_53i_find_a_number_in_a_sorted_array_s2/sfo_53i_find_a_number_in_a_sorted_array_s2.java +++ b/sword_for_offer/codes/java/sfo_53i_find_a_number_in_a_sorted_array_s2/sfo_53i_find_a_number_in_a_sorted_array_s2.java @@ -14,12 +14,15 @@ class Solution { public int search(int[] nums, int target) { return helper(nums, target) - helper(nums, target - 1); } + int helper(int[] nums, int tar) { int i = 0, j = nums.length - 1; - while(i <= j) { + while (i <= j) { int m = (i + j) / 2; - if(nums[m] <= tar) i = m + 1; - else j = m - 1; + if (nums[m] <= tar) + i = m + 1; + else + j = m - 1; } return i; } diff --git a/sword_for_offer/codes/java/sfo_53ii_the_missing_number_from_0_to_n1_s1/sfo_53ii_the_missing_number_from_0_to_n1_s1.java b/sword_for_offer/codes/java/sfo_53ii_the_missing_number_from_0_to_n1_s1/sfo_53ii_the_missing_number_from_0_to_n1_s1.java index 10d408d..89ac756 100644 --- a/sword_for_offer/codes/java/sfo_53ii_the_missing_number_from_0_to_n1_s1/sfo_53ii_the_missing_number_from_0_to_n1_s1.java +++ b/sword_for_offer/codes/java/sfo_53ii_the_missing_number_from_0_to_n1_s1/sfo_53ii_the_missing_number_from_0_to_n1_s1.java @@ -13,10 +13,12 @@ class Solution { public int missingNumber(int[] nums) { int i = 0, j = nums.length - 1; - while(i <= j) { + while (i <= j) { int m = (i + j) / 2; - if(nums[m] == m) i = m + 1; - else j = m - 1; + if (nums[m] == m) + i = m + 1; + else + j = m - 1; } return i; } diff --git a/sword_for_offer/codes/java/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1.java b/sword_for_offer/codes/java/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1.java index a074bd3..ad4576d 100644 --- a/sword_for_offer/codes/java/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1.java +++ b/sword_for_offer/codes/java/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1.java @@ -12,16 +12,21 @@ // ===== Solution Code ===== class Solution { int res, k; + public int kthLargest(TreeNode root, int k) { this.k = k; dfs(root); return res; } + void dfs(TreeNode root) { - if(root == null) return; + if (root == null) + return; dfs(root.right); - if(k == 0) return; - if(--k == 0) res = root.val; + if (k == 0) + return; + if (--k == 0) + res = root.val; dfs(root.left); } } diff --git a/sword_for_offer/codes/java/sfo_55i_depth_of_a_binary_tree_s1/sfo_55i_depth_of_a_binary_tree_s1.java b/sword_for_offer/codes/java/sfo_55i_depth_of_a_binary_tree_s1/sfo_55i_depth_of_a_binary_tree_s1.java index 35da883..98bdc8a 100644 --- a/sword_for_offer/codes/java/sfo_55i_depth_of_a_binary_tree_s1/sfo_55i_depth_of_a_binary_tree_s1.java +++ b/sword_for_offer/codes/java/sfo_55i_depth_of_a_binary_tree_s1/sfo_55i_depth_of_a_binary_tree_s1.java @@ -12,7 +12,8 @@ // ===== Solution Code ===== class Solution { public int maxDepth(TreeNode root) { - if(root == null) return 0; + if (root == null) + return 0; return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1; } } diff --git a/sword_for_offer/codes/java/sfo_55i_depth_of_a_binary_tree_s2/sfo_55i_depth_of_a_binary_tree_s2.java b/sword_for_offer/codes/java/sfo_55i_depth_of_a_binary_tree_s2/sfo_55i_depth_of_a_binary_tree_s2.java index c4b65b8..ac221a2 100644 --- a/sword_for_offer/codes/java/sfo_55i_depth_of_a_binary_tree_s2/sfo_55i_depth_of_a_binary_tree_s2.java +++ b/sword_for_offer/codes/java/sfo_55i_depth_of_a_binary_tree_s2/sfo_55i_depth_of_a_binary_tree_s2.java @@ -12,14 +12,21 @@ // ===== Solution Code ===== class Solution { public int maxDepth(TreeNode root) { - if(root == null) return 0; - List queue = new LinkedList() {{ add(root); }}, tmp; + if (root == null) + return 0; + List queue = new LinkedList() { + { + add(root); + } + }, tmp; int res = 0; - while(!queue.isEmpty()) { + while (!queue.isEmpty()) { tmp = new LinkedList<>(); - for(TreeNode node : queue) { - if(node.left != null) tmp.add(node.left); - if(node.right != null) tmp.add(node.right); + for (TreeNode node : queue) { + if (node.left != null) + tmp.add(node.left); + if (node.right != null) + tmp.add(node.right); } queue = tmp; res++; diff --git a/sword_for_offer/codes/java/sfo_55ii_balanced_binary_tree_s1/sfo_55ii_balanced_binary_tree_s1.java b/sword_for_offer/codes/java/sfo_55ii_balanced_binary_tree_s1/sfo_55ii_balanced_binary_tree_s1.java index e56430e..edf8841 100644 --- a/sword_for_offer/codes/java/sfo_55ii_balanced_binary_tree_s1/sfo_55ii_balanced_binary_tree_s1.java +++ b/sword_for_offer/codes/java/sfo_55ii_balanced_binary_tree_s1/sfo_55ii_balanced_binary_tree_s1.java @@ -16,11 +16,14 @@ public boolean isBalanced(TreeNode root) { } private int recur(TreeNode root) { - if (root == null) return 0; + if (root == null) + return 0; int left = recur(root.left); - if(left == -1) return -1; + if (left == -1) + return -1; int right = recur(root.right); - if(right == -1) return -1; + if (right == -1) + return -1; return Math.abs(left - right) < 2 ? Math.max(left, right) + 1 : -1; } } diff --git a/sword_for_offer/codes/java/sfo_55ii_balanced_binary_tree_s2/sfo_55ii_balanced_binary_tree_s2.java b/sword_for_offer/codes/java/sfo_55ii_balanced_binary_tree_s2/sfo_55ii_balanced_binary_tree_s2.java index 4a56d19..33af271 100644 --- a/sword_for_offer/codes/java/sfo_55ii_balanced_binary_tree_s2/sfo_55ii_balanced_binary_tree_s2.java +++ b/sword_for_offer/codes/java/sfo_55ii_balanced_binary_tree_s2/sfo_55ii_balanced_binary_tree_s2.java @@ -12,12 +12,14 @@ // ===== Solution Code ===== class Solution { public boolean isBalanced(TreeNode root) { - if (root == null) return true; + if (root == null) + return true; return Math.abs(depth(root.left) - depth(root.right)) <= 1 && isBalanced(root.left) && isBalanced(root.right); } private int depth(TreeNode root) { - if (root == null) return 0; + if (root == null) + return 0; return Math.max(depth(root.left), depth(root.right)) + 1; } } diff --git a/sword_for_offer/codes/java/sfo_56i_single_number_i_s1/sfo_56i_single_number_i_s1.java b/sword_for_offer/codes/java/sfo_56i_single_number_i_s1/sfo_56i_single_number_i_s1.java index ffa7108..1537dff 100644 --- a/sword_for_offer/codes/java/sfo_56i_single_number_i_s1/sfo_56i_single_number_i_s1.java +++ b/sword_for_offer/codes/java/sfo_56i_single_number_i_s1/sfo_56i_single_number_i_s1.java @@ -13,15 +13,17 @@ class Solution { public int[] singleNumbers(int[] nums) { int x = 0, y = 0, n = 0, m = 1; - for(int num : nums) // 1. 遍历异或 + for (int num : nums) // 1. 遍历异或 n ^= num; - while((n & m) == 0) // 2. 循环左移，计算 m + while ((n & m) == 0) // 2. 循环左移，计算 m m <<= 1; - for(int num: nums) { // 3. 遍历 nums 分组 - if((num & m) != 0) x ^= num; // 4. 当 num & m != 0 - else y ^= num; // 4. 当 num & m == 0 + for (int num : nums) { // 3. 遍历 nums 分组 + if ((num & m) != 0) + x ^= num; // 4. 当 num & m != 0 + else + y ^= num; // 4. 当 num & m == 0 } - return new int[] {x, y}; // 5. 返回出现一次的数字 + return new int[] { x, y }; // 5. 返回出现一次的数字 } } diff --git a/sword_for_offer/codes/java/sfo_56ii_single_number_ii_s1/sfo_56ii_single_number_ii_s1.java b/sword_for_offer/codes/java/sfo_56ii_single_number_ii_s1/sfo_56ii_single_number_ii_s1.java index 06afa44..7699d21 100644 --- a/sword_for_offer/codes/java/sfo_56ii_single_number_ii_s1/sfo_56ii_single_number_ii_s1.java +++ b/sword_for_offer/codes/java/sfo_56ii_single_number_ii_s1/sfo_56ii_single_number_ii_s1.java @@ -13,7 +13,7 @@ class Solution { public int singleNumber(int[] nums) { int ones = 0, twos = 0; - for(int num : nums){ + for (int num : nums) { ones = ones ^ num & ~twos; twos = twos ^ num & ~ones; } diff --git a/sword_for_offer/codes/java/sfo_56ii_single_number_ii_s2/sfo_56ii_single_number_ii_s2.java b/sword_for_offer/codes/java/sfo_56ii_single_number_ii_s2/sfo_56ii_single_number_ii_s2.java index ae40946..a75d6ad 100644 --- a/sword_for_offer/codes/java/sfo_56ii_single_number_ii_s2/sfo_56ii_single_number_ii_s2.java +++ b/sword_for_offer/codes/java/sfo_56ii_single_number_ii_s2/sfo_56ii_single_number_ii_s2.java @@ -13,16 +13,16 @@ class Solution { public int singleNumber(int[] nums) { int[] counts = new int[32]; - for(int num : nums) { - for(int i = 0; i < 32; i++) { + for (int num : nums) { + for (int i = 0; i < 32; i++) { counts[i] += num & 1; // 更新第 i 位 1 的个数之和 - num >>= 1; // 第 i 位 --> 第 i 位 + num >>= 1; // 第 i 位 --> 第 i 位 } } int res = 0, m = 3; - for(int i = 31; i >= 0; i--) { + for (int i = 31; i >= 0; i--) { res <<= 1; - res |= counts[i] % m; // 恢复第 i 位 + res |= counts[i] % m; // 恢复第 i 位 } return res; } diff --git a/sword_for_offer/codes/java/sfo_57_two_numbers_with_sum_s_s1/sfo_57_two_numbers_with_sum_s_s1.java b/sword_for_offer/codes/java/sfo_57_two_numbers_with_sum_s_s1/sfo_57_two_numbers_with_sum_s_s1.java index 6a0b88a..54dacf5 100644 --- a/sword_for_offer/codes/java/sfo_57_two_numbers_with_sum_s_s1/sfo_57_two_numbers_with_sum_s_s1.java +++ b/sword_for_offer/codes/java/sfo_57_two_numbers_with_sum_s_s1/sfo_57_two_numbers_with_sum_s_s1.java @@ -13,11 +13,14 @@ class Solution { public int[] twoSum(int[] nums, int target) { int i = 0, j = nums.length - 1; - while(i < j) { + while (i < j) { int s = nums[i] + nums[j]; - if(s < target) i++; - else if(s > target) j--; - else return new int[] { nums[i], nums[j] }; + if (s < target) + i++; + else if (s > target) + j--; + else + return new int[] { nums[i], nums[j] }; } return new int[0]; } diff --git a/sword_for_offer/codes/java/sfo_57ii_consecutive_numbers_with_sum_s_s1/sfo_57ii_consecutive_numbers_with_sum_s_s1.java b/sword_for_offer/codes/java/sfo_57ii_consecutive_numbers_with_sum_s_s1/sfo_57ii_consecutive_numbers_with_sum_s_s1.java index 80e1fd6..eae2556 100644 --- a/sword_for_offer/codes/java/sfo_57ii_consecutive_numbers_with_sum_s_s1/sfo_57ii_consecutive_numbers_with_sum_s_s1.java +++ b/sword_for_offer/codes/java/sfo_57ii_consecutive_numbers_with_sum_s_s1/sfo_57ii_consecutive_numbers_with_sum_s_s1.java @@ -15,11 +15,11 @@ public int[][] findContinuousSequence(int target) { int i = 1; double j = 2.0; List res = new ArrayList<>(); - while(i < j) { + while (i < j) { j = (-1 + Math.sqrt(1 + 4 * (2 * target + (long) i * i - i))) / 2; - if(i < j && j == (int)j) { - int[] ans = new int[(int)j - i + 1]; - for(int k = i; k <= (int)j; k++) + if (i < j && j == (int) j) { + int[] ans = new int[(int) j - i + 1]; + for (int k = i; k <= (int) j; k++) ans[k - i] = k; res.add(ans); } diff --git a/sword_for_offer/codes/java/sfo_57ii_consecutive_numbers_with_sum_s_s2/sfo_57ii_consecutive_numbers_with_sum_s_s2.java b/sword_for_offer/codes/java/sfo_57ii_consecutive_numbers_with_sum_s_s2/sfo_57ii_consecutive_numbers_with_sum_s_s2.java index d9d60b4..2bd5ea6 100644 --- a/sword_for_offer/codes/java/sfo_57ii_consecutive_numbers_with_sum_s_s2/sfo_57ii_consecutive_numbers_with_sum_s_s2.java +++ b/sword_for_offer/codes/java/sfo_57ii_consecutive_numbers_with_sum_s_s2/sfo_57ii_consecutive_numbers_with_sum_s_s2.java @@ -14,14 +14,14 @@ class Solution { public int[][] findContinuousSequence(int target) { int i = 1, j = 2, s = 3; List res = new ArrayList<>(); - while(i < j) { - if(s == target) { + while (i < j) { + if (s == target) { int[] ans = new int[j - i + 1]; - for(int k = i; k <= j; k++) + for (int k = i; k <= j; k++) ans[k - i] = k; res.add(ans); } - if(s >= target) { + if (s >= target) { s -= i; i++; } else { diff --git a/sword_for_offer/codes/java/sfo_58i_reverse_order_of_words_s1/sfo_58i_reverse_order_of_words_s1.java b/sword_for_offer/codes/java/sfo_58i_reverse_order_of_words_s1/sfo_58i_reverse_order_of_words_s1.java index 3bb039f..10e6feb 100644 --- a/sword_for_offer/codes/java/sfo_58i_reverse_order_of_words_s1/sfo_58i_reverse_order_of_words_s1.java +++ b/sword_for_offer/codes/java/sfo_58i_reverse_order_of_words_s1/sfo_58i_reverse_order_of_words_s1.java @@ -12,16 +12,18 @@ // ===== Solution Code ===== class Solution { public String reverseWords(String s) { - s = s.trim(); // 删除首尾空格 + s = s.trim(); // 删除首尾空格 int j = s.length() - 1, i = j; StringBuilder res = new StringBuilder(); while (i >= 0) { - while (i >= 0 && s.charAt(i) != ' ') i--; // 搜索首个空格 + while (i >= 0 && s.charAt(i) != ' ') + i--; // 搜索首个空格 res.append(s.substring(i + 1, j + 1) + " "); // 添加单词 - while (i >= 0 && s.charAt(i) == ' ') i--; // 跳过单词间空格 - j = i; // j 指向下个单词的尾字符 + while (i >= 0 && s.charAt(i) == ' ') + i--; // 跳过单词间空格 + j = i; // j 指向下个单词的尾字符 } - return res.toString().trim(); // 转化为字符串并返回 + return res.toString().trim(); // 转化为字符串并返回 } } diff --git a/sword_for_offer/codes/java/sfo_58i_reverse_order_of_words_s2/sfo_58i_reverse_order_of_words_s2.java b/sword_for_offer/codes/java/sfo_58i_reverse_order_of_words_s2/sfo_58i_reverse_order_of_words_s2.java index 7258f91..dd4a137 100644 --- a/sword_for_offer/codes/java/sfo_58i_reverse_order_of_words_s2/sfo_58i_reverse_order_of_words_s2.java +++ b/sword_for_offer/codes/java/sfo_58i_reverse_order_of_words_s2/sfo_58i_reverse_order_of_words_s2.java @@ -12,13 +12,14 @@ // ===== Solution Code ===== class Solution { public String reverseWords(String s) { - String[] strs = s.trim().split(" "); // 删除首尾空格，分割字符串 + String[] strs = s.trim().split(" "); // 删除首尾空格，分割字符串 StringBuilder res = new StringBuilder(); for (int i = strs.length - 1; i >= 0; i--) { // 倒序遍历单词列表 - if(strs[i].equals("")) continue; // 遇到空单词则跳过 - res.append(strs[i] + " "); // 将单词拼接至 StringBuilder + if (strs[i].equals("")) + continue; // 遇到空单词则跳过 + res.append(strs[i] + " "); // 将单词拼接至 StringBuilder } - return res.toString().trim(); // 转化为字符串，删除尾部空格，并返回 + return res.toString().trim(); // 转化为字符串，删除尾部空格，并返回 } } diff --git a/sword_for_offer/codes/java/sfo_58ii_left_rotation_of_a_string_s2/sfo_58ii_left_rotation_of_a_string_s2.java b/sword_for_offer/codes/java/sfo_58ii_left_rotation_of_a_string_s2/sfo_58ii_left_rotation_of_a_string_s2.java index 9cea490..ba9f815 100644 --- a/sword_for_offer/codes/java/sfo_58ii_left_rotation_of_a_string_s2/sfo_58ii_left_rotation_of_a_string_s2.java +++ b/sword_for_offer/codes/java/sfo_58ii_left_rotation_of_a_string_s2/sfo_58ii_left_rotation_of_a_string_s2.java @@ -13,9 +13,9 @@ class Solution { public String reverseLeftWords(String s, int n) { StringBuilder res = new StringBuilder(); - for(int i = n; i < s.length(); i++) + for (int i = n; i < s.length(); i++) res.append(s.charAt(i)); - for(int i = 0; i < n; i++) + for (int i = 0; i < n; i++) res.append(s.charAt(i)); return res.toString(); } diff --git a/sword_for_offer/codes/java/sfo_58ii_left_rotation_of_a_string_s3/sfo_58ii_left_rotation_of_a_string_s3.java b/sword_for_offer/codes/java/sfo_58ii_left_rotation_of_a_string_s3/sfo_58ii_left_rotation_of_a_string_s3.java index bd34629..297aced 100644 --- a/sword_for_offer/codes/java/sfo_58ii_left_rotation_of_a_string_s3/sfo_58ii_left_rotation_of_a_string_s3.java +++ b/sword_for_offer/codes/java/sfo_58ii_left_rotation_of_a_string_s3/sfo_58ii_left_rotation_of_a_string_s3.java @@ -13,7 +13,7 @@ class Solution { public String reverseLeftWords(String s, int n) { StringBuilder res = new StringBuilder(); - for(int i = n; i < n + s.length(); i++) + for (int i = n; i < n + s.length(); i++) res.append(s.charAt(i % s.length())); return res.toString(); } diff --git a/sword_for_offer/codes/java/sfo_58ii_left_rotation_of_a_string_s4/sfo_58ii_left_rotation_of_a_string_s4.java b/sword_for_offer/codes/java/sfo_58ii_left_rotation_of_a_string_s4/sfo_58ii_left_rotation_of_a_string_s4.java index 6c4b398..4645c5a 100644 --- a/sword_for_offer/codes/java/sfo_58ii_left_rotation_of_a_string_s4/sfo_58ii_left_rotation_of_a_string_s4.java +++ b/sword_for_offer/codes/java/sfo_58ii_left_rotation_of_a_string_s4/sfo_58ii_left_rotation_of_a_string_s4.java @@ -13,9 +13,9 @@ class Solution { public String reverseLeftWords(String s, int n) { String res = ""; - for(int i = n; i < s.length(); i++) + for (int i = n; i < s.length(); i++) res += s.charAt(i); - for(int i = 0; i < n; i++) + for (int i = 0; i < n; i++) res += s.charAt(i); return res; } diff --git a/sword_for_offer/codes/java/sfo_58ii_left_rotation_of_a_string_s5/sfo_58ii_left_rotation_of_a_string_s5.java b/sword_for_offer/codes/java/sfo_58ii_left_rotation_of_a_string_s5/sfo_58ii_left_rotation_of_a_string_s5.java index 9071803..e2cd05f 100644 --- a/sword_for_offer/codes/java/sfo_58ii_left_rotation_of_a_string_s5/sfo_58ii_left_rotation_of_a_string_s5.java +++ b/sword_for_offer/codes/java/sfo_58ii_left_rotation_of_a_string_s5/sfo_58ii_left_rotation_of_a_string_s5.java @@ -13,7 +13,7 @@ class Solution { public String reverseLeftWords(String s, int n) { String res = ""; - for(int i = n; i < n + s.length(); i++) + for (int i = n; i < n + s.length(); i++) res += s.charAt(i % s.length()); return res; } diff --git a/sword_for_offer/codes/java/sfo_59i_sliding_window_maximum_s1/sfo_59i_sliding_window_maximum_s1.java b/sword_for_offer/codes/java/sfo_59i_sliding_window_maximum_s1/sfo_59i_sliding_window_maximum_s1.java index 7ea0bcc..de0131e 100644 --- a/sword_for_offer/codes/java/sfo_59i_sliding_window_maximum_s1/sfo_59i_sliding_window_maximum_s1.java +++ b/sword_for_offer/codes/java/sfo_59i_sliding_window_maximum_s1/sfo_59i_sliding_window_maximum_s1.java @@ -12,19 +12,20 @@ // ===== Solution Code ===== class Solution { public int[] maxSlidingWindow(int[] nums, int k) { - if(nums.length == 0 || k == 0) return new int[0]; + if (nums.length == 0 || k == 0) + return new int[0]; Deque deque = new LinkedList<>(); int[] res = new int[nums.length - k + 1]; - for(int j = 0, i = 1 - k; j < nums.length; i++, j++) { + for (int j = 0, i = 1 - k; j < nums.length; i++, j++) { // 删除 deque 中对应的 nums[i-1] - if(i > 0 && deque.peekFirst() == nums[i - 1]) + if (i > 0 && deque.peekFirst() == nums[i - 1]) deque.removeFirst(); // 保持 deque 递减 - while(!deque.isEmpty() && deque.peekLast() < nums[j]) + while (!deque.isEmpty() && deque.peekLast() < nums[j]) deque.removeLast(); deque.addLast(nums[j]); // 记录窗口最大值 - if(i >= 0) + if (i >= 0) res[i] = deque.peekFirst(); } return res; diff --git a/sword_for_offer/codes/java/sfo_59i_sliding_window_maximum_s2/sfo_59i_sliding_window_maximum_s2.java b/sword_for_offer/codes/java/sfo_59i_sliding_window_maximum_s2/sfo_59i_sliding_window_maximum_s2.java index f567135..0bcaf47 100644 --- a/sword_for_offer/codes/java/sfo_59i_sliding_window_maximum_s2/sfo_59i_sliding_window_maximum_s2.java +++ b/sword_for_offer/codes/java/sfo_59i_sliding_window_maximum_s2/sfo_59i_sliding_window_maximum_s2.java @@ -12,21 +12,22 @@ // ===== Solution Code ===== class Solution { public int[] maxSlidingWindow(int[] nums, int k) { - if(nums.length == 0 || k == 0) return new int[0]; + if (nums.length == 0 || k == 0) + return new int[0]; Deque deque = new LinkedList<>(); int[] res = new int[nums.length - k + 1]; // 未形成窗口 - for(int i = 0; i < k; i++) { - while(!deque.isEmpty() && deque.peekLast() < nums[i]) + for (int i = 0; i < k; i++) { + while (!deque.isEmpty() && deque.peekLast() < nums[i]) deque.removeLast(); deque.addLast(nums[i]); } res[0] = deque.peekFirst(); // 形成窗口后 - for(int i = k; i < nums.length; i++) { - if(deque.peekFirst() == nums[i - k]) + for (int i = k; i < nums.length; i++) { + if (deque.peekFirst() == nums[i - k]) deque.removeFirst(); - while(!deque.isEmpty() && deque.peekLast() < nums[i]) + while (!deque.isEmpty() && deque.peekLast() < nums[i]) deque.removeLast(); deque.addLast(nums[i]); res[i - k + 1] = deque.peekFirst(); diff --git a/sword_for_offer/codes/java/sfo_59ii_max_queue_s1/sfo_59ii_max_queue_s1.java b/sword_for_offer/codes/java/sfo_59ii_max_queue_s1/sfo_59ii_max_queue_s1.java index 2dabf6a..034aa53 100644 --- a/sword_for_offer/codes/java/sfo_59ii_max_queue_s1/sfo_59ii_max_queue_s1.java +++ b/sword_for_offer/codes/java/sfo_59ii_max_queue_s1/sfo_59ii_max_queue_s1.java @@ -13,22 +13,27 @@ class MaxQueue { Queue queue; Deque deque; + public MaxQueue() { queue = new LinkedList<>(); deque = new LinkedList<>(); } + public int max_value() { return deque.isEmpty() ? -1 : deque.peekFirst(); } + public void push_back(int value) { queue.offer(value); - while(!deque.isEmpty() && deque.peekLast() < value) + while (!deque.isEmpty() && deque.peekLast() < value) deque.pollLast(); deque.offerLast(value); } + public int pop_front() { - if(queue.isEmpty()) return -1; - if(queue.peek().equals(deque.peekFirst())) + if (queue.isEmpty()) + return -1; + if (queue.peek().equals(deque.peekFirst())) deque.pollFirst(); return queue.poll(); } diff --git a/sword_for_offer/codes/java/sfo_61_straight_in_poker_s1/sfo_61_straight_in_poker_s1.java b/sword_for_offer/codes/java/sfo_61_straight_in_poker_s1/sfo_61_straight_in_poker_s1.java index fa049df..0d4b137 100644 --- a/sword_for_offer/codes/java/sfo_61_straight_in_poker_s1/sfo_61_straight_in_poker_s1.java +++ b/sword_for_offer/codes/java/sfo_61_straight_in_poker_s1/sfo_61_straight_in_poker_s1.java @@ -14,11 +14,13 @@ class Solution { public boolean isStraight(int[] nums) { Set repeat = new HashSet<>(); int max = 0, min = 14; - for(int num : nums) { - if(num == 0) continue; // 跳过大小王 + for (int num : nums) { + if (num == 0) + continue; // 跳过大小王 max = Math.max(max, num); // 最大牌 min = Math.min(min, num); // 最小牌 - if(repeat.contains(num)) return false; // 若有重复，提前返回 false + if (repeat.contains(num)) + return false; // 若有重复，提前返回 false repeat.add(num); // 添加此牌至 Set } return max - min < 5; // 最大牌 - 最小牌 < 5 则可构成顺子 diff --git a/sword_for_offer/codes/java/sfo_61_straight_in_poker_s2/sfo_61_straight_in_poker_s2.java b/sword_for_offer/codes/java/sfo_61_straight_in_poker_s2/sfo_61_straight_in_poker_s2.java index f177a8c..46cbe2c 100644 --- a/sword_for_offer/codes/java/sfo_61_straight_in_poker_s2/sfo_61_straight_in_poker_s2.java +++ b/sword_for_offer/codes/java/sfo_61_straight_in_poker_s2/sfo_61_straight_in_poker_s2.java @@ -14,9 +14,11 @@ class Solution { public boolean isStraight(int[] nums) { int joker = 0; Arrays.sort(nums); // 数组排序 - for(int i = 0; i < 4; i++) { - if(nums[i] == 0) joker++; // 统计大小王数量 - else if(nums[i] == nums[i + 1]) return false; // 若有重复，提前返回 false + for (int i = 0; i < 4; i++) { + if (nums[i] == 0) + joker++; // 统计大小王数量 + else if (nums[i] == nums[i + 1]) + return false; // 若有重复，提前返回 false } return nums[4] - nums[joker] < 5; // 最大牌 - 最小牌 < 5 则可构成顺子 } diff --git a/sword_for_offer/codes/java/sfo_63_the_maximum_profit_of_stocks_s1/sfo_63_the_maximum_profit_of_stocks_s1.java b/sword_for_offer/codes/java/sfo_63_the_maximum_profit_of_stocks_s1/sfo_63_the_maximum_profit_of_stocks_s1.java index ea4a6db..7e282a6 100644 --- a/sword_for_offer/codes/java/sfo_63_the_maximum_profit_of_stocks_s1/sfo_63_the_maximum_profit_of_stocks_s1.java +++ b/sword_for_offer/codes/java/sfo_63_the_maximum_profit_of_stocks_s1/sfo_63_the_maximum_profit_of_stocks_s1.java @@ -13,7 +13,7 @@ class Solution { public int maxProfit(int[] prices) { int cost = Integer.MAX_VALUE, profit = 0; - for(int price : prices) { + for (int price : prices) { cost = Math.min(cost, price); profit = Math.max(profit, price - cost); } diff --git a/sword_for_offer/codes/java/sfo_64_solve_1_2___n_s1/sfo_64_solve_1_2___n_s1.java b/sword_for_offer/codes/java/sfo_64_solve_1_2___n_s1/sfo_64_solve_1_2___n_s1.java index ce9d927..232279a 100644 --- a/sword_for_offer/codes/java/sfo_64_solve_1_2___n_s1/sfo_64_solve_1_2___n_s1.java +++ b/sword_for_offer/codes/java/sfo_64_solve_1_2___n_s1/sfo_64_solve_1_2___n_s1.java @@ -12,6 +12,7 @@ // ===== Solution Code ===== class Solution { int res = 0; + public int sumNums(int n) { boolean x = n > 1 && sumNums(n - 1) > 0; res += n; diff --git a/sword_for_offer/codes/java/sfo_65_implement_addition_operation_without_arithmetic_operators_s1/sfo_65_implement_addition_operation_without_arithmetic_operators_s1.java b/sword_for_offer/codes/java/sfo_65_implement_addition_operation_without_arithmetic_operators_s1/sfo_65_implement_addition_operation_without_arithmetic_operators_s1.java index 9083b9a..5616dac 100644 --- a/sword_for_offer/codes/java/sfo_65_implement_addition_operation_without_arithmetic_operators_s1/sfo_65_implement_addition_operation_without_arithmetic_operators_s1.java +++ b/sword_for_offer/codes/java/sfo_65_implement_addition_operation_without_arithmetic_operators_s1/sfo_65_implement_addition_operation_without_arithmetic_operators_s1.java @@ -12,8 +12,8 @@ // ===== Solution Code ===== class Solution { public int add(int a, int b) { - while(b != 0) { // 当进位为 0 时跳出 - int c = (a & b) << 1; // c = 进位 + while (b != 0) { // 当进位为 0 时跳出 + int c = (a & b) << 1; // c = 进位 a ^= b; // a = 非进位和 b = c; // b = 进位 } diff --git a/sword_for_offer/codes/java/sfo_66_a_product_array_puzzle_s1/sfo_66_a_product_array_puzzle_s1.java b/sword_for_offer/codes/java/sfo_66_a_product_array_puzzle_s1/sfo_66_a_product_array_puzzle_s1.java index 105a996..b13b0d8 100644 --- a/sword_for_offer/codes/java/sfo_66_a_product_array_puzzle_s1/sfo_66_a_product_array_puzzle_s1.java +++ b/sword_for_offer/codes/java/sfo_66_a_product_array_puzzle_s1/sfo_66_a_product_array_puzzle_s1.java @@ -13,14 +13,15 @@ class Solution { public int[] constructArr(int[] a) { int len = a.length; - if(len == 0) return new int[0]; + if (len == 0) + return new int[0]; int[] b = new int[len]; b[0] = 1; int tmp = 1; - for(int i = 1; i < len; i++) { + for (int i = 1; i < len; i++) { b[i] = b[i - 1] * a[i - 1]; } - for(int i = len - 2; i >= 0; i--) { + for (int i = len - 2; i >= 0; i--) { tmp *= a[i + 1]; b[i] *= tmp; } diff --git a/sword_for_offer/codes/java/sfo_67_convert_string_to_int_s1/sfo_67_convert_string_to_int_s1.java b/sword_for_offer/codes/java/sfo_67_convert_string_to_int_s1/sfo_67_convert_string_to_int_s1.java index 2552d74..605c0d0 100644 --- a/sword_for_offer/codes/java/sfo_67_convert_string_to_int_s1/sfo_67_convert_string_to_int_s1.java +++ b/sword_for_offer/codes/java/sfo_67_convert_string_to_int_s1/sfo_67_convert_string_to_int_s1.java @@ -13,14 +13,19 @@ class Solution { public int strToInt(String str) { char[] c = str.trim().toCharArray(); - if(c.length == 0) return 0; + if (c.length == 0) + return 0; int res = 0, bndry = Integer.MAX_VALUE / 10; int i = 1, sign = 1; - if(c[0] == '-') sign = -1; - else if(c[0] != '+') i = 0; - for(int j = i; j < c.length; j++) { - if(c[j] < '0' || c[j] > '9') break; - if(res > bndry || res == bndry && c[j] > '7') return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE; + if (c[0] == '-') + sign = -1; + else if (c[0] != '+') + i = 0; + for (int j = i; j < c.length; j++) { + if (c[j] < '0' || c[j] > '9') + break; + if (res > bndry || res == bndry && c[j] > '7') + return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE; res = res * 10 + (c[j] - '0'); } return sign * res; diff --git a/sword_for_offer/codes/java/sfo_67_convert_string_to_int_s2/sfo_67_convert_string_to_int_s2.java b/sword_for_offer/codes/java/sfo_67_convert_string_to_int_s2/sfo_67_convert_string_to_int_s2.java index 20ca8ac..44eb0b9 100644 --- a/sword_for_offer/codes/java/sfo_67_convert_string_to_int_s2/sfo_67_convert_string_to_int_s2.java +++ b/sword_for_offer/codes/java/sfo_67_convert_string_to_int_s2/sfo_67_convert_string_to_int_s2.java @@ -14,14 +14,19 @@ class Solution { public int strToInt(String str) { int res = 0, bndry = Integer.MAX_VALUE / 10; int i = 0, sign = 1, length = str.length(); - if(length == 0) return 0; - while(str.charAt(i) == ' ') - if(++i == length) return 0; - if(str.charAt(i) == '-') sign = -1; - if(str.charAt(i) == '-' || str.charAt(i) == '+') i++; - for(int j = i; j < length; j++) { - if(str.charAt(j) < '0' || str.charAt(j) > '9') break; - if(res > bndry || res == bndry && str.charAt(j) > '7') + if (length == 0) + return 0; + while (str.charAt(i) == ' ') + if (++i == length) + return 0; + if (str.charAt(i) == '-') + sign = -1; + if (str.charAt(i) == '-' || str.charAt(i) == '+') + i++; + for (int j = i; j < length; j++) { + if (str.charAt(j) < '0' || str.charAt(j) > '9') + break; + if (res > bndry || res == bndry && str.charAt(j) > '7') return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE; res = res * 10 + (str.charAt(j) - '0'); } diff --git a/sword_for_offer/codes/java/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1.java b/sword_for_offer/codes/java/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1.java index 1171c3b..b048ca3 100644 --- a/sword_for_offer/codes/java/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1.java +++ b/sword_for_offer/codes/java/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1.java @@ -12,12 +12,13 @@ // ===== Solution Code ===== class Solution { public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { - while(root != null) { - if(root.val < p.val && root.val < q.val) // p,q 都在 root 的右子树中 + while (root != null) { + if (root.val < p.val && root.val < q.val) // p,q 都在 root 的右子树中 root = root.right; // 遍历至右子节点 - else if(root.val > p.val && root.val > q.val) // p,q 都在 root 的左子树中 + else if (root.val > p.val && root.val > q.val) // p,q 都在 root 的左子树中 root = root.left; // 遍历至左子节点 - else break; + else + break; } return root; } @@ -26,7 +27,8 @@ else if(root.val > p.val && root.val > q.val) // p,q 都在 root 的左子树中 public class sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1 { public static void main(String[] args) { // ======= Test Case ======= - TreeNode root = TreeNode.arrToTree(new Integer[] { 6, 2, 8, 0, 4, 7, 9, null, null, 3, 5, null, null, null, null, null, null, null, null }); + TreeNode root = TreeNode.arrToTree(new Integer[] { 6, 2, 8, 0, 4, 7, 9, null, null, 3, 5, null, null, null, + null, null, null, null, null }); TreeNode p = TreeNode.getTreeNode(root, 2); TreeNode q = TreeNode.getTreeNode(root, 8); // ====== Driver Code ====== diff --git a/sword_for_offer/codes/java/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2.java b/sword_for_offer/codes/java/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2.java index 506dc47..01d1fad 100644 --- a/sword_for_offer/codes/java/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2.java +++ b/sword_for_offer/codes/java/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2.java @@ -12,17 +12,18 @@ // ===== Solution Code ===== class Solution { public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { - if(p.val > q.val) { // 保证 p.val < q.val + if (p.val > q.val) { // 保证 p.val < q.val TreeNode tmp = p; p = q; q = tmp; } - while(root != null) { - if(root.val < p.val) // p,q 都在 root 的右子树中 + while (root != null) { + if (root.val < p.val) // p,q 都在 root 的右子树中 root = root.right; // 遍历至右子节点 - else if(root.val > q.val) // p,q 都在 root 的左子树中 + else if (root.val > q.val) // p,q 都在 root 的左子树中 root = root.left; // 遍历至左子节点 - else break; + else + break; } return root; } @@ -31,7 +32,8 @@ else if(root.val > q.val) // p,q 都在 root 的左子树中 public class sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2 { public static void main(String[] args) { // ======= Test Case ======= - TreeNode root = TreeNode.arrToTree(new Integer[] { 6, 2, 8, 0, 4, 7, 9, null, null, 3, 5, null, null, null, null, null, null, null, null }); + TreeNode root = TreeNode.arrToTree(new Integer[] { 6, 2, 8, 0, 4, 7, 9, null, null, 3, 5, null, null, null, + null, null, null, null, null }); TreeNode p = TreeNode.getTreeNode(root, 2); TreeNode q = TreeNode.getTreeNode(root, 8); // ====== Driver Code ====== diff --git a/sword_for_offer/codes/java/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3.java b/sword_for_offer/codes/java/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3.java index 09dac21..b108d91 100644 --- a/sword_for_offer/codes/java/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3.java +++ b/sword_for_offer/codes/java/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3.java @@ -12,9 +12,9 @@ // ===== Solution Code ===== class Solution { public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { - if(root.val < p.val && root.val < q.val) + if (root.val < p.val && root.val < q.val) return lowestCommonAncestor(root.right, p, q); - if(root.val > p.val && root.val > q.val) + if (root.val > p.val && root.val > q.val) return lowestCommonAncestor(root.left, p, q); return root; } @@ -23,7 +23,8 @@ public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { public class sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3 { public static void main(String[] args) { // ======= Test Case ======= - TreeNode root = TreeNode.arrToTree(new Integer[] { 6, 2, 8, 0, 4, 7, 9, null, null, 3, 5, null, null, null, null, null, null, null, null }); + TreeNode root = TreeNode.arrToTree(new Integer[] { 6, 2, 8, 0, 4, 7, 9, null, null, 3, 5, null, null, null, + null, null, null, null, null }); TreeNode p = TreeNode.getTreeNode(root, 2); TreeNode q = TreeNode.getTreeNode(root, 8); // ====== Driver Code ====== diff --git a/sword_for_offer/codes/java/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1.java b/sword_for_offer/codes/java/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1.java index 038eb99..f3bd533 100644 --- a/sword_for_offer/codes/java/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1.java +++ b/sword_for_offer/codes/java/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1.java @@ -12,11 +12,14 @@ // ===== Solution Code ===== class Solution { public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { - if(root == null || root == p || root == q) return root; + if (root == null || root == p || root == q) + return root; TreeNode left = lowestCommonAncestor(root.left, p, q); TreeNode right = lowestCommonAncestor(root.right, p, q); - if(left == null) return right; - if(right == null) return left; + if (left == null) + return right; + if (right == null) + return left; return root; } } @@ -24,7 +27,8 @@ public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { public class sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1 { public static void main(String[] args) { // ======= Test Case ======= - TreeNode root = TreeNode.arrToTree(new Integer[] { 3, 5, 1, 6, 2, 0, 8, null, null, 7, 4, null, null, null, null, null, null, null, null }); + TreeNode root = TreeNode.arrToTree(new Integer[] { 3, 5, 1, 6, 2, 0, 8, null, null, 7, 4, null, null, null, + null, null, null, null, null }); TreeNode p = TreeNode.getTreeNode(root, 5); TreeNode q = TreeNode.getTreeNode(root, 1); // ====== Driver Code ====== diff --git a/sword_for_offer/codes/java/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2.java b/sword_for_offer/codes/java/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2.java index e53a916..eea06cc 100644 --- a/sword_for_offer/codes/java/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2.java +++ b/sword_for_offer/codes/java/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2.java @@ -12,12 +12,16 @@ // ===== Solution Code ===== class Solution { public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { - if(root == null || root == p || root == q) return root; + if (root == null || root == p || root == q) + return root; TreeNode left = lowestCommonAncestor(root.left, p, q); TreeNode right = lowestCommonAncestor(root.right, p, q); - if(left == null && right == null) return null; // 1. - if(left == null) return right; // 3. - if(right == null) return left; // 4. + if (left == null && right == null) + return null; // 1. + if (left == null) + return right; // 3. + if (right == null) + return left; // 4. return root; // 2. if(left != null and right != null) } } @@ -25,7 +29,8 @@ public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { public class sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2 { public static void main(String[] args) { // ======= Test Case ======= - TreeNode root = TreeNode.arrToTree(new Integer[] { 3, 5, 1, 6, 2, 0, 8, null, null, 7, 4, null, null, null, null, null, null, null, null }); + TreeNode root = TreeNode.arrToTree(new Integer[] { 3, 5, 1, 6, 2, 0, 8, null, null, 7, 4, null, null, null, + null, null, null, null, null }); TreeNode p = TreeNode.getTreeNode(root, 5); TreeNode q = TreeNode.getTreeNode(root, 1); // ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/include/__init__.py b/sword_for_offer/codes/python/include/__init__.py index aacc319..815faf5 100644 --- a/sword_for_offer/codes/python/include/__init__.py +++ b/sword_for_offer/codes/python/include/__init__.py @@ -3,6 +3,11 @@ import functools import collections from typing import List -from .linked_list import ListNode, list_to_linked_list, linked_list_to_list, get_list_node +from .linked_list import ( + ListNode, + list_to_linked_list, + linked_list_to_list, + get_list_node, +) from .binary_tree import TreeNode, list_to_tree, tree_to_list, get_tree_node -from .print_util import print_matrix, print_linked_list, print_tree \ No newline at end of file +from .print_util import print_matrix, print_linked_list, print_tree diff --git a/sword_for_offer/codes/python/include/binary_tree.py b/sword_for_offer/codes/python/include/binary_tree.py index d905d82..85bcc51 100644 --- a/sword_for_offer/codes/python/include/binary_tree.py +++ b/sword_for_offer/codes/python/include/binary_tree.py @@ -1,19 +1,21 @@ -''' +""" File: binary_tree.py Created Time: 2021-12-11 Author: Krahets (krahets@163.com) -''' +""" import collections + class TreeNode: - """Definition for a binary tree node - """ + """Definition for a binary tree node""" + def __init__(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right + def list_to_tree(arr): """Generate a binary tree with a list @@ -41,6 +43,7 @@ def list_to_tree(arr): i += 1 return root + def tree_to_list(root): """Serialize a tree into an array @@ -49,8 +52,9 @@ def tree_to_list(root): Returns: [type]: [description] - """ - if not root: return [] + """ + if not root: + return [] queue = collections.deque() queue.append(root) res = [] @@ -60,9 +64,11 @@ def tree_to_list(root): res.append(node.val) queue.append(node.left) queue.append(node.right) - else: res.append(None) + else: + res.append(None) return res + def get_tree_node(root, val): """Get a tree node with specific value in a binary tree @@ -72,7 +78,7 @@ def get_tree_node(root, val): Returns: [type]: [description] - """ + """ if not root: return if root.val == val: diff --git a/sword_for_offer/codes/python/include/linked_list.py b/sword_for_offer/codes/python/include/linked_list.py index f6773c6..7155e00 100644 --- a/sword_for_offer/codes/python/include/linked_list.py +++ b/sword_for_offer/codes/python/include/linked_list.py @@ -1,16 +1,18 @@ -''' +""" File: linked_list.py Created Time: 2021-12-11 Author: Krahets (krahets@163.com) -''' +""" + class ListNode: - """Definition for a singly-linked list node - """ + """Definition for a singly-linked list node""" + def __init__(self, val=0, next=None): self.val = val self.next = next + def list_to_linked_list(arr): """Generate a linked list with a list @@ -19,7 +21,7 @@ def list_to_linked_list(arr): Returns: [type]: [description] - """ + """ dum = head = ListNode(0) for a in arr: node = ListNode(a) @@ -27,6 +29,7 @@ def list_to_linked_list(arr): head = head.next return dum.next + def linked_list_to_list(head): """Serialize a linked list into an array @@ -35,13 +38,14 @@ def linked_list_to_list(head): Returns: [type]: [description] - """ + """ arr = [] while head: arr.append(head.val) head = head.next return arr + def get_list_node(head, val): """Get a list node with specific value from a linked list @@ -51,7 +55,7 @@ def get_list_node(head, val): Returns: [type]: [description] - """ + """ while head and head.val != val: head = head.next return head diff --git a/sword_for_offer/codes/python/include/print_util.py b/sword_for_offer/codes/python/include/print_util.py index b7c2ba2..5dba2fa 100644 --- a/sword_for_offer/codes/python/include/print_util.py +++ b/sword_for_offer/codes/python/include/print_util.py @@ -1,45 +1,49 @@ -''' +""" File: print_util.py Created Time: 2021-12-11 Author: Krahets (krahets@163.com) -''' +""" from .binary_tree import TreeNode, tree_to_list from .linked_list import ListNode, linked_list_to_list + def print_matrix(mat): """Print a matrix Args: mat ([type]): [description] - """ + """ pstr = [] for arr in mat: - pstr.append(' ' + str(arr)) + pstr.append(" " + str(arr)) + + print("[\n" + ",\n".join(pstr) + "\n]") - print('[\n' + ',\n'.join(pstr) + '\n]') def print_linked_list(head): """Print a linked list Args: head ([type]): [description] - """ + """ arr = linked_list_to_list(head) - print(' -> '.join([str(a) for a in arr])) + print(" -> ".join([str(a) for a in arr])) class Trunk: def __init__(self, prev=None, str=None): self.prev = prev self.str = str - + + def showTrunks(p): if p is None: return showTrunks(p.prev) - print(p.str, end='') - + print(p.str, end="") + + def print_tree(root, prev=None, isLeft=False): """Print a binary tree This tree printer is borrowed from TECHIE DELIGHT @@ -51,23 +55,23 @@ def print_tree(root, prev=None, isLeft=False): """ if root is None: return - - prev_str = ' ' + + prev_str = " " trunk = Trunk(prev, prev_str) print_tree(root.right, trunk, True) - + if prev is None: - trunk.str = '———' + trunk.str = "———" elif isLeft: - trunk.str = '/———' - prev_str = ' |' + trunk.str = "/———" + prev_str = " |" else: - trunk.str = '\———' + trunk.str = "\———" prev.str = prev_str - + showTrunks(trunk) - print(' ' + str(root.val)) + print(" " + str(root.val)) if prev: prev.str = prev_str - trunk.str = ' |' + trunk.str = " |" print_tree(root.left, trunk, False) diff --git a/sword_for_offer/codes/python/sfo_03_find_duplicate_numbers_in_an_array_s1.py b/sword_for_offer/codes/python/sfo_03_find_duplicate_numbers_in_an_array_s1.py index 9860934..852561d 100644 --- a/sword_for_offer/codes/python/sfo_03_find_duplicate_numbers_in_an_array_s1.py +++ b/sword_for_offer/codes/python/sfo_03_find_duplicate_numbers_in_an_array_s1.py @@ -1,20 +1,23 @@ -''' +""" File: sfo_03_find_duplicate_numbers_in_an_array_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def findRepeatNumber(self, nums: List[int]) -> int: dic = set() for num in nums: - if num in dic: return num + if num in dic: + return num dic.add(num) return -1 + # ======= Test Case ======= nums = [2, 3, 1, 0, 2, 5, 3] # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_03_find_duplicate_numbers_in_an_array_s2.py b/sword_for_offer/codes/python/sfo_03_find_duplicate_numbers_in_an_array_s2.py index 0a38433..2e48b8c 100644 --- a/sword_for_offer/codes/python/sfo_03_find_duplicate_numbers_in_an_array_s2.py +++ b/sword_for_offer/codes/python/sfo_03_find_duplicate_numbers_in_an_array_s2.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_03_find_duplicate_numbers_in_an_array_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def findRepeatNumber(self, nums: List[int]) -> int: @@ -14,10 +15,12 @@ def findRepeatNumber(self, nums: List[int]) -> int: if nums[i] == i: i += 1 continue - if nums[nums[i]] == nums[i]: return nums[i] + if nums[nums[i]] == nums[i]: + return nums[i] nums[nums[i]], nums[i] = nums[i], nums[nums[i]] return -1 + # ======= Test Case ======= nums = [2, 3, 1, 0, 2, 5, 3] # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_04_find_a_number_in_2d_matrix_s1.py b/sword_for_offer/codes/python/sfo_04_find_a_number_in_2d_matrix_s1.py index 7ef68af..ddd2e2f 100644 --- a/sword_for_offer/codes/python/sfo_04_find_a_number_in_2d_matrix_s1.py +++ b/sword_for_offer/codes/python/sfo_04_find_a_number_in_2d_matrix_s1.py @@ -1,28 +1,33 @@ -''' +""" File: sfo_04_find_a_number_in_2d_matrix_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def findNumberIn2DArray(self, matrix: List[List[int]], target: int) -> bool: i, j = len(matrix) - 1, 0 while i >= 0 and j < len(matrix[0]): - if matrix[i][j] > target: i -= 1 - elif matrix[i][j] < target: j += 1 - else: return True + if matrix[i][j] > target: + i -= 1 + elif matrix[i][j] < target: + j += 1 + else: + return True return False + # ======= Test Case ======= matrix = [ [1, 4, 7, 11, 15], [2, 5, 8, 12, 19], [3, 6, 9, 16, 22], [10, 13, 14, 17, 24], - [18, 21, 23, 26, 30] + [18, 21, 23, 26, 30], ] target = 5 # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_05_replace_spaces_s1.py b/sword_for_offer/codes/python/sfo_05_replace_spaces_s1.py index 399288c..4e3fe0b 100644 --- a/sword_for_offer/codes/python/sfo_05_replace_spaces_s1.py +++ b/sword_for_offer/codes/python/sfo_05_replace_spaces_s1.py @@ -1,20 +1,24 @@ -''' +""" File: sfo_05_replace_spaces_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def replaceSpace(self, s: str) -> str: res = [] for c in s: - if c == ' ': res.append("%20") - else: res.append(c) + if c == " ": + res.append("%20") + else: + res.append(c) return "".join(res) + # ======= Test Case ======= s = "We are happy." # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_06_print_a_linked_list_in_reverse_order_s1.py b/sword_for_offer/codes/python/sfo_06_print_a_linked_list_in_reverse_order_s1.py index 77a034a..aa41b0c 100644 --- a/sword_for_offer/codes/python/sfo_06_print_a_linked_list_in_reverse_order_s1.py +++ b/sword_for_offer/codes/python/sfo_06_print_a_linked_list_in_reverse_order_s1.py @@ -1,16 +1,18 @@ -''' +""" File: sfo_06_print_a_linked_list_in_reverse_order_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def reversePrint(self, head: ListNode) -> List[int]: return self.reversePrint(head.next) + [head.val] if head else [] + # ======= Test Case ======= head = list_to_linked_list([1, 3, 2]) # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_06_print_a_linked_list_in_reverse_order_s2.py b/sword_for_offer/codes/python/sfo_06_print_a_linked_list_in_reverse_order_s2.py index 13baf7a..20b2ac0 100644 --- a/sword_for_offer/codes/python/sfo_06_print_a_linked_list_in_reverse_order_s2.py +++ b/sword_for_offer/codes/python/sfo_06_print_a_linked_list_in_reverse_order_s2.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_06_print_a_linked_list_in_reverse_order_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def reversePrint(self, head: ListNode) -> List[int]: @@ -15,6 +16,7 @@ def reversePrint(self, head: ListNode) -> List[int]: head = head.next return stack[::-1] + # ======= Test Case ======= head = list_to_linked_list([1, 3, 2]) # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_07_reconstruct_binary_tree_s1.py b/sword_for_offer/codes/python/sfo_07_reconstruct_binary_tree_s1.py index 327f355..22b7dcf 100644 --- a/sword_for_offer/codes/python/sfo_07_reconstruct_binary_tree_s1.py +++ b/sword_for_offer/codes/python/sfo_07_reconstruct_binary_tree_s1.py @@ -1,27 +1,30 @@ -''' +""" File: sfo_07_reconstruct_binary_tree_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode: def recur(root, left, right): - if left > right: return # 递归终止 - node = TreeNode(preorder[root]) # 建立根节点 - i = dic[preorder[root]] # 划分根节点、左子树、右子树 - node.left = recur(root + 1, left, i - 1) # 开启左子树递归 - node.right = recur(i - left + root + 1, i + 1, right) # 开启右子树递归 - return node # 回溯返回根节点 + if left > right: + return # 递归终止 + node = TreeNode(preorder[root]) # 建立根节点 + i = dic[preorder[root]] # 划分根节点、左子树、右子树 + node.left = recur(root + 1, left, i - 1) # 开启左子树递归 + node.right = recur(i - left + root + 1, i + 1, right) # 开启右子树递归 + return node # 回溯返回根节点 dic, preorder = {}, preorder for i in range(len(inorder)): dic[inorder[i]] = i return recur(0, 0, len(inorder) - 1) + # ======= Test Case ======= preorder = [3, 9, 20, 15, 7] inorder = [9, 3, 15, 20, 7] diff --git a/sword_for_offer/codes/python/sfo_09_implement_a_queue_using_two_stacks_s1.py b/sword_for_offer/codes/python/sfo_09_implement_a_queue_using_two_stacks_s1.py index 739e34b..8442e64 100644 --- a/sword_for_offer/codes/python/sfo_09_implement_a_queue_using_two_stacks_s1.py +++ b/sword_for_offer/codes/python/sfo_09_implement_a_queue_using_two_stacks_s1.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_09_implement_a_queue_using_two_stacks_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class CQueue: def __init__(self): @@ -15,12 +16,15 @@ def appendTail(self, value: int) -> None: self.A.append(value) def deleteHead(self) -> int: - if self.B: return self.B.pop() - if not self.A: return -1 + if self.B: + return self.B.pop() + if not self.A: + return -1 while self.A: self.B.append(self.A.pop()) return self.B.pop() + # ====== Driver Code ====== c_queue = CQueue() res = [ @@ -28,6 +32,6 @@ def deleteHead(self) -> int: c_queue.appendTail(5), c_queue.appendTail(2), c_queue.deleteHead(), - c_queue.deleteHead() + c_queue.deleteHead(), ] print(res) diff --git a/sword_for_offer/codes/python/sfo_10i_fibonacci_numbers_s1.py b/sword_for_offer/codes/python/sfo_10i_fibonacci_numbers_s1.py index 6c5f638..9a60f72 100644 --- a/sword_for_offer/codes/python/sfo_10i_fibonacci_numbers_s1.py +++ b/sword_for_offer/codes/python/sfo_10i_fibonacci_numbers_s1.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_10-i_fibonacci_numbers_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def fib(self, n: int) -> int: @@ -14,6 +15,7 @@ def fib(self, n: int) -> int: a, b = b, (a + b) % 1000000007 return a + # ======= Test Case ======= n = 5 # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_10i_fibonacci_numbers_s2.py b/sword_for_offer/codes/python/sfo_10i_fibonacci_numbers_s2.py index 11a3d66..0e4d1eb 100644 --- a/sword_for_offer/codes/python/sfo_10i_fibonacci_numbers_s2.py +++ b/sword_for_offer/codes/python/sfo_10i_fibonacci_numbers_s2.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_10-i_fibonacci_numbers_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== # 不考虑大数越界问题 class Solution: @@ -15,6 +16,7 @@ def fib(self, n: int) -> int: a, b = b, a + b return a % 1000000007 + # ======= Test Case ======= n = 5 # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_10ii_frog_jump_s1.py b/sword_for_offer/codes/python/sfo_10ii_frog_jump_s1.py index 664d4bf..bf1b87c 100644 --- a/sword_for_offer/codes/python/sfo_10ii_frog_jump_s1.py +++ b/sword_for_offer/codes/python/sfo_10ii_frog_jump_s1.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_10-ii_frog_jump_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def numWays(self, n: int) -> int: @@ -14,6 +15,7 @@ def numWays(self, n: int) -> int: a, b = b, (a + b) % 1000000007 return a + # ======= Test Case ======= n = 7 # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_10ii_frog_jump_s2.py b/sword_for_offer/codes/python/sfo_10ii_frog_jump_s2.py index d903790..3bb8049 100644 --- a/sword_for_offer/codes/python/sfo_10ii_frog_jump_s2.py +++ b/sword_for_offer/codes/python/sfo_10ii_frog_jump_s2.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_10-ii_frog_jump_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== # 不考虑大数越界问题 class Solution: @@ -15,6 +16,7 @@ def numWays(self, n: int) -> int: a, b = b, a + b return a % 1000000007 + # ======= Test Case ======= n = 7 # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_11_find_minimum_in_rotated_sorted_array_s1.py b/sword_for_offer/codes/python/sfo_11_find_minimum_in_rotated_sorted_array_s1.py index 165d7df..80b56c5 100644 --- a/sword_for_offer/codes/python/sfo_11_find_minimum_in_rotated_sorted_array_s1.py +++ b/sword_for_offer/codes/python/sfo_11_find_minimum_in_rotated_sorted_array_s1.py @@ -1,22 +1,27 @@ -''' +""" File: sfo_11_find_minimum_in_rotated_sorted_array_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def minArray(self, numbers: List[int]) -> int: i, j = 0, len(numbers) - 1 while i < j: m = (i + j) // 2 - if numbers[m] > numbers[j]: i = m + 1 - elif numbers[m] < numbers[j]: j = m - else: j -= 1 + if numbers[m] > numbers[j]: + i = m + 1 + elif numbers[m] < numbers[j]: + j = m + else: + j -= 1 return numbers[i] + # ======= Test Case ======= numbers = [3, 4, 5, 1, 2] # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_11_find_minimum_in_rotated_sorted_array_s2.py b/sword_for_offer/codes/python/sfo_11_find_minimum_in_rotated_sorted_array_s2.py index 623549c..32494a6 100644 --- a/sword_for_offer/codes/python/sfo_11_find_minimum_in_rotated_sorted_array_s2.py +++ b/sword_for_offer/codes/python/sfo_11_find_minimum_in_rotated_sorted_array_s2.py @@ -1,22 +1,27 @@ -''' +""" File: sfo_11_find_minimum_in_rotated_sorted_array_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def minArray(self, numbers: List[int]) -> int: i, j = 0, len(numbers) - 1 while i < j: m = (i + j) // 2 - if numbers[m] > numbers[j]: i = m + 1 - elif numbers[m] < numbers[j]: j = m - else: return min(numbers[i:j]) + if numbers[m] > numbers[j]: + i = m + 1 + elif numbers[m] < numbers[j]: + j = m + else: + return min(numbers[i:j]) return numbers[i] + # ======= Test Case ======= numbers = [3, 4, 5, 1, 2] # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_12_word_search_s1.py b/sword_for_offer/codes/python/sfo_12_word_search_s1.py index 1ba89de..5b1a36d 100644 --- a/sword_for_offer/codes/python/sfo_12_word_search_s1.py +++ b/sword_for_offer/codes/python/sfo_12_word_search_s1.py @@ -1,33 +1,43 @@ -''' +""" File: sfo_12_word_search_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def exist(self, board: List[List[str]], word: str) -> bool: def dfs(i, j, k): - if not 0 <= i < len(board) or not 0 <= j < len(board[0]) or board[i][j] != word[k]: return False - if k == len(word) - 1: return True - board[i][j] = '' - res = dfs(i + 1, j, k + 1) or dfs(i - 1, j, k + 1) or dfs(i, j + 1, k + 1) or dfs(i, j - 1, k + 1) + if ( + not 0 <= i < len(board) + or not 0 <= j < len(board[0]) + or board[i][j] != word[k] + ): + return False + if k == len(word) - 1: + return True + board[i][j] = "" + res = ( + dfs(i + 1, j, k + 1) + or dfs(i - 1, j, k + 1) + or dfs(i, j + 1, k + 1) + or dfs(i, j - 1, k + 1) + ) board[i][j] = word[k] return res for i in range(len(board)): for j in range(len(board[0])): - if dfs(i, j, 0): return True + if dfs(i, j, 0): + return True return False + # ======= Test Case ======= -board = [ - ['A', 'B', 'C', 'E'], - ['S', 'F', 'C', 'S'], - ['A', 'D', 'E', 'E'] -] +board = [["A", "B", "C", "E"], ["S", "F", "C", "S"], ["A", "D", "E", "E"]] word = "ABCCED" # ====== Driver Code ====== slt = Solution() diff --git a/sword_for_offer/codes/python/sfo_13_range_of_motion_of_a_robot_s1.py b/sword_for_offer/codes/python/sfo_13_range_of_motion_of_a_robot_s1.py index d8c5a94..7a91e35 100644 --- a/sword_for_offer/codes/python/sfo_13_range_of_motion_of_a_robot_s1.py +++ b/sword_for_offer/codes/python/sfo_13_range_of_motion_of_a_robot_s1.py @@ -1,22 +1,29 @@ -''' +""" File: sfo_13_range_of_motion_of_a_robot_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def movingCount(self, m: int, n: int, k: int) -> int: def dfs(i, j, si, sj): - if i >= m or j >= n or k < si + sj or (i, j) in visited: return 0 - visited.add((i,j)) - return 1 + dfs(i + 1, j, si + 1 if (i + 1) % 10 else si - 8, sj) + dfs(i, j + 1, si, sj + 1 if (j + 1) % 10 else sj - 8) + if i >= m or j >= n or k < si + sj or (i, j) in visited: + return 0 + visited.add((i, j)) + return ( + 1 + + dfs(i + 1, j, si + 1 if (i + 1) % 10 else si - 8, sj) + + dfs(i, j + 1, si, sj + 1 if (j + 1) % 10 else sj - 8) + ) visited = set() return dfs(0, 0, 0, 0) + # ======= Test Case ======= m = 2 n = 3 diff --git a/sword_for_offer/codes/python/sfo_13_range_of_motion_of_a_robot_s2.py b/sword_for_offer/codes/python/sfo_13_range_of_motion_of_a_robot_s2.py index c422b6b..066f4b7 100644 --- a/sword_for_offer/codes/python/sfo_13_range_of_motion_of_a_robot_s2.py +++ b/sword_for_offer/codes/python/sfo_13_range_of_motion_of_a_robot_s2.py @@ -1,23 +1,26 @@ -''' +""" File: sfo_13_range_of_motion_of_a_robot_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def movingCount(self, m: int, n: int, k: int) -> int: queue, visited = [(0, 0, 0, 0)], set() while queue: i, j, si, sj = queue.pop(0) - if i >= m or j >= n or k < si + sj or (i, j) in visited: continue - visited.add((i,j)) + if i >= m or j >= n or k < si + sj or (i, j) in visited: + continue + visited.add((i, j)) queue.append((i + 1, j, si + 1 if (i + 1) % 10 else si - 8, sj)) queue.append((i, j + 1, si, sj + 1 if (j + 1) % 10 else sj - 8)) return len(visited) + # ======= Test Case ======= m = 2 n = 3 diff --git a/sword_for_offer/codes/python/sfo_14i_cut_the_rope_i_s1.py b/sword_for_offer/codes/python/sfo_14i_cut_the_rope_i_s1.py index d112cd1..31a9de3 100644 --- a/sword_for_offer/codes/python/sfo_14i_cut_the_rope_i_s1.py +++ b/sword_for_offer/codes/python/sfo_14i_cut_the_rope_i_s1.py @@ -1,20 +1,25 @@ -''' +""" File: sfo_14-i_cut_the_rope_i_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def cuttingRope(self, n: int) -> int: - if n <= 3: return n - 1 + if n <= 3: + return n - 1 a, b = n // 3, n % 3 - if b == 0: return int(math.pow(3, a)) - if b == 1: return int(math.pow(3, a - 1) * 4) + if b == 0: + return int(math.pow(3, a)) + if b == 1: + return int(math.pow(3, a - 1) * 4) return int(math.pow(3, a) * 2) + # ======= Test Case ======= n = 10 # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_14ii_cut_the_rope_ii_s1.py b/sword_for_offer/codes/python/sfo_14ii_cut_the_rope_ii_s1.py index a2b6e41..4edb028 100644 --- a/sword_for_offer/codes/python/sfo_14ii_cut_the_rope_ii_s1.py +++ b/sword_for_offer/codes/python/sfo_14ii_cut_the_rope_ii_s1.py @@ -1,23 +1,29 @@ -''' +""" File: sfo_14-ii_cut_the_rope_ii_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def cuttingRope(self, n: int) -> int: - if n <= 3: return n - 1 - a, b, p, x, rem = n // 3 - 1, n % 3, 1000000007, 3 , 1 + if n <= 3: + return n - 1 + a, b, p, x, rem = n // 3 - 1, n % 3, 1000000007, 3, 1 while a > 0: - if a % 2: rem = (rem * x) % p - x = x ** 2 % p + if a % 2: + rem = (rem * x) % p + x = x**2 % p a //= 2 - if b == 0: return (rem * 3) % p # = 3^(a+1) % p - if b == 1: return (rem * 4) % p # = 3^a * 4 % p - return (rem * 6) % p # = 3^(a+1) * 2 % p + if b == 0: + return (rem * 3) % p # = 3^(a+1) % p + if b == 1: + return (rem * 4) % p # = 3^a * 4 % p + return (rem * 6) % p # = 3^(a+1) * 2 % p + # ======= Test Case ======= n = 10 diff --git a/sword_for_offer/codes/python/sfo_14ii_cut_the_rope_ii_s2.py b/sword_for_offer/codes/python/sfo_14ii_cut_the_rope_ii_s2.py index 95c9b87..8414149 100644 --- a/sword_for_offer/codes/python/sfo_14ii_cut_the_rope_ii_s2.py +++ b/sword_for_offer/codes/python/sfo_14ii_cut_the_rope_ii_s2.py @@ -1,20 +1,25 @@ -''' +""" File: sfo_14-ii_cut_the_rope_ii_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== # 由于语言特性，Python 可以不考虑大数越界问题 class Solution: def cuttingRope(self, n: int) -> int: - if n <= 3: return n - 1 + if n <= 3: + return n - 1 a, b, p = n // 3, n % 3, 1000000007 - if b == 0: return 3 ** a % p - if b == 1: return 3 ** (a - 1) * 4 % p - return 3 ** a * 2 % p + if b == 0: + return 3**a % p + if b == 1: + return 3 ** (a - 1) * 4 % p + return 3**a * 2 % p + # ======= Test Case ======= n = 10 diff --git a/sword_for_offer/codes/python/sfo_15_number_of_1_bits_s1.py b/sword_for_offer/codes/python/sfo_15_number_of_1_bits_s1.py index 622f11f..8513dab 100644 --- a/sword_for_offer/codes/python/sfo_15_number_of_1_bits_s1.py +++ b/sword_for_offer/codes/python/sfo_15_number_of_1_bits_s1.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_15_number_of_1_bits_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def hammingWeight(self, n: int) -> int: @@ -15,6 +16,7 @@ def hammingWeight(self, n: int) -> int: n >>= 1 return res + # ======= Test Case ======= n = 11 # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_15_number_of_1_bits_s2.py b/sword_for_offer/codes/python/sfo_15_number_of_1_bits_s2.py index 67198c4..b08a141 100644 --- a/sword_for_offer/codes/python/sfo_15_number_of_1_bits_s2.py +++ b/sword_for_offer/codes/python/sfo_15_number_of_1_bits_s2.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_15_number_of_1_bits_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def hammingWeight(self, n: int) -> int: @@ -15,6 +16,7 @@ def hammingWeight(self, n: int) -> int: n &= n - 1 return res + # ======= Test Case ======= n = 11 # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_16_powers_of_integers_s1.py b/sword_for_offer/codes/python/sfo_16_powers_of_integers_s1.py index 73d3597..52186af 100644 --- a/sword_for_offer/codes/python/sfo_16_powers_of_integers_s1.py +++ b/sword_for_offer/codes/python/sfo_16_powers_of_integers_s1.py @@ -1,23 +1,28 @@ -''' +""" File: sfo_16_powers_of_integers_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def myPow(self, x: float, n: int) -> float: - if x == 0.0: return 0.0 + if x == 0.0: + return 0.0 res = 1 - if n < 0: x, n = 1 / x, -n + if n < 0: + x, n = 1 / x, -n while n: - if n & 1: res *= x + if n & 1: + res *= x x *= x n >>= 1 return res + # ======= Test Case ======= x = 2.0 n = 10 diff --git a/sword_for_offer/codes/python/sfo_17_print_from_1_to_the_largest_n_digits_s1.py b/sword_for_offer/codes/python/sfo_17_print_from_1_to_the_largest_n_digits_s1.py index b624ffc..b8e3f80 100644 --- a/sword_for_offer/codes/python/sfo_17_print_from_1_to_the_largest_n_digits_s1.py +++ b/sword_for_offer/codes/python/sfo_17_print_from_1_to_the_largest_n_digits_s1.py @@ -1,19 +1,21 @@ -''' +""" File: sfo_17_print_from_1_to_the_largest_n_digits_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def printNumbers(self, n: int) -> List[int]: res = [] - for i in range(1, 10 ** n): + for i in range(1, 10**n): res.append(i) return res + # ======= Test Case ======= n = 1 # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_17_print_from_1_to_the_largest_n_digits_s2.py b/sword_for_offer/codes/python/sfo_17_print_from_1_to_the_largest_n_digits_s2.py index 71034ca..958cbe2 100644 --- a/sword_for_offer/codes/python/sfo_17_print_from_1_to_the_largest_n_digits_s2.py +++ b/sword_for_offer/codes/python/sfo_17_print_from_1_to_the_largest_n_digits_s2.py @@ -1,26 +1,28 @@ -''' +""" File: sfo_17_print_from_1_to_the_largest_n_digits_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def printNumbers(self, n: int) -> List[int]: def dfs(x): - if x == n: # 终止条件：已固定完所有位 - res.append(''.join(num)) # 拼接 num 并添加至 res 尾部 + if x == n: # 终止条件：已固定完所有位 + res.append("".join(num)) # 拼接 num 并添加至 res 尾部 return - for i in range(10): # 遍历 0 - 9 - num[x] = str(i) # 固定第 x 位为 i - dfs(x + 1) # 开启固定第 x + 1 位 - - num = ['0'] * n # 起始数字定义为 n 个 0 组成的字符列表 - res = [] # 数字字符串列表 - dfs(0) # 开启全排列递归 - return ','.join(res) # 拼接所有数字字符串，使用逗号隔开，并返回 + for i in range(10): # 遍历 0 - 9 + num[x] = str(i) # 固定第 x 位为 i + dfs(x + 1) # 开启固定第 x + 1 位 + + num = ["0"] * n # 起始数字定义为 n 个 0 组成的字符列表 + res = [] # 数字字符串列表 + dfs(0) # 开启全排列递归 + return ",".join(res) # 拼接所有数字字符串，使用逗号隔开，并返回 + # ======= Test Case ======= n = 1 diff --git a/sword_for_offer/codes/python/sfo_17_print_from_1_to_the_largest_n_digits_s3.py b/sword_for_offer/codes/python/sfo_17_print_from_1_to_the_largest_n_digits_s3.py index aceb8f2..fb3151c 100644 --- a/sword_for_offer/codes/python/sfo_17_print_from_1_to_the_largest_n_digits_s3.py +++ b/sword_for_offer/codes/python/sfo_17_print_from_1_to_the_largest_n_digits_s3.py @@ -1,31 +1,36 @@ -''' +""" File: sfo_17_print_from_1_to_the_largest_n_digits_s3.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def printNumbers(self, n: int) -> List[int]: def dfs(x): if x == n: - s = ''.join(num[self.start:]) - if s != '0': res.append(s) - if n - self.start == self.nine: self.start -= 1 + s = "".join(num[self.start :]) + if s != "0": + res.append(s) + if n - self.start == self.nine: + self.start -= 1 return for i in range(10): - if i == 9: self.nine += 1 + if i == 9: + self.nine += 1 num[x] = str(i) dfs(x + 1) self.nine -= 1 - - num, res = ['0'] * n, [] + + num, res = ["0"] * n, [] self.nine = 0 self.start = n - 1 dfs(0) - return ','.join(res) + return ",".join(res) + # ======= Test Case ======= n = 1 diff --git a/sword_for_offer/codes/python/sfo_17_print_from_1_to_the_largest_n_digits_s4.py b/sword_for_offer/codes/python/sfo_17_print_from_1_to_the_largest_n_digits_s4.py index dfd0281..febd883 100644 --- a/sword_for_offer/codes/python/sfo_17_print_from_1_to_the_largest_n_digits_s4.py +++ b/sword_for_offer/codes/python/sfo_17_print_from_1_to_the_largest_n_digits_s4.py @@ -1,32 +1,37 @@ -''' +""" File: sfo_17_print_from_1_to_the_largest_n_digits_s4.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def printNumbers(self, n: int) -> List[int]: def dfs(x): if x == n: - s = ''.join(num[self.start:]) - if s != '0': res.append(int(s)) - if n - self.start == self.nine: self.start -= 1 + s = "".join(num[self.start :]) + if s != "0": + res.append(int(s)) + if n - self.start == self.nine: + self.start -= 1 return for i in range(10): - if i == 9: self.nine += 1 + if i == 9: + self.nine += 1 num[x] = str(i) dfs(x + 1) self.nine -= 1 - - num, res = ['0'] * n, [] + + num, res = ["0"] * n, [] self.nine = 0 self.start = n - 1 dfs(0) return res + # ======= Test Case ======= n = 1 # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_18_delete_a_node_from_a_linked_list_s1.py b/sword_for_offer/codes/python/sfo_18_delete_a_node_from_a_linked_list_s1.py index 8d51d64..ed0644e 100644 --- a/sword_for_offer/codes/python/sfo_18_delete_a_node_from_a_linked_list_s1.py +++ b/sword_for_offer/codes/python/sfo_18_delete_a_node_from_a_linked_list_s1.py @@ -1,21 +1,25 @@ -''' +""" File: sfo_18_delete_a_node_from_a_linked_list_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def deleteNode(self, head: ListNode, val: int) -> ListNode: - if head.val == val: return head.next + if head.val == val: + return head.next pre, cur = head, head.next while cur and cur.val != val: pre, cur = cur, cur.next - if cur: pre.next = cur.next + if cur: + pre.next = cur.next return head + # ======= Test Case ======= head = list_to_linked_list([4, 5, 1, 9]) val = 5 diff --git a/sword_for_offer/codes/python/sfo_19_regular_expression_matching_s1.py b/sword_for_offer/codes/python/sfo_19_regular_expression_matching_s1.py index 391b4f7..25eaf5c 100644 --- a/sword_for_offer/codes/python/sfo_19_regular_expression_matching_s1.py +++ b/sword_for_offer/codes/python/sfo_19_regular_expression_matching_s1.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_19_regular_expression_matching_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def isMatch(self, s: str, p: str) -> bool: @@ -13,14 +14,19 @@ def isMatch(self, s: str, p: str) -> bool: dp = [[False] * n for _ in range(m)] dp[0][0] = True for j in range(2, n, 2): - dp[0][j] = dp[0][j - 2] and p[j - 1] == '*' + dp[0][j] = dp[0][j - 2] and p[j - 1] == "*" for i in range(1, m): for j in range(1, n): - dp[i][j] = dp[i][j - 2] or dp[i - 1][j] and (s[i - 1] == p[j - 2] or p[j - 2] == '.') \ - if p[j - 1] == '*' else \ - dp[i - 1][j - 1] and (p[j - 1] == '.' or s[i - 1] == p[j - 1]) + dp[i][j] = ( + dp[i][j - 2] + or dp[i - 1][j] + and (s[i - 1] == p[j - 2] or p[j - 2] == ".") + if p[j - 1] == "*" + else dp[i - 1][j - 1] and (p[j - 1] == "." or s[i - 1] == p[j - 1]) + ) return dp[-1][-1] + # ======= Test Case ======= s = "mississippi" p = "mis*is*p*." diff --git a/sword_for_offer/codes/python/sfo_19_regular_expression_matching_s2.py b/sword_for_offer/codes/python/sfo_19_regular_expression_matching_s2.py index d3fb4e0..9e92f1f 100644 --- a/sword_for_offer/codes/python/sfo_19_regular_expression_matching_s2.py +++ b/sword_for_offer/codes/python/sfo_19_regular_expression_matching_s2.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_19_regular_expression_matching_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def isMatch(self, s: str, p: str) -> bool: @@ -14,19 +15,25 @@ def isMatch(self, s: str, p: str) -> bool: dp[0][0] = True # 初始化首行 for j in range(2, n, 2): - dp[0][j] = dp[0][j - 2] and p[j - 1] == '*' + dp[0][j] = dp[0][j - 2] and p[j - 1] == "*" # 状态转移 for i in range(1, m): for j in range(1, n): - if p[j - 1] == '*': - if dp[i][j - 2]: dp[i][j] = True # 1. - elif dp[i - 1][j] and s[i - 1] == p[j - 2]: dp[i][j] = True # 2. - elif dp[i - 1][j] and p[j - 2] == '.': dp[i][j] = True # 3. + if p[j - 1] == "*": + if dp[i][j - 2]: + dp[i][j] = True # 1. + elif dp[i - 1][j] and s[i - 1] == p[j - 2]: + dp[i][j] = True # 2. + elif dp[i - 1][j] and p[j - 2] == ".": + dp[i][j] = True # 3. else: - if dp[i - 1][j - 1] and s[i - 1] == p[j - 1]: dp[i][j] = True # 1. - elif dp[i - 1][j - 1] and p[j - 1] == '.': dp[i][j] = True # 2. + if dp[i - 1][j - 1] and s[i - 1] == p[j - 1]: + dp[i][j] = True # 1. + elif dp[i - 1][j - 1] and p[j - 1] == ".": + dp[i][j] = True # 2. return dp[-1][-1] + # ======= Test Case ======= s = "mississippi" p = "mis*is*p*." diff --git a/sword_for_offer/codes/python/sfo_20_a_string_representing_a_numeric_value_s1.py b/sword_for_offer/codes/python/sfo_20_a_string_representing_a_numeric_value_s1.py index f74ca64..c5f82b9 100644 --- a/sword_for_offer/codes/python/sfo_20_a_string_representing_a_numeric_value_s1.py +++ b/sword_for_offer/codes/python/sfo_20_a_string_representing_a_numeric_value_s1.py @@ -1,36 +1,44 @@ -''' +""" File: sfo_20_a_string_representing_a_numeric_value_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def isNumber(self, s: str) -> bool: states = [ - { ' ': 0, 's': 1, 'd': 2, '.': 4 }, # 0. start with 'blank' - { 'd': 2, '.': 4 } , # 1. 'sign' before 'e' - { 'd': 2, '.': 3, 'e': 5, ' ': 8 }, # 2. 'digit' before 'dot' - { 'd': 3, 'e': 5, ' ': 8 }, # 3. 'digit' after 'dot' - { 'd': 3 }, # 4. 'digit' after 'dot' (‘blank’ before 'dot') - { 's': 6, 'd': 7 }, # 5. 'e' - { 'd': 7 }, # 6. 'sign' after 'e' - { 'd': 7, ' ': 8 }, # 7. 'digit' after 'e' - { ' ': 8 } # 8. end with 'blank' + {" ": 0, "s": 1, "d": 2, ".": 4}, # 0. start with 'blank' + {"d": 2, ".": 4}, # 1. 'sign' before 'e' + {"d": 2, ".": 3, "e": 5, " ": 8}, # 2. 'digit' before 'dot' + {"d": 3, "e": 5, " ": 8}, # 3. 'digit' after 'dot' + {"d": 3}, # 4. 'digit' after 'dot' (‘blank’ before 'dot') + {"s": 6, "d": 7}, # 5. 'e' + {"d": 7}, # 6. 'sign' after 'e' + {"d": 7, " ": 8}, # 7. 'digit' after 'e' + {" ": 8}, # 8. end with 'blank' ] - p = 0 # start with state 0 + p = 0 # start with state 0 for c in s: - if '0' <= c <= '9': t = 'd' # digit - elif c in "+-": t = 's' # sign - elif c in "eE": t = 'e' # e or E - elif c in ". ": t = c # dot, blank - else: t = '?' # unknown - if t not in states[p]: return False + if "0" <= c <= "9": + t = "d" # digit + elif c in "+-": + t = "s" # sign + elif c in "eE": + t = "e" # e or E + elif c in ". ": + t = c # dot, blank + else: + t = "?" # unknown + if t not in states[p]: + return False p = states[p][t] return p in (2, 3, 7, 8) + # ======= Test Case ======= s = " .1 " # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_21_adjust_the_order_of_numbers_in_an_array_s1.py b/sword_for_offer/codes/python/sfo_21_adjust_the_order_of_numbers_in_an_array_s1.py index 1b99687..f12c8a6 100644 --- a/sword_for_offer/codes/python/sfo_21_adjust_the_order_of_numbers_in_an_array_s1.py +++ b/sword_for_offer/codes/python/sfo_21_adjust_the_order_of_numbers_in_an_array_s1.py @@ -1,21 +1,25 @@ -''' +""" File: sfo_21_adjust_the_order_of_numbers_in_an_array_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def exchange(self, nums: List[int]) -> List[int]: i, j = 0, len(nums) - 1 while i < j: - while i < j and nums[i] & 1 == 1: i += 1 - while i < j and nums[j] & 1 == 0: j -= 1 + while i < j and nums[i] & 1 == 1: + i += 1 + while i < j and nums[j] & 1 == 0: + j -= 1 nums[i], nums[j] = nums[j], nums[i] return nums + # ======= Test Case ======= nums = [1, 2, 3, 4] # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1.py b/sword_for_offer/codes/python/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1.py index 6af418a..2b22689 100644 --- a/sword_for_offer/codes/python/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1.py +++ b/sword_for_offer/codes/python/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s1.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_22_the_k-th_node_from_the_end_of_a_linked_list_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def getKthFromEnd(self, head: ListNode, k: int) -> ListNode: @@ -16,6 +17,7 @@ def getKthFromEnd(self, head: ListNode, k: int) -> ListNode: former, latter = former.next, latter.next return latter + # ======= Test Case ======= head = list_to_linked_list([1, 2, 3, 4, 5]) k = 2 diff --git a/sword_for_offer/codes/python/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2.py b/sword_for_offer/codes/python/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2.py index ce7d207..94f036b 100644 --- a/sword_for_offer/codes/python/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2.py +++ b/sword_for_offer/codes/python/sfo_22_the_kth_node_from_the_end_of_a_linked_list_s2.py @@ -1,22 +1,25 @@ -''' +""" File: sfo_22_the_k-th_node_from_the_end_of_a_linked_list_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def getKthFromEnd(self, head: ListNode, k: int) -> ListNode: former, latter = head, head for _ in range(k): - if not former: return + if not former: + return former = former.next while former: former, latter = former.next, latter.next return latter + # ======= Test Case ======= head = list_to_linked_list([1, 2, 3, 4, 5]) k = 2 diff --git a/sword_for_offer/codes/python/sfo_24_reverse_a_linked_list_s1.py b/sword_for_offer/codes/python/sfo_24_reverse_a_linked_list_s1.py index 0ebfbe8..2c199ea 100644 --- a/sword_for_offer/codes/python/sfo_24_reverse_a_linked_list_s1.py +++ b/sword_for_offer/codes/python/sfo_24_reverse_a_linked_list_s1.py @@ -1,22 +1,24 @@ -''' +""" File: sfo_24_reverse_a_linked_list_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def reverseList(self, head: ListNode) -> ListNode: cur, pre = head, None while cur: - tmp = cur.next # 暂存后继节点 cur.next - cur.next = pre # 修改 next 引用指向 - pre = cur # pre 暂存 cur - cur = tmp # cur 访问下一节点 + tmp = cur.next # 暂存后继节点 cur.next + cur.next = pre # 修改 next 引用指向 + pre = cur # pre 暂存 cur + cur = tmp # cur 访问下一节点 return pre + # ======= Test Case ======= head = list_to_linked_list([1, 2, 3, 4, 5]) # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_24_reverse_a_linked_list_s2.py b/sword_for_offer/codes/python/sfo_24_reverse_a_linked_list_s2.py index 1bbb4d5..73f4eca 100644 --- a/sword_for_offer/codes/python/sfo_24_reverse_a_linked_list_s2.py +++ b/sword_for_offer/codes/python/sfo_24_reverse_a_linked_list_s2.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_24_reverse_a_linked_list_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def reverseList(self, head: ListNode) -> ListNode: @@ -14,6 +15,7 @@ def reverseList(self, head: ListNode) -> ListNode: cur.next, pre, cur = pre, cur, cur.next return pre + # ======= Test Case ======= head = list_to_linked_list([1, 2, 3, 4, 5]) # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_24_reverse_a_linked_list_s3.py b/sword_for_offer/codes/python/sfo_24_reverse_a_linked_list_s3.py index 68c30a8..49cf3ce 100644 --- a/sword_for_offer/codes/python/sfo_24_reverse_a_linked_list_s3.py +++ b/sword_for_offer/codes/python/sfo_24_reverse_a_linked_list_s3.py @@ -1,21 +1,24 @@ -''' +""" File: sfo_24_reverse_a_linked_list_s3.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def reverseList(self, head: ListNode) -> ListNode: def recur(cur, pre): - if not cur: return pre # 终止条件 - res = recur(cur.next, cur) # 递归后继节点 - cur.next = pre # 修改节点引用指向 - return res # 返回反转链表的头节点 - - return recur(head, None) # 调用递归并返回 + if not cur: + return pre # 终止条件 + res = recur(cur.next, cur) # 递归后继节点 + cur.next = pre # 修改节点引用指向 + return res # 返回反转链表的头节点 + + return recur(head, None) # 调用递归并返回 + # ======= Test Case ======= head = list_to_linked_list([1, 2, 3, 4, 5]) diff --git a/sword_for_offer/codes/python/sfo_25_combine_two_sorted_linked_lists_s1.py b/sword_for_offer/codes/python/sfo_25_combine_two_sorted_linked_lists_s1.py index 96bf01e..58a317a 100644 --- a/sword_for_offer/codes/python/sfo_25_combine_two_sorted_linked_lists_s1.py +++ b/sword_for_offer/codes/python/sfo_25_combine_two_sorted_linked_lists_s1.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_25_combine_two_sorted_linked_lists_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode: @@ -19,6 +20,7 @@ def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode: cur.next = l1 if l1 else l2 return dum.next + # ======= Test Case ======= l1 = list_to_linked_list([1, 2, 4]) l2 = list_to_linked_list([1, 3, 4]) diff --git a/sword_for_offer/codes/python/sfo_26_substructure_of_a_binary_tree_s1.py b/sword_for_offer/codes/python/sfo_26_substructure_of_a_binary_tree_s1.py index ce9f016..c9cf5d6 100644 --- a/sword_for_offer/codes/python/sfo_26_substructure_of_a_binary_tree_s1.py +++ b/sword_for_offer/codes/python/sfo_26_substructure_of_a_binary_tree_s1.py @@ -1,20 +1,28 @@ -''' +""" File: sfo_26_substructure_of_a_binary_tree_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def isSubStructure(self, A: TreeNode, B: TreeNode) -> bool: def recur(A, B): - if not B: return True - if not A or A.val != B.val: return False + if not B: + return True + if not A or A.val != B.val: + return False return recur(A.left, B.left) and recur(A.right, B.right) - return bool(A and B) and (recur(A, B) or self.isSubStructure(A.left, B) or self.isSubStructure(A.right, B)) + return bool(A and B) and ( + recur(A, B) + or self.isSubStructure(A.left, B) + or self.isSubStructure(A.right, B) + ) + # ======= Test Case ======= A = list_to_tree([3, 4, 5, 1, 2, None, None, None, None, None, None]) diff --git a/sword_for_offer/codes/python/sfo_27_mirror_of_a_binary_tree_s1.py b/sword_for_offer/codes/python/sfo_27_mirror_of_a_binary_tree_s1.py index d3d3310..90b6bb3 100644 --- a/sword_for_offer/codes/python/sfo_27_mirror_of_a_binary_tree_s1.py +++ b/sword_for_offer/codes/python/sfo_27_mirror_of_a_binary_tree_s1.py @@ -1,20 +1,25 @@ -''' +""" File: sfo_27_mirror_of_a_binary_tree_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def mirrorTree(self, root: TreeNode) -> TreeNode: - if not root: return + if not root: + return root.left, root.right = self.mirrorTree(root.right), self.mirrorTree(root.left) return root + # ======= Test Case ======= -root = list_to_tree([4, 2, 7, 1, 3, 6, 9, None, None, None, None, None, None, None, None]) +root = list_to_tree( + [4, 2, 7, 1, 3, 6, 9, None, None, None, None, None, None, None, None] +) # ====== Driver Code ====== slt = Solution() res = slt.mirrorTree(root) diff --git a/sword_for_offer/codes/python/sfo_27_mirror_of_a_binary_tree_s2.py b/sword_for_offer/codes/python/sfo_27_mirror_of_a_binary_tree_s2.py index 92f4b55..b991f0c 100644 --- a/sword_for_offer/codes/python/sfo_27_mirror_of_a_binary_tree_s2.py +++ b/sword_for_offer/codes/python/sfo_27_mirror_of_a_binary_tree_s2.py @@ -1,22 +1,27 @@ -''' +""" File: sfo_27_mirror_of_a_binary_tree_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def mirrorTree(self, root: TreeNode) -> TreeNode: - if not root: return + if not root: + return tmp = root.left root.left = self.mirrorTree(root.right) root.right = self.mirrorTree(tmp) return root + # ======= Test Case ======= -root = list_to_tree([4, 2, 7, 1, 3, 6, 9, None, None, None, None, None, None, None, None]) +root = list_to_tree( + [4, 2, 7, 1, 3, 6, 9, None, None, None, None, None, None, None, None] +) # ====== Driver Code ====== slt = Solution() res = slt.mirrorTree(root) diff --git a/sword_for_offer/codes/python/sfo_27_mirror_of_a_binary_tree_s3.py b/sword_for_offer/codes/python/sfo_27_mirror_of_a_binary_tree_s3.py index 4b50d29..ac1786f 100644 --- a/sword_for_offer/codes/python/sfo_27_mirror_of_a_binary_tree_s3.py +++ b/sword_for_offer/codes/python/sfo_27_mirror_of_a_binary_tree_s3.py @@ -1,25 +1,32 @@ -''' +""" File: sfo_27_mirror_of_a_binary_tree_s3.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def mirrorTree(self, root: TreeNode) -> TreeNode: - if not root: return + if not root: + return stack = [root] while stack: node = stack.pop() - if node.left: stack.append(node.left) - if node.right: stack.append(node.right) + if node.left: + stack.append(node.left) + if node.right: + stack.append(node.right) node.left, node.right = node.right, node.left return root + # ======= Test Case ======= -root = list_to_tree([4, 2, 7, 1, 3, 6, 9, None, None, None, None, None, None, None, None]) +root = list_to_tree( + [4, 2, 7, 1, 3, 6, 9, None, None, None, None, None, None, None, None] +) # ====== Driver Code ====== slt = Solution() res = slt.mirrorTree(root) diff --git a/sword_for_offer/codes/python/sfo_28_symmetric_binary_tree_s1.py b/sword_for_offer/codes/python/sfo_28_symmetric_binary_tree_s1.py index 181cbde..fda24e6 100644 --- a/sword_for_offer/codes/python/sfo_28_symmetric_binary_tree_s1.py +++ b/sword_for_offer/codes/python/sfo_28_symmetric_binary_tree_s1.py @@ -1,23 +1,29 @@ -''' +""" File: sfo_28_symmetric_binary_tree_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def isSymmetric(self, root: TreeNode) -> bool: def recur(L, R): - if not L and not R: return True - if not L or not R or L.val != R.val: return False + if not L and not R: + return True + if not L or not R or L.val != R.val: + return False return recur(L.left, R.right) and recur(L.right, R.left) return not root or recur(root.left, root.right) + # ======= Test Case ======= -root = list_to_tree([1, 2, 2, 3, 4, 4, 3, None, None, None, None, None, None, None, None]) +root = list_to_tree( + [1, 2, 2, 3, 4, 4, 3, None, None, None, None, None, None, None, None] +) # ====== Driver Code ====== slt = Solution() res = slt.isSymmetric(root) diff --git a/sword_for_offer/codes/python/sfo_29_print_a_given_matrix_in_spiral_form_s1.py b/sword_for_offer/codes/python/sfo_29_print_a_given_matrix_in_spiral_form_s1.py index 243f22e..432742e 100644 --- a/sword_for_offer/codes/python/sfo_29_print_a_given_matrix_in_spiral_form_s1.py +++ b/sword_for_offer/codes/python/sfo_29_print_a_given_matrix_in_spiral_form_s1.py @@ -1,37 +1,44 @@ -''' +""" File: sfo_29_print_a_given_matrix_in_spiral_form_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def spiralOrder(self, matrix: List[List[int]]) -> List[int]: - if not matrix: return [] + if not matrix: + return [] l, r, t, b, res = 0, len(matrix[0]) - 1, 0, len(matrix) - 1, [] while True: - for i in range(l, r + 1): res.append(matrix[t][i]) # left to right + for i in range(l, r + 1): + res.append(matrix[t][i]) # left to right t += 1 - if t > b: break - for i in range(t, b + 1): res.append(matrix[i][r]) # top to bottom + if t > b: + break + for i in range(t, b + 1): + res.append(matrix[i][r]) # top to bottom r -= 1 - if l > r: break - for i in range(r, l - 1, -1): res.append(matrix[b][i]) # right to left + if l > r: + break + for i in range(r, l - 1, -1): + res.append(matrix[b][i]) # right to left b -= 1 - if t > b: break - for i in range(b, t - 1, -1): res.append(matrix[i][l]) # bottom to top + if t > b: + break + for i in range(b, t - 1, -1): + res.append(matrix[i][l]) # bottom to top l += 1 - if l > r: break + if l > r: + break return res + # ======= Test Case ======= -matrix = [ - [1, 2, 3], - [4, 5, 6], - [7, 8, 9] -] +matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]] # ====== Driver Code ====== slt = Solution() res = slt.spiralOrder(matrix) diff --git a/sword_for_offer/codes/python/sfo_30_min_stack_s1.py b/sword_for_offer/codes/python/sfo_30_min_stack_s1.py index 8c9b245..9499f0c 100644 --- a/sword_for_offer/codes/python/sfo_30_min_stack_s1.py +++ b/sword_for_offer/codes/python/sfo_30_min_stack_s1.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_30_min_stack_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class MinStack: def __init__(self): @@ -26,6 +27,7 @@ def top(self) -> int: def min(self) -> int: return self.B[-1] + # ====== Driver Code ====== min_stack = MinStack() res = [ @@ -35,6 +37,6 @@ def min(self) -> int: min_stack.min(), min_stack.pop(), min_stack.top(), - min_stack.min() + min_stack.min(), ] print(res) diff --git a/sword_for_offer/codes/python/sfo_31_validate_stack_sequences_s1.py b/sword_for_offer/codes/python/sfo_31_validate_stack_sequences_s1.py index f89697a..5041242 100644 --- a/sword_for_offer/codes/python/sfo_31_validate_stack_sequences_s1.py +++ b/sword_for_offer/codes/python/sfo_31_validate_stack_sequences_s1.py @@ -1,22 +1,24 @@ -''' +""" File: sfo_31_validate_stack_sequences_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def validateStackSequences(self, pushed: List[int], popped: List[int]) -> bool: stack, i = [], 0 for num in pushed: - stack.append(num) # num 入栈 - while stack and stack[-1] == popped[i]: # 循环判断与出栈 + stack.append(num) # num 入栈 + while stack and stack[-1] == popped[i]: # 循环判断与出栈 stack.pop() i += 1 return not stack + # ======= Test Case ======= pushed = [1, 2, 3, 4, 5] popped = [4, 5, 3, 2, 1] diff --git a/sword_for_offer/codes/python/sfo_32i_print_a_binary_tree_topbottom_i_s1.py b/sword_for_offer/codes/python/sfo_32i_print_a_binary_tree_topbottom_i_s1.py index 8e38e0b..9c7204e 100644 --- a/sword_for_offer/codes/python/sfo_32i_print_a_binary_tree_topbottom_i_s1.py +++ b/sword_for_offer/codes/python/sfo_32i_print_a_binary_tree_topbottom_i_s1.py @@ -1,24 +1,29 @@ -''' +""" File: sfo_32-i_print_a_binary_tree_top-bottom_i_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def levelOrder(self, root: TreeNode) -> List[int]: - if not root: return [] + if not root: + return [] res, queue = [], collections.deque() queue.append(root) while queue: node = queue.popleft() res.append(node.val) - if node.left: queue.append(node.left) - if node.right: queue.append(node.right) + if node.left: + queue.append(node.left) + if node.right: + queue.append(node.right) return res + # ======= Test Case ======= root = list_to_tree([3, 9, 20, None, None, 15, 7, None, None, None, None]) # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_32ii_print_a_binary_tree_topbottom_ii_s1.py b/sword_for_offer/codes/python/sfo_32ii_print_a_binary_tree_topbottom_ii_s1.py index b509990..d2fa243 100644 --- a/sword_for_offer/codes/python/sfo_32ii_print_a_binary_tree_topbottom_ii_s1.py +++ b/sword_for_offer/codes/python/sfo_32ii_print_a_binary_tree_topbottom_ii_s1.py @@ -1,15 +1,17 @@ -''' +""" File: sfo_32-ii_print_a_binary_tree_top-bottom_ii_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def levelOrder(self, root: TreeNode) -> List[List[int]]: - if not root: return [] + if not root: + return [] res, queue = [], collections.deque() queue.append(root) while queue: @@ -17,11 +19,14 @@ def levelOrder(self, root: TreeNode) -> List[List[int]]: for _ in range(len(queue)): node = queue.popleft() tmp.append(node.val) - if node.left: queue.append(node.left) - if node.right: queue.append(node.right) + if node.left: + queue.append(node.left) + if node.right: + queue.append(node.right) res.append(tmp) return res + # ======= Test Case ======= root = list_to_tree([3, 9, 20, None, None, 15, 7, None, None, None, None]) # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_32iii_print_a_binary_tree_topbottom_iii_s1.py b/sword_for_offer/codes/python/sfo_32iii_print_a_binary_tree_topbottom_iii_s1.py index 6e07ea1..3088639 100644 --- a/sword_for_offer/codes/python/sfo_32iii_print_a_binary_tree_topbottom_iii_s1.py +++ b/sword_for_offer/codes/python/sfo_32iii_print_a_binary_tree_topbottom_iii_s1.py @@ -1,27 +1,34 @@ -''' +""" File: sfo_32-iii_print_a_binary_tree_top-bottom_iii_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def levelOrder(self, root: TreeNode) -> List[List[int]]: - if not root: return [] + if not root: + return [] res, deque = [], collections.deque([root]) while deque: tmp = collections.deque() for _ in range(len(deque)): node = deque.popleft() - if len(res) % 2 == 0: tmp.append(node.val) # 奇数层 -> 插入队列尾部 - else: tmp.appendleft(node.val) # 偶数层 -> 插入队列头部 - if node.left: deque.append(node.left) - if node.right: deque.append(node.right) + if len(res) % 2 == 0: + tmp.append(node.val) # 奇数层 -> 插入队列尾部 + else: + tmp.appendleft(node.val) # 偶数层 -> 插入队列头部 + if node.left: + deque.append(node.left) + if node.right: + deque.append(node.right) res.append(list(tmp)) return res + # ======= Test Case ======= root = list_to_tree([3, 9, 20, None, None, 15, 7, None, None, None, None]) # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_32iii_print_a_binary_tree_topbottom_iii_s2.py b/sword_for_offer/codes/python/sfo_32iii_print_a_binary_tree_topbottom_iii_s2.py index 5d10b84..79b934e 100644 --- a/sword_for_offer/codes/python/sfo_32iii_print_a_binary_tree_topbottom_iii_s2.py +++ b/sword_for_offer/codes/python/sfo_32iii_print_a_binary_tree_topbottom_iii_s2.py @@ -1,15 +1,17 @@ -''' +""" File: sfo_32-iii_print_a_binary_tree_top-bottom_iii_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def levelOrder(self, root: TreeNode) -> List[List[int]]: - if not root: return [] + if not root: + return [] res, deque = [], collections.deque() deque.append(root) while deque: @@ -20,10 +22,13 @@ def levelOrder(self, root: TreeNode) -> List[List[int]]: node = deque.popleft() tmp.append(node.val) # 先左后右加入下层节点 - if node.left: deque.append(node.left) - if node.right: deque.append(node.right) + if node.left: + deque.append(node.left) + if node.right: + deque.append(node.right) res.append(tmp) - if not deque: break # 若为空则提前跳出 + if not deque: + break # 若为空则提前跳出 # 打印偶数层 tmp = [] for _ in range(len(deque)): @@ -31,11 +36,14 @@ def levelOrder(self, root: TreeNode) -> List[List[int]]: node = deque.pop() tmp.append(node.val) # 先右后左加入下层节点 - if node.right: deque.appendleft(node.right) - if node.left: deque.appendleft(node.left) + if node.right: + deque.appendleft(node.right) + if node.left: + deque.appendleft(node.left) res.append(tmp) return res + # ======= Test Case ======= root = list_to_tree([3, 9, 20, None, None, 15, 7, None, None, None, None]) # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_32iii_print_a_binary_tree_topbottom_iii_s3.py b/sword_for_offer/codes/python/sfo_32iii_print_a_binary_tree_topbottom_iii_s3.py index 93a0efc..45c495d 100644 --- a/sword_for_offer/codes/python/sfo_32iii_print_a_binary_tree_topbottom_iii_s3.py +++ b/sword_for_offer/codes/python/sfo_32iii_print_a_binary_tree_topbottom_iii_s3.py @@ -1,15 +1,17 @@ -''' +""" File: sfo_32-iii_print_a_binary_tree_top-bottom_iii_s3.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def levelOrder(self, root: TreeNode) -> List[List[int]]: - if not root: return [] + if not root: + return [] res, queue = [], collections.deque() queue.append(root) while queue: @@ -17,11 +19,14 @@ def levelOrder(self, root: TreeNode) -> List[List[int]]: for _ in range(len(queue)): node = queue.popleft() tmp.append(node.val) - if node.left: queue.append(node.left) - if node.right: queue.append(node.right) + if node.left: + queue.append(node.left) + if node.right: + queue.append(node.right) res.append(tmp[::-1] if len(res) % 2 else tmp) return res + # ======= Test Case ======= root = list_to_tree([3, 9, 20, None, None, 15, 7, None, None, None, None]) # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_33_postorder_traversal_of_a_binary_search_tree_s1.py b/sword_for_offer/codes/python/sfo_33_postorder_traversal_of_a_binary_search_tree_s1.py index 57befdf..62a3d28 100644 --- a/sword_for_offer/codes/python/sfo_33_postorder_traversal_of_a_binary_search_tree_s1.py +++ b/sword_for_offer/codes/python/sfo_33_postorder_traversal_of_a_binary_search_tree_s1.py @@ -1,24 +1,29 @@ -''' +""" File: sfo_33_post-order_traversal_of_a_binary_search_tree_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def verifyPostorder(self, postorder: List[int]) -> bool: def recur(i, j): - if i >= j: return True + if i >= j: + return True p = i - while postorder[p] < postorder[j]: p += 1 + while postorder[p] < postorder[j]: + p += 1 m = p - while postorder[p] > postorder[j]: p += 1 + while postorder[p] > postorder[j]: + p += 1 return p == j and recur(i, m - 1) and recur(m, j - 1) return recur(0, len(postorder) - 1) + # ======= Test Case ======= postorder = [1, 6, 3, 2, 5] # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_33_postorder_traversal_of_a_binary_search_tree_s2.py b/sword_for_offer/codes/python/sfo_33_postorder_traversal_of_a_binary_search_tree_s2.py index 3adac64..893745f 100644 --- a/sword_for_offer/codes/python/sfo_33_postorder_traversal_of_a_binary_search_tree_s2.py +++ b/sword_for_offer/codes/python/sfo_33_postorder_traversal_of_a_binary_search_tree_s2.py @@ -1,22 +1,25 @@ -''' +""" File: sfo_33_post-order_traversal_of_a_binary_search_tree_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def verifyPostorder(self, postorder: List[int]) -> bool: stack, root = [], float("+inf") for i in range(len(postorder) - 1, -1, -1): - if postorder[i] > root: return False - while(stack and postorder[i] < stack[-1]): + if postorder[i] > root: + return False + while stack and postorder[i] < stack[-1]: root = stack.pop() stack.append(postorder[i]) return True + # ======= Test Case ======= postorder = [1, 6, 3, 2, 5] # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_34_all_xsum_paths_in_a_binary_tree_s1.py b/sword_for_offer/codes/python/sfo_34_all_xsum_paths_in_a_binary_tree_s1.py index 226d3b2..f0ac844 100644 --- a/sword_for_offer/codes/python/sfo_34_all_xsum_paths_in_a_binary_tree_s1.py +++ b/sword_for_offer/codes/python/sfo_34_all_xsum_paths_in_a_binary_tree_s1.py @@ -1,17 +1,20 @@ -''' +""" File: sfo_34_all_x-sum_paths_in_a_binary_tree_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]: res, path = [], [] + def recur(root, tar): - if not root: return + if not root: + return path.append(root.val) tar -= root.val if tar == 0 and not root.left and not root.right: @@ -23,8 +26,33 @@ def recur(root, tar): recur(root, sum) return res + # ======= Test Case ======= -root = list_to_tree([5, 4, 8, 11, None, 13, 4, 7, 2, None, None, 5, 1, None, None, None, None, None, None, None, None]) +root = list_to_tree( + [ + 5, + 4, + 8, + 11, + None, + 13, + 4, + 7, + 2, + None, + None, + 5, + 1, + None, + None, + None, + None, + None, + None, + None, + None, + ] +) sum = 22 # ====== Driver Code ====== slt = Solution() diff --git a/sword_for_offer/codes/python/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1.py b/sword_for_offer/codes/python/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1.py index 92b1271..81c74f2 100644 --- a/sword_for_offer/codes/python/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1.py +++ b/sword_for_offer/codes/python/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1.py @@ -1,22 +1,25 @@ -''' +""" File: sfo_35_clone_a_linked_list_with_next_and_random_pointer_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # Definition for a Node. class Node: - def __init__(self, x: int, next: 'Node' = None, random: 'Node' = None): + def __init__(self, x: int, next: "Node" = None, random: "Node" = None): self.val = int(x) self.next = next self.random = random + # ===== Solution Code ===== class Solution: - def copyRandomList(self, head: 'Node') -> 'Node': - if not head: return + def copyRandomList(self, head: "Node") -> "Node": + if not head: + return dic = {} # 3. 复制各节点，并建立 “原节点 -> 新节点” 的 Map 映射 cur = head @@ -32,10 +35,11 @@ def copyRandomList(self, head: 'Node') -> 'Node': # 5. 返回新链表的头节点 return dic[head] + # ======= Test Case ======= test_case = [[7, None], [13, 0], [11, 4], [10, 2], [1, 0]] # Construct nodes -node_list = [Node(val) for val,_ in test_case] +node_list = [Node(val) for val, _ in test_case] # Build next reference for i in range(len(test_case) - 1): node_list[i].next = node_list[i + 1] @@ -52,5 +56,10 @@ def copyRandomList(self, head: 'Node') -> 'Node': node_list_new = [] while res: node_list_new.append(res) - res = res.next -print([[node.val, node_list_new.index(node.random) if node.random else None] for node in node_list_new]) + res = res.next +print( + [ + [node.val, node_list_new.index(node.random) if node.random else None] + for node in node_list_new + ] +) diff --git a/sword_for_offer/codes/python/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2.py b/sword_for_offer/codes/python/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2.py index e934774..59940c4 100644 --- a/sword_for_offer/codes/python/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2.py +++ b/sword_for_offer/codes/python/sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2.py @@ -1,22 +1,25 @@ -''' +""" File: sfo_35_clone_a_linked_list_with_next_and_random_pointer_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # Definition for a Node. class Node: - def __init__(self, x: int, next: 'Node' = None, random: 'Node' = None): + def __init__(self, x: int, next: "Node" = None, random: "Node" = None): self.val = int(x) self.next = next self.random = random + # ===== Solution Code ===== class Solution: - def copyRandomList(self, head: 'Node') -> 'Node': - if not head: return + def copyRandomList(self, head: "Node") -> "Node": + if not head: + return cur = head # 1. 复制各节点，并构建拼接链表 while cur: @@ -38,13 +41,14 @@ def copyRandomList(self, head: 'Node') -> 'Node': cur.next = cur.next.next pre = pre.next cur = cur.next - pre.next = None # 单独处理原链表尾节点 - return res # 返回新链表头节点 + pre.next = None # 单独处理原链表尾节点 + return res # 返回新链表头节点 + # ======= Test Case ======= test_case = [[7, None], [13, 0], [11, 4], [10, 2], [1, 0]] # Construct nodes -node_list = [Node(val) for val,_ in test_case] +node_list = [Node(val) for val, _ in test_case] # Build next reference for i in range(len(test_case) - 1): node_list[i].next = node_list[i + 1] @@ -61,5 +65,10 @@ def copyRandomList(self, head: 'Node') -> 'Node': node_list_new = [] while res: node_list_new.append(res) - res = res.next -print([[node.val, node_list_new.index(node.random) if node.random else None] for node in node_list_new]) + res = res.next +print( + [ + [node.val, node_list_new.index(node.random) if node.random else None] + for node in node_list_new + ] +) diff --git a/sword_for_offer/codes/python/sfo_36_binary_search_tree_and_doubly_linked_list_s1.py b/sword_for_offer/codes/python/sfo_36_binary_search_tree_and_doubly_linked_list_s1.py index 79af898..6a190c9 100644 --- a/sword_for_offer/codes/python/sfo_36_binary_search_tree_and_doubly_linked_list_s1.py +++ b/sword_for_offer/codes/python/sfo_36_binary_search_tree_and_doubly_linked_list_s1.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_36_binary_search_tree_and_doubly_linked_list_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # Definition for a Node. class Node: def __init__(self, val, left=None, right=None): @@ -13,25 +14,29 @@ def __init__(self, val, left=None, right=None): self.left = left self.right = right + # ===== Solution Code ===== class Solution: - def treeToDoublyList(self, root: 'Node') -> 'Node': + def treeToDoublyList(self, root: "Node") -> "Node": def dfs(cur): - if not cur: return - dfs(cur.left) # 递归左子树 - if self.pre: # 修改节点引用 + if not cur: + return + dfs(cur.left) # 递归左子树 + if self.pre: # 修改节点引用 self.pre.right, cur.left = cur, self.pre - else: # 记录头节点 + else: # 记录头节点 self.head = cur - self.pre = cur # 保存 cur - dfs(cur.right) # 递归右子树 - - if not root: return + self.pre = cur # 保存 cur + dfs(cur.right) # 递归右子树 + + if not root: + return self.pre = None dfs(root) self.head.left, self.pre.right = self.pre, self.head return self.head + # ======= Test Case ======= node_list = [Node(i) for i in range(1, 6)] node_list[3].left = node_list[1] @@ -47,4 +52,4 @@ def dfs(cur): for _ in range(len(node_list) + 1): nodes_val.append(str(res.val)) res = res.right -print(' <-> '.join(nodes_val)) +print(" <-> ".join(nodes_val)) diff --git a/sword_for_offer/codes/python/sfo_37_serialize_and_deserialize_a_binary_tree_s1.py b/sword_for_offer/codes/python/sfo_37_serialize_and_deserialize_a_binary_tree_s1.py index e7f4ce0..fe74f93 100644 --- a/sword_for_offer/codes/python/sfo_37_serialize_and_deserialize_a_binary_tree_s1.py +++ b/sword_for_offer/codes/python/sfo_37_serialize_and_deserialize_a_binary_tree_s1.py @@ -1,15 +1,17 @@ -''' +""" File: sfo_37_serialize_and_deserialize_a_binary_tree_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Codec: def serialize(self, root: TreeNode) -> str: - if not root: return "[]" + if not root: + return "[]" queue = collections.deque() queue.append(root) res = [] @@ -19,12 +21,14 @@ def serialize(self, root: TreeNode) -> str: res.append(str(node.val)) queue.append(node.left) queue.append(node.right) - else: res.append("null") - return '[' + ','.join(res) + ']' + else: + res.append("null") + return "[" + ",".join(res) + "]" def deserialize(self, data: str) -> TreeNode: - if data == "[]": return - vals, i = data[1:-1].split(','), 1 + if data == "[]": + return + vals, i = data[1:-1].split(","), 1 root = TreeNode(int(vals[0])) queue = collections.deque() queue.append(root) @@ -40,6 +44,7 @@ def deserialize(self, data: str) -> TreeNode: i += 1 return root + # ======= Test Case ======= data = "[1,2,3,null,null,4,5,null,null,null,null]" # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_38_all_permutations_of_a_string_s1.py b/sword_for_offer/codes/python/sfo_38_all_permutations_of_a_string_s1.py index 4058033..83ecc34 100644 --- a/sword_for_offer/codes/python/sfo_38_all_permutations_of_a_string_s1.py +++ b/sword_for_offer/codes/python/sfo_38_all_permutations_of_a_string_s1.py @@ -1,29 +1,34 @@ -''' +""" File: sfo_38_all_permutations_of_a_string_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def permutation(self, s: str) -> List[str]: c, res = list(s), [] + def dfs(x): if x == len(c) - 1: - res.append(''.join(c)) # 添加排列方案 + res.append("".join(c)) # 添加排列方案 return dic = set() for i in range(x, len(c)): - if c[i] in dic: continue # 重复，因此剪枝 + if c[i] in dic: + continue # 重复，因此剪枝 dic.add(c[i]) c[i], c[x] = c[x], c[i] # 交换，将 c[i] 固定在第 x 位 - dfs(x + 1) # 开启固定第 x + 1 位字符 + dfs(x + 1) # 开启固定第 x + 1 位字符 c[i], c[x] = c[x], c[i] # 恢复交换 + dfs(0) return res + # ======= Test Case ======= s = "abc" # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_39_the_majority_element_in_an_array_s1.py b/sword_for_offer/codes/python/sfo_39_the_majority_element_in_an_array_s1.py index f58152c..c18e823 100644 --- a/sword_for_offer/codes/python/sfo_39_the_majority_element_in_an_array_s1.py +++ b/sword_for_offer/codes/python/sfo_39_the_majority_element_in_an_array_s1.py @@ -1,20 +1,23 @@ -''' +""" File: sfo_39_the_majority_element_in_an_array_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def majorityElement(self, nums: List[int]) -> int: votes = 0 for num in nums: - if votes == 0: x = num + if votes == 0: + x = num votes += 1 if num == x else -1 return x + # ======= Test Case ======= nums = [1, 2, 3, 2, 2, 2, 5, 4, 2] # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_39_the_majority_element_in_an_array_s2.py b/sword_for_offer/codes/python/sfo_39_the_majority_element_in_an_array_s2.py index a248aa4..8c7f33b 100644 --- a/sword_for_offer/codes/python/sfo_39_the_majority_element_in_an_array_s2.py +++ b/sword_for_offer/codes/python/sfo_39_the_majority_element_in_an_array_s2.py @@ -1,22 +1,26 @@ -''' +""" File: sfo_39_the_majority_element_in_an_array_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def majorityElement(self, nums: List[int]) -> int: votes, count = 0, 0 for num in nums: - if votes == 0: x = num + if votes == 0: + x = num votes += 1 if num == x else -1 # 验证 x 是否为众数 for num in nums: - if num == x: count += 1 - return x if count > len(nums) // 2 else 0 # 当无众数时返回 0 + if num == x: + count += 1 + return x if count > len(nums) // 2 else 0 # 当无众数时返回 0 + # ======= Test Case ======= nums = [1, 2, 3, 2, 2, 2, 5, 4, 2] diff --git a/sword_for_offer/codes/python/sfo_40_the_smallest_k_numbers_s1.py b/sword_for_offer/codes/python/sfo_40_the_smallest_k_numbers_s1.py index b094794..3473ccf 100644 --- a/sword_for_offer/codes/python/sfo_40_the_smallest_k_numbers_s1.py +++ b/sword_for_offer/codes/python/sfo_40_the_smallest_k_numbers_s1.py @@ -1,31 +1,36 @@ -''' +""" File: sfo_40_the_smallest_k_numbers_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def getLeastNumbers(self, arr: List[int], k: int) -> List[int]: def quick_sort(arr, l, r): # 子数组长度为 1 时终止递归 - if l >= r: return + if l >= r: + return # 哨兵划分操作（以 arr[l] 作为基准数） i, j = l, r while i < j: - while i < j and arr[j] >= arr[l]: j -= 1 - while i < j and arr[i] <= arr[l]: i += 1 + while i < j and arr[j] >= arr[l]: + j -= 1 + while i < j and arr[i] <= arr[l]: + i += 1 arr[i], arr[j] = arr[j], arr[i] arr[l], arr[i] = arr[i], arr[l] # 递归左（右）子数组执行哨兵划分 quick_sort(arr, l, i - 1) quick_sort(arr, i + 1, r) - + quick_sort(arr, 0, len(arr) - 1) return arr[:k] + # ======= Test Case ======= arr = [3, 2, 1] k = 2 diff --git a/sword_for_offer/codes/python/sfo_40_the_smallest_k_numbers_s2.py b/sword_for_offer/codes/python/sfo_40_the_smallest_k_numbers_s2.py index 7c2eb5d..831c8e5 100644 --- a/sword_for_offer/codes/python/sfo_40_the_smallest_k_numbers_s2.py +++ b/sword_for_offer/codes/python/sfo_40_the_smallest_k_numbers_s2.py @@ -1,28 +1,36 @@ -''' +""" File: sfo_40_the_smallest_k_numbers_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def getLeastNumbers(self, arr: List[int], k: int) -> List[int]: - if k >= len(arr): return arr + if k >= len(arr): + return arr + def quick_sort(l, r): i, j = l, r while i < j: - while i < j and arr[j] >= arr[l]: j -= 1 - while i < j and arr[i] <= arr[l]: i += 1 + while i < j and arr[j] >= arr[l]: + j -= 1 + while i < j and arr[i] <= arr[l]: + i += 1 arr[i], arr[j] = arr[j], arr[i] arr[l], arr[i] = arr[i], arr[l] - if k < i: return quick_sort(l, i - 1) - if k > i: return quick_sort(i + 1, r) + if k < i: + return quick_sort(l, i - 1) + if k > i: + return quick_sort(i + 1, r) return arr[:k] - + return quick_sort(0, len(arr) - 1) + # ======= Test Case ======= arr = [3, 2, 1] k = 2 diff --git a/sword_for_offer/codes/python/sfo_41_find_median_from_data_stream_s1.py b/sword_for_offer/codes/python/sfo_41_find_median_from_data_stream_s1.py index 8d664dd..10c92c6 100644 --- a/sword_for_offer/codes/python/sfo_41_find_median_from_data_stream_s1.py +++ b/sword_for_offer/codes/python/sfo_41_find_median_from_data_stream_s1.py @@ -1,18 +1,19 @@ -''' +""" File: sfo_41_find_median_from_data_stream_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * # ===== Solution Code ===== from heapq import * + class MedianFinder: def __init__(self): - self.A = [] # 小顶堆，保存较大的一半 - self.B = [] # 大顶堆，保存较小的一半 + self.A = [] # 小顶堆，保存较大的一半 + self.B = [] # 大顶堆，保存较小的一半 def addNum(self, num: int) -> None: if len(self.A) != len(self.B): @@ -23,7 +24,10 @@ def addNum(self, num: int) -> None: heappush(self.A, -heappop(self.B)) def findMedian(self) -> float: - return self.A[0] if len(self.A) != len(self.B) else (self.A[0] - self.B[0]) / 2.0 + return ( + self.A[0] if len(self.A) != len(self.B) else (self.A[0] - self.B[0]) / 2.0 + ) + # ====== Driver Code ====== median_finder = MedianFinder() @@ -32,6 +36,6 @@ def findMedian(self) -> float: median_finder.addNum(2), median_finder.findMedian(), median_finder.addNum(3), - median_finder.findMedian() + median_finder.findMedian(), ] print(res) diff --git a/sword_for_offer/codes/python/sfo_41_find_median_from_data_stream_s2.py b/sword_for_offer/codes/python/sfo_41_find_median_from_data_stream_s2.py index c9faa94..bcc7649 100644 --- a/sword_for_offer/codes/python/sfo_41_find_median_from_data_stream_s2.py +++ b/sword_for_offer/codes/python/sfo_41_find_median_from_data_stream_s2.py @@ -1,18 +1,19 @@ -''' +""" File: sfo_41_find_median_from_data_stream_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * # ===== Solution Code ===== from heapq import * + class MedianFinder: def __init__(self): - self.A = [] # 小顶堆，保存较大的一半 - self.B = [] # 大顶堆，保存较小的一半 + self.A = [] # 小顶堆，保存较大的一半 + self.B = [] # 大顶堆，保存较小的一半 def addNum(self, num: int) -> None: if len(self.A) != len(self.B): @@ -21,7 +22,10 @@ def addNum(self, num: int) -> None: heappush(self.A, -heappushpop(self.B, -num)) def findMedian(self) -> float: - return self.A[0] if len(self.A) != len(self.B) else (self.A[0] - self.B[0]) / 2.0 + return ( + self.A[0] if len(self.A) != len(self.B) else (self.A[0] - self.B[0]) / 2.0 + ) + # ====== Driver Code ====== median_finder = MedianFinder() @@ -30,6 +34,6 @@ def findMedian(self) -> float: median_finder.addNum(2), median_finder.findMedian(), median_finder.addNum(3), - median_finder.findMedian() + median_finder.findMedian(), ] print(res) diff --git a/sword_for_offer/codes/python/sfo_42_largest_sum_contiguous_subarray_s1.py b/sword_for_offer/codes/python/sfo_42_largest_sum_contiguous_subarray_s1.py index f0a9843..b804e97 100644 --- a/sword_for_offer/codes/python/sfo_42_largest_sum_contiguous_subarray_s1.py +++ b/sword_for_offer/codes/python/sfo_42_largest_sum_contiguous_subarray_s1.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_42_largest_sum_contiguous_subarray_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def maxSubArray(self, nums: List[int]) -> int: @@ -13,6 +14,7 @@ def maxSubArray(self, nums: List[int]) -> int: nums[i] += max(nums[i - 1], 0) return max(nums) + # ======= Test Case ======= nums = [-2, 1, -3, 4, -1, 2, 1, -5, 4] # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1.py b/sword_for_offer/codes/python/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1.py index df8f0ce..f68514b 100644 --- a/sword_for_offer/codes/python/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1.py +++ b/sword_for_offer/codes/python/sfo_43_total_number_of_1_in_integers_from_1_to_n_s1.py @@ -1,26 +1,31 @@ -''' +""" File: sfo_43_total_number_of_1_in_integers_from_1_to_n_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def countDigitOne(self, n: int) -> int: digit, res = 1, 0 high, cur, low = n // 10, n % 10, 0 while high != 0 or cur != 0: - if cur == 0: res += high * digit - elif cur == 1: res += high * digit + low + 1 - else: res += (high + 1) * digit + if cur == 0: + res += high * digit + elif cur == 1: + res += high * digit + low + 1 + else: + res += (high + 1) * digit low += cur * digit cur = high % 10 high //= 10 digit *= 10 return res + # ======= Test Case ======= n = 12 # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_44_nth_digit_s1.py b/sword_for_offer/codes/python/sfo_44_nth_digit_s1.py index c754d4a..52434a8 100644 --- a/sword_for_offer/codes/python/sfo_44_nth_digit_s1.py +++ b/sword_for_offer/codes/python/sfo_44_nth_digit_s1.py @@ -1,22 +1,24 @@ -''' +""" File: sfo_44_nth_digit_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def findNthDigit(self, n: int) -> int: digit, start, count = 1, 1, 9 - while n > count: # 1. + while n > count: # 1. n -= count start *= 10 digit += 1 count = 9 * start * digit - num = start + (n - 1) // digit # 2. - return int(str(num)[(n - 1) % digit]) # 3. + num = start + (n - 1) // digit # 2. + return int(str(num)[(n - 1) % digit]) # 3. + # ======= Test Case ======= n = 3 diff --git a/sword_for_offer/codes/python/sfo_45_arrange_an_array_into_the_smallest_number_s1.py b/sword_for_offer/codes/python/sfo_45_arrange_an_array_into_the_smallest_number_s1.py index b8cc155..aa7ee1a 100644 --- a/sword_for_offer/codes/python/sfo_45_arrange_an_array_into_the_smallest_number_s1.py +++ b/sword_for_offer/codes/python/sfo_45_arrange_an_array_into_the_smallest_number_s1.py @@ -1,28 +1,33 @@ -''' +""" File: sfo_45_arrange_an_array_into_the_smallest_number_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def minNumber(self, nums: List[int]) -> str: - def quick_sort(l , r): - if l >= r: return + def quick_sort(l, r): + if l >= r: + return i, j = l, r while i < j: - while strs[j] + strs[l] >= strs[l] + strs[j] and i < j: j -= 1 - while strs[i] + strs[l] <= strs[l] + strs[i] and i < j: i += 1 + while strs[j] + strs[l] >= strs[l] + strs[j] and i < j: + j -= 1 + while strs[i] + strs[l] <= strs[l] + strs[i] and i < j: + i += 1 strs[i], strs[j] = strs[j], strs[i] strs[i], strs[l] = strs[l], strs[i] quick_sort(l, i - 1) quick_sort(i + 1, r) - + strs = [str(num) for num in nums] quick_sort(0, len(strs) - 1) - return ''.join(strs) + return "".join(strs) + # ======= Test Case ======= nums = [3, 30, 34, 5, 9] diff --git a/sword_for_offer/codes/python/sfo_45_arrange_an_array_into_the_smallest_number_s2.py b/sword_for_offer/codes/python/sfo_45_arrange_an_array_into_the_smallest_number_s2.py index 2fb7fa0..b6fa2d4 100644 --- a/sword_for_offer/codes/python/sfo_45_arrange_an_array_into_the_smallest_number_s2.py +++ b/sword_for_offer/codes/python/sfo_45_arrange_an_array_into_the_smallest_number_s2.py @@ -1,23 +1,28 @@ -''' +""" File: sfo_45_arrange_an_array_into_the_smallest_number_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def minNumber(self, nums: List[int]) -> str: def sort_rule(x, y): a, b = x + y, y + x - if a > b: return 1 - elif a < b: return -1 - else: return 0 - + if a > b: + return 1 + elif a < b: + return -1 + else: + return 0 + strs = [str(num) for num in nums] - strs.sort(key = functools.cmp_to_key(sort_rule)) - return ''.join(strs) + strs.sort(key=functools.cmp_to_key(sort_rule)) + return "".join(strs) + # ======= Test Case ======= nums = [3, 30, 34, 5, 9] diff --git a/sword_for_offer/codes/python/sfo_46_translate_numbers_into_strings_s1.py b/sword_for_offer/codes/python/sfo_46_translate_numbers_into_strings_s1.py index 82a58cb..aade548 100644 --- a/sword_for_offer/codes/python/sfo_46_translate_numbers_into_strings_s1.py +++ b/sword_for_offer/codes/python/sfo_46_translate_numbers_into_strings_s1.py @@ -1,20 +1,22 @@ -''' +""" File: sfo_46_translate_numbers_into_strings_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def translateNum(self, num: int) -> int: s = str(num) a = b = 1 for i in range(2, len(s) + 1): - a, b = (a + b if "10" <= s[i - 2:i] <= "25" else a), a + a, b = (a + b if "10" <= s[i - 2 : i] <= "25" else a), a return a + # ======= Test Case ======= num = 12258 # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_46_translate_numbers_into_strings_s2.py b/sword_for_offer/codes/python/sfo_46_translate_numbers_into_strings_s2.py index 5602423..f3af450 100644 --- a/sword_for_offer/codes/python/sfo_46_translate_numbers_into_strings_s2.py +++ b/sword_for_offer/codes/python/sfo_46_translate_numbers_into_strings_s2.py @@ -1,23 +1,25 @@ -''' +""" File: sfo_46_translate_numbers_into_strings_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def translateNum(self, num: int) -> int: s = str(num) a = b = 1 for i in range(2, len(s) + 1): - tmp = s[i - 2:i] + tmp = s[i - 2 : i] c = a + b if "10" <= tmp <= "25" else a b = a a = c return a + # ======= Test Case ======= num = 12258 # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_46_translate_numbers_into_strings_s3.py b/sword_for_offer/codes/python/sfo_46_translate_numbers_into_strings_s3.py index 5418a75..7e3641a 100644 --- a/sword_for_offer/codes/python/sfo_46_translate_numbers_into_strings_s3.py +++ b/sword_for_offer/codes/python/sfo_46_translate_numbers_into_strings_s3.py @@ -1,20 +1,22 @@ -''' +""" File: sfo_46_translate_numbers_into_strings_s3.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def translateNum(self, num: int) -> int: s = str(num) a = b = 1 for i in range(len(s) - 2, -1, -1): - a, b = (a + b if "10" <= s[i:i + 2] <= "25" else a), a + a, b = (a + b if "10" <= s[i : i + 2] <= "25" else a), a return a + # ======= Test Case ======= num = 12258 # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_46_translate_numbers_into_strings_s4.py b/sword_for_offer/codes/python/sfo_46_translate_numbers_into_strings_s4.py index c66cb0f..68fb160 100644 --- a/sword_for_offer/codes/python/sfo_46_translate_numbers_into_strings_s4.py +++ b/sword_for_offer/codes/python/sfo_46_translate_numbers_into_strings_s4.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_46_translate_numbers_into_strings_s4.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def translateNum(self, num: int) -> int: @@ -18,6 +19,7 @@ def translateNum(self, num: int) -> int: y = x return a + # ======= Test Case ======= num = 12258 # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_46_translate_numbers_into_strings_s5.py b/sword_for_offer/codes/python/sfo_46_translate_numbers_into_strings_s5.py index 869b510..b5177f2 100644 --- a/sword_for_offer/codes/python/sfo_46_translate_numbers_into_strings_s5.py +++ b/sword_for_offer/codes/python/sfo_46_translate_numbers_into_strings_s5.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_46_translate_numbers_into_strings_s5.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def translateNum(self, num: int) -> int: @@ -20,6 +21,7 @@ def translateNum(self, num: int) -> int: y = x return a + # ======= Test Case ======= num = 12258 # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_47_the_maximum_value_of_gifts_s1.py b/sword_for_offer/codes/python/sfo_47_the_maximum_value_of_gifts_s1.py index 0b7f514..a97d19f 100644 --- a/sword_for_offer/codes/python/sfo_47_the_maximum_value_of_gifts_s1.py +++ b/sword_for_offer/codes/python/sfo_47_the_maximum_value_of_gifts_s1.py @@ -1,28 +1,30 @@ -''' +""" File: sfo_47_the_maximum_value_of_gifts_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def maxValue(self, grid: List[List[int]]) -> int: for i in range(len(grid)): for j in range(len(grid[0])): - if i == 0 and j == 0: continue - if i == 0: grid[i][j] += grid[i][j - 1] - elif j == 0: grid[i][j] += grid[i - 1][j] - else: grid[i][j] += max(grid[i][j - 1], grid[i - 1][j]) + if i == 0 and j == 0: + continue + if i == 0: + grid[i][j] += grid[i][j - 1] + elif j == 0: + grid[i][j] += grid[i - 1][j] + else: + grid[i][j] += max(grid[i][j - 1], grid[i - 1][j]) return grid[-1][-1] + # ======= Test Case ======= -grid = [ - [1, 3, 1], - [1, 5, 1], - [4, 2, 1] -] +grid = [[1, 3, 1], [1, 5, 1], [4, 2, 1]] # ====== Driver Code ====== slt = Solution() res = slt.maxValue(grid) diff --git a/sword_for_offer/codes/python/sfo_47_the_maximum_value_of_gifts_s2.py b/sword_for_offer/codes/python/sfo_47_the_maximum_value_of_gifts_s2.py index 58c8b85..bc681f0 100644 --- a/sword_for_offer/codes/python/sfo_47_the_maximum_value_of_gifts_s2.py +++ b/sword_for_offer/codes/python/sfo_47_the_maximum_value_of_gifts_s2.py @@ -1,30 +1,28 @@ -''' +""" File: sfo_47_the_maximum_value_of_gifts_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def maxValue(self, grid: List[List[int]]) -> int: m, n = len(grid), len(grid[0]) - for j in range(1, n): # 初始化第一行 + for j in range(1, n): # 初始化第一行 grid[0][j] += grid[0][j - 1] - for i in range(1, m): # 初始化第一列 + for i in range(1, m): # 初始化第一列 grid[i][0] += grid[i - 1][0] for i in range(1, m): for j in range(1, n): grid[i][j] += max(grid[i][j - 1], grid[i - 1][j]) return grid[-1][-1] + # ======= Test Case ======= -grid = [ - [1, 3, 1], - [1, 5, 1], - [4, 2, 1] -] +grid = [[1, 3, 1], [1, 5, 1], [4, 2, 1]] # ====== Driver Code ====== slt = Solution() res = slt.maxValue(grid) diff --git a/sword_for_offer/codes/python/sfo_48_the_longest_substring_without_repeated_characters_s1.py b/sword_for_offer/codes/python/sfo_48_the_longest_substring_without_repeated_characters_s1.py index 5d01d6e..566e1c2 100644 --- a/sword_for_offer/codes/python/sfo_48_the_longest_substring_without_repeated_characters_s1.py +++ b/sword_for_offer/codes/python/sfo_48_the_longest_substring_without_repeated_characters_s1.py @@ -1,23 +1,25 @@ -''' +""" File: sfo_48_the_longest_substring_without_repeated_characters_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def lengthOfLongestSubstring(self, s: str) -> int: dic = {} res = tmp = 0 for j in range(len(s)): - i = dic.get(s[j], -1) # 获取索引 i - dic[s[j]] = j # 更新哈希表 - tmp = tmp + 1 if tmp < j - i else j - i # dp[j - 1] -> dp[j] - res = max(res, tmp) # max(dp[j - 1], dp[j]) + i = dic.get(s[j], -1) # 获取索引 i + dic[s[j]] = j # 更新哈希表 + tmp = tmp + 1 if tmp < j - i else j - i # dp[j - 1] -> dp[j] + res = max(res, tmp) # max(dp[j - 1], dp[j]) return res + # ======= Test Case ======= s = "abcabcbb" # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_48_the_longest_substring_without_repeated_characters_s2.py b/sword_for_offer/codes/python/sfo_48_the_longest_substring_without_repeated_characters_s2.py index 6d75027..cb7117f 100644 --- a/sword_for_offer/codes/python/sfo_48_the_longest_substring_without_repeated_characters_s2.py +++ b/sword_for_offer/codes/python/sfo_48_the_longest_substring_without_repeated_characters_s2.py @@ -1,22 +1,25 @@ -''' +""" File: sfo_48_the_longest_substring_without_repeated_characters_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def lengthOfLongestSubstring(self, s: str) -> int: res = tmp = i = 0 for j in range(len(s)): i = j - 1 - while i >= 0 and s[i] != s[j]: i -= 1 # 线性查找 i - tmp = tmp + 1 if tmp < j - i else j - i # dp[j - 1] -> dp[j] - res = max(res, tmp) # max(dp[j - 1], dp[j]) + while i >= 0 and s[i] != s[j]: + i -= 1 # 线性查找 i + tmp = tmp + 1 if tmp < j - i else j - i # dp[j - 1] -> dp[j] + res = max(res, tmp) # max(dp[j - 1], dp[j]) return res + # ======= Test Case ======= s = "abcabcbb" # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_48_the_longest_substring_without_repeated_characters_s3.py b/sword_for_offer/codes/python/sfo_48_the_longest_substring_without_repeated_characters_s3.py index 8f3f2b2..7619d4f 100644 --- a/sword_for_offer/codes/python/sfo_48_the_longest_substring_without_repeated_characters_s3.py +++ b/sword_for_offer/codes/python/sfo_48_the_longest_substring_without_repeated_characters_s3.py @@ -1,22 +1,24 @@ -''' +""" File: sfo_48_the_longest_substring_without_repeated_characters_s3.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def lengthOfLongestSubstring(self, s: str) -> int: dic, res, i = {}, 0, -1 for j in range(len(s)): if s[j] in dic: - i = max(dic[s[j]], i) # 更新左指针 i - dic[s[j]] = j # 哈希表记录 - res = max(res, j - i) # 更新结果 + i = max(dic[s[j]], i) # 更新左指针 i + dic[s[j]] = j # 哈希表记录 + res = max(res, j - i) # 更新结果 return res + # ======= Test Case ======= s = "abcabcbb" # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_49_ugly_numbers_s1.py b/sword_for_offer/codes/python/sfo_49_ugly_numbers_s1.py index 1b11dca..cba1d81 100644 --- a/sword_for_offer/codes/python/sfo_49_ugly_numbers_s1.py +++ b/sword_for_offer/codes/python/sfo_49_ugly_numbers_s1.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_49_ugly_numbers_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def nthUglyNumber(self, n: int) -> int: @@ -13,11 +14,15 @@ def nthUglyNumber(self, n: int) -> int: for i in range(1, n): n2, n3, n5 = dp[a] * 2, dp[b] * 3, dp[c] * 5 dp[i] = min(n2, n3, n5) - if dp[i] == n2: a += 1 - if dp[i] == n3: b += 1 - if dp[i] == n5: c += 1 + if dp[i] == n2: + a += 1 + if dp[i] == n3: + b += 1 + if dp[i] == n5: + c += 1 return dp[-1] + # ======= Test Case ======= n = 10 # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1.py b/sword_for_offer/codes/python/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1.py index 999e1bd..60ce984 100644 --- a/sword_for_offer/codes/python/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1.py +++ b/sword_for_offer/codes/python/sfo_50_find_the_first_nonrepeating_character_in_a_string_s1.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_50_find_the_first_non-repeating_character_in_a_string_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def firstUniqChar(self, s: str) -> str: @@ -13,8 +14,10 @@ def firstUniqChar(self, s: str) -> str: for c in s: dic[c] = not c in dic for c in s: - if dic[c]: return c - return ' ' + if dic[c]: + return c + return " " + # ======= Test Case ======= s = "abaccdeff" diff --git a/sword_for_offer/codes/python/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2.py b/sword_for_offer/codes/python/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2.py index 781a11d..8490ec6 100644 --- a/sword_for_offer/codes/python/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2.py +++ b/sword_for_offer/codes/python/sfo_50_find_the_first_nonrepeating_character_in_a_string_s2.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_50_find_the_first_non-repeating_character_in_a_string_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def firstUniqChar(self, s: str) -> str: @@ -13,8 +14,10 @@ def firstUniqChar(self, s: str) -> str: for c in s: dic[c] = not c in dic for k, v in dic.items(): - if v: return k - return ' ' + if v: + return k + return " " + # ======= Test Case ======= s = "abaccdeff" diff --git a/sword_for_offer/codes/python/sfo_50_find_the_first_nonrepeating_character_in_a_string_s3.py b/sword_for_offer/codes/python/sfo_50_find_the_first_nonrepeating_character_in_a_string_s3.py index b8f4e55..578f661 100644 --- a/sword_for_offer/codes/python/sfo_50_find_the_first_nonrepeating_character_in_a_string_s3.py +++ b/sword_for_offer/codes/python/sfo_50_find_the_first_nonrepeating_character_in_a_string_s3.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_50_find_the_first_non-repeating_character_in_a_string_s3.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def firstUniqChar(self, s: str) -> str: @@ -13,8 +14,10 @@ def firstUniqChar(self, s: str) -> str: for c in s: dic[c] = not c in dic for k, v in dic.items(): - if v: return k - return ' ' + if v: + return k + return " " + # ======= Test Case ======= s = "abaccdeff" diff --git a/sword_for_offer/codes/python/sfo_51_reversed_pairs_in_an_array_s1.py b/sword_for_offer/codes/python/sfo_51_reversed_pairs_in_an_array_s1.py index 4f041ff..534bdfa 100644 --- a/sword_for_offer/codes/python/sfo_51_reversed_pairs_in_an_array_s1.py +++ b/sword_for_offer/codes/python/sfo_51_reversed_pairs_in_an_array_s1.py @@ -1,23 +1,25 @@ -''' +""" File: sfo_51_reversed_pairs_in_an_array_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def reversePairs(self, nums: List[int]) -> int: def merge_sort(l, r): # 终止条件 - if l >= r: return 0 + if l >= r: + return 0 # 递归划分 m = (l + r) // 2 res = merge_sort(l, m) + merge_sort(m + 1, r) # 合并阶段 i, j = l, m + 1 - tmp[l:r + 1] = nums[l:r + 1] + tmp[l : r + 1] = nums[l : r + 1] for k in range(l, r + 1): if i == m + 1: nums[k] = tmp[j] @@ -28,12 +30,13 @@ def merge_sort(l, r): else: nums[k] = tmp[j] j += 1 - res += m - i + 1 # 统计逆序对 + res += m - i + 1 # 统计逆序对 return res - + tmp = [0] * len(nums) return merge_sort(0, len(nums) - 1) + # ======= Test Case ======= nums = [7, 5, 6, 4] # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_52_the_first_common_node_in_two_linked_lists_s1.py b/sword_for_offer/codes/python/sfo_52_the_first_common_node_in_two_linked_lists_s1.py index acaa700..6d8b14d 100644 --- a/sword_for_offer/codes/python/sfo_52_the_first_common_node_in_two_linked_lists_s1.py +++ b/sword_for_offer/codes/python/sfo_52_the_first_common_node_in_two_linked_lists_s1.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_52_the_first_common_node_in_two_linked_lists_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode: @@ -15,6 +16,7 @@ def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode: B = B.next if B else headA return A + # ======= Test Case ======= # Build two linked lists with intersection of 8 # headA = 4 -> 1 -> 8 -> 4 -> 5 @@ -24,7 +26,7 @@ def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode: headB = list_to_linked_list([5, 0, 1, 8, 4, 5]) intersectA = get_list_node(head=headA, val=8) intersectB = get_list_node(head=headB, val=1) -intersectB.next = intersectA # Concatenate the two lists +intersectB.next = intersectA # Concatenate the two lists # ====== Driver Code ====== slt = Solution() res = slt.getIntersectionNode(headA, headB) diff --git a/sword_for_offer/codes/python/sfo_53i_find_a_number_in_a_sorted_array_s1.py b/sword_for_offer/codes/python/sfo_53i_find_a_number_in_a_sorted_array_s1.py index 89d2c3a..96abc7f 100644 --- a/sword_for_offer/codes/python/sfo_53i_find_a_number_in_a_sorted_array_s1.py +++ b/sword_for_offer/codes/python/sfo_53i_find_a_number_in_a_sorted_array_s1.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_53-i_find_a_number_in_a_sorted_array_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def search(self, nums: List[int], target: int) -> int: @@ -13,20 +14,26 @@ def search(self, nums: List[int], target: int) -> int: i, j = 0, len(nums) - 1 while i <= j: m = (i + j) // 2 - if nums[m] <= target: i = m + 1 - else: j = m - 1 + if nums[m] <= target: + i = m + 1 + else: + j = m - 1 right = i # 若数组中无 target ，则提前返回 - if j >= 0 and nums[j] != target: return 0 + if j >= 0 and nums[j] != target: + return 0 # 搜索左边界 left i = 0 while i <= j: m = (i + j) // 2 - if nums[m] < target: i = m + 1 - else: j = m - 1 + if nums[m] < target: + i = m + 1 + else: + j = m - 1 left = j return right - left - 1 + # ======= Test Case ======= nums = [5, 7, 7, 8, 8, 10] target = 8 diff --git a/sword_for_offer/codes/python/sfo_53i_find_a_number_in_a_sorted_array_s2.py b/sword_for_offer/codes/python/sfo_53i_find_a_number_in_a_sorted_array_s2.py index 6e2630b..9978b4b 100644 --- a/sword_for_offer/codes/python/sfo_53i_find_a_number_in_a_sorted_array_s2.py +++ b/sword_for_offer/codes/python/sfo_53i_find_a_number_in_a_sorted_array_s2.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_53-i_find_a_number_in_a_sorted_array_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def search(self, nums: List[int], target: int) -> int: @@ -13,11 +14,15 @@ def helper(tar): i, j = 0, len(nums) - 1 while i <= j: m = (i + j) // 2 - if nums[m] <= tar: i = m + 1 - else: j = m - 1 + if nums[m] <= tar: + i = m + 1 + else: + j = m - 1 return i + return helper(target) - helper(target - 1) + # ======= Test Case ======= nums = [5, 7, 7, 8, 8, 10] target = 8 diff --git a/sword_for_offer/codes/python/sfo_53ii_the_missing_number_from_0_to_n1_s1.py b/sword_for_offer/codes/python/sfo_53ii_the_missing_number_from_0_to_n1_s1.py index 1caabf3..09eea1b 100644 --- a/sword_for_offer/codes/python/sfo_53ii_the_missing_number_from_0_to_n1_s1.py +++ b/sword_for_offer/codes/python/sfo_53ii_the_missing_number_from_0_to_n1_s1.py @@ -1,21 +1,25 @@ -''' +""" File: sfo_53-ii_the_missing_number_from_0_to_n-1_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def missingNumber(self, nums: List[int]) -> int: i, j = 0, len(nums) - 1 while i <= j: m = (i + j) // 2 - if nums[m] == m: i = m + 1 - else: j = m - 1 + if nums[m] == m: + i = m + 1 + else: + j = m - 1 return i + # ======= Test Case ======= nums = [0, 1, 3] # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1.py b/sword_for_offer/codes/python/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1.py index faf982d..3c3504a 100644 --- a/sword_for_offer/codes/python/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1.py +++ b/sword_for_offer/codes/python/sfo_54_the_kth_largest_node_of_a_binary_search_tree_s1.py @@ -1,26 +1,31 @@ -''' +""" File: sfo_54_the_k-th_largest_node_of_a_binary_search_tree_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def kthLargest(self, root: TreeNode, k: int) -> int: def dfs(root): - if not root: return + if not root: + return dfs(root.right) - if self.k == 0: return + if self.k == 0: + return self.k -= 1 - if self.k == 0: self.res = root.val + if self.k == 0: + self.res = root.val dfs(root.left) self.k = k dfs(root) return self.res + # ======= Test Case ======= root = list_to_tree([3, 1, 4, None, 2, None, None, None, None]) k = 1 diff --git a/sword_for_offer/codes/python/sfo_55i_depth_of_a_binary_tree_s1.py b/sword_for_offer/codes/python/sfo_55i_depth_of_a_binary_tree_s1.py index 936c7bb..de6a2c1 100644 --- a/sword_for_offer/codes/python/sfo_55i_depth_of_a_binary_tree_s1.py +++ b/sword_for_offer/codes/python/sfo_55i_depth_of_a_binary_tree_s1.py @@ -1,17 +1,20 @@ -''' +""" File: sfo_55-i_depth_of_a_binary_tree_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def maxDepth(self, root: TreeNode) -> int: - if not root: return 0 + if not root: + return 0 return max(self.maxDepth(root.left), self.maxDepth(root.right)) + 1 + # ======= Test Case ======= root = list_to_tree([3, 9, 20, None, None, 15, 7, None, None, None, None]) # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_55i_depth_of_a_binary_tree_s2.py b/sword_for_offer/codes/python/sfo_55i_depth_of_a_binary_tree_s2.py index 2d133e8..6853140 100644 --- a/sword_for_offer/codes/python/sfo_55i_depth_of_a_binary_tree_s2.py +++ b/sword_for_offer/codes/python/sfo_55i_depth_of_a_binary_tree_s2.py @@ -1,25 +1,30 @@ -''' +""" File: sfo_55-i_depth_of_a_binary_tree_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def maxDepth(self, root: TreeNode) -> int: - if not root: return 0 + if not root: + return 0 queue, res = [root], 0 while queue: tmp = [] for node in queue: - if node.left: tmp.append(node.left) - if node.right: tmp.append(node.right) + if node.left: + tmp.append(node.left) + if node.right: + tmp.append(node.right) queue = tmp res += 1 return res + # ======= Test Case ======= root = list_to_tree([3, 9, 20, None, None, 15, 7, None, None, None, None]) # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_55ii_balanced_binary_tree_s1.py b/sword_for_offer/codes/python/sfo_55ii_balanced_binary_tree_s1.py index a31efff..6286c3f 100644 --- a/sword_for_offer/codes/python/sfo_55ii_balanced_binary_tree_s1.py +++ b/sword_for_offer/codes/python/sfo_55ii_balanced_binary_tree_s1.py @@ -1,24 +1,29 @@ -''' +""" File: sfo_55-ii_balanced_binary_tree_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def isBalanced(self, root: TreeNode) -> bool: def recur(root): - if not root: return 0 + if not root: + return 0 left = recur(root.left) - if left == -1: return -1 + if left == -1: + return -1 right = recur(root.right) - if right == -1: return -1 + if right == -1: + return -1 return max(left, right) + 1 if abs(left - right) <= 1 else -1 return recur(root) != -1 + # ======= Test Case ======= root = list_to_tree([3, 9, 20, None, None, 15, 7, None, None, None, None]) # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_55ii_balanced_binary_tree_s2.py b/sword_for_offer/codes/python/sfo_55ii_balanced_binary_tree_s2.py index c95d67d..ffdc19d 100644 --- a/sword_for_offer/codes/python/sfo_55ii_balanced_binary_tree_s2.py +++ b/sword_for_offer/codes/python/sfo_55ii_balanced_binary_tree_s2.py @@ -1,22 +1,29 @@ -''' +""" File: sfo_55-ii_balanced_binary_tree_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def isBalanced(self, root: TreeNode) -> bool: - if not root: return True - return abs(self.depth(root.left) - self.depth(root.right)) <= 1 and \ - self.isBalanced(root.left) and self.isBalanced(root.right) + if not root: + return True + return ( + abs(self.depth(root.left) - self.depth(root.right)) <= 1 + and self.isBalanced(root.left) + and self.isBalanced(root.right) + ) def depth(self, root): - if not root: return 0 + if not root: + return 0 return max(self.depth(root.left), self.depth(root.right)) + 1 + # ======= Test Case ======= root = list_to_tree([3, 9, 20, None, None, 15, 7, None, None, None, None]) # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_56i_single_number_i_s1.py b/sword_for_offer/codes/python/sfo_56i_single_number_i_s1.py index e7c017f..192b3f1 100644 --- a/sword_for_offer/codes/python/sfo_56i_single_number_i_s1.py +++ b/sword_for_offer/codes/python/sfo_56i_single_number_i_s1.py @@ -1,23 +1,27 @@ -''' +""" File: sfo_56-i_single_number_i_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def singleNumbers(self, nums: List[int]) -> List[int]: x, y, n, m = 0, 0, 0, 1 - for num in nums: # 1. 遍历异或 + for num in nums: # 1. 遍历异或 n ^= num - while n & m == 0: # 2. 循环左移，计算 m - m <<= 1 - for num in nums: # 3. 遍历 nums 分组 - if num & m: x ^= num # 4. 当 num & m != 0 - else: y ^= num # 4. 当 num & m == 0 - return x, y # 5. 返回出现一次的数字 + while n & m == 0: # 2. 循环左移，计算 m + m <<= 1 + for num in nums: # 3. 遍历 nums 分组 + if num & m: + x ^= num # 4. 当 num & m != 0 + else: + y ^= num # 4. 当 num & m == 0 + return x, y # 5. 返回出现一次的数字 + # ======= Test Case ======= nums = [4, 1, 4, 6] diff --git a/sword_for_offer/codes/python/sfo_56ii_single_number_ii_s1.py b/sword_for_offer/codes/python/sfo_56ii_single_number_ii_s1.py index bd4d484..479e959 100644 --- a/sword_for_offer/codes/python/sfo_56ii_single_number_ii_s1.py +++ b/sword_for_offer/codes/python/sfo_56ii_single_number_ii_s1.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_56-ii_single_number_ii_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def singleNumber(self, nums: List[int]) -> int: @@ -15,6 +16,7 @@ def singleNumber(self, nums: List[int]) -> int: twos = twos ^ num & ~ones return ones + # ======= Test Case ======= nums = [3, 4, 3, 3] # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_56ii_single_number_ii_s2.py b/sword_for_offer/codes/python/sfo_56ii_single_number_ii_s2.py index 15ac9b3..b041fbc 100644 --- a/sword_for_offer/codes/python/sfo_56ii_single_number_ii_s2.py +++ b/sword_for_offer/codes/python/sfo_56ii_single_number_ii_s2.py @@ -1,24 +1,26 @@ -''' +""" File: sfo_56-ii_single_number_ii_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def singleNumber(self, nums: List[int]) -> int: counts = [0] * 32 for num in nums: for i in range(32): - counts[i] += num & 1 # 更新第 i 位 1 的个数之和 - num >>= 1 # 第 i 位 --> 第 i 位 + counts[i] += num & 1 # 更新第 i 位 1 的个数之和 + num >>= 1 # 第 i 位 --> 第 i 位 res, m = 0, 3 for i in range(31, -1, -1): res <<= 1 - res |= counts[i] % m # 恢复第 i 位 - return res if counts[31] % m == 0 else ~(res ^ 0xffffffff) + res |= counts[i] % m # 恢复第 i 位 + return res if counts[31] % m == 0 else ~(res ^ 0xFFFFFFFF) + # ======= Test Case ======= nums = [3, 4, 3, 3] diff --git a/sword_for_offer/codes/python/sfo_57_two_numbers_with_sum_s_s1.py b/sword_for_offer/codes/python/sfo_57_two_numbers_with_sum_s_s1.py index 5395a8c..c016462 100644 --- a/sword_for_offer/codes/python/sfo_57_two_numbers_with_sum_s_s1.py +++ b/sword_for_offer/codes/python/sfo_57_two_numbers_with_sum_s_s1.py @@ -1,22 +1,27 @@ -''' +""" File: sfo_57_two_numbers_with_sum_s_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: i, j = 0, len(nums) - 1 while i < j: s = nums[i] + nums[j] - if s > target: j -= 1 - elif s < target: i += 1 - else: return nums[i], nums[j] + if s > target: + j -= 1 + elif s < target: + i += 1 + else: + return nums[i], nums[j] return [] + # ======= Test Case ======= nums = [2, 7, 11, 15] target = 9 diff --git a/sword_for_offer/codes/python/sfo_57ii_consecutive_numbers_with_sum_s_s1.py b/sword_for_offer/codes/python/sfo_57ii_consecutive_numbers_with_sum_s_s1.py index 8c87a6c..ef5dab7 100644 --- a/sword_for_offer/codes/python/sfo_57ii_consecutive_numbers_with_sum_s_s1.py +++ b/sword_for_offer/codes/python/sfo_57ii_consecutive_numbers_with_sum_s_s1.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_57-ii_consecutive_numbers_with_sum_s_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def findContinuousSequence(self, target: int) -> List[List[int]]: @@ -17,6 +18,7 @@ def findContinuousSequence(self, target: int) -> List[List[int]]: i += 1 return res + # ======= Test Case ======= target = 9 # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_57ii_consecutive_numbers_with_sum_s_s2.py b/sword_for_offer/codes/python/sfo_57ii_consecutive_numbers_with_sum_s_s2.py index 20ee985..5a3ecfe 100644 --- a/sword_for_offer/codes/python/sfo_57ii_consecutive_numbers_with_sum_s_s2.py +++ b/sword_for_offer/codes/python/sfo_57ii_consecutive_numbers_with_sum_s_s2.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_57-ii_consecutive_numbers_with_sum_s_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def findContinuousSequence(self, target: int) -> List[List[int]]: @@ -21,6 +22,7 @@ def findContinuousSequence(self, target: int) -> List[List[int]]: s += j return res + # ======= Test Case ======= target = 9 # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_58i_reverse_order_of_words_s1.py b/sword_for_offer/codes/python/sfo_58i_reverse_order_of_words_s1.py index 6ad06c3..e83e542 100644 --- a/sword_for_offer/codes/python/sfo_58i_reverse_order_of_words_s1.py +++ b/sword_for_offer/codes/python/sfo_58i_reverse_order_of_words_s1.py @@ -1,23 +1,27 @@ -''' +""" File: sfo_58-i_reverse_order_of_words_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def reverseWords(self, s: str) -> str: - s = s.strip() # 删除首尾空格 + s = s.strip() # 删除首尾空格 i = j = len(s) - 1 res = [] while i >= 0: - while i >= 0 and s[i] != ' ': i -= 1 # 搜索首个空格 - res.append(s[i + 1: j + 1]) # 添加单词 - while i >= 0 and s[i] == ' ': i -= 1 # 跳过单词间空格 - j = i # j 指向下个单词的尾字符 - return ' '.join(res) # 拼接并返回 + while i >= 0 and s[i] != " ": + i -= 1 # 搜索首个空格 + res.append(s[i + 1 : j + 1]) # 添加单词 + while i >= 0 and s[i] == " ": + i -= 1 # 跳过单词间空格 + j = i # j 指向下个单词的尾字符 + return " ".join(res) # 拼接并返回 + # ======= Test Case ======= s = "the sky is blue" diff --git a/sword_for_offer/codes/python/sfo_58i_reverse_order_of_words_s2.py b/sword_for_offer/codes/python/sfo_58i_reverse_order_of_words_s2.py index f910316..e9a6ee0 100644 --- a/sword_for_offer/codes/python/sfo_58i_reverse_order_of_words_s2.py +++ b/sword_for_offer/codes/python/sfo_58i_reverse_order_of_words_s2.py @@ -1,18 +1,20 @@ -''' +""" File: sfo_58-i_reverse_order_of_words_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def reverseWords(self, s: str) -> str: - s = s.strip() # 删除首尾空格 - strs = s.split() # 分割字符串 - strs.reverse() # 翻转单词列表 - return ' '.join(strs) # 拼接为字符串并返回 + s = s.strip() # 删除首尾空格 + strs = s.split() # 分割字符串 + strs.reverse() # 翻转单词列表 + return " ".join(strs) # 拼接为字符串并返回 + # ======= Test Case ======= s = "the sky is blue" diff --git a/sword_for_offer/codes/python/sfo_58i_reverse_order_of_words_s3.py b/sword_for_offer/codes/python/sfo_58i_reverse_order_of_words_s3.py index 179ce64..210c392 100644 --- a/sword_for_offer/codes/python/sfo_58i_reverse_order_of_words_s3.py +++ b/sword_for_offer/codes/python/sfo_58i_reverse_order_of_words_s3.py @@ -1,15 +1,17 @@ -''' +""" File: sfo_58-i_reverse_order_of_words_s3.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def reverseWords(self, s: str) -> str: - return ' '.join(s.strip().split()[::-1]) + return " ".join(s.strip().split()[::-1]) + # ======= Test Case ======= s = "the sky is blue" diff --git a/sword_for_offer/codes/python/sfo_58ii_left_rotation_of_a_string_s1.py b/sword_for_offer/codes/python/sfo_58ii_left_rotation_of_a_string_s1.py index 72c2c68..02841a3 100644 --- a/sword_for_offer/codes/python/sfo_58ii_left_rotation_of_a_string_s1.py +++ b/sword_for_offer/codes/python/sfo_58ii_left_rotation_of_a_string_s1.py @@ -1,16 +1,18 @@ -''' +""" File: sfo_58-ii_left_rotation_of_a_string_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def reverseLeftWords(self, s: str, n: int) -> str: return s[n:] + s[:n] + # ======= Test Case ======= s = "abcdefg" n = 2 diff --git a/sword_for_offer/codes/python/sfo_58ii_left_rotation_of_a_string_s2.py b/sword_for_offer/codes/python/sfo_58ii_left_rotation_of_a_string_s2.py index 952e596..f0b5cdc 100644 --- a/sword_for_offer/codes/python/sfo_58ii_left_rotation_of_a_string_s2.py +++ b/sword_for_offer/codes/python/sfo_58ii_left_rotation_of_a_string_s2.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_58-ii_left_rotation_of_a_string_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def reverseLeftWords(self, s: str, n: int) -> str: @@ -14,7 +15,8 @@ def reverseLeftWords(self, s: str, n: int) -> str: res.append(s[i]) for i in range(n): res.append(s[i]) - return ''.join(res) + return "".join(res) + # ======= Test Case ======= s = "abcdefg" diff --git a/sword_for_offer/codes/python/sfo_58ii_left_rotation_of_a_string_s3.py b/sword_for_offer/codes/python/sfo_58ii_left_rotation_of_a_string_s3.py index d6960a2..5ab7e1b 100644 --- a/sword_for_offer/codes/python/sfo_58ii_left_rotation_of_a_string_s3.py +++ b/sword_for_offer/codes/python/sfo_58ii_left_rotation_of_a_string_s3.py @@ -1,18 +1,20 @@ -''' +""" File: sfo_58-ii_left_rotation_of_a_string_s3.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def reverseLeftWords(self, s: str, n: int) -> str: res = [] for i in range(n, n + len(s)): res.append(s[i % len(s)]) - return ''.join(res) + return "".join(res) + # ======= Test Case ======= s = "abcdefg" diff --git a/sword_for_offer/codes/python/sfo_58ii_left_rotation_of_a_string_s4.py b/sword_for_offer/codes/python/sfo_58ii_left_rotation_of_a_string_s4.py index aa55c6b..9102e0e 100644 --- a/sword_for_offer/codes/python/sfo_58ii_left_rotation_of_a_string_s4.py +++ b/sword_for_offer/codes/python/sfo_58ii_left_rotation_of_a_string_s4.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_58-ii_left_rotation_of_a_string_s4.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def reverseLeftWords(self, s: str, n: int) -> str: @@ -16,6 +17,7 @@ def reverseLeftWords(self, s: str, n: int) -> str: res += s[i] return res + # ======= Test Case ======= s = "abcdefg" n = 2 diff --git a/sword_for_offer/codes/python/sfo_58ii_left_rotation_of_a_string_s5.py b/sword_for_offer/codes/python/sfo_58ii_left_rotation_of_a_string_s5.py index a5915fc..b6c94e8 100644 --- a/sword_for_offer/codes/python/sfo_58ii_left_rotation_of_a_string_s5.py +++ b/sword_for_offer/codes/python/sfo_58ii_left_rotation_of_a_string_s5.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_58-ii_left_rotation_of_a_string_s5.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def reverseLeftWords(self, s: str, n: int) -> str: @@ -14,6 +15,7 @@ def reverseLeftWords(self, s: str, n: int) -> str: res += s[i % len(s)] return res + # ======= Test Case ======= s = "abcdefg" n = 2 diff --git a/sword_for_offer/codes/python/sfo_59i_sliding_window_maximum_s1.py b/sword_for_offer/codes/python/sfo_59i_sliding_window_maximum_s1.py index c25d44f..44300cb 100644 --- a/sword_for_offer/codes/python/sfo_59i_sliding_window_maximum_s1.py +++ b/sword_for_offer/codes/python/sfo_59i_sliding_window_maximum_s1.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_59-i_sliding_window_maximum_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]: @@ -24,6 +25,7 @@ def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]: res.append(deque[0]) return res + # ======= Test Case ======= nums = [1, 3, -1, -3, 5, 3, 6, 7] k = 3 diff --git a/sword_for_offer/codes/python/sfo_59i_sliding_window_maximum_s2.py b/sword_for_offer/codes/python/sfo_59i_sliding_window_maximum_s2.py index 9cbcb0c..8eaf583 100644 --- a/sword_for_offer/codes/python/sfo_59i_sliding_window_maximum_s2.py +++ b/sword_for_offer/codes/python/sfo_59i_sliding_window_maximum_s2.py @@ -1,15 +1,17 @@ -''' +""" File: sfo_59-i_sliding_window_maximum_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]: - if not nums or k == 0: return [] + if not nums or k == 0: + return [] deque = collections.deque() # 未形成窗口 for i in range(k): @@ -27,6 +29,7 @@ def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]: res.append(deque[0]) return res + # ======= Test Case ======= nums = [1, 3, -1, -3, 5, 3, 6, 7] k = 3 diff --git a/sword_for_offer/codes/python/sfo_59ii_max_queue_s1.py b/sword_for_offer/codes/python/sfo_59ii_max_queue_s1.py index c29fbf3..e27023f 100644 --- a/sword_for_offer/codes/python/sfo_59ii_max_queue_s1.py +++ b/sword_for_offer/codes/python/sfo_59ii_max_queue_s1.py @@ -1,14 +1,15 @@ -''' +""" File: sfo_59-ii_max_queue_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * # ===== Solution Code ===== import queue + class MaxQueue: def __init__(self): self.queue = queue.Queue() @@ -24,12 +25,14 @@ def push_back(self, value: int) -> None: self.deque.append(value) def pop_front(self) -> int: - if self.queue.empty(): return -1 + if self.queue.empty(): + return -1 val = self.queue.get() if val == self.deque[0]: self.deque.popleft() return val + # ====== Driver Code ====== max_queue = MaxQueue() res = [ @@ -37,6 +40,6 @@ def pop_front(self) -> int: max_queue.push_back(2), max_queue.max_value(), max_queue.pop_front(), - max_queue.max_value() + max_queue.max_value(), ] print(res) diff --git a/sword_for_offer/codes/python/sfo_60_probabilities_for_rolling_n_dices_s1.py b/sword_for_offer/codes/python/sfo_60_probabilities_for_rolling_n_dices_s1.py index 551ee63..1b0cf95 100644 --- a/sword_for_offer/codes/python/sfo_60_probabilities_for_rolling_n_dices_s1.py +++ b/sword_for_offer/codes/python/sfo_60_probabilities_for_rolling_n_dices_s1.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_60_probabilities_for_rolling_n_dices_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def dicesProbability(self, n: int) -> List[float]: @@ -18,6 +19,7 @@ def dicesProbability(self, n: int) -> List[float]: dp = tmp return dp + # ======= Test Case ======= n = 2 # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_61_straight_in_poker_s1.py b/sword_for_offer/codes/python/sfo_61_straight_in_poker_s1.py index 8fedc91..8aad293 100644 --- a/sword_for_offer/codes/python/sfo_61_straight_in_poker_s1.py +++ b/sword_for_offer/codes/python/sfo_61_straight_in_poker_s1.py @@ -1,23 +1,27 @@ -''' +""" File: sfo_61_straight_in_poker_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def isStraight(self, nums: List[int]) -> bool: repeat = set() ma, mi = 0, 14 for num in nums: - if num == 0: continue # 跳过大小王 - ma = max(ma, num) # 最大牌 - mi = min(mi, num) # 最小牌 - if num in repeat: return False # 若有重复，提前返回 false - repeat.add(num) # 添加牌至 Set - return ma - mi < 5 # 最大牌 - 最小牌 < 5 则可构成顺子 + if num == 0: + continue # 跳过大小王 + ma = max(ma, num) # 最大牌 + mi = min(mi, num) # 最小牌 + if num in repeat: + return False # 若有重复，提前返回 false + repeat.add(num) # 添加牌至 Set + return ma - mi < 5 # 最大牌 - 最小牌 < 5 则可构成顺子 + # ======= Test Case ======= nums = [0, 0, 1, 2, 5] diff --git a/sword_for_offer/codes/python/sfo_61_straight_in_poker_s2.py b/sword_for_offer/codes/python/sfo_61_straight_in_poker_s2.py index 5eb68c9..b7bed8b 100644 --- a/sword_for_offer/codes/python/sfo_61_straight_in_poker_s2.py +++ b/sword_for_offer/codes/python/sfo_61_straight_in_poker_s2.py @@ -1,20 +1,24 @@ -''' +""" File: sfo_61_straight_in_poker_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def isStraight(self, nums: List[int]) -> bool: joker = 0 - nums.sort() # 数组排序 + nums.sort() # 数组排序 for i in range(4): - if nums[i] == 0: joker += 1 # 统计大小王数量 - elif nums[i] == nums[i + 1]: return False # 若有重复，提前返回 false - return nums[4] - nums[joker] < 5 # 最大牌 - 最小牌 < 5 则可构成顺子 + if nums[i] == 0: + joker += 1 # 统计大小王数量 + elif nums[i] == nums[i + 1]: + return False # 若有重复，提前返回 false + return nums[4] - nums[joker] < 5 # 最大牌 - 最小牌 < 5 则可构成顺子 + # ======= Test Case ======= nums = [0, 0, 1, 2, 5] diff --git a/sword_for_offer/codes/python/sfo_62_josephus_problem_s1.py b/sword_for_offer/codes/python/sfo_62_josephus_problem_s1.py index b6836ca..6e427ca 100644 --- a/sword_for_offer/codes/python/sfo_62_josephus_problem_s1.py +++ b/sword_for_offer/codes/python/sfo_62_josephus_problem_s1.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_62_josephus_problem_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def lastRemaining(self, n: int, m: int) -> int: @@ -14,6 +15,7 @@ def lastRemaining(self, n: int, m: int) -> int: x = (x + m) % i return x + # ======= Test Case ======= n = 5 m = 3 diff --git a/sword_for_offer/codes/python/sfo_63_the_maximum_profit_of_stocks_s1.py b/sword_for_offer/codes/python/sfo_63_the_maximum_profit_of_stocks_s1.py index 9e1a2db..517c2a7 100644 --- a/sword_for_offer/codes/python/sfo_63_the_maximum_profit_of_stocks_s1.py +++ b/sword_for_offer/codes/python/sfo_63_the_maximum_profit_of_stocks_s1.py @@ -1,11 +1,12 @@ -''' +""" File: sfo_63_the_maximum_profit_of_stocks_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def maxProfit(self, prices: List[int]) -> int: @@ -15,6 +16,7 @@ def maxProfit(self, prices: List[int]) -> int: profit = max(profit, price - cost) return profit + # ======= Test Case ======= prices = [7, 1, 5, 3, 6, 4] # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_64_solve_1_2___n_s1.py b/sword_for_offer/codes/python/sfo_64_solve_1_2___n_s1.py index cdca0a0..b8c81e1 100644 --- a/sword_for_offer/codes/python/sfo_64_solve_1_2___n_s1.py +++ b/sword_for_offer/codes/python/sfo_64_solve_1_2___n_s1.py @@ -1,20 +1,23 @@ -''' +""" File: sfo_64_solve_1+2+...+n_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def __init__(self): self.res = 0 + def sumNums(self, n: int) -> int: n > 1 and self.sumNums(n - 1) self.res += n return self.res + # ======= Test Case ======= n = 3 # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_65_implement_addition_operation_without_arithmetic_operators_s1.py b/sword_for_offer/codes/python/sfo_65_implement_addition_operation_without_arithmetic_operators_s1.py index 0f4879b..750ee38 100644 --- a/sword_for_offer/codes/python/sfo_65_implement_addition_operation_without_arithmetic_operators_s1.py +++ b/sword_for_offer/codes/python/sfo_65_implement_addition_operation_without_arithmetic_operators_s1.py @@ -1,19 +1,21 @@ -''' +""" File: sfo_65_implement_addition_operation_without_arithmetic_operators_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def add(self, a: int, b: int) -> int: - x = 0xffffffff + x = 0xFFFFFFFF a, b = a & x, b & x while b != 0: a, b = (a ^ b), (a & b) << 1 & x - return a if a <= 0x7fffffff else ~(a ^ x) + return a if a <= 0x7FFFFFFF else ~(a ^ x) + # ======= Test Case ======= a = 1 diff --git a/sword_for_offer/codes/python/sfo_66_a_product_array_puzzle_s1.py b/sword_for_offer/codes/python/sfo_66_a_product_array_puzzle_s1.py index e09aed9..8c94bee 100644 --- a/sword_for_offer/codes/python/sfo_66_a_product_array_puzzle_s1.py +++ b/sword_for_offer/codes/python/sfo_66_a_product_array_puzzle_s1.py @@ -1,22 +1,24 @@ -''' +""" File: sfo_66_a_product_array_puzzle_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def constructArr(self, a: List[int]) -> List[int]: b, tmp = [1] * len(a), 1 for i in range(1, len(a)): - b[i] = b[i - 1] * a[i - 1] # 下三角 + b[i] = b[i - 1] * a[i - 1] # 下三角 for i in range(len(a) - 2, -1, -1): - tmp *= a[i + 1] # 上三角 - b[i] *= tmp # 下三角 * 上三角 + tmp *= a[i + 1] # 上三角 + b[i] *= tmp # 下三角 * 上三角 return b + # ======= Test Case ======= a = [1, 2, 3, 4, 5] # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_67_convert_string_to_int_s1.py b/sword_for_offer/codes/python/sfo_67_convert_string_to_int_s1.py index 209159d..e03b37f 100644 --- a/sword_for_offer/codes/python/sfo_67_convert_string_to_int_s1.py +++ b/sword_for_offer/codes/python/sfo_67_convert_string_to_int_s1.py @@ -1,26 +1,33 @@ -''' +""" File: sfo_67_convert_string_to_int_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def strToInt(self, str: str) -> int: - str = str.strip() # 删除首尾空格 - if not str: return 0 # 字符串为空则直接返回 + str = str.strip() # 删除首尾空格 + if not str: + return 0 # 字符串为空则直接返回 res, i, sign = 0, 1, 1 - int_max, int_min, bndry = 2 ** 31 - 1, -2 ** 31, 2 ** 31 // 10 - if str[0] == '-': sign = -1 # 保存负号 - elif str[0] != '+': i = 0 # 若无符号位，则需从 i = 0 开始数字拼接 + int_max, int_min, bndry = 2**31 - 1, -(2**31), 2**31 // 10 + if str[0] == "-": + sign = -1 # 保存负号 + elif str[0] != "+": + i = 0 # 若无符号位，则需从 i = 0 开始数字拼接 for c in str[i:]: - if not '0' <= c <= '9' : break # 遇到非数字的字符则跳出 - if res > bndry or res == bndry and c > '7': return int_max if sign == 1 else int_min # 数字越界处理 - res = 10 * res + ord(c) - ord('0') # 数字拼接 + if not "0" <= c <= "9": + break # 遇到非数字的字符则跳出 + if res > bndry or res == bndry and c > "7": + return int_max if sign == 1 else int_min # 数字越界处理 + res = 10 * res + ord(c) - ord("0") # 数字拼接 return sign * res + # ======= Test Case ======= str = "42" # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_67_convert_string_to_int_s2.py b/sword_for_offer/codes/python/sfo_67_convert_string_to_int_s2.py index ef72d07..54aa850 100644 --- a/sword_for_offer/codes/python/sfo_67_convert_string_to_int_s2.py +++ b/sword_for_offer/codes/python/sfo_67_convert_string_to_int_s2.py @@ -1,29 +1,36 @@ -''' +""" File: sfo_67_convert_string_to_int_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: def strToInt(self, str: str) -> int: res, i, sign, length = 0, 0, 1, len(str) - int_max, int_min, bndry = 2 ** 31 - 1, -2 ** 31, 2 ** 31 // 10 - if not str: return 0 # 空字符串，提前返回 - while str[i] == ' ': + int_max, int_min, bndry = 2**31 - 1, -(2**31), 2**31 // 10 + if not str: + return 0 # 空字符串，提前返回 + while str[i] == " ": + i += 1 + if i == length: + return 0 # 字符串全为空格，提前返回 + if str[i] == "-": + sign = -1 + if str[i] in "+-": i += 1 - if i == length: return 0 # 字符串全为空格，提前返回 - if str[i] == '-': sign = -1 - if str[i] in '+-': i += 1 for j in range(i, length): - if not '0' <= str[j] <= '9' : break - if res > bndry or res == bndry and str[j] > '7': + if not "0" <= str[j] <= "9": + break + if res > bndry or res == bndry and str[j] > "7": return int_max if sign == 1 else int_min - res = 10 * res + ord(str[j]) - ord('0') + res = 10 * res + ord(str[j]) - ord("0") return sign * res + # ======= Test Case ======= str = "42" # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1.py b/sword_for_offer/codes/python/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1.py index fc31bdf..3613836 100644 --- a/sword_for_offer/codes/python/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1.py +++ b/sword_for_offer/codes/python/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s1.py @@ -1,24 +1,51 @@ -''' +""" File: sfo_68-i_the_nearest_common_ancestor_of_a_binary_search_tree_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: - def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode': + def lowestCommonAncestor( + self, root: "TreeNode", p: "TreeNode", q: "TreeNode" + ) -> "TreeNode": while root: - if root.val < p.val and root.val < q.val: # p,q 都在 root 的右子树中 - root = root.right # 遍历至右子节点 - elif root.val > p.val and root.val > q.val: # p,q 都在 root 的左子树中 - root = root.left # 遍历至左子节点 - else: break + if root.val < p.val and root.val < q.val: # p,q 都在 root 的右子树中 + root = root.right # 遍历至右子节点 + elif root.val > p.val and root.val > q.val: # p,q 都在 root 的左子树中 + root = root.left # 遍历至左子节点 + else: + break return root + # ======= Test Case ======= -root = list_to_tree([6, 2, 8, 0, 4, 7, 9, None, None, 3, 5, None, None, None, None, None, None, None, None]) +root = list_to_tree( + [ + 6, + 2, + 8, + 0, + 4, + 7, + 9, + None, + None, + 3, + 5, + None, + None, + None, + None, + None, + None, + None, + None, + ] +) p = get_tree_node(root=root, val=2) q = get_tree_node(root=root, val=8) # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2.py b/sword_for_offer/codes/python/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2.py index e5ca00f..2f0b9fd 100644 --- a/sword_for_offer/codes/python/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2.py +++ b/sword_for_offer/codes/python/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s2.py @@ -1,25 +1,53 @@ -''' +""" File: sfo_68-i_the_nearest_common_ancestor_of_a_binary_search_tree_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: - def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode': - if p.val > q.val: p, q = q, p # 保证 p.val < q.val + def lowestCommonAncestor( + self, root: "TreeNode", p: "TreeNode", q: "TreeNode" + ) -> "TreeNode": + if p.val > q.val: + p, q = q, p # 保证 p.val < q.val while root: - if root.val < p.val: # p,q 都在 root 的右子树中 - root = root.right # 遍历至右子节点 - elif root.val > q.val: # p,q 都在 root 的左子树中 - root = root.left # 遍历至左子节点 - else: break + if root.val < p.val: # p,q 都在 root 的右子树中 + root = root.right # 遍历至右子节点 + elif root.val > q.val: # p,q 都在 root 的左子树中 + root = root.left # 遍历至左子节点 + else: + break return root + # ======= Test Case ======= -root = list_to_tree([6, 2, 8, 0, 4, 7, 9, None, None, 3, 5, None, None, None, None, None, None, None, None]) +root = list_to_tree( + [ + 6, + 2, + 8, + 0, + 4, + 7, + 9, + None, + None, + 3, + 5, + None, + None, + None, + None, + None, + None, + None, + None, + ] +) p = get_tree_node(root=root, val=2) q = get_tree_node(root=root, val=8) # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3.py b/sword_for_offer/codes/python/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3.py index 4373e96..b8668f6 100644 --- a/sword_for_offer/codes/python/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3.py +++ b/sword_for_offer/codes/python/sfo_68i_the_nearest_common_ancestor_of_a_binary_search_tree_s3.py @@ -1,22 +1,48 @@ -''' +""" File: sfo_68-i_the_nearest_common_ancestor_of_a_binary_search_tree_s3.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: - def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode': + def lowestCommonAncestor( + self, root: "TreeNode", p: "TreeNode", q: "TreeNode" + ) -> "TreeNode": if root.val < p.val and root.val < q.val: return self.lowestCommonAncestor(root.right, p, q) if root.val > p.val and root.val > q.val: return self.lowestCommonAncestor(root.left, p, q) return root + # ======= Test Case ======= -root = list_to_tree([6, 2, 8, 0, 4, 7, 9, None, None, 3, 5, None, None, None, None, None, None, None, None]) +root = list_to_tree( + [ + 6, + 2, + 8, + 0, + 4, + 7, + 9, + None, + None, + 3, + 5, + None, + None, + None, + None, + None, + None, + None, + None, + ] +) p = get_tree_node(root=root, val=2) q = get_tree_node(root=root, val=8) # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1.py b/sword_for_offer/codes/python/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1.py index 4b23d33..a3a8d40 100644 --- a/sword_for_offer/codes/python/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1.py +++ b/sword_for_offer/codes/python/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s1.py @@ -1,23 +1,52 @@ -''' +""" File: sfo_68-ii_the_nearest_common_ancestor_of_a_binary_tree_s1.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: - def lowestCommonAncestor(self, root: TreeNode, p: TreeNode, q: TreeNode) -> TreeNode: - if not root or root == p or root == q: return root + def lowestCommonAncestor( + self, root: TreeNode, p: TreeNode, q: TreeNode + ) -> TreeNode: + if not root or root == p or root == q: + return root left = self.lowestCommonAncestor(root.left, p, q) right = self.lowestCommonAncestor(root.right, p, q) - if not left: return right - if not right: return left + if not left: + return right + if not right: + return left return root + # ======= Test Case ======= -root = list_to_tree([3, 5, 1, 6, 2, 0, 8, None, None, 7, 4, None, None, None, None, None, None, None, None]) +root = list_to_tree( + [ + 3, + 5, + 1, + 6, + 2, + 0, + 8, + None, + None, + 7, + 4, + None, + None, + None, + None, + None, + None, + None, + None, + ] +) p = get_tree_node(root=root, val=5) q = get_tree_node(root=root, val=1) # ====== Driver Code ====== diff --git a/sword_for_offer/codes/python/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2.py b/sword_for_offer/codes/python/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2.py index 962492a..882f0a9 100644 --- a/sword_for_offer/codes/python/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2.py +++ b/sword_for_offer/codes/python/sfo_68ii_the_nearest_common_ancestor_of_a_binary_tree_s2.py @@ -1,24 +1,54 @@ -''' +""" File: sfo_68-ii_the_nearest_common_ancestor_of_a_binary_tree_s2.py Created Time: 2021-12-09 Author: Krahets (krahets@163.com) -''' +""" from include import * + # ===== Solution Code ===== class Solution: - def lowestCommonAncestor(self, root: TreeNode, p: TreeNode, q: TreeNode) -> TreeNode: - if not root or root == p or root == q: return root + def lowestCommonAncestor( + self, root: TreeNode, p: TreeNode, q: TreeNode + ) -> TreeNode: + if not root or root == p or root == q: + return root left = self.lowestCommonAncestor(root.left, p, q) right = self.lowestCommonAncestor(root.right, p, q) - if not left and not right: return # 1. - if not left: return right # 3. - if not right: return left # 4. - return root # 2. if left and right: + if not left and not right: + return # 1. + if not left: + return right # 3. + if not right: + return left # 4. + return root # 2. if left and right: + # ======= Test Case ======= -root = list_to_tree([3, 5, 1, 6, 2, 0, 8, None, None, 7, 4, None, None, None, None, None, None, None, None]) +root = list_to_tree( + [ + 3, + 5, + 1, + 6, + 2, + 0, + 8, + None, + None, + 7, + 4, + None, + None, + None, + None, + None, + None, + None, + None, + ] +) p = get_tree_node(root=root, val=5) q = get_tree_node(root=root, val=1) # ====== Driver Code ====== diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 03. \346\225\260\347\273\204\344\270\255\351\207\215\345\244\215\347\232\204\346\225\260\345\255\227.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 03. \346\225\260\347\273\204\344\270\255\351\207\215\345\244\215\347\232\204\346\225\260\345\255\227.md" new file mode 100644 index 0000000..d0f232c --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 03. \346\225\260\347\273\204\344\270\255\351\207\215\345\244\215\347\232\204\346\225\260\345\255\227.md" @@ -0,0 +1,138 @@ +## 方法一：哈希表 / Set + +利用数据结构特点，容易想到使用哈希表（Set）记录数组的各个数字，当查找到重复数字则直接返回。 + +### 算法流程： + +1. 初始化： 新建 HashSet ，记为 $dic$ ； +2. 遍历数组 $nums$ 中的每个数字 $num$ ： + 1. 当 $num$ 在 $dic$ 中，说明重复，直接返回 $num$ ； + 2. 将 $num$ 添加至 $dic$ 中； +3. 返回 $-1$ 。本题中一定有重复数字，因此这里返回多少都可以。 + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 遍历数组使用 $O(N)$ ，HashSet 添加与查找元素皆为 $O(1)$ 。 +- **空间复杂度 $O(N)$ ：** HashSet 占用 $O(N)$ 大小的额外空间。 + + + +### 代码： + +```Python [] +class Solution: + def findRepeatNumber(self, nums: [int]) -> int: + dic = set() + for num in nums: + if num in dic: return num + dic.add(num) + return -1 +``` + +```Java [] +class Solution { + public int findRepeatNumber(int[] nums) { + Set dic = new HashSet<>(); + for(int num : nums) { + if(dic.contains(num)) return num; + dic.add(num); + } + return -1; + } +} +``` + +```C++ [] +class Solution { +public: + int findRepeatNumber(vector& nums) { + unordered_map map; + for(int num : nums) { + if(map[num]) return num; + map[num] = true; + } + return -1; + } +}; +``` + +## 方法二：原地交换 + +题目说明尚未被充分使用，即 `在一个长度为 n 的数组 nums 里的所有数字都在 0 ~ n-1 的范围内` 。 此说明含义：数组元素的 **索引** 和 **值** 是 **一对多** 的关系。 +因此，可遍历数组并通过交换操作，使元素的 **索引** 与 **值** 一一对应（即 $nums[i] = i$ ）。因而，就能通过索引映射对应的值，起到与字典等价的作用。 + +![Picture0.png](https://pic.leetcode-cn.com/1618146573-bOieFQ-Picture0.png){:width=500} + +遍历中，第一次遇到数字 $x$ 时，将其交换至索引 $x$ 处；而当第二次遇到数字 $x$ 时，一定有 $nums[x] = x$ ，此时即可得到一组重复数字。 + +### 算法流程： + +1. 遍历数组 $nums$ ，设索引初始值为 $i = 0$ : + 1. **若 $nums[i] = i$ ：** 说明此数字已在对应索引位置，无需交换，因此跳过； + 2. **若 $nums[nums[i]] = nums[i]$ ：** 代表索引 $nums[i]$ 处和索引 $i$ 处的元素值都为 $nums[i]$ ，即找到一组重复值，返回此值 $nums[i]$ ； + 3. **否则：** 交换索引为 $i$ 和 $nums[i]$ 的元素值，将此数字交换至对应索引位置。 + +2. 若遍历完毕尚未返回，则返回 $-1$ 。 + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 遍历数组使用 $O(N)$ ，每轮遍历的判断和交换操作使用 $O(1)$ 。 +- **空间复杂度 $O(1)$ ：** 使用常数复杂度的额外空间。 + + + +### 代码： + +Python 中， $a, b = c, d$ 操作的原理是先暂存元组 $(c, d)$ ，然后 “按左右顺序” 赋值给 a 和 b 。 +因此，若写为 $nums[i], nums[nums[i]] = nums[nums[i]], nums[i]$ ，则 $nums[i]$ 会先被赋值，之后 $nums[nums[i]]$ 指向的元素则会出错。 + +```Python [] +class Solution: + def findRepeatNumber(self, nums: List[int]) -> int: + i = 0 + while i < len(nums): + if nums[i] == i: + i += 1 + continue + if nums[nums[i]] == nums[i]: return nums[i] + nums[nums[i]], nums[i] = nums[i], nums[nums[i]] + return -1 +``` + +```Java [] +class Solution { + public int findRepeatNumber(int[] nums) { + int i = 0; + while(i < nums.length) { + if(nums[i] == i) { + i++; + continue; + } + if(nums[nums[i]] == nums[i]) return nums[i]; + int tmp = nums[i]; + nums[i] = nums[tmp]; + nums[tmp] = tmp; + } + return -1; + } +} +``` + +```C++ [] +class Solution { +public: + int findRepeatNumber(vector& nums) { + int i = 0; + while(i < nums.size()) { + if(nums[i] == i) { + i++; + continue; + } + if(nums[nums[i]] == nums[i]) + return nums[i]; + swap(nums[i],nums[nums[i]]); + } + return -1; + } +}; +``` \ No newline at end of file diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 04. \344\272\214\347\273\264\346\225\260\347\273\204\344\270\255\347\232\204\346\237\245\346\211\276.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 04. \344\272\214\347\273\264\346\225\260\347\273\204\344\270\255\347\232\204\346\237\245\346\211\276.md" new file mode 100644 index 0000000..dc5465d --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 04. \344\272\214\347\273\264\346\225\260\347\273\204\344\270\255\347\232\204\346\237\245\346\211\276.md" @@ -0,0 +1,73 @@ +## 解题思路： + +> 若使用暴力法遍历矩阵 `matrix` ，则时间复杂度为 $O(NM)$ 。暴力法未利用矩阵 **“从上到下递增、从左到右递增”** 的特点，显然不是最优解法。 + +如下图所示，我们将矩阵逆时针旋转 45° ，并将其转化为图形式，发现其类似于 **二叉搜索树** ，即对于每个元素，其左分支元素更小、右分支元素更大。因此，通过从 “根节点” 开始搜索，遇到比 `target` 大的元素就向左，反之向右，即可找到目标值 `target` 。 + +![Picture1.png](https://pic.leetcode-cn.com/6584ea93812d27112043d203ea90e4b0950117d45e0452d0c630fcb247fbc4af-Picture1.png){:width=450} + +“根节点” 对应的是矩阵的 “左下角” 和 “右上角” 元素，本文称之为 **标志数** ，以 `matrix` 中的 **左下角元素** 为标志数 `flag` ，则有: + +1. 若 `flag > target` ，则 `target` 一定在 `flag` 所在 **行的上方** ，即 `flag` 所在行可被消去。 +2. 若 `flag < target` ，则 `target` 一定在 `flag` 所在 **列的右方** ，即 `flag` 所在列可被消去。 + +### **算法流程：** + +1. 从矩阵 `matrix` 左下角元素（索引设为 `(i, j)` ）开始遍历，并与目标值对比： + - 当 `matrix[i][j] > target` 时，执行 `i--` ，即消去第 `i` 行元素； + - 当 `matrix[i][j] < target` 时，执行 `j++` ，即消去第 `j` 列元素； + - 当 `matrix[i][j] = target` 时，返回 $true$ ，代表找到目标值。 +2. 若行索引或列索引越界，则代表矩阵中无目标值，返回 $false$ 。 + +> 每轮 `i` 或 `j` 移动后，相当于生成了“消去一行（列）的新矩阵”， 索引`(i,j)` 指向新矩阵的左下角元素（标志数），因此可重复使用以上性质消去行（列）。 + +### 复杂度分析： + +- 时间复杂度 $O(M+N)$ ：其中，$N$ 和 $M$ 分别为矩阵行数和列数，此算法最多循环 $M+N$ 次。 +- 空间复杂度 $O(1)$ : `i`, `j` 指针使用常数大小额外空间。 + + + +## 代码： + +```Python [] +class Solution: + def findNumberIn2DArray(self, matrix: List[List[int]], target: int) -> bool: + i, j = len(matrix) - 1, 0 + while i >= 0 and j < len(matrix[0]): + if matrix[i][j] > target: i -= 1 + elif matrix[i][j] < target: j += 1 + else: return True + return False +``` + +```Java [] +class Solution { + public boolean findNumberIn2DArray(int[][] matrix, int target) { + int i = matrix.length - 1, j = 0; + while(i >= 0 && j < matrix[0].length) + { + if(matrix[i][j] > target) i--; + else if(matrix[i][j] < target) j++; + else return true; + } + return false; + } +} +``` + +```C++ [] +class Solution { +public: + bool findNumberIn2DArray(vector>& matrix, int target) { + int i = matrix.size() - 1, j = 0; + while(i >= 0 && j < matrix[0].size()) + { + if(matrix[i][j] > target) i--; + else if(matrix[i][j] < target) j++; + else return true; + } + return false; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 05. \346\233\277\346\215\242\347\251\272\346\240\274.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 05. \346\233\277\346\215\242\347\251\272\346\240\274.md" new file mode 100644 index 0000000..27deabe --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 05. \346\233\277\346\215\242\347\251\272\346\240\274.md" @@ -0,0 +1,96 @@ +## 方法一：遍历添加 + +在 Python 和 Java 等语言中，字符串都被设计成「不可变」的类型，即无法直接修改字符串的某一位字符，需要新建一个字符串实现。 + +### 算法流程： + +1. 初始化一个 list (Python) / StringBuilder (Java) ，记为 `res` ； +2. 遍历列表 `s` 中的每个字符 `c` ： + - 当 `c` 为空格时：向 `res` 后添加字符串 "%20" ； + - 当 `c` 不为空格时：向 `res` 后添加字符 `c` ； +3. 将列表 `res` 转化为字符串并返回。 + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 遍历使用 $O(N)$ ，每轮添加（修改）字符操作使用 $O(1)$ ； +- **空间复杂度 $O(N)$ ：** Python 新建的 list 和 Java 新建的 StringBuilder 都使用了线性大小的额外空间。 + + + +## 代码： + +```Python [] +class Solution: + def replaceSpace(self, s: str) -> str: + res = [] + for c in s: + if c == ' ': res.append("%20") + else: res.append(c) + return "".join(res) +``` + +```Java [] +class Solution { + public String replaceSpace(String s) { + StringBuilder res = new StringBuilder(); + for(Character c : s.toCharArray()) + { + if(c == ' ') res.append("%20"); + else res.append(c); + } + return res.toString(); + } +} +``` + +## 方法二：原地修改 + +在 C++ 语言中， string 被设计成「可变」的类型（[参考资料](https://stackoverflow.com/questions/28442719/are-c-strings-mutable-unlike-java-strings)），因此可以在不新建字符串的情况下实现原地修改。 + +由于需要将空格替换为 "%20" ，字符串的总字符数增加，因此需要扩展原字符串 s 的长度，计算公式为：`新字符串长度 = 原字符串长度 + 2 * 空格个数` ，示例如下图所示。 + +![Picture6.png](https://pic.leetcode-cn.com/1599931882-pPgkor-Picture6.png){:width=450} + +### 算法流程： + +1. 初始化：空格数量 `count` ，字符串 `s` 的长度 `len` ； +2. 统计空格数量：遍历 `s` ，遇空格则 `count++` ； +3. 修改 `s` 长度：添加完 "%20" 后的字符串长度应为 `len + 2 * count` ； +4. 倒序遍历修改：`i` 指向原字符串尾部元素， `j` 指向新字符串尾部元素；当 `i = j` 时跳出（代表左方已没有空格，无需继续遍历）； + - 当 `s[i]` 不为空格时：执行 `s[j] = s[i]` ； + - 当 `s[i]` 为空格时：将字符串闭区间 `[j-2, j]` 的元素修改为 "%20" ；由于修改了 3 个元素，因此需要 `j -= 2` ； +5. 返回值：已修改的字符串 `s` ； + + + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 遍历统计、遍历修改皆使用 $O(N)$ 时间。 +- **空间复杂度 $O(1)$ ：** 由于是原地扩展 `s` 长度，因此使用 $O(1)$ 额外空间。 + +```C++ [] +class Solution { +public: + string replaceSpace(string s) { + int count = 0, len = s.size(); + // 统计空格数量 + for (char c : s) { + if (c == ' ') count++; + } + // 修改 s 长度 + s.resize(len + 2 * count); + // 倒序遍历修改 + for(int i = len - 1, j = s.size() - 1; i < j; i--, j--) { + if (s[i] != ' ') + s[j] = s[i]; + else { + s[j - 2] = '%'; + s[j - 1] = '2'; + s[j] = '0'; + j -= 2; + } + } + return s; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 06. \344\273\216\345\260\276\345\210\260\345\244\264\346\211\223\345\215\260\351\223\276\350\241\250.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 06. \344\273\216\345\260\276\345\210\260\345\244\264\346\211\223\345\215\260\351\223\276\350\241\250.md" new file mode 100644 index 0000000..32398f5 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 06. \344\273\216\345\260\276\345\210\260\345\244\264\346\211\223\345\215\260\351\223\276\350\241\250.md" @@ -0,0 +1,130 @@ +## 方法一：递归法 + +利用递归，先递推至链表末端；回溯时，依次将节点值加入列表，即可实现链表值的倒序输出。 + +### 递归解析： + +1. **终止条件：** 当 `head == None` 时，代表越过了链表尾节点，则返回空列表； +2. **递推工作：** 访问下一节点 `head.next` ； +3. **回溯阶段：** + - **Python：** 返回 `当前 list + 当前节点值 [head.val]` ； + - **Java / C++：** 将当前节点值 `head.val` 加入列表 `tmp` ； + +### 复杂度分析： + +- **时间复杂度 $O(N)$：** 遍历链表，递归 $N$ 次。 +- **空间复杂度 $O(N)$：** 系统递归需要使用 $O(N)$ 的栈空间。 + +> 图解以 Python 代码为例。 + + + +### 代码： + +```Python [] +class Solution: + def reversePrint(self, head: ListNode) -> List[int]: + return self.reversePrint(head.next) + [head.val] if head else [] +``` + +```Java [] +class Solution { + ArrayList tmp = new ArrayList(); + public int[] reversePrint(ListNode head) { + recur(head); + int[] res = new int[tmp.size()]; + for(int i = 0; i < res.length; i++) + res[i] = tmp.get(i); + return res; + } + void recur(ListNode head) { + if(head == null) return; + recur(head.next); + tmp.add(head.val); + } +} +``` + +```C++ [] +class Solution { +public: + vector reversePrint(ListNode* head) { + recur(head); + return res; + } +private: + vector res; + void recur(ListNode* head) { + if(head == nullptr) return; + recur(head->next); + res.push_back(head->val); + } +}; +``` + +## 方法二：辅助栈法 + +链表只能 **从前至后** 访问每个节点，而题目要求 **倒序输出** 各节点值，这种 **先入后出** 的需求可以借助 **栈** 来实现。 + +### 算法流程： + +1. **入栈：** 遍历链表，将各节点值 `push` 入栈。 +2. **出栈：** 将各节点值 `pop` 出栈，存储于数组并返回。 + +### 复杂度分析： + +- **时间复杂度 $O(N)$：** 入栈和出栈共使用 $O(N)$ 时间。 +- **空间复杂度 $O(N)$：** 辅助栈 `stack` 和数组 `res` 共使用 $O(N)$ 的额外空间。 + +> 图解以 Java 代码为例，Python 无需将 `stack` 转移至 `res`，而是直接返回倒序数组。 + + + +### 代码： + +Java 数组长度不可变，因此使用 List 先存储，再转为数组并返回。 + +```Python [] +class Solution: + def reversePrint(self, head: ListNode) -> List[int]: + stack = [] + while head: + stack.append(head.val) + head = head.next + return stack[::-1] +``` + +```Java [] +class Solution { + public int[] reversePrint(ListNode head) { + LinkedList stack = new LinkedList(); + while(head != null) { + stack.addLast(head.val); + head = head.next; + } + int[] res = new int[stack.size()]; + for(int i = 0; i < res.length; i++) + res[i] = stack.removeLast(); + return res; + } +} +``` + +```C++ [] +class Solution { +public: + vector reversePrint(ListNode* head) { + stack stk; + while(head != nullptr) { + stk.push(head->val); + head = head->next; + } + vector res; + while(!stk.empty()) { + res.push_back(stk.top()); + stk.pop(); + } + return res; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 07. \351\207\215\345\273\272\344\272\214\345\217\211\346\240\221.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 07. \351\207\215\345\273\272\344\272\214\345\217\211\346\240\221.md" new file mode 100644 index 0000000..c7f4c9d --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 07. \351\207\215\345\273\272\344\272\214\345\217\211\346\240\221.md" @@ -0,0 +1,117 @@ +### 解题思路： + +前序遍历性质： 节点按照 `[ 根节点 | 左子树 | 右子树 ]` 排序。 +中序遍历性质： 节点按照 `[ 左子树 | 根节点 | 右子树 ]` 排序。 + +> 以题目示例为例： +> +> - 前序遍历划分 `[ 3 | 9 | 20 15 7 ]` +> - 中序遍历划分 `[ 9 | 3 | 15 20 7 ]` + +根据以上性质，可得出以下推论： + +1. 前序遍历的首元素 为 树的根节点 `node` 的值。 +2. 在中序遍历中搜索根节点 `node` 的索引 ，可将 中序遍历 划分为 `[ 左子树 | 根节点 | 右子树 ]` 。 +3. 根据中序遍历中的左（右）子树的节点数量，可将 前序遍历 划分为 `[ 根节点 | 左子树 | 右子树 ] ` 。 + +![Picture1.png](https://pic.leetcode-cn.com/1629825510-roByLr-Picture1.png){:width=550} + +通过以上三步，可确定 **三个节点** ：1.树的根节点、2.左子树根节点、3.右子树根节点。 + +根据「分治算法」思想，对于树的左、右子树，仍可复用以上方法划分子树的左右子树。 + +## 分治算法解析： + +- **递推参数：** 根节点在前序遍历的索引 `root` 、子树在中序遍历的左边界 `left` 、子树在中序遍历的右边界 `right` ； + +- **终止条件：** 当 `left > right` ，代表已经越过叶节点，此时返回 $null$ ； + +- **递推工作：** + + 1. **建立根节点 `node` ：** 节点值为 `preorder[root]` ； + 2. **划分左右子树：** 查找根节点在中序遍历 `inorder` 中的索引 `i` ； + + > 为了提升效率，本文使用哈希表 `dic` 存储中序遍历的值与索引的映射，查找操作的时间复杂度为 $O(1)$ ； + + 3. **构建左右子树：** 开启左右子树递归； + + | | 根节点索引 | 中序遍历左边界 | 中序遍历右边界 | + | ---------- | --------------------- | -------------- | -------------- | + | **左子树** | `root + 1` | `left` | `i - 1` | + | **右子树** | `i - left + root + 1` | `i + 1` | `right` | + + > **TIPS：** `i - left + root + 1`含义为 `根节点索引 + 左子树长度 + 1` + +- **返回值：** 回溯返回 `node` ，作为上一层递归中根节点的左 / 右子节点； + + + +## 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为树的节点数量。初始化 HashMap 需遍历 `inorder` ，占用 $O(N)$ 。递归共建立 $N$ 个节点，每层递归中的节点建立、搜索操作占用 $O(1)$ ，因此使用 $O(N)$ 时间。 +- **空间复杂度 $O(N)$ ：** HashMap 使用 $O(N)$ 额外空间；最差情况下（输入二叉树为链表时），递归深度达到 $N$ ，占用 $O(N)$ 的栈帧空间；因此总共使用 $O(N)$ 空间。 + +### 代码： + +> 注意：本文方法只适用于 “无重复节点值” 的二叉树。 + +```Python [] +class Solution: + def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode: + def recur(root, left, right): + if left > right: return # 递归终止 + node = TreeNode(preorder[root]) # 建立根节点 + i = dic[preorder[root]] # 划分根节点、左子树、右子树 + node.left = recur(root + 1, left, i - 1) # 开启左子树递归 + node.right = recur(i - left + root + 1, i + 1, right) # 开启右子树递归 + return node # 回溯返回根节点 + + dic, preorder = {}, preorder + for i in range(len(inorder)): + dic[inorder[i]] = i + return recur(0, 0, len(inorder) - 1) +``` + +```Java [] +class Solution { + int[] preorder; + HashMap dic = new HashMap<>(); + public TreeNode buildTree(int[] preorder, int[] inorder) { + this.preorder = preorder; + for(int i = 0; i < inorder.length; i++) + dic.put(inorder[i], i); + return recur(0, 0, inorder.length - 1); + } + TreeNode recur(int root, int left, int right) { + if(left > right) return null; // 递归终止 + TreeNode node = new TreeNode(preorder[root]); // 建立根节点 + int i = dic.get(preorder[root]); // 划分根节点、左子树、右子树 + node.left = recur(root + 1, left, i - 1); // 开启左子树递归 + node.right = recur(root + i - left + 1, i + 1, right); // 开启右子树递归 + return node; // 回溯返回根节点 + } +} +``` + +```C++ [] +class Solution { +public: + TreeNode* buildTree(vector& preorder, vector& inorder) { + this->preorder = preorder; + for(int i = 0; i < inorder.size(); i++) + dic[inorder[i]] = i; + return recur(0, 0, inorder.size() - 1); + } +private: + vector preorder; + unordered_map dic; + TreeNode* recur(int root, int left, int right) { + if(left > right) return nullptr; // 递归终止 + TreeNode* node = new TreeNode(preorder[root]); // 建立根节点 + int i = dic[preorder[root]]; // 划分根节点、左子树、右子树 + node->left = recur(root + 1, left, i - 1); // 开启左子树递归 + node->right = recur(root + i - left + 1, i + 1, right); // 开启右子树递归 + return node; // 回溯返回根节点 + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 09. \347\224\250\344\270\244\344\270\252\346\240\210\345\256\236\347\216\260\351\230\237\345\210\227.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 09. \347\224\250\344\270\244\344\270\252\346\240\210\345\256\236\347\216\260\351\230\237\345\210\227.md" new file mode 100644 index 0000000..cbbb833 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 09. \347\224\250\344\270\244\344\270\252\346\240\210\345\256\236\347\216\260\351\230\237\345\210\227.md" @@ -0,0 +1,95 @@ +## 解题思路： + +栈实现队列的出队操作效率低下：栈底元素（对应队首元素）无法直接删除，需要将上方所有元素出栈。 + +列表倒序操作可使用双栈实现：设有含三个元素的栈 `A = [1,2,3]` 和空栈 `B = []` 。若循环执行 `A` 元素出栈并添加入栈 `B` ，直到栈 `A` 为空，则 `A = []` , `B = [3,2,1]` ，即栈 `B` 元素为栈 `A` 元素倒序。 + +利用栈 `B` 删除队首元素：倒序后，`B` 执行出栈则相当于删除了 `A` 的栈底元素，即对应队首元素。 + +![Picture1.png](https://pic.leetcode-cn.com/1599286207-HnnMhX-Picture1.png){:width=500} + +题目要求实现 **加入队尾**`appendTail()` 和 **删除队首**`deleteHead()` 两个函数的正常工作。因此，可以设计栈 `A` 用于加入队尾操作，栈 `B` 用于将元素倒序，从而实现删除队首元素。 + +### 函数设计： + +1. **加入队尾 `appendTail()` ：** 将数字 `val` 加入栈 `A` 即可。 +2. **删除队首`deleteHead()` ：** 有以下三种情况。 + 1. **当栈 `B` 不为空：** `B`中仍有已完成倒序的元素，因此直接返回 `B` 的栈顶元素。 + 2. **否则，当 `A` 为空：** 即两个栈都为空，无元素，因此返回 -1 。 + 3. **否则：** 将栈 `A` 元素全部转移至栈 `B` 中，实现元素倒序，并返回栈 `B` 的栈顶元素。 + + + +### 复杂度分析： + +> 以下分析仅满足添加 $N$ 个元素并删除 $N$ 个元素，即栈初始和结束状态下都为空的情况。 + +- **时间复杂度：** `appendTail()`函数为 $O(1)$ ；`deleteHead()` 函数在 $N$ 次队首元素删除操作中总共需完成 $N$ 个元素的倒序。 +- **空间复杂度 $O(N)$ ：** 最差情况下，栈 `A` 和 `B` 共保存 $N$ 个元素。 + +## 代码： + +Python 和 Java 的栈的 `pop()` 函数返回栈顶元素，而 C++ 不返回；因此对于 C++ ，需要先使用 `top()` 方法暂存栈顶元素，再执行 `pop()` 出栈操作。 + +```Python [] +class CQueue: + def __init__(self): + self.A, self.B = [], [] + + def appendTail(self, value: int) -> None: + self.A.append(value) + + def deleteHead(self) -> int: + if self.B: return self.B.pop() + if not self.A: return -1 + while self.A: + self.B.append(self.A.pop()) + return self.B.pop() +``` + +```Java [] +class CQueue { + LinkedList A, B; + public CQueue() { + A = new LinkedList(); + B = new LinkedList(); + } + public void appendTail(int value) { + A.addLast(value); + } + public int deleteHead() { + if(!B.isEmpty()) return B.removeLast(); + if(A.isEmpty()) return -1; + while(!A.isEmpty()) + B.addLast(A.removeLast()); + return B.removeLast(); + } +} +``` + +```C++ [] +class CQueue { +public: + stack A, B; + CQueue() {} + void appendTail(int value) { + A.push(value); + } + int deleteHead() { + if(!B.empty()) { + int tmp = B.top(); + B.pop(); + return tmp; + } + if(A.empty()) return -1; + while(!A.empty()) { + int tmp = A.top(); + A.pop(); + B.push(tmp); + } + int tmp = B.top(); + B.pop(); + return tmp; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 10- I. \346\226\220\346\263\242\351\202\243\345\245\221\346\225\260\345\210\227.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 10- I. \346\226\220\346\263\242\351\202\243\345\245\221\346\225\260\345\210\227.md" new file mode 100644 index 0000000..67135ff --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 10- I. \346\226\220\346\263\242\351\202\243\345\245\221\346\225\260\345\210\227.md" @@ -0,0 +1,98 @@ +## 解题思路： + +斐波那契数列的定义是 $f(n + 1) = f(n) + f(n - 1)$ ，生成第 $n$ 项的做法有以下几种： + +1. **递归法：** + - **原理：** 把 $f(n)$ 问题的计算拆分成 $f(n-1)$ 和 $f(n-2)$ 两个子问题的计算，并递归，以 $f(0)$ 和 $f(1)$ 为终止条件。 + - **缺点：** 大量重复的递归计算，例如 $f(n)$ 和 $f(n - 1)$ 两者向下递归需要 **各自计算** $f(n - 2)$ 的值。 +2. **记忆化递归法：** + - **原理：** 在递归法的基础上，新建一个长度为 $n$ 的数组，用于在递归时存储 $f(0)$ 至 $f(n)$ 的数字值，重复遇到某数字则直接从数组取用，避免了重复的递归计算。 + - **缺点：** 记忆化存储需要使用 $O(N)$ 的额外空间。 +3. **动态规划：** + - **原理：** 以斐波那契数列性质 $f(n + 1) = f(n) + f(n - 1)$ 为转移方程。 + - 从计算效率、空间复杂度上看，动态规划是本题的最佳解法。 + +> 下图帮助理解递归法的 “重复计算” 概念。 + +![Picture1.png](https://pic.leetcode-cn.com/1599882883-mtYecf-Picture1.png){:width=500} + +## 动态规划解析： + +- **状态定义：** 设 $dp$ 为一维数组，其中 $dp[i]$ 的值代表 斐波那契数列第 $i$ 个数字 。 +- **转移方程：** $dp[i + 1] = dp[i] + dp[i - 1]$ ，即对应数列定义 $f(n + 1) = f(n) + f(n - 1)$ ； +- **初始状态：** $dp[0] = 0$, $dp[1] = 1$ ，即初始化前两个数字； +- **返回值：** $dp[n]$ ，即斐波那契数列的第 $n$ 个数字。 + +## 空间复杂度降低： + +> 若新建长度为 $n$ 的 $dp$ 列表，则空间复杂度为 $O(N)$ 。 + +- 由于 $dp$ 列表第 $i$ 项只与第 $i-1$ 和第 $i-2$ 项有关，因此只需要初始化三个整形变量 `sum`, `a`, `b` ，利用辅助变量 $sum$ 使 $a, b$ 两数字交替前进即可 *（具体实现见代码）* 。 +- 节省了 $dp$ 列表空间，因此空间复杂度降至 $O(1)$ 。 + +## 循环求余法： + +> **大数越界：** 随着 $n$ 增大, $f(n)$ 会超过 `Int32` 甚至 `Int64` 的取值范围，导致最终的返回值错误。 + +- **求余运算规则：** 设正整数 $x, y, p$ ，求余符号为 $\odot$ ，则有 $(x + y) \odot p = (x \odot p + y \odot p) \odot p$ 。 + +- **解析：** 根据以上规则，可推出 $f(n) \odot p = [f(n-1) \odot p + f(n-2) \odot p] \odot p$ ，从而可以在循环过程中每次计算 $sum = (a + b) \odot 1000000007$ ，此操作与最终返回前取余等价。 + + + +## 复杂度分析： + +- **时间复杂度 $O(n)$ ：** 计算 $f(n)$ 需循环 $n$ 次，每轮循环内计算操作使用 $O(1)$ 。 +- **空间复杂度 $O(1)$ ：** 几个标志变量使用常数大小的额外空间。 + +## 代码： + +```Python [] +class Solution: + def fib(self, n: int) -> int: + a, b = 0, 1 + for _ in range(n): + a, b = b, (a + b) % 1000000007 + return a +``` + +```Java [] +class Solution { + public int fib(int n) { + int a = 0, b = 1, sum; + for(int i = 0; i < n; i++){ + sum = (a + b) % 1000000007; + a = b; + b = sum; + } + return a; + } +} +``` + +```C++ [] +class Solution { +public: + int fib(int n) { + int a = 0, b = 1, sum; + for(int i = 0; i < n; i++){ + sum = (a + b) % 1000000007; + a = b; + b = sum; + } + return a; + } +}; +``` + +由于 Python 中整形数字的大小限制取决计算机的内存（可理解为无限大），因此也可不考虑大数越界问题；但当数字很大时，加法运算的效率也会降低，因此不推荐此方法。 + +```Python [] +# 不考虑大数越界问题 +class Solution: + def fib(self, n: int) -> int: + a, b = 0, 1 + for _ in range(n): + a, b = b, a + b + return a % 1000000007 +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 10- II. \351\235\222\350\233\231\350\267\263\345\217\260\351\230\266\351\227\256\351\242\230.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 10- II. \351\235\222\350\233\231\350\267\263\345\217\260\351\230\266\351\227\256\351\242\230.md" new file mode 100644 index 0000000..a6cfc76 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 10- II. \351\235\222\350\233\231\350\267\263\345\217\260\351\230\266\351\227\256\351\242\230.md" @@ -0,0 +1,105 @@ +## 解题思路： + +设跳上 $n$ 级台阶有 $f(n)$ 种跳法。在所有跳法中，青蛙的最后一步只有两种情况： **跳上 $1$ 级或 $2$ 级台阶**。 + +1. **当为 $1$ 级台阶：** 剩 $n-1$ 个台阶，此情况共有 $f(n-1)$ 种跳法； +2. **当为 $2$ 级台阶：** 剩 $n-2$ 个台阶，此情况共有 $f(n-2)$ 种跳法。 + +即 $f(n)$ 为以上两种情况之和，即 $f(n)=f(n-1)+f(n-2)$ ，以上递推性质为斐波那契数列。因此，本题可转化为 **求斐波那契数列第 $n$ 项的值** ，与 [剑指Offer 10- I. 斐波那契数列](https://leetcode-cn.com/problems/fei-bo-na-qi-shu-lie-lcof/solution/mian-shi-ti-10-i-fei-bo-na-qi-shu-lie-dong-tai-gui/) 等价，唯一的不同在于起始数字不同。 + +- 青蛙跳台阶问题： $f(0)=1$ , $f(1)=1$ , $f(2)=2$ ； +- 斐波那契数列问题： $f(0)=0$ , $f(1)=1$ , $f(2)=1$ 。 + +![Picture1.png](https://pic.leetcode-cn.com/1599883153-UckfTw-Picture1.png){:width=500} + +斐波那契数列的定义是 $f(n + 1) = f(n) + f(n - 1)$ ，生成第 $n$ 项的做法有以下几种： + +1. **递归法：** + - **原理：** 把 $f(n)$ 问题的计算拆分成 $f(n-1)$ 和 $f(n-2)$ 两个子问题的计算，并递归，以 $f(0)$ 和 $f(1)$ 为终止条件。 + - **缺点：** 大量重复的递归计算，例如 $f(n)$ 和 $f(n - 1)$ 两者向下递归都需要计算 $f(n - 2)$ 的值。 +2. **记忆化递归法：** + - **原理：** 在递归法的基础上，新建一个长度为 $n$ 的数组，用于在递归时存储 $f(0)$ 至 $f(n)$ 的数字值，重复遇到某数字时则直接从数组取用，避免了重复的递归计算。 + - **缺点：** 记忆化存储的数组需要使用 $O(N)$ 的额外空间。 +3. **动态规划：** + - **原理：** 以斐波那契数列性质 $f(n + 1) = f(n) + f(n - 1)$ 为转移方程。 + - 从计算效率、空间复杂度上看，动态规划是本题的最佳解法。 + +**动态规划解析：** + +- **状态定义：** 设 $dp$ 为一维数组，其中 $dp[i]$ 的值代表斐波那契数列的第 $i$ 个数字。 +- **转移方程：** $dp[i + 1] = dp[i] + dp[i - 1]$ ，即对应数列定义 $f(n + 1) = f(n) + f(n - 1)$ ； +- **初始状态：** $dp[0] = 1$, $dp[1] = 1$ ，即初始化前两个数字； +- **返回值：** $dp[n]$ ，即斐波那契数列的第 $n$ 个数字。 + +**空间复杂度降低：** + +> 若新建长度为 $n$ 的 $dp$ 列表，则空间复杂度为 $O(N)$ 。 + +- 由于 $dp$ 列表第 $i$ 项只与第 $i-1$ 和第 $i-2$ 项有关，因此只需要初始化三个整形变量 `sum`, `a`, `b` ，利用辅助变量 $sum$ 使 $a, b$ 两数字交替前进即可 *（具体实现见代码）* 。 +- 因为节省了 $dp$ 列表空间，因此空间复杂度降至 $O(1)$ 。 + +**循环求余法：** + +> **大数越界：** 随着 $n$ 增大, $f(n)$ 会超过 `Int32` 甚至 `Int64` 的取值范围，导致最终的返回值错误。 + +- **求余运算规则：** 设正整数 $x, y, p$ ，求余符号为 $\odot$ ，则有 $(x + y) \odot p = (x \odot p + y \odot p) \odot p$ 。 +- **解析：** 根据以上规则，可推出 $f(n) \odot p = [f(n-1) \odot p + f(n-2) \odot p] \odot p$ ，从而可以在循环过程中每次计算 $sum = a + b \odot 1000000007$ ，此操作与最终返回前取余等价。 + + + +**复杂度分析：** + +- **时间复杂度 $O(n)$ ：** 计算 $f(n)$ 需循环 $n$ 次，每轮循环内计算操作使用 $O(1)$ 。 +- **空间复杂度 $O(1)$ ：** 几个标志变量使用常数大小的额外空间。 + +## 代码： + +```Python [] +class Solution: + def numWays(self, n: int) -> int: + a, b = 1, 1 + for _ in range(n): + a, b = b, (a + b) % 1000000007 + return a +``` + +```Java [] +class Solution { + public int numWays(int n) { + int a = 1, b = 1, sum; + for(int i = 0; i < n; i++){ + sum = (a + b) % 1000000007; + a = b; + b = sum; + } + return a; + } +} +``` + +```C++ [] +class Solution { +public: + int numWays(int n) { + int a = 1, b = 1, sum; + for(int i = 0; i < n; i++){ + sum = (a + b) % 1000000007; + a = b; + b = sum; + } + return a; + } +}; +``` + +由于 Python 中整形数字的大小限制取决计算机的内存（可理解为无限大），因此也可不考虑大数越界问题；但当数字很大时，加法运算的效率也会降低，因此不推荐此方法。 + +```Python [] +# 不考虑大数越界问题 +class Solution: + def numWays(self, n: int) -> int: + a, b = 1, 1 + for _ in range(n): + a, b = b, a + b + return a % 1000000007 +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 11. \346\227\213\350\275\254\346\225\260\347\273\204\347\232\204\346\234\200\345\260\217\346\225\260\345\255\227.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 11. \346\227\213\350\275\254\346\225\260\347\273\204\347\232\204\346\234\200\345\260\217\346\225\260\345\255\227.md" new file mode 100644 index 0000000..359f522 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 11. \346\227\213\350\275\254\346\225\260\347\273\204\347\232\204\346\234\200\345\260\217\346\225\260\345\255\227.md" @@ -0,0 +1,161 @@ +## 解题思路： + +> 为精简篇幅，本文将数组 $numbers$ 缩写为 $nums$ 。 + +如下图所示，寻找旋转数组的最小元素即为寻找 **右排序数组** 的首个元素 $nums[x]$ ，称 $x$ 为 **旋转点** 。 + +![Picture1.png](https://pic.leetcode-cn.com/1599404042-JMvjtL-Picture1.png){:width=450} + +排序数组的查找问题首先考虑使用 **二分法** 解决，其可将 **遍历法** 的 **线性级别** 时间复杂度降低至 **对数级别** 。 + +### 算法流程： + +1. **初始化：** 声明 $i$, $j$ 双指针分别指向 $nums$ 数组左右两端； +2. **循环二分：** 设 $m = (i + j) / 2$ 为每次二分的中点（ "`/`" 代表向下取整除法，因此恒有 $i \leq m < j$ ），可分为以下三种情况： + 1. **当 $nums[m] > nums[j]$ 时：** $m$ 一定在 左排序数组 中，即旋转点 $x$ 一定在 $[m + 1, j]$ 闭区间内，因此执行 $i = m + 1$； + 2. **当 $nums[m] < nums[j]$ 时：** $m$ 一定在 右排序数组 中，即旋转点 $x$ 一定在$[i, m]$ 闭区间内，因此执行 $j = m$； + 3. **当 $nums[m] = nums[j]$ 时：** 无法判断 $m$ 在哪个排序数组中，即无法判断旋转点 $x$ 在 $[i, m]$ 还是 $[m + 1, j]$ 区间中。**解决方案：** 执行 $j = j - 1$ 缩小判断范围，分析见下文。 +3. **返回值：** 当 $i = j$ 时跳出二分循环，并返回 **旋转点的值** $nums[i]$ 即可。 + +### 正确性证明： + +当 $nums[m] = nums[j]$ 时，无法判定 $m$ 在左（右）排序数组，自然也无法通过二分法安全地缩小区间，因为其会导致旋转点 $x$ 不在区间 $[i, j]$ 内。举例如下： + +> 设以下两个旋转点值为 $0$ 的示例数组，则当 $i = 0$, $j = 4$ 时 $m = 2$ ，两示例结果不同。 +> 示例一 $[1, 0, 1, 1, 1]$ ：旋转点 $x = 1$ ，因此 $m = 2$ 在 **右排序数组** 中。 +> 示例二 $[1, 1, 1, 0, 1]$ ：旋转点 $x = 3$ ，因此 $m = 2$ 在 **左排序数组** 中。 + +而证明 $j = j - 1$ 正确（缩小区间安全性），需分为两种情况： + +1. **当 $x < j$ 时：** 易得执行 $j = j - 1$ 后，旋转点 $x$ 仍在区间 $[i, j]$ 内。 +2. **当 $x = j$ 时：** 执行 $j = j - 1$ 后越过（丢失）了旋转点 $x$ ，但最终返回的元素值 $nums[i]$ 仍等于旋转点值 $nums[x]$ 。 + + 1. 由于 $x = j$ ，因此 $nums[x] = nums[j] = nums[m] \leq number[i]$ ; + 2. 又由于 $i \leq m 综上所述，此方法可以保证返回值 $nums[i]$ 等于旋转点值 $nums[x]$ ，但在少数特例下 $i \ne x$ ；而本题目只要求返回 “旋转点的值” ，因此本方法正确。 + +**补充思考：** 为什么本题二分法不用 $nums[m]$ 和 $nums[i]$ 作比较？ + +二分目的是判断 $m$ 在哪个排序数组中，从而缩小区间。而在 $nums[m] > nums[i]$情况下，无法判断 $m$ 在哪个排序数组中。本质上是由于 $j$ 初始值肯定在右排序数组中； $i$ 初始值无法确定在哪个排序数组中。举例如下： + +> 对于以下两示例，当 $i = 0, j = 4, m = 2$ 时，有 `nums[m] > nums[i]` ，而结果不同。 +> $[1, 2, 3, 4 ,5]$ 旋转点 $x = 0$ ： $m$ 在右排序数组（此示例只有右排序数组）； +> $[3, 4, 5, 1 ,2]$ 旋转点 $x = 3$ ： $m$ 在左排序数组。 + + + +### 复杂度分析： + +- **时间复杂度 $O(\log_2 N)$ ：** 在特例情况下（例如 $[1, 1, 1, 1]$），会退化到 $O(N)$。 +- **空间复杂度 $O(1)$ ：** $i$ , $j$ , $m$ 变量使用常数大小的额外空间。 + +## 代码： + +```Python [] +class Solution: + def minArray(self, numbers: [int]) -> int: + i, j = 0, len(numbers) - 1 + while i < j: + m = (i + j) // 2 + if numbers[m] > numbers[j]: i = m + 1 + elif numbers[m] < numbers[j]: j = m + else: j -= 1 + return numbers[i] +``` + +```Java [] +class Solution { + public int minArray(int[] numbers) { + int i = 0, j = numbers.length - 1; + while (i < j) { + int m = (i + j) / 2; + if (numbers[m] > numbers[j]) i = m + 1; + else if (numbers[m] < numbers[j]) j = m; + else j--; + } + return numbers[i]; + } +} +``` + +```C++ [] +class Solution { +public: + int minArray(vector& numbers) { + int i = 0, j = numbers.size() - 1; + while (i < j) { + int m = (i + j) / 2; + if (numbers[m] > numbers[j]) i = m + 1; + else if (numbers[m] < numbers[j]) j = m; + else j--; + } + return numbers[i]; + } +}; +``` + +实际上，当出现 $nums[m] = nums[j]$ 时，一定有区间 $[i, m]$ 内所有元素相等 或 区间 $[m, j]$ 内所有元素相等（或两者皆满足）。对于寻找此类数组的最小值问题，可直接放弃二分查找，而使用线性查找替代。 + +```Python [] +class Solution: + def minArray(self, numbers: [int]) -> int: + i, j = 0, len(numbers) - 1 + while i < j: + m = (i + j) // 2 + if numbers[m] > numbers[j]: i = m + 1 + elif numbers[m] < numbers[j]: j = m + else: return min(numbers[i:j]) + return numbers[i] +``` + +```Java [] +class Solution { + public int minArray(int[] numbers) { + int i = 0, j = numbers.length - 1; + while (i < j) { + int m = (i + j) / 2; + if (numbers[m] > numbers[j]) i = m + 1; + else if (numbers[m] < numbers[j]) j = m; + else { + int x = i; + for(int k = i + 1; k < j; k++) { + if(numbers[k] < numbers[x]) x = k; + } + return numbers[x]; + } + } + return numbers[i]; + } +} +``` + +```C++ [] +class Solution { +public: + int minArray(vector& numbers) { + int i = 0, j = numbers.size() - 1; + while (i < j) { + int m = (i + j) / 2; + if (numbers[m] > numbers[j]) i = m + 1; + else if (numbers[m] < numbers[j]) j = m; + else { + int x = i; + for(int k = i + 1; k < j; k++) { + if(numbers[k] < numbers[x]) x = k; + } + return numbers[x]; + } + } + return numbers[i]; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 12. \347\237\251\351\230\265\344\270\255\347\232\204\350\267\257\345\276\204.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 12. \347\237\251\351\230\265\344\270\255\347\232\204\350\267\257\345\276\204.md" new file mode 100644 index 0000000..93ddeb2 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 12. \347\237\251\351\230\265\344\270\255\347\232\204\350\267\257\345\276\204.md" @@ -0,0 +1,101 @@ +## 解题思路： + +本问题是典型的矩阵搜索问题，可使用 **深度优先搜索（DFS）+ 剪枝** 解决。 + +- **深度优先搜索：** 可以理解为暴力法遍历矩阵中所有字符串可能性。DFS 通过递归，先朝一个方向搜到底，再回溯至上个节点，沿另一个方向搜索，以此类推。 +- **剪枝：** 在搜索中，遇到 `这条路不可能和目标字符串匹配成功` 的情况（*例如：此矩阵元素和目标字符不同、此元素已被访问）*，则应立即返回，称之为 `可行性剪枝` 。 + +![Picture0.png](https://pic.leetcode-cn.com/1604944042-glmqJO-Picture0.png){:width=500} + +### DFS 解析： + +- **递归参数：** 当前元素在矩阵 `board` 中的行列索引 `i` 和 `j` ，当前目标字符在 `word` 中的索引 `k` 。 +- **终止条件：** + 1. 返回 $false$ ： (1) 行或列索引越界 **或** (2) 当前矩阵元素与目标字符不同 **或** (3) 当前矩阵元素已访问过 （ (3) 可合并至 (2) ） 。 + 2. 返回 $true$ ： `k = len(word) - 1` ，即字符串 `word` 已全部匹配。 +- **递推工作：** + 1. 标记当前矩阵元素： 将 `board[i][j]` 修改为 **空字符** `''` ，代表此元素已访问过，防止之后搜索时重复访问。 + 2. 搜索下一单元格： 朝当前元素的 **上、下、左、右** 四个方向开启下层递归，使用 `或` 连接 （代表只需找到一条可行路径就直接返回，不再做后续 DFS ），并记录结果至 `res` 。 + 3. 还原当前矩阵元素： 将 `board[i][j]` 元素还原至初始值，即 `word[k]` 。 +- **返回值：** 返回布尔量 `res` ，代表是否搜索到目标字符串。 + +> 使用空字符（Python: `''` , Java/C++: `'\0'` ）做标记是为了防止标记字符与矩阵原有字符重复。当存在重复时，此算法会将矩阵原有字符认作标记字符，从而出现错误。 + + + +### 复杂度分析： + +> $M, N$ 分别为矩阵行列大小， $K$ 为字符串 `word` 长度。 + +- **时间复杂度 $O(3^KMN)$ ：** 最差情况下，需要遍历矩阵中长度为 $K$ 字符串的所有方案，时间复杂度为 $O(3^K)$；矩阵中共有 $MN$ 个起点，时间复杂度为 $O(MN)$ 。 + - **方案数计算：** 设字符串长度为 $K$ ，搜索中每个字符有上、下、左、右四个方向可以选择，舍弃回头（上个字符）的方向，剩下 $3$ 种选择，因此方案数的复杂度为 $O(3^K)$ 。 +- **空间复杂度 $O(K)$ ：** 搜索过程中的递归深度不超过 $K$ ，因此系统因函数调用累计使用的栈空间占用 $O(K)$ （因为函数返回后，系统调用的[栈空间会释放](https://leetcode-cn.com/explore/orignial/card/recursion-i/259/complexity-analysis/1223/)）。最坏情况下 $K = MN$ ，递归深度为 $MN$ ，此时系统栈使用 $O(MN)$ 的额外空间。 + +## 代码： + +```Python [] +class Solution: + def exist(self, board: List[List[str]], word: str) -> bool: + def dfs(i, j, k): + if not 0 <= i < len(board) or not 0 <= j < len(board[0]) or board[i][j] != word[k]: return False + if k == len(word) - 1: return True + board[i][j] = '' + res = dfs(i + 1, j, k + 1) or dfs(i - 1, j, k + 1) or dfs(i, j + 1, k + 1) or dfs(i, j - 1, k + 1) + board[i][j] = word[k] + return res + + for i in range(len(board)): + for j in range(len(board[0])): + if dfs(i, j, 0): return True + return False +``` + +```Java [] +class Solution { + public boolean exist(char[][] board, String word) { + char[] words = word.toCharArray(); + for(int i = 0; i < board.length; i++) { + for(int j = 0; j < board[0].length; j++) { + if(dfs(board, words, i, j, 0)) return true; + } + } + return false; + } + boolean dfs(char[][] board, char[] word, int i, int j, int k) { + if(i >= board.length || i < 0 || j >= board[0].length || j < 0 || board[i][j] != word[k]) return false; + if(k == word.length - 1) return true; + board[i][j] = '\0'; + boolean res = dfs(board, word, i + 1, j, k + 1) || dfs(board, word, i - 1, j, k + 1) || + dfs(board, word, i, j + 1, k + 1) || dfs(board, word, i , j - 1, k + 1); + board[i][j] = word[k]; + return res; + } +} +``` + +```C++ [] +class Solution { +public: + bool exist(vector>& board, string word) { + rows = board.size(); + cols = board[0].size(); + for(int i = 0; i < rows; i++) { + for(int j = 0; j < cols; j++) { + if(dfs(board, word, i, j, 0)) return true; + } + } + return false; + } +private: + int rows, cols; + bool dfs(vector>& board, string word, int i, int j, int k) { + if(i >= rows || i < 0 || j >= cols || j < 0 || board[i][j] != word[k]) return false; + if(k == word.size() - 1) return true; + board[i][j] = '\0'; + bool res = dfs(board, word, i + 1, j, k + 1) || dfs(board, word, i - 1, j, k + 1) || + dfs(board, word, i, j + 1, k + 1) || dfs(board, word, i , j - 1, k + 1); + board[i][j] = word[k]; + return res; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 13. \346\234\272\345\231\250\344\272\272\347\232\204\350\277\220\345\212\250\350\214\203\345\233\264.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 13. \346\234\272\345\231\250\344\272\272\347\232\204\350\277\220\345\212\250\350\214\203\345\233\264.md" new file mode 100644 index 0000000..a47df7a --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 13. \346\234\272\345\231\250\344\272\272\347\232\204\350\277\220\345\212\250\350\214\203\345\233\264.md" @@ -0,0 +1,240 @@ +## 解题思路： + +> 本题与 [矩阵中的路径](https://leetcode-cn.com/problems/ju-zhen-zhong-de-lu-jing-lcof/solution/mian-shi-ti-12-ju-zhen-zhong-de-lu-jing-shen-du-yo/) 类似，是典型的搜索 & 回溯问题。在介绍回溯算法算法前，为提升计算效率，首先讲述两项前置工作： **数位之和计算** 、 **可达解分析** 。 + +### 数位之和计算： + +设一数字 $x$ ，向下取整除法符号 $//$ ，求余符号 $\odot$ ，则有： + +- $x \odot 10$ ：得到 $x$ 的个位数字； +- $x // 10$ ： 令 $x$ 的十进制数向右移动一位，即删除个位数字。 + +因此，可通过循环求得数位和 $s$ ，数位和计算的封装函数如下所示： + +```Python [] +def sums(x): + s = 0 + while x != 0: + s += x % 10 + x = x // 10 + return s +``` + +```Java [] +int sums(int x) + int s = 0; + while(x != 0) { + s += x % 10; + x = x / 10; + } + return s; +``` + +```C++ [] +int sums(int x) + int s = 0; + while(x != 0) { + s += x % 10; + x = x / 10; + } + return s; +``` + +由于机器人每次只能移动一格（即只能从 $x$ 运动至 $x \pm 1$），因此每次只需计算 $x$ 到 $x \pm 1$ 的**数位和增量**。本题说明 $1 \leq n,m \leq 100$ ，以下公式仅在此范围适用。 + +**数位和增量公式：** 设 $x$ 的数位和为 $s_x$ ， $x+1$ 的数位和为 $s_{x+1}$ ； + +1. **当 $(x + 1) \odot 10 = 0$ 时：** $s_{x+1} = s_x - 8$ ，例如 $19, 20$ 的数位和分别为 $10, 2$ ； +2. **当 $(x + 1) \odot 10 \neq 0$ 时：** $s_{x+1} = s_x + 1$ ，例如 $1, 2$ 的数位和分别为 $1, 2$ 。 + +> 以下代码为增量公式的三元表达式写法，将整合入最终代码中。 + +```Python [] +s_x + 1 if (x + 1) % 10 else s_x - 8 +``` + +```Java [] +(x + 1) % 10 != 0 ? s_x + 1 : s_x - 8; +``` + +```C++ [] +(x + 1) % 10 != 0 ? s_x + 1 : s_x - 8; +``` + +### 可达解分析： + +根据数位和增量公式得知，数位和每逢 **进位** 突变一次。根据此特点，矩阵中 **满足数位和的解** 构成的几何形状形如多个 **等腰直角三角形** ，每个三角形的直角顶点位于 $0, 10, 20, ...$ 等数位和突变的矩阵索引处 。 + +三角形内的解虽然都满足数位和要求，但由于机器人每步只能走一个单元格，而三角形间不一定是连通的，因此机器人不一定能到达，称之为 **不可达解** ；同理，可到达的解称为 **可达解** *（本题求此解）* 。 + +> 图例展示了 $n,m = 20$ ， $k \in [6, 19]$ 的可达解、不可达解、非解，以及连通性的变化。 + + + +根据可达解的结构和连通性，易推出机器人可 **仅通过向右和向下移动，访问所有可达解** 。 + +- **三角形内部：** 全部连通，易证； +- **两三角形连通处：** 若某三角形内的解为可达解，则必与其左边或上边的三角形连通（即相交），即机器人必可从左边或上边走进此三角形。 + +![Picture9.png](https://pic.leetcode-cn.com/1603024999-XMpudY-Picture9.png){:width=500} + +## 方法一：深度优先遍历 DFS + +- **深度优先搜索：** 可以理解为暴力法模拟机器人在矩阵中的所有路径。DFS 通过递归，先朝一个方向搜到底，再回溯至上个节点，沿另一个方向搜索，以此类推。 +- **剪枝：** 在搜索中，遇到数位和超出目标值、此元素已访问，则应立即返回，称之为 `可行性剪枝` 。 + +### **算法解析：** + +- **递归参数：** 当前元素在矩阵中的行列索引 `i` 和 `j` ，两者的数位和 `si`, `sj` 。 +- **终止条件：** 当 ① 行列索引越界 **或** ② 数位和超出目标值 `k` **或** ③ 当前元素已访问过 时，返回 $0$ ，代表不计入可达解。 +- **递推工作：** + 1. **标记当前单元格** ：将索引 `(i, j)` 存入 Set `visited` 中，代表此单元格已被访问过。 + 2. **搜索下一单元格：** 计算当前元素的 **下、右** 两个方向元素的数位和，并开启下层递归 。 +- **回溯返回值：** 返回 `1 + 右方搜索的可达解总数 + 下方搜索的可达解总数`，代表从本单元格递归搜索的可达解总数。 + + + +### 复杂度分析： + +> 设矩阵行列数分别为 $M, N$ 。 + +- **时间复杂度 $O(MN)$ ：** 最差情况下，机器人遍历矩阵所有单元格，此时时间复杂度为 $O(MN)$ 。 +- **空间复杂度 $O(MN)$ ：** 最差情况下，Set `visited` 内存储矩阵所有单元格的索引，使用 $O(MN)$ 的额外空间。 + +### 代码： + +> Java/C++ 代码中 `visited` 为辅助矩阵，Python 中为 Set 。 + +```Python [] +class Solution: + def movingCount(self, m: int, n: int, k: int) -> int: + def dfs(i, j, si, sj): + if i >= m or j >= n or k < si + sj or (i, j) in visited: return 0 + visited.add((i,j)) + return 1 + dfs(i + 1, j, si + 1 if (i + 1) % 10 else si - 8, sj) + dfs(i, j + 1, si, sj + 1 if (j + 1) % 10 else sj - 8) + + visited = set() + return dfs(0, 0, 0, 0) +``` + +```Java [] +class Solution { + int m, n, k; + boolean[][] visited; + public int movingCount(int m, int n, int k) { + this.m = m; this.n = n; this.k = k; + this.visited = new boolean[m][n]; + return dfs(0, 0, 0, 0); + } + public int dfs(int i, int j, int si, int sj) { + if(i >= m || j >= n || k < si + sj || visited[i][j]) return 0; + visited[i][j] = true; + return 1 + dfs(i + 1, j, (i + 1) % 10 != 0 ? si + 1 : si - 8, sj) + dfs(i, j + 1, si, (j + 1) % 10 != 0 ? sj + 1 : sj - 8); + } +} +``` + +```C++ [] +class Solution { +public: + int movingCount(int m, int n, int k) { + vector> visited(m, vector(n, 0)); + return dfs(0, 0, 0, 0, visited, m, n, k); + } +private: + int dfs(int i, int j, int si, int sj, vector> &visited, int m, int n, int k) { + if(i >= m || j >= n || k < si + sj || visited[i][j]) return 0; + visited[i][j] = true; + return 1 + dfs(i + 1, j, (i + 1) % 10 != 0 ? si + 1 : si - 8, sj, visited, m, n, k) + + dfs(i, j + 1, si, (j + 1) % 10 != 0 ? sj + 1 : sj - 8, visited, m, n, k); + } +}; +``` + +## 方法二：广度优先遍历 BFS + +- **BFS/DFS ：** 两者目标都是遍历整个矩阵，不同点在于搜索顺序不同。DFS 是朝一个方向走到底，再回退，以此类推；BFS 则是按照“平推”的方式向前搜索。 +- **BFS 实现：** 通常利用队列实现广度优先遍历。 + +### 算法解析： + +- **初始化：** 将机器人初始点 $(0, 0)$ 加入队列 `queue` ； +- **迭代终止条件：** `queue` 为空。代表已遍历完所有可达解。 +- **迭代工作：** + 1. **单元格出队：** 将队首单元格的 索引、数位和 弹出，作为当前搜索单元格。 + 2. **判断是否跳过：** 若 ① 行列索引越界 **或** ② 数位和超出目标值 `k` **或** ③ 当前元素已访问过 时，执行 `continue` 。 + 3. **标记当前单元格** ：将单元格索引 `(i, j)` 存入 Set `visited` 中，代表此单元格 **已被访问过** 。 + 4. **单元格入队：** 将当前元素的 **下方、右方** 单元格的 **索引、数位和** 加入 `queue` 。 +- **返回值：** Set `visited` 的长度 `len(visited)` ，即可达解的数量。 + +> Java/C++ 使用了辅助变量 `res` 统计可达解数量； Python 直接返回 Set 的元素数 `len(visited)` 即可。 + + + +### 复杂度分析： + +> 设矩阵行列数分别为 $M, N$ 。 + +- **时间复杂度 $O(MN)$ ：** 最差情况下，机器人遍历矩阵所有单元格，此时时间复杂度为 $O(MN)$ 。 +- **空间复杂度 $O(MN)$ ：** 最差情况下，Set `visited` 内存储矩阵所有单元格的索引，使用 $O(MN)$ 的额外空间。 + +### 代码： + +> Java/C++ 代码中 `visited` 为辅助矩阵，Python 中为 Set 。 + +```Python [] +class Solution: + def movingCount(self, m: int, n: int, k: int) -> int: + queue, visited = [(0, 0, 0, 0)], set() + while queue: + i, j, si, sj = queue.pop(0) + if i >= m or j >= n or k < si + sj or (i, j) in visited: continue + visited.add((i,j)) + queue.append((i + 1, j, si + 1 if (i + 1) % 10 else si - 8, sj)) + queue.append((i, j + 1, si, sj + 1 if (j + 1) % 10 else sj - 8)) + return len(visited) +``` + +```Java [] +class Solution { + public int movingCount(int m, int n, int k) { + boolean[][] visited = new boolean[m][n]; + int res = 0; + Queue queue= new LinkedList(); + queue.add(new int[] { 0, 0, 0, 0 }); + while(queue.size() > 0) { + int[] x = queue.poll(); + int i = x[0], j = x[1], si = x[2], sj = x[3]; + if(i >= m || j >= n || k < si + sj || visited[i][j]) continue; + visited[i][j] = true; + res ++; + queue.add(new int[] { i + 1, j, (i + 1) % 10 != 0 ? si + 1 : si - 8, sj }); + queue.add(new int[] { i, j + 1, si, (j + 1) % 10 != 0 ? sj + 1 : sj - 8 }); + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int movingCount(int m, int n, int k) { + vector> visited(m, vector(n, 0)); + int res = 0; + queue> que; + que.push({ 0, 0, 0, 0 }); + while(que.size() > 0) { + vector x = que.front(); + que.pop(); + int i = x[0], j = x[1], si = x[2], sj = x[3]; + if(i >= m || j >= n || k < si + sj || visited[i][j]) continue; + visited[i][j] = true; + res++; + que.push({ i + 1, j, (i + 1) % 10 != 0 ? si + 1 : si - 8, sj }); + que.push({ i, j + 1, si, (j + 1) % 10 != 0 ? sj + 1 : sj - 8 }); + } + return res; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 14- I. \345\211\252\347\273\263\345\255\220.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 14- I. \345\211\252\347\273\263\345\255\220.md" new file mode 100644 index 0000000..5aacae5 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 14- I. \345\211\252\347\273\263\345\255\220.md" @@ -0,0 +1,164 @@ +## 解题思路： + +- 设将长度为 $n$ 的绳子切为 $a$ 段： + +$$ +n = n_1 + n_2 + ... + n_a +$$ + +- 本题等价于求解： + +$$ +\max(n_1 \times n_2 \times ... \times n_a) +$$ + +> 以下数学推导总体分为两步：① 当所有绳段长度相等时，乘积最大。② 最优的绳段长度为 $3$ 。 + +### 数学推导： + +- 以下公式为“算术几何均值不等式” ，等号当且仅当 $n_1 = n_2 = ... = n_a$ 时成立。 + +$$ +\frac{n_1 + n_2 + ... + n_a}{a} \geq \sqrt[a]{n_1 n_2 ... n_a} +$$ + +> **推论一：** 将绳子 **以相等的长度等分为多段** ，得到的乘积最大。 + +- 设将绳子按照 $x$ 长度等分为 $a$ 段，即 $n = ax$ ，则乘积为 $x^a$ 。观察以下公式，由于 $n$ 为常数，因此当 $x^{\frac{1}{x}}$ 取最大值时， 乘积达到最大值。 + +$$ +x^a = x^{\frac{n}{x}} = (x^{\frac{1}{x}})^n +$$ + +- 根据分析，可将问题转化为求 $y = x^{\frac{1}{x}}$ 的极大值，因此对 $x$ 求导数。 + +$$ +\begin{aligned} + \ln y & = \frac{1}{x} \ln x & \text{取对数} \\ + \frac{1}{y} \dot {y} & = \frac{1}{x^2} - \frac{1}{x^2} \ln x & \text{对 $x$ 求导} \\ + & = \frac{1 - \ln x}{x^2} \\ + \dot {y} & = \frac{1 - \ln x}{x^2} x^{\frac{1}{x}} & \text{整理得} +\end{aligned} +$$ + +- 令 $\dot {y} = 0$ ，则 $1 - \ln x = 0$ ，易得驻点为 $x_0 = e \approx 2.7$ ；根据以下公式，可知 $x_0$ 为极大值点。 + +$$ +\dot {y} +\begin{cases} + > 0 & , x \in [- \infty, e) \\ + < 0 & , x \in (e, \infty] \\ +\end{cases} +$$ + +- 由于切分长度 $x$ 必须为整数，最接近 $e$ 的整数为 $2$ 或 $3$ 。如下式所示，代入 $x = 2$ 和 $x = 3$ ，得出 $x = 3$ 时，乘积达到最大。 + +$$ +y(3) = 3^{1/3} \approx 1.44 \\ +y(2) = 2^{1/2} \approx 1.41 +$$ + +- 口算对比方法：给两数字同时取 $6$ 次方，再对比。 + +$$ +[y(3)]^6 = (3^{1/3})^6 = 9 \\ +[y(2)]^6 = (2^{1/2})^6 = 8 +$$ + +> **推论二：** 尽可能将绳子以长度 $3$ 等分为多段时，乘积最大。 + +### 切分规则： + +1. **最优：** $3$ 。把绳子尽可能切为多个长度为 $3$ 的片段，留下的最后一段绳子的长度可能为 $0,1,2$ 三种情况。 +2. **次优：** $2$ 。若最后一段绳子长度为 $2$ ；则保留，不再拆为 $1+1$ 。 +3. **最差：** $1$ 。若最后一段绳子长度为 $1$ ；则应把一份 $3 + 1$ 替换为 $2 + 2$，因为 $2 \times 2 > 3 \times 1$。 + +### 算法流程： + +1. 当 $n \leq 3$ 时，按照规则应不切分，但由于题目要求必须剪成 $m>1$ 段，因此必须剪出一段长度为 $1$ 的绳子，即返回 $n - 1$ 。 +2. 当 $n>3$ 时，求 $n$ 除以 $3$ 的 整数部分 $a$ 和 余数部分 $b$ （即 $n = 3a + b$ ），并分为以下三种情况： + - 当 $b = 0$ 时，直接返回 $3^a$； + - 当 $b = 1$ 时，要将一个 $1 + 3$ 转换为 $2+2$，因此返回 $3^{a-1} \times 4$； + - 当 $b = 2$ 时，返回 $3^a \times 2$。 + +![Picture1.png](https://pic.leetcode-cn.com/1f9adeaa7b9fff0ab19c9d29e3a8f98749011d22dc162d67bdbe223f1a38119f-Picture1.png){:width=600} + +### 复杂度分析： + +- **时间复杂度 $O(1)$ ：** 仅有求整、求余、次方运算。 + - [求整和求余运算](https://stackoverflow.com/questions/35189851/time-complexity-of-modulo-operator-in-python)：资料提到不超过机器数的整数可以看作是 $O(1)$ ； + - [幂运算](https://stackoverflow.com/questions/32418731/java-math-powa-b-time-complexity)：查阅资料，提到浮点取幂为 $O(1)$ 。 +- **空间复杂度 $O(1)$ ：** 变量 `a` 和 `b` 使用常数大小额外空间。 + +## 代码： + +> Python 中常见有三种幂计算函数： **`*`** 和 **`pow()`** 的时间复杂度均为 $O(\log a)$ ；而 **`math.pow()`** 始终调用 C 库的 `pow()` 函数，其执行浮点取幂，时间复杂度为 $O(1)$ 。 + +```Python [] +class Solution: + def cuttingRope(self, n: int) -> int: + if n <= 3: return n - 1 + a, b = n // 3, n % 3 + if b == 0: return int(math.pow(3, a)) + if b == 1: return int(math.pow(3, a - 1) * 4) + return int(math.pow(3, a) * 2) +``` + +```Java [] +class Solution { + public int cuttingRope(int n) { + if(n <= 3) return n - 1; + int a = n / 3, b = n % 3; + if(b == 0) return (int)Math.pow(3, a); + if(b == 1) return (int)Math.pow(3, a - 1) * 4; + return (int)Math.pow(3, a) * 2; + } +} +``` + +```C++ [] +class Solution { +public: + int cuttingRope(int n) { + if(n <= 3) return n - 1; + int a = n / 3, b = n % 3; + if(b == 0) return pow(3, a); + if(b == 1) return pow(3, a - 1) * 4; + return pow(3, a) * 2; + } +}; +``` + +数学推导需要一定的知识基础。下面分享一种基于贪心算法的思路，个人认为适合于时间有限情况下的快速解题。 + +### 贪心算法思路： + +> 设一绳子长度为 $n$ ( $n>1$ )，则其必可被切分为两段 $n=n_1+n_2$ 。 +> 根据经验推测，切分的两数字乘积往往原数字更大，即往往有 $n_1 \times n_2 > n_1 + n_2 = n$ 。 +> +> - **例如绳子长度为 $6$ ：** $6 = 3 + 3 < 3 \times 3 = 9$ ； +> - **也有少数反例，例如 $2$ ：** $2 = 1 + 1 > 1 \times 1 = 1$ 。 + +- **推论一：** 合理的切分方案可以带来更大的乘积。 + +> 设一绳子长度为 $n$ ( $n>1$ )，**切分为两段** $n=n_1+n_2$ ，**切分为三段** $n=n_1+n_2+n_3$ 。 +> 根据经验推测，**三段** 的乘积往往更大，即往往有 $n_1 n_2 n_3 > n_1 n_2$ 。 +> +> - **例如绳子长度为 $9$ ：** 两段 $9=4+5$ 和 三段 $9=3+3+3$，则有 $4 \times 5 < 3 \times 3 \times 3$ 。 +> - **也有少数反例，例如 $6$ ：** 两段 $6=3+3$ 和 三段 $6=2+2+2$，则有 $3 \times 3 > 2 \times 2 \times 2$ 。 + +- **推论二：** 若切分方案合理，绳子段切分的越多，乘积越大。 + +> 总体上看，貌似长绳子切分为越多段乘积越大，但其实到某个长度分界点后，乘积到达最大值，就不应再切分了。 +> **问题转化：** 是否有**优先级最高的长度** $x$ 存在？若有，则应该尽可能把绳子以 $x$ 长度切为多段，以获取最大乘积。 + +- **推论三：** 为使乘积最大，只有长度为 $2$ 和 $3$ 的绳子不应再切分，且 $3$ 比 $2$ 更优 *（详情见下表）* 。 + +| 绳子切分方案 | 乘积 | 结论 | +| ------------- | ------------------------------------------ | ----------------------------------------------------------------- | +| $2 = 1 + 1$ | $1 \times 1 = 1$ | $2$ 不应切分 | +| $3=1+2$ | $1 \times 2 = 2$ | $3$ 不应切分 | +| $4=2+2=1+3$ | $2 \times 2 = 4 > 1 \times 3 = 3$ | $4$ 和 $2$ 等价，且 $2+2$ 比 $1+3$ 更优 | +| $5=2+3=1+4$ | $2 \times 3 = 6 > 1 \times 4 = 4$ | $5$ 应切分为 $2+3$ | +| $6=3+3=2+2+2$ | $3 \times 3 = 9 > 2 \times 2 \times 2 = 8$ | $6$ 应切分为 $3+3$ ，进而**推出 $3$ 比 $2$ 更优** | +| $>7$ | ... | **长绳**（长度>7）可转化为多个**短绳**（长度1~6），因此肯定应切分 | diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 14- II. \345\211\252\347\273\263\345\255\220 II.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 14- II. \345\211\252\347\273\263\345\255\220 II.md" new file mode 100644 index 0000000..a6e571e --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 14- II. \345\211\252\347\273\263\345\255\220 II.md" @@ -0,0 +1,232 @@ +此题与 [剑指 Offer 14- I. 剪绳子](https://leetcode-cn.com/problems/jian-sheng-zi-lcof/) 主体等价，唯一不同在于本题目涉及 **大数越界情况下的求余问题** 。建议先做上一道题，在此基础上再研究此题目的大数求余方法。 + +## 解题思路： + +- 设将长度为 $n$ 的绳子切为 $a$ 段： + +$$ +n = n_1 + n_2 + ... + n_a +$$ + +- 本题等价于求解： + +$$ +\max(n_1 \times n_2 \times ... \times n_a) +$$ + +> 以下数学推导总体分为两步：① 当所有绳段长度相等时，乘积最大。② 最优的绳段长度为 $3$ 。 + +### 数学推导： + +- 以下公式为“算术几何均值不等式” ，等号当且仅当 $n_1 = n_2 = ... = n_a$ 时成立。 + +$$ +\frac{n_1 + n_2 + ... + n_a}{a} \geq \sqrt[a]{n_1 n_2 ... n_a} +$$ + +> **推论一：** 将绳子 **以相等的长度等分为多段** ，得到的乘积最大。 + +- 设将绳子按照 $x$ 长度等分为 $a$ 段，即 $n = ax$ ，则乘积为 $x^a$ 。观察以下公式，由于 $n$ 为常数，因此当 $x^{\frac{1}{x}}$ 取最大值时， 乘积达到最大值。 + +$$ +x^a = x^{\frac{n}{x}} = (x^{\frac{1}{x}})^n +$$ + +- 根据分析，可将问题转化为求 $y = x^{\frac{1}{x}}$ 的极大值，因此对 $x$ 求导数。 + +$$ +\begin{aligned} + \ln y & = \frac{1}{x} \ln x & \text{取对数} \\ + \frac{1}{y} \dot {y} & = \frac{1}{x^2} - \frac{1}{x^2} \ln x & \text{对 $x$ 求导} \\ + & = \frac{1 - \ln x}{x^2} \\ + \dot {y} & = \frac{1 - \ln x}{x^2} x^{\frac{1}{x}} & \text{整理得} +\end{aligned} +$$ + +- 令 $\dot {y} = 0$ ，则 $1 - \ln x = 0$ ，易得驻点为 $x_0 = e \approx 2.7$ ；根据以下公式，可知 $x_0$ 为极大值点。 + +$$ +\dot {y} +\begin{cases} + > 0 & , x \in [- \infty, e) \\ + < 0 & , x \in (e, \infty] \\ +\end{cases} +$$ + +- 由于切分长度 $x$ 必须为整数，最接近 $e$ 的整数为 $2$ 或 $3$ 。如下式所示，代入 $x = 2$ 和 $x = 3$ ，得出 $x = 3$ 时，乘积达到最大。 + +$$ +y(3) = 3^{1/3} \approx 1.44 \\ +y(2) = 2^{1/2} \approx 1.41 +$$ + +- 口算对比方法：给两数字同时取 $6$ 次方，再对比。 + +$$ +[y(3)]^6 = (3^{1/3})^6 = 9 \\ +[y(2)]^6 = (2^{1/2})^6 = 8 +$$ + +> **推论二：** 尽可能将绳子以长度 $3$ 等分为多段时，乘积最大。 + +### 切分规则： + +1. **最优：** $3$ 。把绳子尽可能切为多个长度为 $3$ 的片段，留下的最后一段绳子的长度可能为 $0,1,2$ 三种情况。 +2. **次优：** $2$ 。若最后一段绳子长度为 $2$ ；则保留，不再拆为 $1+1$ 。 +3. **最差：** $1$ 。若最后一段绳子长度为 $1$ ；则应把一份 $3 + 1$ 替换为 $2 + 2$，因为 $2 \times 2 > 3 \times 1$。 + +### 算法流程： + +1. 当 $n \leq 3$ 时，按照规则应不切分，但由于题目要求必须剪成 $m>1$ 段，因此必须剪出一段长度为 $1$ 的绳子，即返回 $n - 1$ 。 +2. 当 $n>3$ 时，求 $n$ 除以 $3$ 的 整数部分 $a$ 和 余数部分 $b$ （即 $n = 3a + b$ ），并分为以下三种情况（设求余操作符号为 "$\odot$" ）： + - 当 $b = 0$ 时，直接返回 $3^a \odot 1000000007$； + - 当 $b = 1$ 时，要将一个 $1 + 3$ 转换为 $2+2$，因此返回 $(3^{a-1} \times 4)\odot 1000000007$； + - 当 $b = 2$ 时，返回 $(3^a \times 2) \odot 1000000007$。 + +![Picture1.png](https://pic.leetcode-cn.com/1f9adeaa7b9fff0ab19c9d29e3a8f98749011d22dc162d67bdbe223f1a38119f-Picture1.png){:width=600} + +### 大数求余解法： + +**大数越界：** 当 $a$ 增大时，最后返回的 $3^a$ 大小以指数级别增长，可能超出 `int32` 甚至 `int64` 的取值范围，导致返回值错误。 +**大数求余问题：** 在仅使用 `int32` 类型存储的前提下，正确计算 $x^a$ 对 $p$ 求余（即 $x^a \odot p$ ）的值。 +**解决方案：** *循环求余* 、 *快速幂求余* ，其中后者的时间复杂度更低，两种方法均基于以下求余运算规则推出： + +$$ +(xy) \odot p = [(x \odot p)(y \odot p)] \odot p +$$ + +### 1. 循环求余： + +- 根据求余运算性质推出（∵ 本题中 $x 0: + if a % 2: rem = (rem * x) % p + x = x ** 2 % p + a //= 2 + return rem +``` + +- **帮助理解：** 根据下表， 初始状态 $rem=1$, $x=3$, $a=19$, $p=1000000007$ ，最后会将 $rem \times (x^a \odot p)$ 化为 $rem \times (x^0 \odot p) = rem \times 1$ 的形式，即 $rem$ 为余数答案。 + +| $n$ | $rem \times (x^a \odot p)$ | $rem_n=rem_{n-1} \times x_{n-1} \odot p$ | $x_n=x_{n-1}^2 \odot p$ | $a_n=a_{n-1}//2$ | +| --- | -----------------------------------------: | ---------------------------------------: | -----------------------------: | :--------------: | +| $1$ | $1 \times (3^{19} \odot p)$ | $1$ | $3$ | $19$ | +| $2$ | $3 \times (9^{9} \odot p)$ | $3=1\times3\odot p$ | $9=3^2 \odot p$ | $9=19//2$ | +| $3$ | $27 \times (81^{4} \odot p)$ | $27 = 3 \times 9 \odot p$ | $81=9^2\odot p$ | $4=9//2$ | +| $4$ | $27 \times (6561^{2} \odot p)$ | $27$ | $6561=81^2 \odot p$ | $2=4//2$ | +| $5$ | $27 \times (43046721^{1} \odot p)$ | $27$ | $43046721=6561^2 \odot p$ | $1=2//2$ | +| $6$ | $162261460 \times (175880701^{0} \odot p)$ | $162261460=27 \times 43046721 \odot p$ | $175880701=43046721^2 \odot p$ | $0=1//2$ | + +### 复杂度分析： + +> 以下为 **二分求余法** 的复杂度。 + +- **时间复杂度 $O(\log_2 N)$ ：** 其中 $N=a$ ，二分法为对数级别复杂度，每轮仅有求整、求余、次方运算。 + - [求整和求余运算](https://stackoverflow.com/questions/35189851/time-complexity-of-modulo-operator-in-python)：资料提到不超过机器数的整数可以看作是 $O(1)$ ； + - [幂运算](https://stackoverflow.com/questions/32418731/java-math-powa-b-time-complexity)：查阅资料，提到浮点取幂为 $O(1)$ 。 +- **空间复杂度 $O(1)$ ：** 变量 `a, b, p, x, rem` 使用常数大小额外空间。 + +### 代码： + +**Python 代码：** 由于语言特性，理论上 Python 中的变量取值范围由系统内存大小决定（无限大），因此在 Python 中其实不用考虑大数越界问题。 +**Java/C++ 代码：** 根据二分法计算原理，至少要保证变量 `x` 和 `rem` 可以正确存储 $1000000007^2$ ，而 $2^{64} > 1000000007^2 > 2^{32}$ ，因此我们选取 `long` 类型。 + +```Python [] +class Solution: + def cuttingRope(self, n: int) -> int: + if n <= 3: return n - 1 + a, b, p, x, rem = n // 3 - 1, n % 3, 1000000007, 3 , 1 + while a > 0: + if a % 2: rem = (rem * x) % p + x = x ** 2 % p + a //= 2 + if b == 0: return (rem * 3) % p # = 3^(a+1) % p + if b == 1: return (rem * 4) % p # = 3^a * 4 % p + return (rem * 6) % p # = 3^(a+1) * 2 % p +``` + +```Java [] +class Solution { + public int cuttingRope(int n) { + if(n <= 3) return n - 1; + int b = n % 3, p = 1000000007; + long rem = 1, x = 3; + for(int a = n / 3 - 1; a > 0; a /= 2) { + if(a % 2 == 1) rem = (rem * x) % p; + x = (x * x) % p; + } + if(b == 0) return (int)(rem * 3 % p); + if(b == 1) return (int)(rem * 4 % p); + return (int)(rem * 6 % p); + } +} +``` + +```C++ [] +class Solution { +public: + int cuttingRope(int n) { + if(n <= 3) return n - 1; + int b = n % 3, p = 1000000007; + long rem = 1, x = 3; + for(int a = n / 3 - 1; a > 0; a /= 2) { + if(a % 2 == 1) rem = (rem * x) % p; + x = (x * x) % p; + } + if(b == 0) return (int)(rem * 3 % p); + if(b == 1) return (int)(rem * 4 % p); + return (int)(rem * 6 % p); + } +}; +``` + +```Python [] +# 由于语言特性，Python 可以不考虑大数越界问题 +class Solution: + def cuttingRope(self, n: int) -> int: + if n <= 3: return n - 1 + a, b, p = n // 3, n % 3, 1000000007 + if b == 0: return 3 ** a % p + if b == 1: return 3 ** (a - 1) * 4 % p + return 3 ** a * 2 % p +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 15. \344\272\214\350\277\233\345\210\266\344\270\255 1 \347\232\204\344\270\252\346\225\260.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 15. \344\272\214\350\277\233\345\210\266\344\270\255 1 \347\232\204\344\270\252\346\225\260.md" new file mode 100644 index 0000000..92febc7 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 15. \344\272\214\350\277\233\345\210\266\344\270\255 1 \347\232\204\344\270\252\346\225\260.md" @@ -0,0 +1,123 @@ +## 方法一：逐位判断 + +- 根据 **与运算** 定义，设二进制数字 $n$ ，则有： + - 若 $n \& 1 = 0$ ，则 $n$ 二进制 **最右一位** 为 $0$ ； + - 若 $n \& 1 = 1$ ，则 $n$ 二进制 **最右一位** 为 $1$ 。 +- 根据以上特点，考虑以下 **循环判断** ： + 1. 判断 $n$ 最右一位是否为 $1$ ，根据结果计数。 + 2. 将 $n$ 右移一位（本题要求把数字 $n$ 看作无符号数，因此使用 **无符号右移** 操作）。 + +### 算法流程： + +1. 初始化数量统计变量 $res = 0$ 。 +2. 循环逐位判断： 当 $n = 0$ 时跳出。 + 1. **`res += n & 1` ：** 若 $n \& 1 = 1$ ，则统计数 $res$ 加一。 + 2. **`n >>= 1` ：** 将二进制数字 $n$ 无符号右移一位（ Java 中无符号右移为 "$>>>$" ） 。 +3. 返回统计数量 $res$ 。 + + + +### 复杂度分析： + +- **时间复杂度 $O(\log_2 n)$ ：** 此算法循环内部仅有 **移位、与、加** 等基本运算，占用 $O(1)$ ；逐位判断需循环 $log_2 n$ 次，其中 $\log_2 n$ 代表数字 $n$ 最高位 $1$ 的所在位数（例如 $\log_2 4 = 2$, $\log_2 16 = 4$）。 +- **空间复杂度 $O(1)$ ：** 变量 $res$ 使用常数大小额外空间。 + +### 代码： + +```Python [] +class Solution: + def hammingWeight(self, n: int) -> int: + res = 0 + while n: + res += n & 1 + n >>= 1 + return res +``` + +```Java [] +public class Solution { + public int hammingWeight(int n) { + int res = 0; + while(n != 0) { + res += n & 1; + n >>>= 1; + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int hammingWeight(uint32_t n) { + unsigned int res = 0; // c++ 使用无符号数 + while(n != 0) { + res += n & 1; + n >>= 1; + } + return res; + } +}; +``` + +## 方法二：巧用 $n \& (n - 1)$ + +- **$(n - 1)$ 解析：** 二进制数字 $n$ 最右边的 $1$ 变成 $0$ ，此 $1$ 右边的 $0$ 都变成 $1$ 。 +- **$n \& (n - 1)$ 解析：** 二进制数字 $n$ 最右边的 $1$ 变成 $0$ ，其余不变。 + +![Picture1.png](https://pic.leetcode-cn.com/f23d9ef4fcfd65d7fbe29e477cbf36110b2f34558020e8cff09a1e13c0275c43-Picture1.png){:width=400} + +### 算法流程： + +1. 初始化数量统计变量 $res$ 。 +2. 循环消去最右边的 $1$ ：当 $n = 0$ 时跳出。 + 1. **`res += 1` ：** 统计变量加 $1$ ； + 2. **`n &= n - 1` ：** 消去数字 $n$ 最右边的 $1$ 。 +3. 返回统计数量 $res$ 。 + + + +### 复杂度分析： + +- **时间复杂度 $O(M)$ ：** $n \& (n - 1)$ 操作仅有减法和与运算，占用 $O(1)$ ；设 $M$ 为二进制数字 $n$ 中 $1$ 的个数，则需循环 $M$ 次（每轮消去一个 $1$ ），占用 $O(M)$ 。 +- **空间复杂度 $O(1)$ ：** 变量 $res$ 使用常数大小额外空间。 + +### 代码： + +```Python [] +class Solution: + def hammingWeight(self, n: int) -> int: + res = 0 + while n: + res += 1 + n &= n - 1 + return res +``` + +```Java [] +public class Solution { + public int hammingWeight(int n) { + int res = 0; + while(n != 0) { + res++; + n &= n - 1; + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int hammingWeight(uint32_t n) { + int res = 0; + while(n != 0) { + res++; + n &= n - 1; + } + return res; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 16. \346\225\260\345\200\274\347\232\204\346\225\264\346\225\260\346\254\241\346\226\271.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 16. \346\225\260\345\200\274\347\232\204\346\225\264\346\225\260\346\254\241\346\226\271.md" new file mode 100644 index 0000000..ae9bbdf --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 16. \346\225\260\345\200\274\347\232\204\346\225\264\346\225\260\346\254\241\346\226\271.md" @@ -0,0 +1,128 @@ +## 解题思路： + +求 $x^n$ 最简单的方法是通过循环将 $n$ 个 $x$ 乘起来，依次求 $x^1, x^2, ..., x^{n-1}, x^n$ ，时间复杂度为 $O(n)$ 。 +**快速幂法** 可将时间复杂度降低至 $O(\log n)$ ，以下从 「分治法」 和 「二进制」 两个角度解析快速幂法。 + +### 快速幂解析（分治法角度）： + +> 快速幂实际上是分治思想的一种应用。 + +- **二分推导：** $x^n = x^{n/2} \times x^{n/2} = (x^2)^{n/2}$ ，令 $n/2$ 为整数，则需要分为奇偶两种情况（设向下取整除法符号为 "$//$" ）： + +$$ +x^n = +\begin{cases} + (x^2)^{n//2} & , n 为偶数 \\ + x(x^2)^{n//2} & , n 为奇数 \\ +\end{cases} +$$ + +> 观察发现，当 $n$ 为奇数时，二分后会多出一项 $x$ 。 + +- **幂结果获取：** + - 根据推导，可通过循环 $x = x^2$ 操作，每次把幂从 $n$ 降至 $n//2$ ，直至将幂降为 $0$ ； + - 设 $res=1$ ，则初始状态 $x^n = x^n \times res$ 。在循环二分时，每当 $n$ 为奇数时，将多出的一项 $x$ 乘入 $res$ ，则最终可化至 $x^n = x^0 \times res = res$ ，返回 $res$ 即可。 + +![Picture2.png](https://pic.leetcode-cn.com/1599885604-YzlkAN-Picture2.png){:width=500} + +- **转化为位运算：** + - 向下整除 $n // 2$ **等价于** 右移一位 $n >> 1$ ； + - 取余数 $n \% 2$ **等价于** 判断二进制最右位 $n \& 1$ ； + +### 快速幂解析（二进制角度）： + +> 利用十进制数字 $n$ 的二进制表示，可对快速幂进行数学化解释。 + +- 对于任何十进制正整数 $n$ ，设其二进制为 "$b_m...b_3b_2b_1$"（ $b_i$ 为二进制某位值，$i \in [1,m]$ ），则有： + - **二进制转十进制：** $n = 1b_1 + 2b_2 + 4b_3 + ... + 2^{m-1}b_m$ *（即二进制转十进制公式）* ； + - **幂的二进制展开：** $x^n = x^{1b_1 + 2b_2 + 4b_3 + ... + 2^{m-1}b_m} = x^{1b_1}x^{2b_2}x^{4b_3}...x^{2^{m-1}b_m}$ ； +- 根据以上推导，可把计算 $x^n$ 转化为解决以下两个问题： + - **计算 $x^1, x^2, x^4, ..., x^{2^{m-1}}$ 的值：** 循环赋值操作 $x = x^2$ 即可； + - **获取二进制各位 $b_1, b_2, b_3, ..., b_m$ 的值：** 循环执行以下操作即可。 + 1. **$n \& 1$ （与操作）：** 判断 $n$ 二进制最右一位是否为 $1$ ； + 2. **$n>>1$ （移位操作）：** $n$ 右移一位（可理解为删除最后一位）。 +- 因此，应用以上操作，可在循环中依次计算 $x^{2^{0}b_1}, x^{2^{1}b_2}, ..., x^{2^{m-1}b_m}$ 的值，并将所有 $x^{2^{i-1}b_i}$ 累计相乘即可，其中： + +$$ +x^{2^{i-1}b_i}= +\begin{cases} + 1 & , b_i = 0 \\ + x^{2^{i-1}} & , b_i = 1 \\ +\end{cases} +$$ + +![Picture1.png](https://pic.leetcode-cn.com/1599885604-yDzVYK-Picture1.png){:width=500} + +### 算法流程： + +1. 当 $x = 0.0$ 时：直接返回 $0.0$ ，以避免后续 $1$ 除以 $0$ 操作报错。**分析：** 数字 $0$ 的正数次幂恒为 $0$ ； $0$ 的 $0$ 次幂和负数次幂没有意义，因此直接返回 $0.0$ 即可。 +2. 初始化 $res = 1$ 。 +3. 当 $n < 0$ 时：把问题转化至 $n \geq 0$ 的范围内，即执行 $x = 1/x$ ，$n = - n$ 。 +4. 循环计算：当 $n = 0$ 时跳出。 + 1. 当 $n \& 1 = 1$ 时：将当前 $x$ 乘入 $res$ （即 $res *= x$ ）。 + 2. 执行 $x = x^2$ （即 $x *= x$ ）。 + 3. 执行 $n$ 右移一位（即 $n >>= 1$）。 +5. 返回 $res$ 。 + +### 复杂度分析： + +- **时间复杂度 $O(\log n)$ ：** 二分的时间复杂度为对数级别。 +- **空间复杂度 $O(1)$ ：** $res$, $b$ 等变量占用常数大小额外空间。 + +## 代码： + +Java 中 int32 变量区间 $n \in [-2147483648, 2147483647]$ ，因此当 $n = -2147483648$ 时执行 $n = -n$ 会因越界而赋值出错。解决方法是先将 $n$ 存入 long 变量 $b$ ，后面用 $b$ 操作即可。 + +```Python [] +class Solution: + def myPow(self, x: float, n: int) -> float: + if x == 0.0: return 0.0 + res = 1 + if n < 0: x, n = 1 / x, -n + while n: + if n & 1: res *= x + x *= x + n >>= 1 + return res +``` + +```Java [] +class Solution { + public double myPow(double x, int n) { + if(x == 0.0f) return 0.0d; + long b = n; + double res = 1.0; + if(b < 0) { + x = 1 / x; + b = -b; + } + while(b > 0) { + if((b & 1) == 1) res *= x; + x *= x; + b >>= 1; + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + double myPow(double x, int n) { + if(x == 0.0f) return 0.0; + long b = n; + double res = 1.0; + if(b < 0) { + x = 1 / x; + b = -b; + } + while(b > 0) { + if((b & 1) == 1) res *= x; + x *= x; + b >>= 1; + } + return res; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 17. \346\211\223\345\215\260\344\273\216 1 \345\210\260\346\234\200\345\244\247\347\232\204 n \344\275\215\346\225\260.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 17. \346\211\223\345\215\260\344\273\216 1 \345\210\260\346\234\200\345\244\247\347\232\204 n \344\275\215\346\225\260.md" new file mode 100644 index 0000000..a8bab59 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 17. \346\211\223\345\215\260\344\273\216 1 \345\210\260\346\234\200\345\244\247\347\232\204 n \344\275\215\346\225\260.md" @@ -0,0 +1,268 @@ +## 解题思路： + +题目要求打印 “从 $1$ 至最大的 $n$ 位数的列表” ，因此需考虑以下两个问题： + +1. **最大的 $n$ 位数（记为 $end$ ）和位数 $n$ 的关系：** 例如最大的 $1$ 位数是 $9$ ，最大的 $2$ 位数是 $99$ ，最大的 $3$ 位数是 $999$ 。则可推出公式： + +$$ +end = 10^n - 1 +$$ + +2. **大数越界问题：** 当 $n$ 较大时，$end$ 会超出 $int32$ 整型的取值范围，超出取值范围的数字无法正常存储。但由于本题要求返回 int 类型数组，相当于默认所有数字都在 int32 整型取值范围内，因此不考虑大数越界问题。 + +因此，只需定义区间 $[1, 10^n - 1]$ 和步长 $1$ ，通过 $for$ 循环生成结果列表 $res$ 并返回即可。 + +### 复杂度分析： + +- **时间复杂度 $O(10^n)$ ：** 生成长度为 $10^n$ 的列表需使用 $O(10^n)$ 时间。 +- **空间复杂度 $O(1)$ ：** 建立列表需使用 $O(1)$ 大小的额外空间（ 列表作为返回结果，不计入额外空间 ）。 + +### 代码： + +```Python [] +class Solution: + def printNumbers(self, n: int) -> List[int]: + res = [] + for i in range(1, 10 ** n): + res.append(i) + return res +``` + +```Java [] +class Solution { + public int[] printNumbers(int n) { + int end = (int)Math.pow(10, n) - 1; + int[] res = new int[end]; + for(int i = 0; i < end; i++) + res[i] = i + 1; + return res; + } +} +``` + +利用 Python 的语言特性，可以简化代码：先使用 `range()` 方法生成可迭代对象，再使用 `list()` 方法转化为列表并返回即可。 + +```Python +class Solution: + def printNumbers(self, n: int) -> List[int]: + return list(range(1, 10 ** n)) +``` + +## 大数打印解法： + +实际上，本题的主要考点是大数越界情况下的打印。需要解决以下三个问题： + +### 1. 表示大数的变量类型： + +- 无论是 short / int / long ... 任意变量类型，数字的取值范围都是有限的。因此，大数的表示应用字符串 String 类型。 + +### 2. 生成数字的字符串集： + +- 使用 int 类型时，每轮可通过 $+1$ 生成下个数字，而此方法无法应用至 String 类型。并且， String 类型的数字的进位操作效率较低，例如 `"9999"` 至 `"10000"` 需要从个位到千位循环判断，进位 4 次。 + +- 观察可知，生成的列表实际上是 $n$ 位 $0$ - $9$ 的 **全排列** ，因此可避开进位操作，通过递归生成数字的 String 列表。 + +### 3. 递归生成全排列： + +- 基于分治算法的思想，先固定高位，向低位递归，当个位已被固定时，添加数字的字符串。例如当 $n = 2$ 时（数字范围 $1 - 99$ ），固定十位为 $0$ - $9$ ，按顺序依次开启递归，固定个位 $0$ - $9$ ，终止递归并添加数字字符串。 + +![Picture1.png](https://pic.leetcode-cn.com/83f4b5930ddc1d42b05c724ea2950ee7f00427b11150c86b45bd88405f8c7c87-Picture1.png){:width=500} + +根据以上方法，可初步编写全排列代码： + +```Python [] +class Solution: + def printNumbers(self, n: int) -> [int]: + def dfs(x): + if x == n: # 终止条件：已固定完所有位 + res.append(''.join(num)) # 拼接 num 并添加至 res 尾部 + return + for i in range(10): # 遍历 0 - 9 + num[x] = str(i) # 固定第 x 位为 i + dfs(x + 1) # 开启固定第 x + 1 位 + + num = ['0'] * n # 起始数字定义为 n 个 0 组成的字符列表 + res = [] # 数字字符串列表 + dfs(0) # 开启全排列递归 + return ','.join(res) # 拼接所有数字字符串，使用逗号隔开，并返回 +``` + +```Java [] +class Solution { + StringBuilder res; + int count = 0, n; + char[] num, loop = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9'}; + public String printNumbers(int n) { + this.n = n; + res = new StringBuilder(); // 数字字符串集 + num = new char[n]; // 定义长度为 n 的字符列表 + dfs(0); // 开启全排列递归 + res.deleteCharAt(res.length() - 1); // 删除最后多余的逗号 + return res.toString(); // 转化为字符串并返回 + } + void dfs(int x) { + if(x == n) { // 终止条件：已固定完所有位 + res.append(String.valueOf(num) + ","); // 拼接 num 并添加至 res 尾部，使用逗号隔开 + return; + } + for(char i : loop) { // 遍历 ‘0‘ - ’9‘ + num[x] = i; // 固定第 x 位为 i + dfs(x + 1); // 开启固定第 x + 1 位 + } + } +} +``` + +在此方法下，各数字字符串被逗号隔开，共同组成长字符串。返回的数字集字符串如下所示： + +```yaml +输入：n = 1 +输出："0,1,2,3,4,5,6,7,8,9" + +输入：n = 2 +输出："00,01,02,...,10,11,12,...,97,98,99" + +输入：n = 3 +输出："000,001,002,...,100,101,102,...,997,998,999" +``` + +观察可知，当前的生成方法仍有以下问题： + +1. 诸如 $00, 01, 02, \cdots$ 应显示为 $0, 1, 2, \cdots$ ，即应 **删除高位多余的 $0$** ; +2. 此方法从 $0$ 开始生成，而题目要求 **列表从 $1$ 开始** ； + +以上两个问题的解决方法如下： + +### 1. 删除高位多余的 $0$ ： + +- **字符串左边界定义：** 声明变量 $start$ 规定字符串的左边界，以保证添加的数字字符串 `num[start:]` 中无高位多余的 $0$ 。例如当 $n = 2$ 时， $1 - 9$ 时 $start = 1$ ， $10 - 99$ 时 $start = 0$ 。 + +- **左边界 $start$ 变化规律：** 观察可知，当输出数字的所有位都是 $9$ 时，则下个数字需要向更高位进 $1$ ，此时左边界 $start$ 需要减 $1$ （即高位多余的 $0$ 减少一个）。例如当 $n = 3$ （数字范围 $1 - 999$ ）时，左边界 $start$ 需要减 $1$ 的情况有： "009" 进位至 "010" ， "099" 进位至 "100" 。设数字各位中 $9$ 的数量为 $nine$ ，所有位都为 $9$ 的判断条件可用以下公式表示： + +$$ +n - start = nine +$$ + +- **统计 $nine$ 的方法：** 固定第 $x$ 位时，当 $i = 9$ 则执行 $nine = nine + 1$ ，并在回溯前恢复 $nine = nine - 1$ 。 + +### 2. 列表从 $1$ 开始： + +- 在以上方法的基础上，添加数字字符串前判断其是否为 `"0"` ，若为 `"0"` 则直接跳过。 + + + +### 复杂度分析： + +- **时间复杂度 $O(10^n)$ ：** 递归的生成的排列的数量为 $10^n$ 。 +- **空间复杂度 $O(10^n)$ ：** 结果列表 $res$ 的长度为 $10^n - 1$ ，各数字字符串的长度区间为 $1, 2, ..., n$ ，因此占用 $O(10^n)$ 大小的额外空间。 + +### 代码： + +为 **正确表示大数** ，以下代码的返回值为数字字符串集拼接而成的长字符串。 + +```Python [] +class Solution: + def printNumbers(self, n: int) -> [int]: + def dfs(x): + if x == n: + s = ''.join(num[self.start:]) + if s != '0': res.append(s) + if n - self.start == self.nine: self.start -= 1 + return + for i in range(10): + if i == 9: self.nine += 1 + num[x] = str(i) + dfs(x + 1) + self.nine -= 1 + + num, res = ['0'] * n, [] + self.nine = 0 + self.start = n - 1 + dfs(0) + return ','.join(res) +``` + +```Java [] +class Solution { + StringBuilder res; + int nine = 0, count = 0, start, n; + char[] num, loop = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9'}; + public String printNumbers(int n) { + this.n = n; + res = new StringBuilder(); + num = new char[n]; + start = n - 1; + dfs(0); + res.deleteCharAt(res.length() - 1); + return res.toString(); + } + void dfs(int x) { + if(x == n) { + String s = String.valueOf(num).substring(start); + if(!s.equals("0")) res.append(s + ","); + if(n - start == nine) start--; + return; + } + for(char i : loop) { + if(i == '9') nine++; + num[x] = i; + dfs(x + 1); + } + nine--; + } +} +``` + +本题要求输出 int 类型数组。为 **运行通过** ，可在添加数字字符串 $s$ 前，将其转化为 int 类型。代码如下所示： + +```Python [] +class Solution: + def printNumbers(self, n: int) -> [int]: + def dfs(x): + if x == n: + s = ''.join(num[self.start:]) + if s != '0': res.append(int(s)) + if n - self.start == self.nine: self.start -= 1 + return + for i in range(10): + if i == 9: self.nine += 1 + num[x] = str(i) + dfs(x + 1) + self.nine -= 1 + + num, res = ['0'] * n, [] + self.nine = 0 + self.start = n - 1 + dfs(0) + return res +``` + +```Java [] +class Solution { + int[] res; + int nine = 0, count = 0, start, n; + char[] num, loop = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9'}; + public int[] printNumbers(int n) { + this.n = n; + res = new int[(int)Math.pow(10, n) - 1]; + num = new char[n]; + start = n - 1; + dfs(0); + return res; + } + void dfs(int x) { + if(x == n) { + String s = String.valueOf(num).substring(start); + if(!s.equals("0")) res[count++] = Integer.parseInt(s); + if(n - start == nine) start--; + return; + } + for(char i : loop) { + if(i == '9') nine++; + num[x] = i; + dfs(x + 1); + } + nine--; + } +} +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 18. \345\210\240\351\231\244\351\223\276\350\241\250\347\232\204\350\212\202\347\202\271.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 18. \345\210\240\351\231\244\351\223\276\350\241\250\347\232\204\350\212\202\347\202\271.md" new file mode 100644 index 0000000..3f2ef03 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 18. \345\210\240\351\231\244\351\223\276\350\241\250\347\232\204\350\212\202\347\202\271.md" @@ -0,0 +1,69 @@ +## 解题思路： + +本题删除值为 `val` 的节点分需为两步：定位节点、修改引用。 + +1. **定位节点：** 遍历链表，直到 `head.val == val` 时跳出，即可定位目标节点。 +2. **修改引用：** 设节点 `cur` 的前驱节点为 `pre` ，后继节点为 `cur.next` ；则执行 `pre.next = cur.next` ，即可实现删除 `cur` 节点。 + +![Picture1.png](https://pic.leetcode-cn.com/1613757478-NBOvjn-Picture1.png){:width=450} + +### 算法流程： + +1. **特例处理：** 当应删除头节点 `head` 时，直接返回 `head.next` 即可。 +2. **初始化：** `pre = head` , `cur = head.next` 。 +3. **定位节点：** 当 `cur` 为空 **或** `cur` 节点值等于 `val` 时跳出。 + 1. 保存当前节点索引，即 `pre = cur` 。 + 2. 遍历下一节点，即 `cur = cur.next` 。 +4. **删除节点：** 若 `cur` 指向某节点，则执行 `pre.next = cur.next` ；若 `cur` 指向 $null$ ，代表链表中不包含值为 `val` 的节点。 +5. **返回值：** 返回链表头部节点 `head` 即可。 + + + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为链表长度，删除操作平均需循环 $N/2$ 次，最差 $N$ 次。 +- **空间复杂度 $O(1)$ ：** `cur`, `pre` 占用常数大小额外空间。 + +## 代码： + +```Python [] +class Solution: + def deleteNode(self, head: ListNode, val: int) -> ListNode: + if head.val == val: return head.next + pre, cur = head, head.next + while cur and cur.val != val: + pre, cur = cur, cur.next + if cur: pre.next = cur.next + return head +``` + +```Java [] +class Solution { + public ListNode deleteNode(ListNode head, int val) { + if(head.val == val) return head.next; + ListNode pre = head, cur = head.next; + while(cur != null && cur.val != val) { + pre = cur; + cur = cur.next; + } + if(cur != null) pre.next = cur.next; + return head; + } +} +``` + +```C++ [] +class Solution { +public: + ListNode* deleteNode(ListNode* head, int val) { + if(head->val == val) return head->next; + ListNode *pre = head, *cur = head->next; + while(cur != nullptr && cur->val != val) { + pre = cur; + cur = cur->next; + } + if(cur != nullptr) pre->next = cur->next; + return head; + } +}; +``` \ No newline at end of file diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 19. \346\255\243\345\210\231\350\241\250\350\276\276\345\274\217\345\214\271\351\205\215.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 19. \346\255\243\345\210\231\350\241\250\350\276\276\345\274\217\345\214\271\351\205\215.md" new file mode 100644 index 0000000..1a9997d --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 19. \346\255\243\345\210\231\350\241\250\350\276\276\345\274\217\345\214\271\351\205\215.md" @@ -0,0 +1,185 @@ +## 解题思路： + +> 设 $s$ 的长度为 $n$ ， $p$ 的长度为 $m$ ；将 $s$ 的第 $i$ 个字符记为 $s_i$ ，$p$ 的第 $j$ 个字符记为 $p_j$ ，将 $s$ 的前 $i$ 个字符组成的子字符串记为 $s[:i]$ , 同理将 $p$ 的前 $j$ 个字符组成的子字符串记为 $p[:j]$ 。 +> +> 因此，本题可转化为求 $s[:n]$ 是否能和 $p[:m]$ 匹配。 + +总体思路是从 $s[:1]$ 和 $p[:1]$ 是否能匹配开始判断，每轮添加一个字符并判断是否能匹配，直至添加完整个字符串 $s$ 和 $p$ 。展开来看，假设 $s[:i]$ 与 $p[:j]$ 可以匹配，那么下一状态有两种： + +1. 添加一个字符 $s_{i+1}$ 后是否能匹配？ +2. 添加字符 $p_{j+1}$ 后是否能匹配？ + +![Picture1.png](https://pic.leetcode-cn.com/1614516402-HyzAil-Picture1.png){:width=500} + +因此，本题的状态共有 $m \times n$ 种，应定义状态矩阵 $dp$ ，$dp[i][j]$ 代表 $s[:i]$ 与 $p[:j]$ 是否可以匹配。 + +做好状态定义，接下来就是根据 「`普通字符`」 , 「`.`」 , 「`*`」三种字符的功能定义，分析出动态规划的转移方程。 + +### 动态规划解析： + +- **状态定义：** 设动态规划矩阵 `dp` ， `dp[i][j]` 代表字符串 `s` 的前 `i` 个字符和 `p` 的前 `j` 个字符能否匹配。 + +- **转移方程：** 需要注意，由于 `dp[0][0]` 代表的是空字符的状态， 因此 `dp[i][j]` 对应的添加字符是 `s[i - 1]` 和 `p[j - 1]` 。 + + - 当 `p[j - 1] = '*'` 时， `dp[i][j]` 在当以下任一情况为 $true$ 时等于 $true$ ： + + 1. **`dp[i][j - 2]`：** 即将字符组合 `p[j - 2] *` 看作出现 0 次时，能否匹配； + 2. **`dp[i - 1][j]` 且 `s[i - 1] = p[j - 2]`:** 即让字符 `p[j - 2]` 多出现 1 次时，能否匹配； + 3. **`dp[i - 1][j]` 且 `p[j - 2] = '.'`:** 即让字符 `'.'` 多出现 1 次时，能否匹配； + + - 当 `p[j - 1] != '*'` 时， `dp[i][j]` 在当以下任一情况为 $true$ 时等于 $true$ ： + + 1. **`dp[i - 1][j - 1]` 且 `s[i - 1] = p[j - 1]`：** 即让字符 `p[j - 1]` 多出现一次时，能否匹配； + 2. **`dp[i - 1][j - 1]` 且 `p[j - 1] = '.'`：** 即将字符 `.` 看作字符 `s[i - 1]` 时，能否匹配； + +- **初始化：** 需要先初始化 `dp` 矩阵首行，以避免状态转移时索引越界。 + + - **`dp[0][0] = true`：** 代表两个空字符串能够匹配。 + - **`dp[0][j] = dp[0][j - 2]` 且 `p[j - 1] = '*'`：** 首行 `s` 为空字符串，因此当 `p` 的偶数位为 `*` 时才能够匹配（即让 `p` 的奇数位出现 0 次，保持 `p` 是空字符串）。因此，循环遍历字符串 `p` ，步长为 2（即只看偶数位）。 + +- **返回值：** `dp` 矩阵右下角字符，代表字符串 `s` 和 `p` 能否匹配。 + + + +### 复杂度分析： + +- **时间复杂度 $O(MN)$ ：** 其中 $M, N$ 分别为 `s` 和 `p` 的长度，状态转移需遍历整个 `dp` 矩阵。 +- **空间复杂度 $O(MN)$ ：** 状态矩阵 `dp` 使用 $O(MN)$ 的额外空间。 + +## 代码： + +```Python [] +class Solution: + def isMatch(self, s: str, p: str) -> bool: + m, n = len(s) + 1, len(p) + 1 + dp = [[False] * n for _ in range(m)] + dp[0][0] = True + for j in range(2, n, 2): + dp[0][j] = dp[0][j - 2] and p[j - 1] == '*' + for i in range(1, m): + for j in range(1, n): + dp[i][j] = dp[i][j - 2] or dp[i - 1][j] and (s[i - 1] == p[j - 2] or p[j - 2] == '.') \ + if p[j - 1] == '*' else \ + dp[i - 1][j - 1] and (p[j - 1] == '.' or s[i - 1] == p[j - 1]) + return dp[-1][-1] +``` + +```Java [] +class Solution { + public boolean isMatch(String s, String p) { + int m = s.length() + 1, n = p.length() + 1; + boolean[][] dp = new boolean[m][n]; + dp[0][0] = true; + for(int j = 2; j < n; j += 2) + dp[0][j] = dp[0][j - 2] && p.charAt(j - 1) == '*'; + for(int i = 1; i < m; i++) { + for(int j = 1; j < n; j++) { + dp[i][j] = p.charAt(j - 1) == '*' ? + dp[i][j - 2] || dp[i - 1][j] && (s.charAt(i - 1) == p.charAt(j - 2) || p.charAt(j - 2) == '.') : + dp[i - 1][j - 1] && (p.charAt(j - 1) == '.' || s.charAt(i - 1) == p.charAt(j - 1)); + } + } + return dp[m - 1][n - 1]; + } +} +``` + +```C++ [] +class Solution { +public: + bool isMatch(string s, string p) { + int m = s.size() + 1, n = p.size() + 1; + vector> dp(m, vector(n, false)); + dp[0][0] = true; + for(int j = 2; j < n; j += 2) + dp[0][j] = dp[0][j - 2] && p[j - 1] == '*'; + for(int i = 1; i < m; i++) { + for(int j = 1; j < n; j++) { + dp[i][j] = p[j - 1] == '*' ? + dp[i][j - 2] || dp[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == '.'): + dp[i - 1][j - 1] && (p[j - 1] == '.' || s[i - 1] == p[j - 1]); + } + } + return dp[m - 1][n - 1]; + } +}; +``` + +以上代码利用布尔运算实现简短长度，若阅读不畅，可先理解以下代码，与文中内容一一对应： + +```Python [] +class Solution: + def isMatch(self, s: str, p: str) -> bool: + m, n = len(s) + 1, len(p) + 1 + dp = [[False] * n for _ in range(m)] + dp[0][0] = True + # 初始化首行 + for j in range(2, n, 2): + dp[0][j] = dp[0][j - 2] and p[j - 1] == '*' + # 状态转移 + for i in range(1, m): + for j in range(1, n): + if p[j - 1] == '*': + if dp[i][j - 2]: dp[i][j] = True # 1. + elif dp[i - 1][j] and s[i - 1] == p[j - 2]: dp[i][j] = True # 2. + elif dp[i - 1][j] and p[j - 2] == '.': dp[i][j] = True # 3. + else: + if dp[i - 1][j - 1] and s[i - 1] == p[j - 1]: dp[i][j] = True # 1. + elif dp[i - 1][j - 1] and p[j - 1] == '.': dp[i][j] = True # 2. + return dp[-1][-1] +``` + +```Java [] +class Solution { + public boolean isMatch(String s, String p) { + int m = s.length() + 1, n = p.length() + 1; + boolean[][] dp = new boolean[m][n]; + dp[0][0] = true; + // 初始化首行 + for(int j = 2; j < n; j += 2) + dp[0][j] = dp[0][j - 2] && p.charAt(j - 1) == '*'; + // 状态转移 + for(int i = 1; i < m; i++) { + for(int j = 1; j < n; j++) { + if(p.charAt(j - 1) == '*') { + if(dp[i][j - 2]) dp[i][j] = true; // 1. + else if(dp[i - 1][j] && s.charAt(i - 1) == p.charAt(j - 2)) dp[i][j] = true; // 2. + else if(dp[i - 1][j] && p.charAt(j - 2) == '.') dp[i][j] = true; // 3. + } else { + if(dp[i - 1][j - 1] && s.charAt(i - 1) == p.charAt(j - 1)) dp[i][j] = true; // 1. + else if(dp[i - 1][j - 1] && p.charAt(j - 1) == '.') dp[i][j] = true; // 2. + } + } + } + return dp[m - 1][n - 1]; + } +} +``` + +```C++ [] +class Solution { +public: + bool isMatch(string s, string p) { + int m = s.size() + 1, n = p.size() + 1; + vector> dp(m, vector(n, false)); + dp[0][0] = true; + // 初始化首行 + for(int j = 2; j < n; j += 2) + dp[0][j] = dp[0][j - 2] && p[j - 1] == '*'; + // 状态转移 + for(int i = 1; i < m; i++) { + for(int j = 1; j < n; j++) { + if(p[j - 1] == '*') { + if(dp[i][j - 2]) dp[i][j] = true; // 1. + else if(dp[i - 1][j] && s[i - 1] == p[j - 2]) dp[i][j] = true; // 2. + else if(dp[i - 1][j] && p[j - 2] == '.') dp[i][j] = true; // 3. + } else { + if(dp[i - 1][j - 1] && s[i - 1] == p[j - 1]) dp[i][j] = true; // 1. + else if(dp[i - 1][j - 1] && p[j - 1] == '.') dp[i][j] = true; // 2. + } + } + } + return dp[m - 1][n - 1]; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 20. \350\241\250\347\244\272\346\225\260\345\200\274\347\232\204\345\255\227\347\254\246\344\270\262.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 20. \350\241\250\347\244\272\346\225\260\345\200\274\347\232\204\345\255\227\347\254\246\344\270\262.md" new file mode 100644 index 0000000..81f022f --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 20. \350\241\250\347\244\272\346\225\260\345\200\274\347\232\204\345\255\227\347\254\246\344\270\262.md" @@ -0,0 +1,112 @@ +## 解题思路： + +本题使用有限状态自动机。根据字符类型和合法数值的特点，先定义状态，再画出状态转移图，最后编写代码即可。 + +**字符类型：** + +空格 「 」、数字「 $0—9$ 」 、正负号 「 $+$, $-$ 」 、小数点 「 $.$ 」 、幂符号 「 $e$, $E$ 」 。 + +**状态定义：** + +按照字符串从左到右的顺序，定义以下 9 种状态。 + +0. 开始的空格 +1. 幂符号前的正负号 +2. 小数点前的数字 +3. 小数点、小数点后的数字 +4. 当小数点前为空格时，小数点、小数点后的数字 +5. 幂符号 +6. 幂符号后的正负号 +7. 幂符号后的数字 +8. 结尾的空格 + +**结束状态：** + +合法的结束状态有 2, 3, 7, 8 。 + +![Picture1.png](https://pic.leetcode-cn.com/1599283151-YmPMis-Picture1.png){:width=650} + +### 算法流程： + +1. **初始化：** + 1. **状态转移表 `states` ：** 设 `states[i]` ，其中 `i` 为所处状态， `states[i]` 使用哈希表存储可转移至的状态。键值对 `(key, value)` 含义：输入字符 `key` ，则从状态 `i` 转移至状态 `value` 。 + 2. **当前状态 `p` ：** 起始状态初始化为 `p = 0` 。 + +2. **状态转移循环：** 遍历字符串 `s` 的每个字符 `c` 。 + 1. **记录字符类型 `t` ：** 分为四种情况。 + - 当 `c` 为正负号时，执行 `t = 's'` ; + - 当 `c` 为数字时，执行 `t = 'd'` ; + - 当 `c` 为 `e` 或 `E` 时，执行 `t = 'e'` ; + - 当 `c` 为 `.` 或 `空格` 时，执行 `t = c` （即用字符本身表示字符类型）; + - 否则，执行 `t = '?'` ，代表为不属于判断范围的非法字符，后续直接返回 $false$ 。 + 2. **终止条件：** 若字符类型 `t` 不在哈希表 `states[p]` 中，说明无法转移至下一状态，因此直接返回 $false$ 。 + 3. **状态转移：** 状态 `p` 转移至 `states[p][t]` 。 + +3. **返回值：** 跳出循环后，若状态 `p` $\in {2, 3, 7, 8}$ ，说明结尾合法，返回 $true$ ，否则返回 $false$ 。 + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为字符串 `s` 的长度，判断需遍历字符串，每轮状态转移的使用 $O(1)$ 时间。 +- **空间复杂度 $O(1)$ ：** `states` 和 `p` 使用常数大小的额外空间。 + + + +## 代码： + +Java 的状态转移表 `states` 使用 Map[] 数组存储。 + +```Python [] +class Solution: + def isNumber(self, s: str) -> bool: + states = [ + { ' ': 0, 's': 1, 'd': 2, '.': 4 }, # 0. start with 'blank' + { 'd': 2, '.': 4 } , # 1. 'sign' before 'e' + { 'd': 2, '.': 3, 'e': 5, ' ': 8 }, # 2. 'digit' before 'dot' + { 'd': 3, 'e': 5, ' ': 8 }, # 3. 'digit' after 'dot' + { 'd': 3 }, # 4. 'digit' after 'dot' (‘blank’ before 'dot') + { 's': 6, 'd': 7 }, # 5. 'e' + { 'd': 7 }, # 6. 'sign' after 'e' + { 'd': 7, ' ': 8 }, # 7. 'digit' after 'e' + { ' ': 8 } # 8. end with 'blank' + ] + p = 0 # start with state 0 + for c in s: + if '0' <= c <= '9': t = 'd' # digit + elif c in "+-": t = 's' # sign + elif c in "eE": t = 'e' # e or E + elif c in ". ": t = c # dot, blank + else: t = '?' # unknown + if t not in states[p]: return False + p = states[p][t] + return p in (2, 3, 7, 8) +``` + +```Java [] +class Solution { + public boolean isNumber(String s) { + Map[] states = { + new HashMap<>() {{ put(' ', 0); put('s', 1); put('d', 2); put('.', 4); }}, // 0. + new HashMap<>() {{ put('d', 2); put('.', 4); }}, // 1. + new HashMap<>() {{ put('d', 2); put('.', 3); put('e', 5); put(' ', 8); }}, // 2. + new HashMap<>() {{ put('d', 3); put('e', 5); put(' ', 8); }}, // 3. + new HashMap<>() {{ put('d', 3); }}, // 4. + new HashMap<>() {{ put('s', 6); put('d', 7); }}, // 5. + new HashMap<>() {{ put('d', 7); }}, // 6. + new HashMap<>() {{ put('d', 7); put(' ', 8); }}, // 7. + new HashMap<>() {{ put(' ', 8); }} // 8. + }; + int p = 0; + char t; + for(char c : s.toCharArray()) { + if(c >= '0' && c <= '9') t = 'd'; + else if(c == '+' || c == '-') t = 's'; + else if(c == 'e' || c == 'E') t = 'e'; + else if(c == '.' || c == ' ') t = c; + else t = '?'; + if(!states[p].containsKey(t)) return false; + p = (int)states[p].get(t); + } + return p == 2 || p == 3 || p == 7 || p == 8; + } +} +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 21. \350\260\203\346\225\264\346\225\260\347\273\204\351\241\272\345\272\217\344\275\277\345\245\207\346\225\260\344\275\215\344\272\216\345\201\266\346\225\260\345\211\215\351\235\242.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 21. \350\260\203\346\225\264\346\225\260\347\273\204\351\241\272\345\272\217\344\275\277\345\245\207\346\225\260\344\275\215\344\272\216\345\201\266\346\225\260\345\211\215\351\235\242.md" new file mode 100644 index 0000000..9cf476d --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 21. \350\260\203\346\225\264\346\225\260\347\273\204\351\241\272\345\272\217\344\275\277\345\245\207\346\225\260\344\275\215\344\272\216\345\201\266\346\225\260\345\211\215\351\235\242.md" @@ -0,0 +1,75 @@ +## 解题思路： + +考虑定义双指针 $i$ , $j$ 分列数组左右两端，循环执行： + +1. 指针 $i$ 从左向右寻找偶数； +2. 指针 $j$ 从右向左寻找奇数； +3. 将 偶数 $nums[i]$ 和 奇数 $nums[j]$ 交换。 + +可始终保证： 指针 $i$ 左边都是奇数，指针 $j$ 右边都是偶数 。 + +![Picture1.png](https://pic.leetcode-cn.com/43e965485da89efa688947bc108232f10b65b5ba5c0dbd6a68227a82c7e451e4-Picture1.png){:width=450} + +### 算法流程： + +1. **初始化：** $i$ , $j$ 双指针，分别指向数组 $nums$ 左右两端； +2. **循环交换：** 当 $i = j$ 时跳出； + 1. 指针 $i$ 遇到奇数则执行 $i = i + 1$ 跳过，直到找到偶数； + 2. 指针 $j$ 遇到偶数则执行 $j = j - 1$ 跳过，直到找到奇数； + 3. 交换 $nums[i]$ 和 $nums[j]$ 值； +3. **返回值：** 返回已修改的 $nums$ 数组。 + + + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为数组 $nums$ 长度，双指针 $i$, $j$ 共同遍历整个数组。 +- **空间复杂度 $O(1)$ ：** 双指针 $i$, $j$ 使用常数大小的额外空间。 + +## 代码： + +$x \& 1$ 位运算 等价于 $x \% 2$ 取余运算，即皆可用于判断数字奇偶性。 + +```Python [] +class Solution: + def exchange(self, nums: List[int]) -> List[int]: + i, j = 0, len(nums) - 1 + while i < j: + while i < j and nums[i] & 1 == 1: i += 1 + while i < j and nums[j] & 1 == 0: j -= 1 + nums[i], nums[j] = nums[j], nums[i] + return nums +``` + +```Java [] +class Solution { + public int[] exchange(int[] nums) { + int i = 0, j = nums.length - 1, tmp; + while(i < j) { + while(i < j && (nums[i] & 1) == 1) i++; + while(i < j && (nums[j] & 1) == 0) j--; + tmp = nums[i]; + nums[i] = nums[j]; + nums[j] = tmp; + } + return nums; + } +} +``` + +```C++ [] +class Solution { +public: + vector exchange(vector& nums) + { + int i = 0, j = nums.size() - 1; + while (i < j) + { + while(i < j && (nums[i] & 1) == 1) i++; + while(i < j && (nums[j] & 1) == 0) j--; + swap(nums[i], nums[j]); + } + return nums; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 22. \351\223\276\350\241\250\344\270\255\345\200\222\346\225\260\347\254\254 k \344\270\252\350\212\202\347\202\271.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 22. \351\223\276\350\241\250\344\270\255\345\200\222\346\225\260\347\254\254 k \344\270\252\350\212\202\347\202\271.md" new file mode 100644 index 0000000..f70e7ae --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 22. \351\223\276\350\241\250\344\270\255\345\200\222\346\225\260\347\254\254 k \344\270\252\350\212\202\347\202\271.md" @@ -0,0 +1,117 @@ +## 解题思路： + +第一时间想到的解法： + +1. 先遍历统计链表长度，记为 $n$ ； +2. 设置一个指针走 $(n-k)$ 步，即可找到链表倒数第 $k$ 个节点； + +使用双指针则可以不用统计链表长度。 + +![Picture1.png](https://pic.leetcode-cn.com/1600794523-AAMvoP-Picture1.png){:width=400} + +### 算法流程： + +1. **初始化：** 前指针 `former` 、后指针 `latter` ，双指针都指向头节点 `head​` 。 +2. **构建双指针距离：** 前指针 `former` 先向前走 $k$ 步（结束后，双指针 `former` 和 `latter` 间相距 $k$ 步）。 +3. **双指针共同移动：** 循环中，双指针 `former` 和 `latter` 每轮都向前走一步，直至 `former` 走过链表 **尾节点** 时跳出（跳出后， `latter` 与尾节点距离为 $k-1$，即 `latter` 指向倒数第 $k$ 个节点）。 +4. **返回值：** 返回 `latter` 即可。 + + + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为链表长度；总体看， `former` 走了 $N$ 步， `latter` 走了 $(N-k)$ 步。 +- **空间复杂度 $O(1)$ ：** 双指针 `former` , `latter` 使用常数大小的额外空间。 + +## 代码： + +```Python [] +class Solution: + def getKthFromEnd(self, head: ListNode, k: int) -> ListNode: + former, latter = head, head + for _ in range(k): + former = former.next + while former: + former, latter = former.next, latter.next + return latter +``` + +```Java [] +class Solution { + public ListNode getKthFromEnd(ListNode head, int k) { + ListNode former = head, latter = head; + for(int i = 0; i < k; i++) + former = former.next; + while(former != null) { + former = former.next; + latter = latter.next; + } + return latter; + } +} +``` + +```C++ [] +class Solution { +public: + ListNode* getKthFromEnd(ListNode* head, int k) { + ListNode *former = head, *latter = head; + for(int i = 0; i < k; i++) + former = former->next; + while(former != nullptr) { + former = former->next; + latter = latter->next; + } + return latter; + } +}; +``` + +本题没有 $k>$ 链表长度的测试样例 ，因此不用考虑越界。考虑越界问题的代码如下： + +```Python [] +class Solution: + def getKthFromEnd(self, head: ListNode, k: int) -> ListNode: + former, latter = head, head + for _ in range(k): + if not former: return + former = former.next + while former: + former, latter = former.next, latter.next + return latter +``` + +```Java [] +class Solution { + public ListNode getKthFromEnd(ListNode head, int k) { + ListNode former = head, latter = head; + for(int i = 0; i < k; i++) { + if(former == null) return null; + former = former.next; + } + while(former != null) { + former = former.next; + latter = latter.next; + } + return latter; + } +} +``` + +```C++ [] +class Solution { +public: + ListNode* getKthFromEnd(ListNode* head, int k) { + ListNode *former = head, *latter = head; + for(int i = 0; i < k; i++) { + if(former == nullptr) return nullptr; + former = former->next; + } + while(former != nullptr) { + former = former->next; + latter = latter->next; + } + return latter; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 24. \345\217\215\350\275\254\351\223\276\350\241\250.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 24. \345\217\215\350\275\254\351\223\276\350\241\250.md" new file mode 100644 index 0000000..08fabdd --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 24. \345\217\215\350\275\254\351\223\276\350\241\250.md" @@ -0,0 +1,138 @@ +## 解题思路： + +如下图所示，题目要求将链表反转。本文介绍迭代（双指针）、递归两种实现方法。 + +![Picture1.png](https://pic.leetcode-cn.com/1604779288-WXygqL-Picture1.png){:width=400} + +## 方法一：迭代（双指针） + +考虑遍历链表，并在访问各节点时修改 `next` 引用指向，算法流程见注释。 + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 遍历链表使用线性大小时间。 +- **空间复杂度 $O(1)$ ：** 变量 `pre` 和 `cur` 使用常数大小额外空间。 + + + +### 代码： + +```Python [] +class Solution: + def reverseList(self, head: ListNode) -> ListNode: + cur, pre = head, None + while cur: + tmp = cur.next # 暂存后继节点 cur.next + cur.next = pre # 修改 next 引用指向 + pre = cur # pre 暂存 cur + cur = tmp # cur 访问下一节点 + return pre +``` + +```Java [] +class Solution { + public ListNode reverseList(ListNode head) { + ListNode cur = head, pre = null; + while(cur != null) { + ListNode tmp = cur.next; // 暂存后继节点 cur.next + cur.next = pre; // 修改 next 引用指向 + pre = cur; // pre 暂存 cur + cur = tmp; // cur 访问下一节点 + } + return pre; + } +} +``` + +```C++ [] +class Solution { +public: + ListNode* reverseList(ListNode* head) { + ListNode *cur = head, *pre = nullptr; + while(cur != nullptr) { + ListNode* tmp = cur->next; // 暂存后继节点 cur.next + cur->next = pre; // 修改 next 引用指向 + pre = cur; // pre 暂存 cur + cur = tmp; // cur 访问下一节点 + } + return pre; + } +}; +``` + +利用 Python 语言的平行赋值语法，可以进一步简化代码（但可读性下降）： + +```Python [] +class Solution: + def reverseList(self, head: ListNode) -> ListNode: + cur, pre = head, None + while cur: + cur.next, pre, cur = pre, cur, cur.next + return pre +``` + +## 方法二：递归 + +考虑使用递归法遍历链表，当越过尾节点后终止递归，在回溯时修改各节点的 `next` 引用指向。 + +### `recur(cur, pre)` 递归函数： + +1. 终止条件：当 `cur` 为空，则返回尾节点 `pre` （即反转链表的头节点）； +2. 递归后继节点，记录返回值（即反转链表的头节点）为 `res` ； +3. 修改当前节点 `cur` 引用指向前驱节点 `pre` ； +4. 返回反转链表的头节点 `res` ； + +### `reverseList(head)` 函数： + +调用并返回 `recur(head, null)` 。传入 `null` 是因为反转链表后， `head` 节点指向 `null` ； + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 遍历链表使用线性大小时间。 +- **空间复杂度 $O(N)$ ：** 遍历链表的递归深度达到 $N$ ，系统使用 $O(N)$ 大小额外空间。 + + + +### 代码： + +```Python [] +class Solution: + def reverseList(self, head: ListNode) -> ListNode: + def recur(cur, pre): + if not cur: return pre # 终止条件 + res = recur(cur.next, cur) # 递归后继节点 + cur.next = pre # 修改节点引用指向 + return res # 返回反转链表的头节点 + + return recur(head, None) # 调用递归并返回 +``` + +```Java [] +class Solution { + public ListNode reverseList(ListNode head) { + return recur(head, null); // 调用递归并返回 + } + private ListNode recur(ListNode cur, ListNode pre) { + if (cur == null) return pre; // 终止条件 + ListNode res = recur(cur.next, cur); // 递归后继节点 + cur.next = pre; // 修改节点引用指向 + return res; // 返回反转链表的头节点 + } +} +``` + +```C++ [] +class Solution { +public: + ListNode* reverseList(ListNode* head) { + return recur(head, nullptr); // 调用递归并返回 + } +private: + ListNode* recur(ListNode* cur, ListNode* pre) { + if (cur == nullptr) return pre; // 终止条件 + ListNode* res = recur(cur->next, cur); // 递归后继节点 + cur->next = pre; // 修改节点引用指向 + return res; // 返回反转链表的头节点 + } +}; +``` \ No newline at end of file diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 25. \345\220\210\345\271\266\344\270\244\344\270\252\346\216\222\345\272\217\347\232\204\351\223\276\350\241\250.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 25. \345\220\210\345\271\266\344\270\244\344\270\252\346\216\222\345\272\217\347\232\204\351\223\276\350\241\250.md" new file mode 100644 index 0000000..990ee1b --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 25. \345\220\210\345\271\266\344\270\244\344\270\252\346\216\222\345\272\217\347\232\204\351\223\276\350\241\250.md" @@ -0,0 +1,88 @@ +## 解题思路： + +- 根据题目描述， 链表 $l_1$ , $l_2$ 是 **递增** 的，因此容易想到使用双指针 $l_1$ 和 $l_2$ 遍历两链表，根据 $l_1.val$ 和 $l_2.val$ 的大小关系确定节点添加顺序，两节点指针交替前进，直至遍历完毕。 + +- **引入伪头节点：** 由于初始状态合并链表中无节点，因此循环第一轮时无法将节点添加到合并链表中。解决方案：初始化一个辅助节点 $dum$ 作为合并链表的伪头节点，将各节点添加至 $dum$ 之后。 + +![Picture1.png](https://pic.leetcode-cn.com/e4c8c97883da50d81498fd1f1e6cdd575429bd65f9f2babb00dc2b709f7ad8b2-Picture1.png){:width=400} + +### 算法流程： + +1. **初始化：** 伪头节点 $dum$ ，节点 $cur$ 指向 $dum$ 。 +2. **循环合并：** 当 $l_1$ 或 $l_2$ 为空时跳出； + 1. 当 $l_1.val < l_2.val$ 时： $cur$ 的后继节点指定为 $l_1$ ，并 $l_1$ 向前走一步； + 2. 当 $l_1.val \geq l_2.val$ 时： $cur$ 的后继节点指定为 $l_2$ ，并 $l_2$ 向前走一步 ； + 3. 节点 $cur$ 向前走一步，即 $cur = cur.next$ 。 +3. **合并剩余尾部：** 跳出时有两种情况，即 $l_1$ 为空 **或** $l_2$ 为空。 + 1. 若 $l_1 \ne null$ ： 将 $l_1$ 添加至节点 $cur$ 之后； + 2. 否则： 将 $l_2$ 添加至节点 $cur$ 之后。 +4. **返回值：** 合并链表在伪头节点 $dum$ 之后，因此返回 $dum.next$ 即可。 + + + +### 复杂度分析： + +- **时间复杂度 $O(M+N)$ ：** $M, N$ 分别为链表 $l_1$, $l_2$ 的长度，合并操作需遍历两链表。 +- **空间复杂度 $O(1)$ ：** 节点引用 $dum$ , $cur$ 使用常数大小的额外空间。 + +## 代码： + +Python 三元表达式写法 `A if x else B` ，代表当 $x = True$ 时执行 $A$ ，否则执行 $B$ 。 + +```Python [] +class Solution: + def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode: + cur = dum = ListNode(0) + while l1 and l2: + if l1.val < l2.val: + cur.next, l1 = l1, l1.next + else: + cur.next, l2 = l2, l2.next + cur = cur.next + cur.next = l1 if l1 else l2 + return dum.next +``` + +```Java [] +class Solution { + public ListNode mergeTwoLists(ListNode l1, ListNode l2) { + ListNode dum = new ListNode(0), cur = dum; + while(l1 != null && l2 != null) { + if(l1.val < l2.val) { + cur.next = l1; + l1 = l1.next; + } + else { + cur.next = l2; + l2 = l2.next; + } + cur = cur.next; + } + cur.next = l1 != null ? l1 : l2; + return dum.next; + } +} +``` + +```C++ [] +class Solution { +public: + ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) { + ListNode* dum = new ListNode(0); + ListNode* cur = dum; + while(l1 != nullptr && l2 != nullptr) { + if(l1->val < l2->val) { + cur->next = l1; + l1 = l1->next; + } + else { + cur->next = l2; + l2 = l2->next; + } + cur = cur->next; + } + cur->next = l1 != nullptr ? l1 : l2; + return dum->next; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 26. \346\240\221\347\232\204\345\255\220\347\273\223\346\236\204.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 26. \346\240\221\347\232\204\345\255\220\347\273\223\346\236\204.md" new file mode 100644 index 0000000..9c18ed7 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 26. \346\240\221\347\232\204\345\255\220\347\273\223\346\236\204.md" @@ -0,0 +1,79 @@ +## 解题思路： + +若树 `B` 是树 `A` 的子结构，则子结构的根节点可能为树 `A` 的任意一个节点。因此，判断树 `B` 是否是树 `A` 的子结构，需完成以下两步工作： + +1. 先序遍历树 `A` 中的每个节点 $n_A$ ；（对应函数 `isSubStructure(A, B)`） +2. 判断树 `A` 中 **以 $n_A$ 为根节点的子树** 是否包含树 `B` 。（对应函数 `recur(A, B)`） + +![Picture1.png](https://pic.leetcode-cn.com/1599290566-VhWsiQ-Picture1.png){:width=500} + +### 算法流程： + +> 本文名词规定：**树 `A`** 的根节点记作 **节点 `A`** ，**树 `B`** 的根节点称为 **节点 `B`** 。 + +**`recur(A, B)` 函数：** + +1. **终止条件：** + 1. 当节点 `B` 为空：说明树 `B` 已匹配完成（越过叶子节点），因此返回 $true$ ； + 2. 当节点 `A` 为空：说明已经越过树 `A` 的叶节点，即匹配失败，返回 $false$ ； + 3. 当节点 `A` 和 `B` 的值不同：说明匹配失败，返回 $false$ ； +2. **返回值：** + 1. 判断 `A` 和 `B` 的 **左子节点** 是否相等，即 `recur(A.left, B.left)` ； + 2. 判断 `A` 和 `B` 的 **右子节点** 是否相等，即 `recur(A.right, B.right)` ； + +**`isSubStructure(A, B)` 函数：** + +1. **特例处理：** 当 树 `A` 为空 **或** 树 `B` 为空 时，直接返回 $false$ ； +2. **返回值：** 若树 `B` 是树 `A` 的子结构，则必满足以下三种情况之一，因此用或 `||` 连接； + 1. 以 **节点 `A` 为根节点的子树** 包含树 `B` ，对应 `recur(A, B)`； + 2. 树 `B` 是 **树 `A` 左子树** 的子结构，对应 `isSubStructure(A.left, B)`； + 3. 树 `B` 是 **树 `A` 右子树** 的子结构，对应 `isSubStructure(A.right, B)`； + > 以上 `2.` `3.` 实质上是在对树 `A` 做 **先序遍历** 。 + + + +### 复杂度分析： + +- **时间复杂度 $O(MN)$ ：** 其中 $M, N$ 分别为树 `A` 和 树 `B` 的节点数量；先序遍历树 `A` 占用 $O(M)$ ，每次调用 `recur(A, B)` 判断占用 $O(N)$ 。 +- **空间复杂度 $O(M)$ ：** 当树 `A` 和树 `B` 都退化为链表时，递归调用深度最大。当 $M \leq N$ 时，遍历树 `A` 与递归判断的总递归深度为 $M$ ；当 $M>N$ 时，最差情况为遍历至树 `A` 的叶节点，此时总递归深度为 $M$。 + +## 代码： + +```Python [] +class Solution: + def isSubStructure(self, A: TreeNode, B: TreeNode) -> bool: + def recur(A, B): + if not B: return True + if not A or A.val != B.val: return False + return recur(A.left, B.left) and recur(A.right, B.right) + + return bool(A and B) and (recur(A, B) or self.isSubStructure(A.left, B) or self.isSubStructure(A.right, B)) +``` + +```Java [] +class Solution { + public boolean isSubStructure(TreeNode A, TreeNode B) { + return (A != null && B != null) && (recur(A, B) || isSubStructure(A.left, B) || isSubStructure(A.right, B)); + } + boolean recur(TreeNode A, TreeNode B) { + if(B == null) return true; + if(A == null || A.val != B.val) return false; + return recur(A.left, B.left) && recur(A.right, B.right); + } +} +``` + +```C++ [] +class Solution { +public: + bool isSubStructure(TreeNode* A, TreeNode* B) { + return (A != nullptr && B != nullptr) && (recur(A, B) || isSubStructure(A->left, B) || isSubStructure(A->right, B)); + } +private: + bool recur(TreeNode* A, TreeNode* B) { + if(B == nullptr) return true; + if(A == nullptr || A->val != B->val) return false; + return recur(A->left, B->left) && recur(A->right, B->right); + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 27. \344\272\214\345\217\211\346\240\221\347\232\204\351\225\234\345\203\217.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 27. \344\272\214\345\217\211\346\240\221\347\232\204\351\225\234\345\203\217.md" new file mode 100644 index 0000000..310ffae --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 27. \344\272\214\345\217\211\346\240\221\347\232\204\351\225\234\345\203\217.md" @@ -0,0 +1,151 @@ +**二叉树镜像定义：** 对于二叉树中任意节点 $root$ ，设其左 / 右子节点分别为 $left, right$ ；则在二叉树的镜像中的对应 $root$ 节点，其左 / 右子节点分别为 $right, left$ 。 + +![Picture1.png](https://pic.leetcode-cn.com/20717714d97fa04d509e4f0525a3089efefc2ea02cc08ce92b77978e9b51f15f-Picture1.png){:width=450} + +## 方法一：递归法 + +- 根据二叉树镜像的定义，考虑递归遍历（dfs）二叉树，交换每个节点的左 / 右子节点，即可生成二叉树的镜像。 + +### 递归解析： + +1. **终止条件：** 当节点 $root$ 为空时（即越过叶节点），则返回 $null$ ； +2. **递推工作：** + 1. 初始化节点 $tmp$ ，用于暂存 $root$ 的左子节点； + 2. 开启递归 **右子节点** $mirrorTree(root.right)$ ，并将返回值作为 $root$ 的 **左子节点** 。 + 3. 开启递归 **左子节点** $mirrorTree(tmp)$ ，并将返回值作为 $root$ 的 **右子节点** 。 +3. **返回值：** 返回当前节点 $root$ ； + +> **Q：** 为何需要暂存 $root$ 的左子节点？ +> **A：** 在递归右子节点 “$root.left = mirrorTree(root.right);$” 执行完毕后， $root.left$ 的值已经发生改变，此时递归左子节点 $mirrorTree(root.left)$ 则会出问题。 + + + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为二叉树的节点数量，建立二叉树镜像需要遍历树的所有节点，占用 $O(N)$ 时间。 +- **空间复杂度 $O(N)$ ：** 最差情况下（当二叉树退化为链表），递归时系统需使用 $O(N)$ 大小的栈空间。 + +## 代码： + +Python 利用平行赋值的写法（即 $a, b = b, a$ ），可省略暂存操作。其原理是先将等号右侧打包成元组 $(b,a)$ ，再序列地分给等号左侧的 $a, b$ 序列。 + +```Java [] +class Solution { + public TreeNode mirrorTree(TreeNode root) { + if(root == null) return null; + TreeNode tmp = root.left; + root.left = mirrorTree(root.right); + root.right = mirrorTree(tmp); + return root; + } +} +``` + +```C++ [] +class Solution { +public: + TreeNode* mirrorTree(TreeNode* root) { + if (root == nullptr) return nullptr; + TreeNode* tmp = root->left; + root->left = mirrorTree(root->right); + root->right = mirrorTree(tmp); + return root; + } +}; +``` + +```Python [] +class Solution: + def mirrorTree(self, root: TreeNode) -> TreeNode: + if not root: return + root.left, root.right = self.mirrorTree(root.right), self.mirrorTree(root.left) + return root +``` + +```Python [] +class Solution: + def mirrorTree(self, root: TreeNode) -> TreeNode: + if not root: return + tmp = root.left + root.left = self.mirrorTree(root.right) + root.right = self.mirrorTree(tmp) + return root +``` + +## 方法二：辅助栈（或队列） + +- 利用栈（或队列）遍历树的所有节点 $node$ ，并交换每个 $node$ 的左 / 右子节点。 + +### 算法流程： + +1. **特例处理：** 当 $root$ 为空时，直接返回 $null$ ； +2. **初始化：** 栈（或队列），本文用栈，并加入根节点 $root$ 。 +3. **循环交换：** 当栈 $stack$ 为空时跳出； + 1. **出栈：** 记为 $node$ ； + 2. **添加子节点：** 将 $node$ 左和右子节点入栈； + 3. **交换：** 交换 $node$ 的左 / 右子节点。 +4. **返回值：** 返回根节点 $root$ 。 + + + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为二叉树的节点数量，建立二叉树镜像需要遍历树的所有节点，占用 $O(N)$ 时间。 +- **空间复杂度 $O(N)$ ：** 如下图所示，最差情况下，栈 $stack$ 最多同时存储 $\frac{N + 1}{2}$ 个节点，占用 $O(N)$ 额外空间。 + +![Picture0.png](https://pic.leetcode-cn.com/1614450330-bTAcyj-Picture0.png){:width=450} + +### 代码： + +```Python [] +class Solution: + def mirrorTree(self, root: TreeNode) -> TreeNode: + if not root: return + stack = [root] + while stack: + node = stack.pop() + if node.left: stack.append(node.left) + if node.right: stack.append(node.right) + node.left, node.right = node.right, node.left + return root +``` + +```Java [] +class Solution { + public TreeNode mirrorTree(TreeNode root) { + if(root == null) return null; + Stack stack = new Stack<>() {{ add(root); }}; + while(!stack.isEmpty()) { + TreeNode node = stack.pop(); + if(node.left != null) stack.add(node.left); + if(node.right != null) stack.add(node.right); + TreeNode tmp = node.left; + node.left = node.right; + node.right = tmp; + } + return root; + } +} +``` + +```C++ [] +class Solution { +public: + TreeNode* mirrorTree(TreeNode* root) { + if(root == nullptr) return nullptr; + stack stack; + stack.push(root); + while (!stack.empty()) + { + TreeNode* node = stack.top(); + stack.pop(); + if (node->left != nullptr) stack.push(node->left); + if (node->right != nullptr) stack.push(node->right); + TreeNode* tmp = node->left; + node->left = node->right; + node->right = tmp; + } + return root; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 28. \345\257\271\347\247\260\347\232\204\344\272\214\345\217\211\346\240\221.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 28. \345\257\271\347\247\260\347\232\204\344\272\214\345\217\211\346\240\221.md" new file mode 100644 index 0000000..99652f4 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 28. \345\257\271\347\247\260\347\232\204\344\272\214\345\217\211\346\240\221.md" @@ -0,0 +1,79 @@ +## 解题思路： + +**对称二叉树定义：** 对于树中 **任意两个对称节点** `L` 和 `R `，一定有： + +- `L.val = R.val` ：即此两对称节点值相等。 +- `L.left.val = R.right.val` ：即 $L$ 的 左子节点 和 $R$ 的 右子节点 对称； +- `L.right.val = R.left.val` ：即 $L$ 的 右子节点 和 $R$ 的 左子节点 对称。 + +根据以上规律，考虑从顶至底递归，判断每对左右节点是否对称，从而判断树是否为对称二叉树。 + +![Picture1.png](https://pic.leetcode-cn.com/1599398062-PbkpuX-Picture1.png){:width=450} + +### 算法流程： + +**`isSymmetric(root)` ：** + +- **特例处理：** 若根节点 `root` 为空，则直接返回 $true$ 。 +- **返回值：** 即 `recur(root.left, root.right)` ; + +**`recur(L, R)` ：** + +- **终止条件：** + - 当 `L` 和 `R` 同时越过叶节点： 此树从顶至底的节点都对称，因此返回 $true$ ； + - 当 `L` 或 `R` 中只有一个越过叶节点： 此树不对称，因此返回 $false$ ； + - 当节点 `L` 值 $\ne$ 节点 `R` 值： 此树不对称，因此返回 $false$ ； +- **递推工作：** + - 判断两节点 `L.left` 和 `R.right` 是否对称，即 `recur(L.left, R.right)` ； + - 判断两节点 `L.right` 和 `R.left` 是否对称，即 `recur(L.right, R.left)` ； +- **返回值：** 两对节点都对称时，才是对称树，因此用与逻辑符 `&&` 连接。 + + + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为二叉树的节点数量，每次执行 `recur()` 可以判断一对节点是否对称，因此最多调用 $N/2$ 次 `recur()` 方法。 +- **空间复杂度 $O(N)$ ：** 如下图所示，最差情况下（二叉树退化为链表），系统使用 $O(N)$ 大小的空间。 + +![Picture2.png](https://pic.leetcode-cn.com/1599398062-LmPbix-Picture2.png){:width=450} + +## 代码： + +```Python [] +class Solution: + def isSymmetric(self, root: TreeNode) -> bool: + def recur(L, R): + if not L and not R: return True + if not L or not R or L.val != R.val: return False + return recur(L.left, R.right) and recur(L.right, R.left) + + return not root or recur(root.left, root.right) +``` + +```Java [] +class Solution { + public boolean isSymmetric(TreeNode root) { + return root == null || recur(root.left, root.right); + } + boolean recur(TreeNode L, TreeNode R) { + if(L == null && R == null) return true; + if(L == null || R == null || L.val != R.val) return false; + return recur(L.left, R.right) && recur(L.right, R.left); + } +} +``` + +```C++ [] +class Solution { +public: + bool isSymmetric(TreeNode* root) { + return root == nullptr || recur(root->left, root->right); + } +private: + bool recur(TreeNode* L, TreeNode* R) { + if(L == nullptr && R == nullptr) return true; + if(L == nullptr || R == nullptr || L->val != R->val) return false; + return recur(L->left, R->right) && recur(L->right, R->left); + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 29. \351\241\272\346\227\266\351\222\210\346\211\223\345\215\260\347\237\251\351\230\265.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 29. \351\241\272\346\227\266\351\222\210\346\211\223\345\215\260\347\237\251\351\230\265.md" new file mode 100644 index 0000000..94a6d71 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 29. \351\241\272\346\227\266\351\222\210\346\211\223\345\215\260\347\237\251\351\230\265.md" @@ -0,0 +1,106 @@ +## 解题思路： + +根据题目示例 `matrix = [[1,2,3],[4,5,6],[7,8,9]]` 的对应输出 `[1,2,3,6,9,8,7,4,5]` 可以发现，顺时针打印矩阵的顺序是 **“从左向右、从上向下、从右向左、从下向上”** 循环。 + +因此，考虑设定矩阵的 “左、上、右、下” 四个边界，模拟以上矩阵遍历顺序。 + +![Picture1.png](https://pic.leetcode-cn.com/7605d807782923e4ad3c7995dc2485f538f202ac326bb330fe997f449123a548-Picture1.png){:width=400} + +### 算法流程： + +1. **空值处理：** 当 `matrix` 为空时，直接返回空列表 `[]` 即可。 +2. **初始化：** 矩阵 左、右、上、下 四个边界 `l` , `r` , `t` , `b` ，用于打印的结果列表 `res` 。 +3. **循环打印：** “从左向右、从上向下、从右向左、从下向上” 四个方向循环打印； + 1. 根据边界打印，即将元素按顺序添加至列表 `res` 尾部； + 2. 边界向内收缩 1 （代表已被打印）； + 3. 判断边界是否相遇（是否打印完毕），若打印完毕则跳出。 +4. **返回值：** 返回 `res` 即可。 + +| 打印方向 | 1. 根据边界打印 | 2. 边界向内收缩 | 3. 是否打印完毕 | +| -------- | ---------------------- | ----------------- | --------------- | +| 从左向右 | 左边界`l` ，右边界 `r` | 上边界 `t` 加 $1$ | 是否 `t > b` | +| 从上向下 | 上边界 `t` ，下边界`b` | 右边界 `r` 减 $1$ | 是否 `l > r` | +| 从右向左 | 右边界 `r` ，左边界`l` | 下边界 `b` 减 $1$ | 是否 `t > b` | +| 从下向上 | 下边界 `b` ，上边界`t` | 左边界 `l` 加 $1$ | 是否 `l > r` | + +### 复杂度分析： + +- **时间复杂度 $O(MN)$ ：** $M, N$ 分别为矩阵行数和列数。 +- **空间复杂度 $O(1)$ ：** 四个边界 `l` , `r` , `t` , `b` 使用常数大小的额外空间。 + + + +## 代码： + +Java/C++ 代码利用了 `++` 操作的便利性，详情可见 [++i 和 i++ 的区别](https://www.jianshu.com/p/b62eac216499) ； + +- `res[x++]` 等价于先给 `res[x]` 赋值，再给 `x` 自增 $1$ ； +- `++t > b` 等价于先给 `t` 自增 $1$ ，再判断 `t > b` 逻辑表达式。 + +> TIPS： 请注意区分数字 `1` 和变量 `l` 。 + +```Python [] +class Solution: + def spiralOrder(self, matrix:[[int]]) -> [int]: + if not matrix: return [] + l, r, t, b, res = 0, len(matrix[0]) - 1, 0, len(matrix) - 1, [] + while True: + for i in range(l, r + 1): res.append(matrix[t][i]) # left to right + t += 1 + if t > b: break + for i in range(t, b + 1): res.append(matrix[i][r]) # top to bottom + r -= 1 + if l > r: break + for i in range(r, l - 1, -1): res.append(matrix[b][i]) # right to left + b -= 1 + if t > b: break + for i in range(b, t - 1, -1): res.append(matrix[i][l]) # bottom to top + l += 1 + if l > r: break + return res +``` + +```Java [] +class Solution { + public int[] spiralOrder(int[][] matrix) { + if(matrix.length == 0) return new int[0]; + int l = 0, r = matrix[0].length - 1, t = 0, b = matrix.length - 1, x = 0; + int[] res = new int[(r + 1) * (b + 1)]; + while(true) { + for(int i = l; i <= r; i++) res[x++] = matrix[t][i]; // left to right + if(++t > b) break; + for(int i = t; i <= b; i++) res[x++] = matrix[i][r]; // top to bottom + if(l > --r) break; + for(int i = r; i >= l; i--) res[x++] = matrix[b][i]; // right to left + if(t > --b) break; + for(int i = b; i >= t; i--) res[x++] = matrix[i][l]; // bottom to top + if(++l > r) break; + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + vector spiralOrder(vector>& matrix) + { + if (matrix.empty()) return {}; + int l = 0, r = matrix[0].size() - 1, t = 0, b = matrix.size() - 1; + vector res; + while(true) + { + for (int i = l; i <= r; i++) res.push_back(matrix[t][i]); // left to right + if (++t > b) break; + for (int i = t; i <= b; i++) res.push_back(matrix[i][r]); // top to bottom + if (l > --r) break; + for (int i = r; i >= l; i--) res.push_back(matrix[b][i]); // right to left + if (t > --b) break; + for (int i = b; i >= t; i--) res.push_back(matrix[i][l]); // bottom to top + if (++l > r) break; + } + return res; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 30. \345\214\205\345\220\253 min \345\207\275\346\225\260\347\232\204\346\240\210.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 30. \345\214\205\345\220\253 min \345\207\275\346\225\260\347\232\204\346\240\210.md" new file mode 100644 index 0000000..d741c51 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 30. \345\214\205\345\220\253 min \345\207\275\346\225\260\347\232\204\346\240\210.md" @@ -0,0 +1,115 @@ +## 解题思路： + +普通栈的 `push()` 和 `pop()` 函数的复杂度为 $O(1)$ ；而获取栈最小值 `min()` 函数需要遍历整个栈，复杂度为 $O(N)$ 。 + +**本题难点：** 将 `min()` 函数复杂度降为 $O(1)$ 。可借助辅助栈实现： + +- **数据栈 `A` ：** 栈 `A` 用于存储所有元素，保证入栈 `push()` 函数、出栈 `pop()` 函数、获取栈顶 `top()` 函数的正常逻辑。 +- **辅助栈 `B` ：** 栈 `B` 中存储栈 `A` 中所有 **非严格降序** 元素的子序列，则栈 `A` 中的最小元素始终对应栈 `B` 的栈顶元素。此时， `min()` 函数只需返回栈 `B` 的栈顶元素即可。 + +因此，只需设法维护好 栈 `B` 的元素，使其保持是栈 `A` 的非严格降序元素的子序列，即可实现 `min()` 函数的 $O(1)$ 复杂度。 + +![Picture1.png](https://pic.leetcode-cn.com/1599880866-aLaPYz-Picture1.png){:width=450} + +### 函数设计： + +**`push(x)` 函数：** 重点为保持栈 `B` 的元素是 **非严格降序** 的； + +1. 执行「元素 `x` 压入栈 `A`」 ； +2. 若「栈 `B` 为空」**或**「`x` $\leq$ 栈 `B` 的栈顶元素」，则执行「元素 `x` 压入栈 `B`」 ； + +**`pop()` 函数：** 重点为保持栈 `A` , `B` 的 **元素一致性** ； + +1. 执行「栈 `A` 元素出栈」，将出栈元素记为 `y` ； +2. 若 「`y` 等于栈 `B` 的栈顶元素」，则执行「栈 `B` 元素出栈」； + +**`top()` 函数：** 直接返回栈 `A` 的栈顶元素，即返回 `A.peek()` ； + +**`min()` 函数：** 直接返回栈 `B` 的栈顶元素，即返回 `B.peek()` ； + + + +### 采用 “非严格” 降序原因： + +在栈 `A` 具有 **重复** 最小值元素时，非严格降序可防止栈 `B` 提前弹出最小值元素，示例如下： + +![Picture2.png](https://pic.leetcode-cn.com/1600086305-BSfBJu-Picture2.png){:width=550} + +### 复杂度分析： + +- **时间复杂度 $O(1)$ ：** `push()`, `pop()`, `top()`, `min()` 四个函数的时间复杂度均为常数级别。 +- **空间复杂度 $O(N)$ ：** 当共有 $N$ 个待入栈元素时，辅助栈 `B` 最差情况下存储 $N$ 个元素，使用 $O(N)$ 额外空间。 + +## 代码： + +Java 代码中，由于 Stack 中存储的是 int 的包装类 Integer ，因此需要使用 `equals()` 代替 `==` ，以比较对象的值。 + +```Python [] +class MinStack: + def __init__(self): + self.A, self.B = [], [] + + def push(self, x: int) -> None: + self.A.append(x) + if not self.B or self.B[-1] >= x: + self.B.append(x) + + def pop(self) -> None: + if self.A.pop() == self.B[-1]: + self.B.pop() + + def top(self) -> int: + return self.A[-1] + + def min(self) -> int: + return self.B[-1] +``` + +```Java [] +class MinStack { + Stack A, B; + public MinStack() { + A = new Stack<>(); + B = new Stack<>(); + } + public void push(int x) { + A.add(x); + if(B.empty() || B.peek() >= x) + B.add(x); + } + public void pop() { + if(A.pop().equals(B.peek())) + B.pop(); + } + public int top() { + return A.peek(); + } + public int min() { + return B.peek(); + } +} +``` + +```C++ [] +class MinStack { +public: + stack A, B; + MinStack() {} + void push(int x) { + A.push(x); + if(B.empty() || B.top() >= x) + B.push(x); + } + void pop() { + if(A.top() == B.top()) + B.pop(); + A.pop(); + } + int top() { + return A.top(); + } + int min() { + return B.top(); + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 31. \346\240\210\347\232\204\345\216\213\345\205\245\343\200\201\345\274\271\345\207\272\345\272\217\345\210\227.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 31. \346\240\210\347\232\204\345\216\213\345\205\245\343\200\201\345\274\271\345\207\272\345\272\217\345\210\227.md" new file mode 100644 index 0000000..47e07a5 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 31. \346\240\210\347\232\204\345\216\213\345\205\245\343\200\201\345\274\271\345\207\272\345\272\217\345\210\227.md" @@ -0,0 +1,82 @@ +## 解题思路： + +如下图所示，给定一个压入序列 $pushed$ 和弹出序列 $popped$ ，则压入 / 弹出操作的顺序（即排列）是 **唯一确定** 的。 + +![Picture1.png](https://pic.leetcode-cn.com/1188474dc6a377fc258004bca84e5a130b663eeb24bf22c4fa4eb998a0249f97-Picture1.png){:width=500} + +如下图所示，栈的数据操作具有 **先入后出** 的特性，因此某些弹出序列是无法实现的。 + +![Picture2.png](https://pic.leetcode-cn.com/3f43b224553bf3a37f9758dbb41655e547795e293524a148380c6f335af315e4-Picture2.png){:width=500} + +考虑借用一个辅助栈 $stack$ ，**模拟** 压入 / 弹出操作的排列。根据是否模拟成功，即可得到结果。 + +- **入栈操作：** 按照压栈序列的顺序执行。 +- **出栈操作：** 每次入栈后，循环判断 “栈顶元素 $=$ 弹出序列的当前元素” 是否成立，将符合弹出序列顺序的栈顶元素全部弹出。 + +> 由于题目规定 `栈的所有数字均不相等` ，因此在循环入栈中，每个元素出栈的位置的可能性是唯一的（若有重复数字，则具有多个可出栈的位置）。因而，在遇到 “栈顶元素 $=$ 弹出序列的当前元素” 就应立即执行出栈。 + +### 算法流程： + +1. **初始化：** 辅助栈 $stack$ ，弹出序列的索引 $i$ ； +2. **遍历压栈序列：** 各元素记为 $num$ ； + 1. 元素 $num$ 入栈； + 2. 循环出栈：若 $stack$ 的栈顶元素 $=$ 弹出序列元素 $popped[i]$ ，则执行出栈与 $i++$ ； +3. **返回值：** 若 $stack$ 为空，则此弹出序列合法。 + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为列表 $pushed$ 的长度；每个元素最多入栈与出栈一次，即最多共 $2N$ 次出入栈操作。 +- **空间复杂度 $O(N)$ ：** 辅助栈 $stack$ 最多同时存储 $N$ 个元素。 + + + +## 代码： + +题目指出 `pushed 是 popped 的排列` 。因此，无需考虑 $pushed$ 和 $popped$ **长度不同** 或 **包含元素不同** 的情况。 + +```Python [] +class Solution: + def validateStackSequences(self, pushed: List[int], popped: List[int]) -> bool: + stack, i = [], 0 + for num in pushed: + stack.append(num) # num 入栈 + while stack and stack[-1] == popped[i]: # 循环判断与出栈 + stack.pop() + i += 1 + return not stack +``` + +```Java [] +class Solution { + public boolean validateStackSequences(int[] pushed, int[] popped) { + Stack stack = new Stack<>(); + int i = 0; + for(int num : pushed) { + stack.push(num); // num 入栈 + while(!stack.isEmpty() && stack.peek() == popped[i]) { // 循环判断与出栈 + stack.pop(); + i++; + } + } + return stack.isEmpty(); + } +} +``` + +```C++ [] +class Solution { +public: + bool validateStackSequences(vector& pushed, vector& popped) { + stack stk; + int i = 0; + for(int num : pushed) { + stk.push(num); // num 入栈 + while(!stk.empty() && stk.top() == popped[i]) { // 循环判断与出栈 + stk.pop(); + i++; + } + } + return stk.empty(); + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 32 - I. \344\273\216\344\270\212\345\210\260\344\270\213\346\211\223\345\215\260\344\272\214\345\217\211\346\240\221.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 32 - I. \344\273\216\344\270\212\345\210\260\344\270\213\346\211\223\345\215\260\344\272\214\345\217\211\346\240\221.md" new file mode 100644 index 0000000..94367ba --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 32 - I. \344\273\216\344\270\212\345\210\260\344\270\213\346\211\223\345\215\260\344\272\214\345\217\211\346\240\221.md" @@ -0,0 +1,81 @@ +## 解题思路： + +题目要求的二叉树的 **从上至下** 打印（即按层打印），又称为二叉树的 **广度优先搜索**（BFS）。 +BFS 通常借助 **队列** 的先入先出特性来实现。 + +![Picture1.png](https://pic.leetcode-cn.com/a872b50fa42011748437ec9123d8f77a104b3d528880efca8b212f91d115f835-Picture1.png){:width=400} + +### 算法流程： + +1. **特例处理：** 当树的根节点为空，则直接返回空列表 `[]` ； +2. **初始化：** 打印结果列表 `res = []` ，包含根节点的队列 `queue = [root]` ； +3. **BFS 循环：** 当队列 `queue` 为空时跳出； + 1. **出队：** 队首元素出队，记为 `node`； + 2. **打印：** 将 `node.val` 添加至列表 `tmp` 尾部； + 3. **添加子节点：** 若 `node` 的左（右）子节点不为空，则将左（右）子节点加入队列 `queue` ； +4. **返回值：** 返回打印结果列表 `res` 即可。 + + + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数量，即 BFS 需循环 $N$ 次。 +- **空间复杂度 $O(N)$ ：** 最差情况下，即当树为平衡二叉树时，最多有 $N/2$ 个树节点**同时**在 `queue` 中，使用 $O(N)$ 大小的额外空间。 + +## 代码： + +Python 中使用 collections 中的双端队列 `deque()` ，其 `popleft()` 方法可达到 $O(1)$ 时间复杂度；列表 list 的 `pop(0)` 方法时间复杂度为 $O(N)$ 。 + +```Python [] +class Solution: + def levelOrder(self, root: TreeNode) -> List[int]: + if not root: return [] + res, queue = [], collections.deque() + queue.append(root) + while queue: + node = queue.popleft() + res.append(node.val) + if node.left: queue.append(node.left) + if node.right: queue.append(node.right) + return res +``` + +```Java [] +class Solution { + public int[] levelOrder(TreeNode root) { + if(root == null) return new int[0]; + Queue queue = new LinkedList<>(){{ add(root); }}; + ArrayList ans = new ArrayList<>(); + while(!queue.isEmpty()) { + TreeNode node = queue.poll(); + ans.add(node.val); + if(node.left != null) queue.add(node.left); + if(node.right != null) queue.add(node.right); + } + int[] res = new int[ans.size()]; + for(int i = 0; i < ans.size(); i++) + res[i] = ans.get(i); + return res; + } +} +``` + +```C++ [] +class Solution { +public: + vector levelOrder(TreeNode* root) { + vector res; + if(!root) return res; + TreeNode*> que; + que.push(root); + while(!que.empty()){ + TreeNode* node = que.front(); + que.pop(); + res.push_back(node->val); + if(node->left) que.push(node->left); + if(node->right) que.push(node->right); + } + return res; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 32 - II. \344\273\216\344\270\212\345\210\260\344\270\213\346\211\223\345\215\260\344\272\214\345\217\211\346\240\221 II.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 32 - II. \344\273\216\344\270\212\345\210\260\344\270\213\346\211\223\345\215\260\344\272\214\345\217\211\346\240\221 II.md" new file mode 100644 index 0000000..6ba20c2 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 32 - II. \344\273\216\344\270\212\345\210\260\344\270\213\346\211\223\345\215\260\344\272\214\345\217\211\346\240\221 II.md" @@ -0,0 +1,94 @@ +## 解题思路： + +> 建议先做 [剑指 Offer 32 - I. 从上到下打印二叉树](https://leetcode-cn.com/problems/cong-shang-dao-xia-da-yin-er-cha-shu-lcof/solution/mian-shi-ti-32-i-cong-shang-dao-xia-da-yin-er-ch-4/) 再做此题，两题仅有微小区别，即本题需将 **每一层打印到一行** 。 + +**I. 按层打印：** 题目要求的二叉树的 **从上至下** 打印（即按层打印），又称为二叉树的 **广度优先搜索**（BFS）。BFS 通常借助 **队列** 的先入先出特性来实现。 + +**II. 每层打印到一行：** 将本层全部节点打印到一行，并将下一层全部节点加入队列，以此类推，即可分为多行打印。 + +![Picture1.png](https://pic.leetcode-cn.com/a58477c74c96779c265ce3028def7625d16042895d2c21f7fb0293df7b213276-Picture1.png){:width=400} + +### 算法流程： + +1. **特例处理：** 当根节点为空，则返回空列表 `[]` ； +2. **初始化：** 打印结果列表 `res = []` ，包含根节点的队列 `queue = [root]` ； +3. **BFS 循环：** 当队列 `queue` 为空时跳出； + 1. 新建一个临时列表 `tmp` ，用于存储当前层打印结果； + 2. **当前层打印循环：** 循环次数为当前层节点数（即队列 `queue` 长度）； + 1. **出队：** 队首元素出队，记为 `node`； + 2. **打印：** 将 `node.val` 添加至 `tmp` 尾部； + 3. **添加子节点：** 若 `node` 的左（右）子节点不为空，则将左（右）子节点加入队列 `queue` ； + 3. 将当前层结果 `tmp` 添加入 `res` 。 +4. **返回值：** 返回打印结果列表 `res` 即可。 + + + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数量，即 BFS 需循环 $N$ 次。 +- **空间复杂度 $O(N)$ ：** 最差情况下，即当树为平衡二叉树时，最多有 $N/2$ 个树节点**同时**在 `queue` 中，使用 $O(N)$ 大小的额外空间。 + +## 代码： + +Python 中使用 collections 中的双端队列 `deque()` ，其 `popleft()` 方法可达到 $O(1)$ 时间复杂度；列表 list 的 `pop(0)` 方法时间复杂度为 $O(N)$ 。 + +```Python [] +class Solution: + def levelOrder(self, root: TreeNode) -> List[List[int]]: + if not root: return [] + res, queue = [], collections.deque() + queue.append(root) + while queue: + tmp = [] + for _ in range(len(queue)): + node = queue.popleft() + tmp.append(node.val) + if node.left: queue.append(node.left) + if node.right: queue.append(node.right) + res.append(tmp) + return res +``` + +```Java [] +class Solution { + public List> levelOrder(TreeNode root) { + Queue queue = new LinkedList<>(); + List> res = new ArrayList<>(); + if(root != null) queue.add(root); + while(!queue.isEmpty()) { + List tmp = new ArrayList<>(); + for(int i = queue.size(); i > 0; i--) { + TreeNode node = queue.poll(); + tmp.add(node.val); + if(node.left != null) queue.add(node.left); + if(node.right != null) queue.add(node.right); + } + res.add(tmp); + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + vector> levelOrder(TreeNode* root) { + queue que; + vector> res; + if(root != NULL) que.push(root); + while(!que.empty()) { + vector tmp; + for(int i = que.size(); i > 0; --i) { + root = que.front(); + que.pop(); + tmp.push_back(root->val); + if(root->left != NULL) que.push(root->left); + if(root->right != NULL) que.push(root->right); + } + res.push_back(tmp); + } + return res; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 32 - III. \344\273\216\344\270\212\345\210\260\344\270\213\346\211\223\345\215\260\344\272\214\345\217\211\346\240\221 III.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 32 - III. \344\273\216\344\270\212\345\210\260\344\270\213\346\211\223\345\215\260\344\272\214\345\217\211\346\240\221 III.md" new file mode 100644 index 0000000..4d9e1d4 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 32 - III. \344\273\216\344\270\212\345\210\260\344\270\213\346\211\223\345\215\260\344\272\214\345\217\211\346\240\221 III.md" @@ -0,0 +1,281 @@ +## 解题思路： + +[剑指 Offer 32 - I. 从上到下打印二叉树](https://leetcode-cn.com/problems/cong-shang-dao-xia-da-yin-er-cha-shu-lcof/solution/mian-shi-ti-32-i-cong-shang-dao-xia-da-yin-er-ch-4/) 主要考察 **树的按层打印** ； +[剑指 Offer 32 - II. 从上到下打印二叉树 II](https://leetcode-cn.com/problems/cong-shang-dao-xia-da-yin-er-cha-shu-ii-lcof/solution/mian-shi-ti-32-ii-cong-shang-dao-xia-da-yin-er-c-5/) 额外要求 **每一层打印到一行** ； +本题额外要求 **打印顺序交替变化**（建议按顺序做此三道题）。 + +![Picture1.png](https://pic.leetcode-cn.com/9513dcb034f5dcdea947a2f667b3d380df4f8732da6397778e00718b77584010-Picture1.png){:width=400} + +## 方法一：层序遍历 + 双端队列 + +- 利用双端队列的两端皆可添加元素的特性，设打印列表（双端队列） `tmp` ，并规定： + - 奇数层 则添加至 `tmp` **尾部** ， + - 偶数层 则添加至 `tmp` **头部** 。 + +### 算法流程： + +1. **特例处理：** 当树的根节点为空，则直接返回空列表 `[]` ； +2. **初始化：** 打印结果空列表 `res` ，包含根节点的双端队列 `deque` ； +3. **BFS 循环：** 当 `deque` 为空时跳出； + 1. 新建列表 `tmp` ，用于临时存储当前层打印结果； + 2. **当前层打印循环：** 循环次数为当前层节点数（即 `deque` 长度）； + 1. **出队：** 队首元素出队，记为 `node`； + 2. **打印：** 若为奇数层，将 `node.val` 添加至 `tmp` 尾部；否则，添加至 `tmp` 头部； + 3. **添加子节点：** 若 `node` 的左（右）子节点不为空，则加入 `deque` ； + 3. 将当前层结果 `tmp` 转化为 list 并添加入 `res` ； +4. **返回值：** 返回打印结果列表 `res` 即可； + + + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数量，即 BFS 需循环 $N$ 次，占用 $O(N)$ ；双端队列的队首和队尾的添加和删除操作的时间复杂度均为 $O(1)$ 。 +- **空间复杂度 $O(N)$ ：** 最差情况下，即当树为满二叉树时，最多有 $N/2$ 个树节点 **同时** 在 `deque` 中，使用 $O(N)$ 大小的额外空间。 + +### 代码： + +> Python 中使用 collections 中的双端队列 `deque()` ，其 `popleft()` 方法可达到 $O(1)$ 时间复杂度；列表 list 的 `pop(0)` 方法时间复杂度为 $O(N)$ 。 +> Java 中将链表 LinkedList 作为双端队列使用。 + +```Python [] +class Solution: + def levelOrder(self, root: TreeNode) -> List[List[int]]: + if not root: return [] + res, deque = [], collections.deque([root]) + while deque: + tmp = collections.deque() + for _ in range(len(deque)): + node = deque.popleft() + if len(res) % 2 == 0: tmp.append(node.val) # 奇数层 -> 插入队列尾部 + else: tmp.appendleft(node.val) # 偶数层 -> 插入队列头部 + if node.left: deque.append(node.left) + if node.right: deque.append(node.right) + res.append(list(tmp)) + return res +``` + +```Java [] +class Solution { + public List> levelOrder(TreeNode root) { + Queue queue = new LinkedList<>(); + List> res = new ArrayList<>(); + if(root != null) queue.add(root); + while(!queue.isEmpty()) { + LinkedList tmp = new LinkedList<>(); + for(int i = queue.size(); i > 0; i--) { + TreeNode node = queue.poll(); + if(res.size() % 2 == 0) tmp.addLast(node.val); + else tmp.addFirst(node.val); + if(node.left != null) queue.add(node.left); + if(node.right != null) queue.add(node.right); + } + res.add(tmp); + } + return res; + } +} +``` + +## 方法二：层序遍历 + 双端队列（奇偶层逻辑分离） + +- 方法一代码简短、容易实现；但需要判断每个节点的所在层奇偶性，即冗余了 $N$ 次判断。 +- 通过将奇偶层逻辑拆分，可以消除冗余的判断。 + +### 算法流程： + +> 与方法一对比，仅 BFS 循环不同。 + +- **BFS 循环：** 循环打印奇 / 偶数层，当 `deque` 为空时跳出； + 1. **打印奇数层：** **从左向右** 打印，**先左后右** 加入下层节点； + 2. 若 `deque` 为空，说明向下无偶数层，则跳出； + 3. **打印偶数层：** **从右向左** 打印，**先右后左** 加入下层节点； + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 同方法一。 +- **空间复杂度 $O(N)$ ：** 同方法一。 + +### 代码： + +```Python [] +class Solution: + def levelOrder(self, root: TreeNode) -> List[List[int]]: + if not root: return [] + res, deque = [], collections.deque() + deque.append(root) + while deque: + tmp = [] + # 打印奇数层 + for _ in range(len(deque)): + # 从左向右打印 + node = deque.popleft() + tmp.append(node.val) + # 先左后右加入下层节点 + if node.left: deque.append(node.left) + if node.right: deque.append(node.right) + res.append(tmp) + if not deque: break # 若为空则提前跳出 + # 打印偶数层 + tmp = [] + for _ in range(len(deque)): + # 从右向左打印 + node = deque.pop() + tmp.append(node.val) + # 先右后左加入下层节点 + if node.right: deque.appendleft(node.right) + if node.left: deque.appendleft(node.left) + res.append(tmp) + return res +``` + +```Java [] +class Solution { + public List> levelOrder(TreeNode root) { + Deque deque = new LinkedList<>(); + List> res = new ArrayList<>(); + if(root != null) deque.add(root); + while(!deque.isEmpty()) { + // 打印奇数层 + List tmp = new ArrayList<>(); + for(int i = deque.size(); i > 0; i--) { + // 从左向右打印 + TreeNode node = deque.removeFirst(); + tmp.add(node.val); + // 先左后右加入下层节点 + if(node.left != null) deque.addLast(node.left); + if(node.right != null) deque.addLast(node.right); + } + res.add(tmp); + if(deque.isEmpty()) break; // 若为空则提前跳出 + // 打印偶数层 + tmp = new ArrayList<>(); + for(int i = deque.size(); i > 0; i--) { + // 从右向左打印 + TreeNode node = deque.removeLast(); + tmp.add(node.val); + // 先右后左加入下层节点 + if(node.right != null) deque.addFirst(node.right); + if(node.left != null) deque.addFirst(node.left); + } + res.add(tmp); + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + vector> levelOrder(TreeNode* root) { + deque deque; + vector> res; + if(root != NULL) deque.push_back(root); + while(!deque.empty()) { + // 打印奇数层 + vector tmp; + for(int i = deque.size(); i > 0; i--) { + // 从左向右打印 + TreeNode* node = deque.front(); + deque.pop_front(); + tmp.push_back(node->val); + // 先左后右加入下层节点 + if(node->left != NULL) deque.push_back(node->left); + if(node->right != NULL) deque.push_back(node->right); + } + res.push_back(tmp); + if(deque.empty()) break; // 若为空则提前跳出 + // 打印偶数层 + tmp.clear(); + for(int i = deque.size(); i > 0; i--) { + // 从右向左打印 + TreeNode* node = deque.back(); + deque.pop_back(); + tmp.push_back(node->val); + // 先右后左加入下层节点 + if(node->right != NULL) deque.push_front(node->right); + if(node->left != NULL) deque.push_front(node->left); + } + res.push_back(tmp); + } + return res; + } +}; +``` + +## 方法三：层序遍历 + 倒序 + +- 此方法的优点是只用列表即可，无需其他数据结构。 +- **偶数层倒序：** 若 `res` 的长度为 **奇数** ，说明当前是偶数层，则对 `tmp` 执行 **倒序** 操作。 + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数量，即 BFS 需循环 $N$ 次，占用 $O(N)$ 。**共完成** 少于 $N$ 个节点的倒序操作，占用 $O(N)$ 。 +- **空间复杂度 $O(N)$ ：** 最差情况下，即当树为满二叉树时，最多有 $N/2$ 个树节点**同时**在 `queue` 中，使用 $O(N)$ 大小的额外空间。 + + + +### 代码： + +```Python [] +class Solution: + def levelOrder(self, root: TreeNode) -> List[List[int]]: + if not root: return [] + res, queue = [], collections.deque() + queue.append(root) + while queue: + tmp = [] + for _ in range(len(queue)): + node = queue.popleft() + tmp.append(node.val) + if node.left: queue.append(node.left) + if node.right: queue.append(node.right) + res.append(tmp[::-1] if len(res) % 2 else tmp) + return res +``` + +```Java [] +class Solution { + public List> levelOrder(TreeNode root) { + Queue queue = new LinkedList<>(); + List> res = new ArrayList<>(); + if(root != null) queue.add(root); + while(!queue.isEmpty()) { + List tmp = new ArrayList<>(); + for(int i = queue.size(); i > 0; i--) { + TreeNode node = queue.poll(); + tmp.add(node.val); + if(node.left != null) queue.add(node.left); + if(node.right != null) queue.add(node.right); + } + if(res.size() % 2 == 1) Collections.reverse(tmp); + res.add(tmp); + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + vector> levelOrder(TreeNode* root) { + queue que; + vector> res; + if(root != NULL) que.push(root); + while(!que.empty()) { + vector tmp; + for(int i = que.size(); i > 0; i--) { + TreeNode* node = que.front(); + que.pop(); + tmp.push_back(node->val); + if(node->left != NULL) que.push(node->left); + if(node->right != NULL) que.push(node->right); + } + if(res.size() % 2 == 1) reverse(tmp.begin(),tmp.end()); + res.push_back(tmp); + } + return res; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 33. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\347\232\204\345\220\216\345\272\217\351\201\215\345\216\206\345\272\217\345\210\227.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 33. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\347\232\204\345\220\216\345\272\217\351\201\215\345\216\206\345\272\217\345\210\227.md" new file mode 100644 index 0000000..496b4cc --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 33. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\347\232\204\345\220\216\345\272\217\351\201\215\345\216\206\345\272\217\345\210\227.md" @@ -0,0 +1,167 @@ +## 解题思路： + +**后序遍历定义：** `[ 左子树 | 右子树 | 根节点 ]` ，即遍历顺序为 “左、右、根” 。 + +**二叉搜索树定义：** 左子树中所有节点的值 $<$ 根节点的值；右子树中所有节点的值 $>$ 根节点的值；其左、右子树也分别为二叉搜索树。 + +![Picture1.png](https://pic.leetcode-cn.com/1599753507-JrFBjm-Picture1.png){:width=500} + +## 方法一：递归分治 + +根据二叉搜索树的定义，可以通过递归，判断所有子树的 **正确性** （即其后序遍历是否满足二叉搜索树的定义） ，若所有子树都正确，则此序列为二叉搜索树的后序遍历。 + +### 递归解析： + +- **终止条件：** 当 $i \geq j$ ，说明此子树节点数量 $\leq 1$ ，无需判别正确性，因此直接返回 $true$ ； +- **递推工作：** + 1. **划分左右子树：** 遍历后序遍历的 $[i, j]$ 区间元素，寻找 **第一个大于根节点** 的节点，索引记为 $m$ 。此时，可划分出左子树区间 $[i,m-1]$ 、右子树区间 $[m, j - 1]$ 、根节点索引 $j$ 。 + 2. **判断是否为二叉搜索树：** + - **左子树区间** $[i, m - 1]$ 内的所有节点都应 $<$ $postorder[j]$ 。而第 `1.划分左右子树` 步骤已经保证左子树区间的正确性，因此只需要判断右子树区间即可。 + - **右子树区间** $[m, j-1]$ 内的所有节点都应 $>$ $postorder[j]$ 。实现方式为遍历，当遇到 $\leq postorder[j]$ 的节点则跳出；则可通过 $p = j$ 判断是否为二叉搜索树。 +- **返回值：** 所有子树都需正确才可判定正确，因此使用 **与逻辑符** $\&\&$ 连接。 + 1. **$p = j$ ：** 判断 **此树** 是否正确。 + 2. **$recur(i, m - 1)$ ：** 判断 **此树的左子树** 是否正确。 + 3. **$recur(m, j - 1)$ ：** 判断 **此树的右子树** 是否正确。 + + + +**复杂度分析：** + +- **时间复杂度 $O(N^2)$ ：** 每次调用 $recur(i,j)$ 减去一个根节点，因此递归占用 $O(N)$ ；最差情况下（即当树退化为链表），每轮递归都需遍历树所有节点，占用 $O(N)$ 。 +- **空间复杂度 $O(N)$ ：** 最差情况下（即当树退化为链表），递归深度将达到 $N$ 。 + +```Python [] +class Solution: + def verifyPostorder(self, postorder: [int]) -> bool: + def recur(i, j): + if i >= j: return True + p = i + while postorder[p] < postorder[j]: p += 1 + m = p + while postorder[p] > postorder[j]: p += 1 + return p == j and recur(i, m - 1) and recur(m, j - 1) + + return recur(0, len(postorder) - 1) +``` + +```Java [] +class Solution { + public boolean verifyPostorder(int[] postorder) { + return recur(postorder, 0, postorder.length - 1); + } + boolean recur(int[] postorder, int i, int j) { + if(i >= j) return true; + int p = i; + while(postorder[p] < postorder[j]) p++; + int m = p; + while(postorder[p] > postorder[j]) p++; + return p == j && recur(postorder, i, m - 1) && recur(postorder, m, j - 1); + } +} +``` + +```C++ [] +class Solution { +public: + bool verifyPostorder(vector& postorder) { + return recur(postorder, 0, postorder.size() - 1); + } +private: + bool recur(vector& postorder, int i, int j) { + if(i >= j) return true; + int p = i; + while(postorder[p] < postorder[j]) p++; + int m = p; + while(postorder[p] > postorder[j]) p++; + return p == j && recur(postorder, i, m - 1) && recur(postorder, m, j - 1); + } +}; +``` + +## 方法二：辅助单调栈 + +**后序遍历倒序：** `[ 根节点 | 右子树 | 左子树 ]` 。类似 **先序遍历的镜像** ，即先序遍历为 “根、左、右” 的顺序，而后序遍历的倒序为 “根、右、左” 顺序。 + +![Picture10.png](https://pic.leetcode-cn.com/1599753507-KaeaWl-Picture10.png){:width=500} + +设后序遍历倒序列表为 $[r_{n}, r_{n-1},...,r_1]$，遍历此列表，设索引为 $i$ ，若为 **二叉搜索树** ，则有： + +- **当节点值 $r_i > r_{i+1}$ 时：** 节点 $r_i$ 一定是节点 $r_{i+1}$ 的右子节点。 +- **当节点值 $r_i < r_{i+1}$ 时：** 节点 $r_i$ 一定是某节点 $root$ 的左子节点，且 $root$ 为节点 $r_{i+1}, r_{i+2},..., r_{n}$ 中值大于且最接近 $r_i$ 的节点（∵ $root$ **直接连接** 左子节点 $r_i$ ）。 + +当遍历时遇到递减节点 $r_i < r_{i+1}$ ，若为二叉搜索树，则对于后序遍历中节点 $r_i$ 右边的任意节点 $r_x \in [r_{i-1}, r_{i-2}, ..., r_1]$ ，必有节点值 $r_x < root$ 。 + +> 节点 $r_x$ 只可能为以下两种情况：① $r_x$ 为 $r_i$ 的左、右子树的各节点；② $r_x$ 为 $root$ 的父节点或更高层父节点的左子树的各节点。在二叉搜索树中，以上节点都应小于 $root$ 。 + +![Picture11.png](https://pic.leetcode-cn.com/1599753507-mgqkoF-Picture11.png){:width=500} + +遍历 “后序遍历的倒序” 会多次遇到递减节点 $r_i$ ，若所有的递减节点 $r_i$ 对应的父节点 $root$ 都满足以上条件，则可判定为二叉搜索树。根据以上特点，考虑借助 **单调栈** 实现： + +1. 借助一个单调栈 $stack$ 存储值递增的节点； +2. 每当遇到值递减的节点 $r_i$ ，则通过出栈来更新节点 $r_i$ 的父节点 $root$ ； +3. 每轮判断 $r_i$ 和 $root$ 的值关系： + 1. 若 $r_i > root$ 则说明不满足二叉搜索树定义，直接返回 $false$ 。 + 2. 若 $r_i < root$ 则说明满足二叉搜索树定义，则继续遍历。 + +### 算法流程： + +1. **初始化：** 单调栈 $stack$ ，父节点值 $root = +\infin$ （初始值为正无穷大，可把树的根节点看为此无穷大节点的左孩子）； +2. **倒序遍历 $postorder$** ：记每个节点为 $r_i$； + 1. **判断：** 若 $r_i>root$ ，说明此后序遍历序列不满足二叉搜索树定义，直接返回 $false$ ； + 2. **更新父节点 $root$ ：** 当栈不为空 **且** $r_i + +**复杂度分析：** + +- **时间复杂度 $O(N)$ ：** 遍历 $postorder$ 所有节点，各节点均入栈 / 出栈一次，使用 $O(N)$ 时间。 +- **空间复杂度 $O(N)$ ：** 最差情况下，单调栈 $stack$ 存储所有节点，使用 $O(N)$ 额外空间。 + +```Python [] +class Solution: + def verifyPostorder(self, postorder: [int]) -> bool: + stack, root = [], float("+inf") + for i in range(len(postorder) - 1, -1, -1): + if postorder[i] > root: return False + while(stack and postorder[i] < stack[-1]): + root = stack.pop() + stack.append(postorder[i]) + return True +``` + +```Java [] +class Solution { + public boolean verifyPostorder(int[] postorder) { + Stack stack = new Stack<>(); + int root = Integer.MAX_VALUE; + for(int i = postorder.length - 1; i >= 0; i--) { + if(postorder[i] > root) return false; + while(!stack.isEmpty() && stack.peek() > postorder[i]) + root = stack.pop(); + stack.add(postorder[i]); + } + return true; + } +} +``` + +```C++ [] +class Solution { +public: + bool verifyPostorder(vector& postorder) { + stack stk; + int root = INT_MAX; + for(int i = postorder.size() - 1; i >= 0; i--) { + if(postorder[i] > root) return false; + while(!stk.empty() && stk.top() > postorder[i]) { + root = stk.top(); + stk.pop(); + } + stk.push(postorder[i]); + } + return true; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 34. \344\272\214\345\217\211\346\240\221\344\270\255\345\222\214\344\270\272\346\237\220\344\270\200\345\200\274\347\232\204\350\267\257\345\276\204.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 34. \344\272\214\345\217\211\346\240\221\344\270\255\345\222\214\344\270\272\346\237\220\344\270\200\345\200\274\347\232\204\350\267\257\345\276\204.md" new file mode 100644 index 0000000..cf10c13 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 34. \344\272\214\345\217\211\346\240\221\344\270\255\345\222\214\344\270\272\346\237\220\344\270\200\345\200\274\347\232\204\350\267\257\345\276\204.md" @@ -0,0 +1,105 @@ +## 解题思路： + +本题是典型的二叉树方案搜索问题，使用回溯法解决，其包含 **先序遍历 + 路径记录** 两部分。 + +- **先序遍历：** 按照 “根、左、右” 的顺序，遍历树的所有节点。 +- **路径记录：** 在先序遍历中，记录从根节点到当前节点的路径。当路径满足 ① 根节点到叶节点形成的路径 **且** ② 各节点值的和等于目标值 `sum` 时，将此路径加入结果列表。 + +![Picture1.png](https://pic.leetcode-cn.com/1599400747-BuGhCT-Picture1.png){:width=500} + +### 算法流程： + +**`pathSum(root, sum)` 函数：** + +- **初始化：** 结果列表 `res` ，路径列表 `path` 。 +- **返回值：** 返回 `res` 即可。 + +**`recur(root, tar) 函数：`** + +- **递推参数：** 当前节点 `root` ，当前目标值 `tar` 。 +- **终止条件：** 若节点 `root` 为空，则直接返回。 +- **递推工作：** + 1. 路径更新： 将当前节点值 `root.val` 加入路径 `path` 。 + 2. 目标值更新： `tar = tar - root.val`（即目标值 `tar` 从 `sum` 减至 $0$ ）。 + 3. 路径记录： 当 ① `root` 为叶节点 **且** ② 路径和等于目标值 ，则将此路径 `path` 加入 `res` 。 + 4. 先序遍历： 递归左 / 右子节点。 + 5. 路径恢复： 向上回溯前，需要将当前节点从路径 `path` 中删除，即执行 `path.pop()` 。 + + + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数，先序遍历需要遍历所有节点。 +- **空间复杂度 $O(N)$ ：** 最差情况下，即树退化为链表时，`path` 存储所有树节点，使用 $O(N)$ 额外空间。 + +## 代码： + +以 Python 语言为例，记录路径时若直接执行 `res.append(path)` ，则是将此 `path` 对象加入了 `res` ；后续 `path` 改变时， `res` 中的 `path` 对象也会随之改变，因此无法实现结果记录。正确做法为： + +- Python: `res.append(list(path))` ； +- Java: `res.add(new LinkedList(path))` ； +- C++: `res.push_back(path)` ； + +> 三者的原理都是避免直接添加 `path` 对象，而是 **拷贝** 了一个 `path` 对象并加入到 `res` 。 + +```Python [] +class Solution: + def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]: + res, path = [], [] + def recur(root, tar): + if not root: return + path.append(root.val) + tar -= root.val + if tar == 0 and not root.left and not root.right: + res.append(list(path)) + recur(root.left, tar) + recur(root.right, tar) + path.pop() + + recur(root, sum) + return res +``` + +```Java [] +class Solution { + LinkedList> res = new LinkedList<>(); + LinkedList path = new LinkedList<>(); + public List> pathSum(TreeNode root, int sum) { + recur(root, sum); + return res; + } + void recur(TreeNode root, int tar) { + if(root == null) return; + path.add(root.val); + tar -= root.val; + if(tar == 0 && root.left == null && root.right == null) + res.add(new LinkedList(path)); + recur(root.left, tar); + recur(root.right, tar); + path.removeLast(); + } +} +``` + +```C++ [] +class Solution { +public: + vector> pathSum(TreeNode* root, int sum) { + recur(root, sum); + return res; + } +private: + vector> res; + vector path; + void recur(TreeNode* root, int tar) { + if(root == nullptr) return; + path.push_back(root->val); + tar -= root->val; + if(tar == 0 && root->left == nullptr && root->right == nullptr) + res.push_back(path); + recur(root->left, tar); + recur(root->right, tar); + path.pop_back(); + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 35. \345\244\215\346\235\202\351\223\276\350\241\250\347\232\204\345\244\215\345\210\266.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 35. \345\244\215\346\235\202\351\223\276\350\241\250\347\232\204\345\244\215\345\210\266.md" new file mode 100644 index 0000000..037bad0 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 35. \345\244\215\346\235\202\351\223\276\350\241\250\347\232\204\345\244\215\345\210\266.md" @@ -0,0 +1,359 @@ +## 解题思路： + +> 普通链表的节点定义如下： + +```Python [] +# Definition for a Node. +class Node: + def __init__(self, x: int, next: 'Node' = None): + self.val = int(x) + self.next = next +``` + +```Java [] +// Definition for a Node. +class Node { + int val; + Node next; + public Node(int val) { + this.val = val; + this.next = null; + } +} +``` + +```C++ [] +// Definition for a Node. +class Node { +public: + int val; + Node* next; + Node(int _val) { + val = _val; + next = NULL; + } +}; +``` + +> 本题链表的节点定义如下： + +```Python [] +# Definition for a Node. +class Node: + def __init__(self, x: int, next: 'Node' = None, random: 'Node' = None): + self.val = int(x) + self.next = next + self.random = random +``` + +```Java [] +// Definition for a Node. +class Node { + int val; + Node next, random; + public Node(int val) { + this.val = val; + this.next = null; + this.random = null; + } +} +``` + +```C++ [] +// Definition for a Node. +class Node { +public: + int val; + Node* next; + Node* random; + Node(int _val) { + val = _val; + next = NULL; + random = NULL; + } +}; +``` + +给定链表的头节点 `head` ，复制普通链表很简单，只需遍历链表，每轮建立新节点 + 构建前驱节点 `pre` 和当前节点 `node` 的引用指向即可。 + +本题链表的节点新增了 `random` 指针，指向链表中的 **任意节点** 或者 $null$ 。这个 `random` 指针意味着在复制过程中，除了构建前驱节点和当前节点的引用指向 `pre.next` ，还要构建前驱节点和其随机节点的引用指向 `pre.random` 。 + +**本题难点：** 在复制链表的过程中构建新链表各节点的 `random` 引用指向。 + +![Picture1.png](https://pic.leetcode-cn.com/1604747285-ELUgCd-Picture1.png){:width=450} + +```Python [] +class Solution: + def copyRandomList(self, head: 'Node') -> 'Node': + cur = head + dum = pre = Node(0) + while cur: + node = Node(cur.val) # 复制节点 cur + pre.next = node # 新链表的 前驱节点 -> 当前节点 + # pre.random = '???' # 新链表的 「 前驱节点 -> 当前节点 」 无法确定 + cur = cur.next # 遍历下一节点 + pre = node # 保存当前新节点 + return dum.next +``` + +```Java [] +class Solution { + public Node copyRandomList(Node head) { + Node cur = head; + Node dum = new Node(0), pre = dum; + while(cur != null) { + Node node = new Node(cur.val); // 复制节点 cur + pre.next = node; // 新链表的 前驱节点 -> 当前节点 + // pre.random = "???"; // 新链表的 「 前驱节点 -> 当前节点 」 无法确定 + cur = cur.next; // 遍历下一节点 + pre = node; // 保存当前新节点 + } + return dum.next; + } +} +``` + +```C++ [] +class Solution { +public: + Node* copyRandomList(Node* head) { + Node* cur = head; + Node* dum = new Node(0), *pre = dum; + while(cur != nullptr) { + Node* node = new Node(cur->val); // 复制节点 cur + pre->next = node; // 新链表的 前驱节点 -> 当前节点 + // pre->random = "???"; // 新链表的 「 前驱节点 -> 当前节点 」 无法确定 + cur = cur->next; // 遍历下一节点 + pre = node; // 保存当前新节点 + } + return dum->next; + } +}; +``` + +> 本文介绍 「哈希表」 ，「拼接 + 拆分」 两种方法。哈希表方法比较直观；拼接 + 拆分方法的空间复杂度更低。 + +## 方法一：哈希表 + +利用哈希表的查询特点，考虑构建 **原链表节点** 和 **新链表对应节点** 的键值对映射关系，再遍历构建新链表各节点的 `next` 和 `random` 引用指向即可。 + +### 算法流程： + +1. 若头节点 `head` 为空节点，直接返回 $null$ ； +2. **初始化：** 哈希表 `dic` ， 节点 `cur` 指向头节点； +3. **复制链表：** + 1. 建立新节点，并向 `dic` 添加键值对 `(原 cur 节点, 新 cur 节点）` ； + 2. `cur` 遍历至原链表下一节点； +4. **构建新链表的引用指向：** + 1. 构建新节点的 `next` 和 `random` 引用指向； + 2. `cur` 遍历至原链表下一节点； +5. **返回值：** 新链表的头节点 `dic[cur]` ； + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 两轮遍历链表，使用 $O(N)$ 时间。 +- **空间复杂度 $O(N)$ ：** 哈希表 `dic` 使用线性大小的额外空间。 + + + +### 代码： + +```Python [] +class Solution: + def copyRandomList(self, head: 'Node') -> 'Node': + if not head: return + dic = {} + # 3. 复制各节点，并建立 “原节点 -> 新节点” 的 Map 映射 + cur = head + while cur: + dic[cur] = Node(cur.val) + cur = cur.next + cur = head + # 4. 构建新节点的 next 和 random 指向 + while cur: + dic[cur].next = dic.get(cur.next) + dic[cur].random = dic.get(cur.random) + cur = cur.next + # 5. 返回新链表的头节点 + return dic[head] +``` + +```Java [] +class Solution { + public Node copyRandomList(Node head) { + if(head == null) return null; + Node cur = head; + Map map = new HashMap<>(); + // 3. 复制各节点，并建立 “原节点 -> 新节点” 的 Map 映射 + while(cur != null) { + map.put(cur, new Node(cur.val)); + cur = cur.next; + } + cur = head; + // 4. 构建新链表的 next 和 random 指向 + while(cur != null) { + map.get(cur).next = map.get(cur.next); + map.get(cur).random = map.get(cur.random); + cur = cur.next; + } + // 5. 返回新链表的头节点 + return map.get(head); + } +} +``` + +```C++ [] +class Solution { +public: + Node* copyRandomList(Node* head) { + if(head == nullptr) return nullptr; + Node* cur = head; + unordered_map map; + // 3. 复制各节点，并建立 “原节点 -> 新节点” 的 Map 映射 + while(cur != nullptr) { + map[cur] = new Node(cur->val); + cur = cur->next; + } + cur = head; + // 4. 构建新链表的 next 和 random 指向 + while(cur != nullptr) { + map[cur]->next = map[cur->next]; + map[cur]->random = map[cur->random]; + cur = cur->next; + } + // 5. 返回新链表的头节点 + return map[head]; + } +}; +``` + +## 方法二：拼接 + 拆分 + +考虑构建 `原节点 1 -> 新节点 1 -> 原节点 2 -> 新节点 2 -> ……` 的拼接链表，如此便可在访问原节点的 `random` 指向节点的同时找到新对应新节点的 `random` 指向节点。 + +### 算法流程： + +1. **复制各节点，构建拼接链表:** + + - 设原链表为 $node1 \rightarrow node2 \rightarrow \cdots$ ，构建的拼接链表如下所示： + +$$ +node1 \rightarrow node1_{new} \rightarrow node2 \rightarrow node2_{new} \rightarrow \cdots +$$ + +2. **构建新链表各节点的 `random` 指向：** + + - 当访问原节点 `cur` 的随机指向节点 `cur.random` 时，对应新节点 `cur.next` 的随机指向节点为 `cur.random.next` 。 + +3. **拆分原 / 新链表：** + + - 设置 `pre` / `cur` 分别指向原 / 新链表头节点，遍历执行 `pre.next = pre.next.next` 和 `cur.next = cur.next.next` 将两链表拆分开。 + +4. 返回新链表的头节点 `res` 即可。 + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 三轮遍历链表，使用 $O(N)$ 时间。 +- **空间复杂度 $O(1)$ ：** 节点引用变量使用常数大小的额外空间。 + + + +### 代码： + +```Python [] +class Solution: + def copyRandomList(self, head: 'Node') -> 'Node': + if not head: return + cur = head + # 1. 复制各节点，并构建拼接链表 + while cur: + tmp = Node(cur.val) + tmp.next = cur.next + cur.next = tmp + cur = tmp.next + # 2. 构建各新节点的 random 指向 + cur = head + while cur: + if cur.random: + cur.next.random = cur.random.next + cur = cur.next.next + # 3. 拆分两链表 + cur = res = head.next + pre = head + while cur.next: + pre.next = pre.next.next + cur.next = cur.next.next + pre = pre.next + cur = cur.next + pre.next = None # 单独处理原链表尾节点 + return res # 返回新链表头节点 +``` + +```Java [] +class Solution { + public Node copyRandomList(Node head) { + if(head == null) return null; + Node cur = head; + // 1. 复制各节点，并构建拼接链表 + while(cur != null) { + Node tmp = new Node(cur.val); + tmp.next = cur.next; + cur.next = tmp; + cur = tmp.next; + } + // 2. 构建各新节点的 random 指向 + cur = head; + while(cur != null) { + if(cur.random != null) + cur.next.random = cur.random.next; + cur = cur.next.next; + } + // 3. 拆分两链表 + cur = head.next; + Node pre = head, res = head.next; + while(cur.next != null) { + pre.next = pre.next.next; + cur.next = cur.next.next; + pre = pre.next; + cur = cur.next; + } + pre.next = null; // 单独处理原链表尾节点 + return res; // 返回新链表头节点 + } +} +``` + +```C++ [] +class Solution { +public: + Node* copyRandomList(Node* head) { + if(head == nullptr) return nullptr; + Node* cur = head; + // 1. 复制各节点，并构建拼接链表 + while(cur != nullptr) { + Node* tmp = new Node(cur->val); + tmp->next = cur->next; + cur->next = tmp; + cur = tmp->next; + } + // 2. 构建各新节点的 random 指向 + cur = head; + while(cur != nullptr) { + if(cur->random != nullptr) + cur->next->random = cur->random->next; + cur = cur->next->next; + } + // 3. 拆分两链表 + cur = head->next; + Node* pre = head, *res = head->next; + while(cur->next != nullptr) { + pre->next = pre->next->next; + cur->next = cur->next->next; + pre = pre->next; + cur = cur->next; + } + pre->next = nullptr; // 单独处理原链表尾节点 + return res; // 返回新链表头节点 + } +}; +``` \ No newline at end of file diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 36. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\344\270\216\345\217\214\345\220\221\351\223\276\350\241\250.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 36. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\344\270\216\345\217\214\345\220\221\351\223\276\350\241\250.md" new file mode 100644 index 0000000..b2efd18 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 36. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\344\270\216\345\217\214\345\220\221\351\223\276\350\241\250.md" @@ -0,0 +1,138 @@ +## 解题思路： + +本文解法基于性质：二叉搜索树的中序遍历为 **递增序列** 。 +将 二叉搜索树 转换成一个 “排序的循环双向链表” ，其中包含三个要素： + +1. **排序链表：** 节点应从小到大排序，因此应使用 **中序遍历** “从小到大”访问树的节点。 +2. **双向链表：** 在构建相邻节点的引用关系时，设前驱节点 `pre` 和当前节点 `cur` ，不仅应构建 `pre.right = cur` ，也应构建 `cur.left = pre` 。 +3. **循环链表：** 设链表头节点 `head` 和尾节点 `tail` ，则应构建 `head.left = tail` 和 `tail.right = head` 。 + +![Picture1.png](https://pic.leetcode-cn.com/1599401091-PKIjds-Picture1.png){:width=500} + +**中序遍历** 为对二叉树作 “左、根、右” 顺序遍历，递归实现如下： + +```Python [] +# 打印中序遍历 +def dfs(root): + if not root: return + dfs(root.left) # 左 + print(root.val) # 根 + dfs(root.right) # 右 +``` + +```Java [] +// 打印中序遍历 +void dfs(Node root) { + if(root == null) return; + dfs(root.left); // 左 + System.out.println(root.val); // 根 + dfs(root.right); // 右 +} +``` + +```C++ [] +// 打印中序遍历 +void dfs(Node* root) { + if(root == nullptr) return; + dfs(root->left); // 左 + cout << root->val << endl; // 根 + dfs(root->right); // 右 +} +``` + +根据以上分析，考虑使用中序遍历访问树的各节点 `cur` ；并在访问每个节点时构建 `cur` 和前驱节点 `pre` 的引用指向；中序遍历完成后，最后构建头节点和尾节点的引用指向即可。 + +### 算法流程： + +**`dfs(cur):`** 递归法中序遍历； + +1. **终止条件：** 当节点 `cur` 为空，代表越过叶节点，直接返回； +2. 递归左子树，即 `dfs(cur.left)` ； +3. **构建链表：** + 1. **当 `pre` 为空时：** 代表正在访问链表头节点，记为 `head` ； + 2. **当 `pre` 不为空时：** 修改双向节点引用，即 `pre.right = cur` ， `cur.left = pre` ； + 3. **保存 `cur` ：** 更新 `pre = cur` ，即节点 `cur` 是后继节点的 `pre` ； +4. 递归右子树，即 `dfs(cur.right)` ； + +**`treeToDoublyList(root)：`** + +1. **特例处理：** 若节点 `root` 为空，则直接返回； +2. **初始化：** 空节点 `pre` ； +3. **转化为双向链表：** 调用 `dfs(root)` ； +4. **构建循环链表：** 中序遍历完成后，`head` 指向头节点， `pre` 指向尾节点，因此修改 `head` 和 `pre` 的双向节点引用即可； +5. **返回值：** 返回链表的头节点 `head` 即可； + + + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数，中序遍历需要访问所有节点。 +- **空间复杂度 $O(N)$ ：** 最差情况下，即树退化为链表时，递归深度达到 $N$，系统使用 $O(N)$ 栈空间。 + +## 代码： + +```Python [] +class Solution: + def treeToDoublyList(self, root: 'Node') -> 'Node': + def dfs(cur): + if not cur: return + dfs(cur.left) # 递归左子树 + if self.pre: # 修改节点引用 + self.pre.right, cur.left = cur, self.pre + else: # 记录头节点 + self.head = cur + self.pre = cur # 保存 cur + dfs(cur.right) # 递归右子树 + + if not root: return + self.pre = None + dfs(root) + self.head.left, self.pre.right = self.pre, self.head + return self.head +``` + +```Java [] +class Solution { + Node pre, head; + public Node treeToDoublyList(Node root) { + if(root == null) return null; + dfs(root); + head.left = pre; + pre.right = head; + return head; + } + void dfs(Node cur) { + if(cur == null) return; + dfs(cur.left); + if(pre != null) pre.right = cur; + else head = cur; + cur.left = pre; + pre = cur; + dfs(cur.right); + } +} +``` + +```C++ [] +class Solution { +public: + Node* treeToDoublyList(Node* root) { + if(root == nullptr) return nullptr; + dfs(root); + head->left = pre; + pre->right = head; + return head; + } +private: + Node *pre, *head; + void dfs(Node* cur) { + if(cur == nullptr) return; + dfs(cur->left); + if(pre != nullptr) pre->right = cur; + else head = cur; + cur->left = pre; + pre = cur; + dfs(cur->right); + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 37. \345\272\217\345\210\227\345\214\226\344\272\214\345\217\211\346\240\221.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 37. \345\272\217\345\210\227\345\214\226\344\272\214\345\217\211\346\240\221.md" new file mode 100644 index 0000000..e9500de --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 37. \345\272\217\345\210\227\345\214\226\344\272\214\345\217\211\346\240\221.md" @@ -0,0 +1,153 @@ +通常使用的前序、中序、后序、层序遍历记录的二叉树的信息不完整，即唯一的输出序列可能对应着多种二叉树可能性。题目要求的 序列化 和 反序列化 是 **可逆操作** 。因此，序列化的字符串应携带 **完整的二叉树信息** 。 + +> 观察题目示例，序列化的字符串实际上是二叉树的 “层序遍历”（BFS）结果，本文也采用层序遍历。 + +为完整表示二叉树，考虑将叶节点下的 `null` 也记录。在此基础上，对于列表中任意某节点 `node` ，其左子节点 `node.left` 和右子节点 `node.right` 在序列中的位置都是 **唯一确定** 的。如下图所示： + +![Picture1.png](https://pic.leetcode-cn.com/1603117385-ehAGsP-Picture1.png){:width=550} + +上图规律可总结为下表： + +| `node.val` | `node` 的索引 | `node.left` 的索引 | `node.right` 的索引 | +| :--------: | :-----------: | :----------------: | :-----------------: | +| $1$ | $0$ | $1$ | $2$ | +| $2$ | $1$ | $3$ | $4$ | +| $3$ | $2$ | $5$ | $6$ | +| $4$ | $5$ | $7$ | $8$ | +| $5$ | $6$ | $9$ | $10$ | + +设 $m$ 为列表区间 $[0, n]$ 中的 `null` 节点个数，则可总结出根节点、左子节点、右子节点的列表索引的递推公式： + +| `node.val` | `node` 的列表索引 | `node.left` 的列表索引 | `node.right` 的列表索引 | +| :----------: | :---------------: | :--------------------: | :---------------------: | +| $\ne$ `null` | $n$ | $2(n-m) + 1$ | $2(n-m) + 2$ | +| $=$ `null` | $n$ | 无 | 无 | + +**序列化** 使用层序遍历实现。**反序列化** 通过以上递推公式反推各节点在序列中的索引，进而实现。 + +## 序列化 Serialize ： + +借助队列，对二叉树做层序遍历，并将越过叶节点的 `null` 也打印出来。 + +### 算法流程： + +1. **特例处理：** 若 `root` 为空，则直接返回空列表 `"[]"` ； +2. **初始化：** 队列 `queue` （包含根节点 `root` ）；序列化列表 `res` ； +3. **层序遍历：** 当 `queue` 为空时跳出； + 1. 节点出队，记为 `node` ； + 2. 若 `node` 不为空：① 打印字符串 `node.val` ，② 将左、右子节点加入 `queue` ； + 3. 否则（若 `node` 为空）：打印字符串 `"null"` ； +4. **返回值：** 拼接列表，用 `','` 隔开，首尾添加中括号； + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数，层序遍历需要访问所有节点，最差情况下需要访问 $N + 1$ 个 `null` ，总体复杂度为 $O(2N + 1) = O(N)$ 。 +- **空间复杂度 $O(N)$ ：** 最差情况下，队列 `queue` 同时存储 $\frac{N + 1}{2}$ 个节点（或 $N+1$ 个 `null` ），使用 $O(N)$ ；列表 `res` 使用 $O(N)$ 。 + + + +## 反序列化 Deserialize ： + +基于本文开始推出的 `node` , `node.left` , `node.right` 在序列化列表中的位置关系，可实现反序列化。 + +利用队列按层构建二叉树，借助一个指针 `i` 指向节点 `node` 的左、右子节点，每构建一个 `node` 的左、右子节点，指针 `i` 就向右移动 $1$ 位。 + +### 算法流程： + +1. **特例处理：** 若 `data` 为空，直接返回 `null` ； +2. **初始化：** 序列化列表 `vals` （先去掉首尾中括号，再用逗号隔开），指针 `i = 1` ，根节点 `root` （值为 `vals[0]` ），队列 `queue`（包含 `root` ）； +3. **按层构建：** 当 `queue` 为空时跳出； + 1. 节点出队，记为 `node` ； + 2. 构建 `node` 的左子节点：`node.left` 的值为 `vals[i]` ，并将 `node.left` 入队； + 3. 执行 `i += 1` ； + 4. 构建 `node` 的右子节点：`node.right` 的值为 `vals[i]` ，并将 `node.right` 入队； + 5. 执行 `i += 1` ； +4. **返回值：** 返回根节点 `root` 即可； + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数，按层构建二叉树需要遍历整个 $vals$ ，其长度最大为 $2N+1$ 。 +- **空间复杂度 $O(N)$ ：** 最差情况下，队列 `queue` 同时存储 $\frac{N + 1}{2}$ 个节点，因此使用 $O(N)$ 额外空间。 + + + +## 代码： + +```Python [] +class Codec: + def serialize(self, root): + if not root: return "[]" + queue = collections.deque() + queue.append(root) + res = [] + while queue: + node = queue.popleft() + if node: + res.append(str(node.val)) + queue.append(node.left) + queue.append(node.right) + else: res.append("null") + return '[' + ','.join(res) + ']' + + def deserialize(self, data): + if data == "[]": return + vals, i = data[1:-1].split(','), 1 + root = TreeNode(int(vals[0])) + queue = collections.deque() + queue.append(root) + while queue: + node = queue.popleft() + if vals[i] != "null": + node.left = TreeNode(int(vals[i])) + queue.append(node.left) + i += 1 + if vals[i] != "null": + node.right = TreeNode(int(vals[i])) + queue.append(node.right) + i += 1 + return root +``` + +```Java [] +public class Codec { + public String serialize(TreeNode root) { + if(root == null) return "[]"; + StringBuilder res = new StringBuilder("["); + Queue queue = new LinkedList<>() {{ add(root); }}; + while(!queue.isEmpty()) { + TreeNode node = queue.poll(); + if(node != null) { + res.append(node.val + ","); + queue.add(node.left); + queue.add(node.right); + } + else res.append("null,"); + } + res.deleteCharAt(res.length() - 1); + res.append("]"); + return res.toString(); + } + + public TreeNode deserialize(String data) { + if(data.equals("[]")) return null; + String[] vals = data.substring(1, data.length() - 1).split(","); + TreeNode root = new TreeNode(Integer.parseInt(vals[0])); + Queue queue = new LinkedList<>() {{ add(root); }}; + int i = 1; + while(!queue.isEmpty()) { + TreeNode node = queue.poll(); + if(!vals[i].equals("null")) { + node.left = new TreeNode(Integer.parseInt(vals[i])); + queue.add(node.left); + } + i++; + if(!vals[i].equals("null")) { + node.right = new TreeNode(Integer.parseInt(vals[i])); + queue.add(node.right); + } + i++; + } + return root; + } +} +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 38. \345\255\227\347\254\246\344\270\262\347\232\204\346\216\222\345\210\227.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 38. \345\255\227\347\254\246\344\270\262\347\232\204\346\216\222\345\210\227.md" new file mode 100644 index 0000000..9deb42b --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 38. \345\255\227\347\254\246\344\270\262\347\232\204\346\216\222\345\210\227.md" @@ -0,0 +1,117 @@ +## 解题思路： + +对于一个长度为 $n$ 的字符串（假设字符互不重复），其排列方案数共有： + +$$ +n \times (n-1) \times (n-2) … \times 2 \times 1 +$$ + +**排列方案的生成：** + +根据字符串排列的特点，考虑深度优先搜索所有排列方案。即通过字符交换，先固定第 $1$ 位字符（ $n$ 种情况）、再固定第 $2$ 位字符（ $n-1$ 种情况）、... 、最后固定第 $n$ 位字符（ $1$ 种情况）。 + +![Picture1.png](https://pic.leetcode-cn.com/1599403497-KXKQcp-Picture1.png){:width=500} + +**重复排列方案与剪枝：** + +当字符串存在重复字符时，排列方案中也存在重复的排列方案。为排除重复方案，需在固定某位字符时，保证 “每种字符只在此位固定一次” ，即遇到重复字符时不交换，直接跳过。从 DFS 角度看，此操作称为 “剪枝” 。 + +![Picture2.png](https://pic.leetcode-cn.com/1599403497-GATdFr-Picture2.png){:width=500} + +### 递归解析： + +1. **终止条件：** 当 `x = len(c) - 1` 时，代表所有位已固定（最后一位只有 $1$ 种情况），则将当前组合 `c` 转化为字符串并加入 `res` ，并返回； +2. **递推参数：** 当前固定位 `x` ； +3. **递推工作：** 初始化一个 Set ，用于排除重复的字符；将第 `x` 位字符与 `i` $\in$ `[x, len(c)]` 字符分别交换，并进入下层递归； + 1. **剪枝：** 若 `c[i]` 在 Set​ 中，代表其是重复字符，因此 “剪枝” ； + 2. 将 `c[i]` 加入 Set​ ，以便之后遇到重复字符时剪枝； + 3. **固定字符：** 将字符 `c[i]` 和 `c[x]` 交换，即固定 `c[i]` 为当前位字符； + 4. **开启下层递归：** 调用 `dfs(x + 1)` ，即开始固定第 `x + 1` 个字符； + 5. **还原交换：** 将字符 `c[i]` 和 `c[x]` 交换（还原之前的交换）； + +> 下图中 `list` 对应文中的列表 `c` 。 + + + +### 复杂度分析： + +- **时间复杂度 $O(N!N)$ ：** $N$ 为字符串 `s` 的长度；时间复杂度和字符串排列的方案数成线性关系，方案数为 $N \times (N-1) \times (N-2) … \times 2 \times 1$ ，即复杂度为 $O(N!)$ ；字符串拼接操作 `join()` 使用 $O(N)$ ；因此总体时间复杂度为 $O(N!N)$ 。 +- **空间复杂度 $O(N^2)$ ：** 全排列的递归深度为 $N$ ，系统累计使用栈空间大小为 $O(N)$ ；递归中辅助 Set 累计存储的字符数量最多为 $N + (N-1) + ... + 2 + 1 = (N+1)N/2$ ，即占用 $O(N^2)$ 的额外空间。 + +## 代码： + +```Python [] +class Solution: + def permutation(self, s: str) -> List[str]: + c, res = list(s), [] + def dfs(x): + if x == len(c) - 1: + res.append(''.join(c)) # 添加排列方案 + return + dic = set() + for i in range(x, len(c)): + if c[i] in dic: continue # 重复，因此剪枝 + dic.add(c[i]) + c[i], c[x] = c[x], c[i] # 交换，将 c[i] 固定在第 x 位 + dfs(x + 1) # 开启固定第 x + 1 位字符 + c[i], c[x] = c[x], c[i] # 恢复交换 + dfs(0) + return res +``` + +```Java [] +class Solution { + List res = new LinkedList<>(); + char[] c; + public String[] permutation(String s) { + c = s.toCharArray(); + dfs(0); + return res.toArray(new String[res.size()]); + } + void dfs(int x) { + if(x == c.length - 1) { + res.add(String.valueOf(c)); // 添加排列方案 + return; + } + HashSet set = new HashSet<>(); + for(int i = x; i < c.length; i++) { + if(set.contains(c[i])) continue; // 重复，因此剪枝 + set.add(c[i]); + swap(i, x); // 交换，将 c[i] 固定在第 x 位 + dfs(x + 1); // 开启固定第 x + 1 位字符 + swap(i, x); // 恢复交换 + } + } + void swap(int a, int b) { + char tmp = c[a]; + c[a] = c[b]; + c[b] = tmp; + } +} +``` + +```C++ [] +class Solution { +public: + vector permutation(string s) { + dfs(s, 0); + return res; + } +private: + vector res; + void dfs(string s, int x) { + if(x == s.size() - 1) { + res.push_back(s); // 添加排列方案 + return; + } + set st; + for(int i = x; i < s.size(); i++) { + if(st.find(s[i]) != st.end()) continue; // 重复，因此剪枝 + st.insert(s[i]); + swap(s[i], s[x]); // 交换，将 s[i] 固定在第 x 位 + dfs(s, x + 1); // 开启固定第 x + 1 位字符 + swap(s[i], s[x]); // 恢复交换 + } + } +}; +``` \ No newline at end of file diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 39. \346\225\260\347\273\204\344\270\255\345\207\272\347\216\260\346\254\241\346\225\260\350\266\205\350\277\207\344\270\200\345\215\212\347\232\204\346\225\260\345\255\227.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 39. \346\225\260\347\273\204\344\270\255\345\207\272\347\216\260\346\254\241\346\225\260\350\266\205\350\277\207\344\270\200\345\215\212\347\232\204\346\225\260\345\255\227.md" new file mode 100644 index 0000000..a3f56e4 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 39. \346\225\260\347\273\204\344\270\255\345\207\272\347\216\260\346\254\241\346\225\260\350\266\205\350\277\207\344\270\200\345\215\212\347\232\204\346\225\260\345\255\227.md" @@ -0,0 +1,134 @@ +## 解题思路： + +> 本文将 “数组中出现次数超过一半的数字” 简称为 **“众数”** 。 +> 需要注意的是，数学中众数的定义为 “数组中出现次数最多的数字” ，与本文定义不同。 + +本题常见的三种解法： + +1. **哈希表统计法：** 遍历数组 `nums` ，用 HashMap 统计各数字的数量，即可找出 众数 。此方法时间和空间复杂度均为 $O(N)$ 。 +2. **数组排序法：** 将数组 `nums` 排序，**数组中点的元素** 一定为众数。 +3. **摩尔投票法：** 核心理念为 **票数正负抵消** 。此方法时间和空间复杂度分别为 $O(N)$ 和 $O(1)$ ，为本题的最佳解法。 + +### 摩尔投票法： + +> 设输入数组 `nums` 的众数为 $x$ ，数组长度为 $n$ 。 + +**推论一：** 若记 **众数** 的票数为 $+1$ ，**非众数** 的票数为 $-1$ ，则一定有所有数字的 **票数和 $> 0$** 。 + +**推论二：** 若数组的前 $a$ 个数字的 **票数和 $= 0$** ，则 数组剩余 $(n-a)$ 个数字的 **票数和一定仍 $>0$** ，即后 $(n-a)$ 个数字的 **众数仍为 $x$** 。 + +![Picture1.png](https://pic.leetcode-cn.com/1603612327-bOQxzq-Picture1.png){:width=500} + +根据以上推论，记数组首个元素为 $n_1$ ，众数为 $x$ ，遍历并统计票数。当发生 **票数和 $= 0$** 时，**剩余数组的众数一定不变** ，这是由于： + +- **当 $n_1 = x$ ：** 抵消的所有数字中，有一半是众数 $x$ 。 +- **当 $n_1 \neq x$ ：** 抵消的所有数字中，众数 $x$ 的数量最少为 0 个，最多为一半。 + +利用此特性，每轮假设发生 **票数和 $= 0$** 都可以 **缩小剩余数组区间** 。当遍历完成时，最后一轮假设的数字即为众数。 + +### 算法流程: + +1. **初始化：** 票数统计 `votes = 0` ， 众数 `x`； +2. **循环：** 遍历数组 `nums` 中的每个数字 `num` ； + 1. 当 票数 `votes` 等于 0 ，则假设当前数字 `num` 是众数； + 2. 当 `num = x` 时，票数 `votes` 自增 1 ；当 `num != x` 时，票数 `votes` 自减 1 ； +3. **返回值：** 返回 `x` 即可； + + + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为数组 `nums` 长度。 +- **空间复杂度 $O(1)$ ：** `votes` 变量使用常数大小的额外空间。 + +## 代码： + +```Python [] +class Solution: + def majorityElement(self, nums: List[int]) -> int: + votes = 0 + for num in nums: + if votes == 0: x = num + votes += 1 if num == x else -1 + return x +``` + +```Java [] +class Solution { + public int majorityElement(int[] nums) { + int x = 0, votes = 0; + for(int num : nums){ + if(votes == 0) x = num; + votes += num == x ? 1 : -1; + } + return x; + } +} +``` + +```C++ [] +class Solution { +public: + int majorityElement(vector& nums) { + int x = 0, votes = 0; + for(int num : nums){ + if(votes == 0) x = num; + votes += num == x ? 1 : -1; + } + return x; + } +}; +``` + +**拓展：** 由于题目说明 `给定的数组总是存在多数元素` ，因此本题不用考虑 **数组不存在众数** 的情况。若考虑，需要加入一个 “验证环节” ，遍历数组 `nums` 统计 `x` 的数量。 + +- 若 `x` 的数量超过数组长度一半，则返回 `x` ； +- 否则，返回未找到众数； + +时间和空间复杂度不变，仍为 $O(N)$ 和 $O(1)$ 。 + +```Python [] +class Solution: + def majorityElement(self, nums: List[int]) -> int: + votes, count = 0, 0 + for num in nums: + if votes == 0: x = num + votes += 1 if num == x else -1 + # 验证 x 是否为众数 + for num in nums: + if num == x: count += 1 + return x if count > len(nums) // 2 else 0 # 当无众数时返回 0 +``` + +```Java [] +class Solution { + public int majorityElement(int[] nums) { + int x = 0, votes = 0, count = 0; + for(int num : nums){ + if(votes == 0) x = num; + votes += num == x ? 1 : -1; + } + // 验证 x 是否为众数 + for(int num : nums) + if(num == x) count++; + return count > nums.length / 2 ? x : 0; // 当无众数时返回 0 + } +} +``` + +```C++ [] +class Solution { +public: + int majorityElement(vector& nums) { + int x = 0, votes = 0, count = 0; + for(int num : nums){ + if(votes == 0) x = num; + votes += num == x ? 1 : -1; + } + // 验证 x 是否为众数 + for(int num : nums) + if(num == x) count++; + return count > nums.size() / 2 ? x : 0; // 当无众数时返回 0 + } +}; +``` \ No newline at end of file diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 40. \346\234\200\345\260\217\347\232\204 k \344\270\252\346\225\260.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 40. \346\234\200\345\260\217\347\232\204 k \344\270\252\346\225\260.md" new file mode 100644 index 0000000..910bae5 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 40. \346\234\200\345\260\217\347\232\204 k \344\270\252\346\225\260.md" @@ -0,0 +1,213 @@ +## 方法一：排序 + +本题使用排序算法解决最直观，对数组 `arr` 执行排序，再返回前 $k$ 个元素即可。使用任意排序算法皆可，本文采用并介绍 **快速排序** ，为下文 **方法二** 做铺垫。 + +### 快速排序原理： + +快速排序算法有两个核心点，分别为 “哨兵划分” 和 “递归” 。 + +**哨兵划分操作：** 以数组某个元素（一般选取首元素）为 **基准数** ，将所有小于基准数的元素移动至其左边，大于基准数的元素移动至其右边。 + +> 如下图所示，为哨兵划分操作流程。通过一轮 **哨兵划分** ，可将数组排序问题拆分为 **两个较短数组的排序问题** （本文称之为左（右）子数组）。 + + + +**递归：** 对 **左子数组** 和 **右子数组** 递归执行 **哨兵划分**，直至子数组长度为 1 时终止递归，即可完成对整个数组的排序。 + +> 如下图所示，为示例数组 `[2,4,1,0,3,5]` 的快速排序流程。观察发现，快速排序和 **二分法** 的原理类似，都是以 $\log$ 时间复杂度实现搜索区间缩小。 + +![Picture1.png](https://pic.leetcode-cn.com/1612615552-rifQwI-Picture1.png){:width=550} + +### 复杂度分析： + +- **时间复杂度 $O(N \log N)$ ：** 库函数、快排等排序算法的平均时间复杂度为 $O(N \log N)$ 。 +- **空间复杂度 $O(N)$ ：** 快速排序的递归深度最好（平均）为 $O(\log N)$ ，最差情况（即输入数组完全倒序）为 $O(N)$。 + +### 代码： + +```Python [] +class Solution: + def getLeastNumbers(self, arr: List[int], k: int) -> List[int]: + def quick_sort(arr, l, r): + # 子数组长度为 1 时终止递归 + if l >= r: return + # 哨兵划分操作（以 arr[l] 作为基准数） + i, j = l, r + while i < j: + while i < j and arr[j] >= arr[l]: j -= 1 + while i < j and arr[i] <= arr[l]: i += 1 + arr[i], arr[j] = arr[j], arr[i] + arr[l], arr[i] = arr[i], arr[l] + # 递归左（右）子数组执行哨兵划分 + quick_sort(arr, l, i - 1) + quick_sort(arr, i + 1, r) + + quick_sort(arr, 0, len(arr) - 1) + return arr[:k] +``` + +```Java [] +class Solution { + public int[] getLeastNumbers(int[] arr, int k) { + quickSort(arr, 0, arr.length - 1); + return Arrays.copyOf(arr, k); + } + private void quickSort(int[] arr, int l, int r) { + // 子数组长度为 1 时终止递归 + if (l >= r) return; + // 哨兵划分操作（以 arr[l] 作为基准数） + int i = l, j = r; + while (i < j) { + while (i < j && arr[j] >= arr[l]) j--; + while (i < j && arr[i] <= arr[l]) i++; + swap(arr, i, j); + } + swap(arr, i, l); + // 递归左（右）子数组执行哨兵划分 + quickSort(arr, l, i - 1); + quickSort(arr, i + 1, r); + } + private void swap(int[] arr, int i, int j) { + int tmp = arr[i]; + arr[i] = arr[j]; + arr[j] = tmp; + } +} +``` + +```C++ [] +class Solution { +public: + vector getLeastNumbers(vector& arr, int k) { + quickSort(arr, 0, arr.size() - 1); + vector res; + res.assign(arr.begin(), arr.begin() + k); + return res; + } +private: + void quickSort(vector& arr, int l, int r) { + // 子数组长度为 1 时终止递归 + if (l >= r) return; + // 哨兵划分操作（以 arr[l] 作为基准数） + int i = l, j = r; + while (i < j) { + while (i < j && arr[j] >= arr[l]) j--; + while (i < j && arr[i] <= arr[l]) i++; + swap(arr[i], arr[j]); + } + swap(arr[i], arr[l]); + // 递归左（右）子数组执行哨兵划分 + quickSort(arr, l, i - 1); + quickSort(arr, i + 1, r); + } +}; +``` + +## 方法二： 基于快速排序的数组划分 + +题目只要求返回最小的 k 个数，对这 k 个数的顺序并没有要求。因此，只需要将数组划分为 **最小的 $k$ 个数** 和 **其他数字** 两部分即可，而快速排序的哨兵划分可完成此目标。 + +根据快速排序原理，如果某次哨兵划分后 **基准数正好是第 $k+1$ 小的数字** ，那么此时基准数左边的所有数字便是题目所求的 **最小的 k 个数** 。 + +根据此思路，考虑在每次哨兵划分后，判断基准数在数组中的索引是否等于 $k$ ，若 $true$ 则直接返回此时数组的前 $k$ 个数字即可。 + +### 算法流程： + +**`getLeastNumbers() 函数：`** + +1. 若 $k$ 大于数组长度，则直接返回整个数组； +2. 执行并返回 `quick_sort()` 即可； + +**`quick_sort() 函数：`** + +> 注意，此时 `quick_sort()` 的功能不是排序整个数组，而是搜索并返回最小的 $k$ 个数。 + +1. **哨兵划分**： + + - 划分完毕后，基准数为 `arr[i]` ，左 / 右子数组区间分别为 $[l, i - 1]$ , $[i + 1, r]$ ； + +2. **递归或返回：** + + - 若 $k < i$ ，代表第 $k + 1$ 小的数字在 **左子数组** 中，则递归左子数组； + - 若 $k > i$ ，代表第 $k + 1$ 小的数字在 **右子数组** 中，则递归右子数组； + - 若 $k = i$ ，代表此时 `arr[k]` 即为第 $k + 1$ 小的数字，则直接返回数组前 $k$ 个数字即可； + + + +### 复杂度分析： + +> 本方法优化时间复杂度的本质是通过判断舍去了不必要的递归（哨兵划分）。 + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为数组元素数量；对于长度为 $N$ 的数组执行哨兵划分操作的时间复杂度为 $O(N)$ ；每轮哨兵划分后根据 $k$ 和 $i$ 的大小关系选择递归，由于 $i$ 分布的随机性，则向下递归子数组的平均长度为 $\frac{N}{2}$ ；因此平均情况下，哨兵划分操作一共有 $N + \frac{N}{2} + \frac{N}{4} + ... + \frac{N}{N} = \frac{N - \frac{1}{2}}{1 - \frac{1}{2}} = 2N - 1$ （等比数列求和），即总体时间复杂度为 $O(N)$ 。 +- **空间复杂度 $O(\log N)$ ：** 划分函数的平均递归深度为 $O(\log N)$ 。 + +## 代码： + +```Python [] +class Solution: + def getLeastNumbers(self, arr: List[int], k: int) -> List[int]: + if k >= len(arr): return arr + def quick_sort(l, r): + i, j = l, r + while i < j: + while i < j and arr[j] >= arr[l]: j -= 1 + while i < j and arr[i] <= arr[l]: i += 1 + arr[i], arr[j] = arr[j], arr[i] + arr[l], arr[i] = arr[i], arr[l] + if k < i: return quick_sort(l, i - 1) + if k > i: return quick_sort(i + 1, r) + return arr[:k] + + return quick_sort(0, len(arr) - 1) +``` + +```Java [] +class Solution { + public int[] getLeastNumbers(int[] arr, int k) { + if (k >= arr.length) return arr; + return quickSort(arr, k, 0, arr.length - 1); + } + private int[] quickSort(int[] arr, int k, int l, int r) { + int i = l, j = r; + while (i < j) { + while (i < j && arr[j] >= arr[l]) j--; + while (i < j && arr[i] <= arr[l]) i++; + swap(arr, i, j); + } + swap(arr, i, l); + if (i > k) return quickSort(arr, k, l, i - 1); + if (i < k) return quickSort(arr, k, i + 1, r); + return Arrays.copyOf(arr, k); + } + private void swap(int[] arr, int i, int j) { + int tmp = arr[i]; + arr[i] = arr[j]; + arr[j] = tmp; + } +} +``` + +```C++ [] +class Solution { +public: + vector getLeastNumbers(vector& arr, int k) { + if (k >= arr.size()) return arr; + return quickSort(arr, k, 0, arr.size() - 1); + } +private: + vector quickSort(vector& arr, int k, int l, int r) { + int i = l, j = r; + while (i < j) { + while (i < j && arr[j] >= arr[l]) j--; + while (i < j && arr[i] <= arr[l]) i++; + swap(arr[i], arr[j]); + } + swap(arr[i], arr[l]); + if (i > k) return quickSort(arr, k, l, i - 1); + if (i < k) return quickSort(arr, k, i + 1, r); + vector res; + res.assign(arr.begin(), arr.begin() + k); + return res; + } +}; +``` \ No newline at end of file diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 41. \346\225\260\346\215\256\346\265\201\344\270\255\347\232\204\344\270\255\344\275\215\346\225\260.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 41. \346\225\260\346\215\256\346\265\201\344\270\255\347\232\204\344\270\255\344\275\215\346\225\260.md" new file mode 100644 index 0000000..ea81d83 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 41. \346\225\260\346\215\256\346\265\201\344\270\255\347\232\204\344\270\255\344\275\215\346\225\260.md" @@ -0,0 +1,134 @@ +## 解题思路： + +> 给定一长度为 $N$ 的无序数组，其中位数的计算方法：首先对数组执行排序（使用 $O(N \log N)$ 时间），然后返回中间元素即可（使用 $O(1)$ 时间）。 + +针对本题，根据以上思路，可以将数据流保存在一个列表中，并在添加元素时 **保持数组有序** 。此方法的时间复杂度为 $O(N)$ ，其中包括： 查找元素插入位置 $O(\log N)$ （二分查找）、向数组某位置插入元素 $O(N)$ （插入位置之后的元素都需要向后移动一位）。 + +> 借助 **堆** 可进一步优化时间复杂度。 + +建立一个 **小顶堆** $A$ 和 **大顶堆** $B$ ，各保存列表的一半元素，且规定： + +- $A$ 保存 **较大** 的一半，长度为 $\frac{N}{2}$（ $N$ 为偶数）或 $\frac{N+1}{2}$（ $N$ 为奇数）； +- $B$ 保存 **较小** 的一半，长度为 $\frac{N}{2}$（ $N$ 为偶数）或 $\frac{N-1}{2}$（ $N$ 为奇数）； + +随后，中位数可仅根据 $A, B$ 的堆顶元素计算得到。 + +![Picture1.png](https://pic.leetcode-cn.com/bcfaca2b1920d2dd6bbb01aeff990698eb36d53830c38ed499ea3239a15296b3-Picture1.png){:width=500} + +### 算法流程： + +> 设元素总数为 $N = m + n$ ，其中 $m$ 和 $n$ 分别为 $A$ 和 $B$ 中的元素个数。 + +**`addNum(num)` 函数：** + +1. 当 $m = n$（即 $N$ 为 **偶数**）：需向 $A$ 添加一个元素。实现方法：将新元素 $num$ 插入至 $B$ ，再将 $B$ 堆顶元素插入至 $A$ ； +2. 当 $m \ne n$（即 $N$ 为 **奇数**）：需向 $B$ 添加一个元素。实现方法：将新元素 $num$ 插入至 $A$ ，再将 $A$ 堆顶元素插入至 $B$ ； + +> 假设插入数字 $num$ 遇到情况 `1.` 。由于 $num$ 可能属于 “较小的一半” （即属于 $B$ ），因此不能将 $nums$ 直接插入至 $A$ 。而应先将 $num$ 插入至 $B$ ，再将 $B$ 堆顶元素插入至 $A$ 。这样就可以始终保持 $A$ 保存较大一半、 $B$ 保存较小一半。 + +**`findMedian()` 函数：** + +1. 当 $m = n$（ $N$ 为 **偶数**）：则中位数为 $($ $A$ 的堆顶元素 + $B$ 的堆顶元素 $)/2$。 +2. 当 $m \ne n$（ $N$ 为 **奇数**）：则中位数为 $A$ 的堆顶元素。 + + + +### 复杂度分析： + +- **时间复杂度：** + - **查找中位数 $O(1)$ ：** 获取堆顶元素使用 $O(1)$ 时间； + - **添加数字 $O(\log N)$ ：** 堆的插入和弹出操作使用 $O(\log N)$ 时间。 +- **空间复杂度 $O(N)$ ：** 其中 $N$ 为数据流中的元素数量，小顶堆 $A$ 和大顶堆 $B$ 最多同时保存 $N$ 个元素。 + +### 代码： + +Python 中 heapq 模块是小顶堆。实现 **大顶堆** 方法： 小顶堆的插入和弹出操作均将元素 **取反** 即可。 +Java 使用 `PriorityQueue<>((x, y) -> (y - x))` 可方便实现大顶堆。 +C++ 中 `greater` 为小顶堆， `less` 为大顶堆。 + +```Python [] +from heapq import * + +class MedianFinder: + def __init__(self): + self.A = [] # 小顶堆，保存较大的一半 + self.B = [] # 大顶堆，保存较小的一半 + + def addNum(self, num: int) -> None: + if len(self.A) != len(self.B): + heappush(self.A, num) + heappush(self.B, -heappop(self.A)) + else: + heappush(self.B, -num) + heappush(self.A, -heappop(self.B)) + + def findMedian(self) -> float: + return self.A[0] if len(self.A) != len(self.B) else (self.A[0] - self.B[0]) / 2.0 +``` + +```Java [] +class MedianFinder { + Queue A, B; + public MedianFinder() { + A = new PriorityQueue<>(); // 小顶堆，保存较大的一半 + B = new PriorityQueue<>((x, y) -> (y - x)); // 大顶堆，保存较小的一半 + } + public void addNum(int num) { + if(A.size() != B.size()) { + A.add(num); + B.add(A.poll()); + } else { + B.add(num); + A.add(B.poll()); + } + } + public double findMedian() { + return A.size() != B.size() ? A.peek() : (A.peek() + B.peek()) / 2.0; + } +} +``` + +```C++ [] +class MedianFinder { +public: + priority_queue, greater> A; // 小顶堆，保存较大的一半 + priority_queue, less> B; // 大顶堆，保存较小的一半 + MedianFinder() { } + void addNum(int num) { + if(A.size() != B.size()) { + A.push(num); + B.push(A.top()); + A.pop(); + } else { + B.push(num); + A.push(B.top()); + B.pop(); + } + } + double findMedian() { + return A.size() != B.size() ? A.top() : (A.top() + B.top()) / 2.0; + } +}; +``` + +> Push item on the heap, then pop and return the smallest item from the heap. The combined action runs more efficiently than heappush() followed by a separate call to heappop(). + +根据以上文档说明，可将 Python 代码优化为： + +```Python [] +from heapq import * + +class MedianFinder: + def __init__(self): + self.A = [] # 小顶堆，保存较大的一半 + self.B = [] # 大顶堆，保存较小的一半 + + def addNum(self, num: int) -> None: + if len(self.A) != len(self.B): + heappush(self.B, -heappushpop(self.A, num)) + else: + heappush(self.A, -heappushpop(self.B, -num)) + + def findMedian(self) -> float: + return self.A[0] if len(self.A) != len(self.B) else (self.A[0] - self.B[0]) / 2.0 +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 42. \350\277\236\347\273\255\345\255\220\346\225\260\347\273\204\347\232\204\346\234\200\345\244\247\345\222\214.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 42. \350\277\236\347\273\255\345\255\220\346\225\260\347\273\204\347\232\204\346\234\200\345\244\247\345\222\214.md" new file mode 100644 index 0000000..948cfb9 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 42. \350\277\236\347\273\255\345\255\220\346\225\260\347\273\204\347\232\204\346\234\200\345\244\247\345\222\214.md" @@ -0,0 +1,79 @@ +## 解题思路： + +| 常见解法 | 时间复杂度 | 空间复杂度 | +| -------- | ------------- | ----------- | +| 暴力搜索 | $O(N^2)$ | $O(1)$ | +| 分治思想 | $O(N \log N)$ | $O(\log N)$ | +| 动态规划 | $O(N)$ | $O(1)$ | + +动态规划是本题的最优解法，以下按照标准流程解题。 + +### 动态规划解析： + +- **状态定义：** 设动态规划列表 $dp$ ，$dp[i]$ 代表以元素 $nums[i]$ 为结尾的连续子数组最大和。 + - 为何定义最大和 $dp[i]$ 中必须包含元素 $nums[i]$ ：保证 $dp[i]$ 递推到 $dp[i+1]$ 的正确性；如果不包含 $nums[i]$ ，递推时则不满足题目的 **连续子数组** 要求。 + +- **转移方程：** 若 $dp[i-1] \leq 0$ ，说明 $dp[i - 1]$ 对 $dp[i]$ 产生负贡献，即 $dp[i-1] + nums[i]$ 还不如 $nums[i]$ 本身大。 + +$$ +dp[i] = +\begin{cases} + dp[i-1] + nums[i] & , dp[i - 1] > 0 \\ + nums[i] & , dp[i - 1] \leq 0 \\ +\end{cases} +$$ + +- **初始状态：** $dp[0] = nums[0]$，即以 $nums[0]$ 结尾的连续子数组最大和为 $nums[0]$ 。 + +- **返回值：** 返回 $dp$ 列表中的最大值，代表全局最大值。 + +![Picture1.png](https://pic.leetcode-cn.com/77d1aa6a444743d3c8606ac951cd7fc38faf68a62064fd2639df517cd666a4d0-Picture1.png){:width=500} + +### 空间复杂度降低： + +- 由于 $dp[i]$ 只与 $dp[i-1]$ 和 $nums[i]$ 有关系，因此可以将原数组 $nums$ 用作 $dp$ 列表，即直接在 $nums$ 上修改即可。 +- 由于省去 $dp$ 列表使用的额外空间，因此空间复杂度从 $O(N)$ 降至 $O(1)$ 。 + +**复杂度分析：** + +- **时间复杂度 $O(N)$ ：** 线性遍历数组 $nums$ 即可获得结果，使用 $O(N)$ 时间。 +- **空间复杂度 $O(1)$ ：** 使用常数大小的额外空间。 + + + +### 代码： + +```Python [] +class Solution: + def maxSubArray(self, nums: List[int]) -> int: + for i in range(1, len(nums)): + nums[i] += max(nums[i - 1], 0) + return max(nums) +``` + +```Java [] +class Solution { + public int maxSubArray(int[] nums) { + int res = nums[0]; + for(int i = 1; i < nums.length; i++) { + nums[i] += Math.max(nums[i - 1], 0); + res = Math.max(res, nums[i]); + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int maxSubArray(vector& nums) { + int res = nums[0]; + for(int i = 1; i < nums.size(); i++) { + if(nums[i - 1] > 0) nums[i] += nums[i - 1]; + if(nums[i] > res) res = nums[i]; + } + return res; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 43. 1\357\275\236n \346\225\264\346\225\260\344\270\255 1 \345\207\272\347\216\260\347\232\204\346\254\241\346\225\260.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 43. 1\357\275\236n \346\225\264\346\225\260\344\270\255 1 \345\207\272\347\216\260\347\232\204\346\254\241\346\225\260.md" new file mode 100644 index 0000000..e88bcf7 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 43. 1\357\275\236n \346\225\264\346\225\260\344\270\255 1 \345\207\272\347\216\260\347\232\204\346\254\241\346\225\260.md" @@ -0,0 +1,161 @@ +## 解题思路： + +将 $1$ ~ $n$ 的个位、十位、百位、...的 $1$ 出现次数相加，即为 $1$ 出现的总次数。 + +设数字 $n$ 是个 $x$ 位数，记 $n$ 的第 $i$ 位为 $n_i$ ，则可将 $n$ 写为 $n_{x} n_{x-1} \cdots n_{2} n_{1}$ ；本文名词规定如下： + +- 称 「 $n_i$ 」称为 **当前位** ，记为 $cur$ ； +- 将 「 $n_{i-1} n_{i-2} \cdots n_{2} n_{1}$ 」称为 **低位** ，记为 $low$ ； +- 将 「 $n_{x} n_{x-1} \cdots n_{i+2} n_{i+1}$ 」称为 **高位** ，记为 $high$ ； +- 将 「 $10^i$ 」称为 **位因子** ，记为 $digit$ ； + +### 某位中 $1$ 出现次数的计算方法： + +根据当前位 $cur$ 值的不同，分为以下三种情况： + +1. 当 **$cur = 0$ 时：** 此位 $1$ 的出现次数只由高位 $high$ 决定，计算公式为： + +$$ +high \times digit +$$ + +> 如下图所示，以 $n = 2304$ 为例，求 $digit = 10$ （即十位）的 $1$ 出现次数。 + +![Picture1.png](https://pic.leetcode-cn.com/1599887431-cVmcVA-Picture1.png){:width=450} + +2. 当 **$cur = 1$ 时：** 此位 $1$ 的出现次数由高位 $high$ 和低位 $low$ 决定，计算公式为： + +$$ +high \times digit + low + 1 +$$ + +> 如下图所示，以 $n = 2314$ 为例，求 $digit = 10$ （即十位）的 $1$ 出现次数。 + +![Picture2.png](https://pic.leetcode-cn.com/1599887431-HAAvVp-Picture2.png){:width=450} + +3. 当 **$cur = 2, 3, \cdots, 9$ 时：** 此位 $1$ 的出现次数只由高位 $high$ 决定，计算公式为： + +$$ +(high + 1) \times digit +$$ + +> 如下图所示，以 $n = 2324$ 为例，求 $digit = 10$ （即十位）的 $1$ 出现次数。 + +![Picture3.png](https://pic.leetcode-cn.com/1599887431-djUZTe-Picture3.png){:width=450} + +### 变量递推公式： + +设计按照 “个位、十位、...” 的顺序计算，则 $high / cur / low / digit$ 应初始化为： + +```Python [] +high = n // 10 +cur = n % 10 +low = 0 +digit = 1 # 个位 +``` + +```Java [] +int high = n / 10; +int cur = n % 10; +int low = 0; +int digit = 1; // 个位 +``` + +```C++ [] +int high = n / 10; +int cur = n % 10; +int low = 0; +int digit = 1; // 个位 +``` + +因此，从个位到最高位的变量递推公式为： + +```Python [] +while high != 0 or cur != 0: # 当 high 和 cur 同时为 0 时，说明已经越过最高位，因此跳出 + low += cur * digit # 将 cur 加入 low ，组成下轮 low + cur = high % 10 # 下轮 cur 是本轮 high 的最低位 + high //= 10 # 将本轮 high 最低位删除，得到下轮 high + digit *= 10 # 位因子每轮 × 10 +``` + +```Java [] +while(high != 0 || cur != 0) { // 当 high 和 cur 同时为 0 时，说明已经越过最高位，因此跳出 + low += cur * digit; // 将 cur 加入 low ，组成下轮 low + cur = high % 10; // 下轮 cur 是本轮 high 的最低位 + high /= 10; // 将本轮 high 最低位删除，得到下轮 high + digit *= 10; // 位因子每轮 × 10 +} +``` + +```C++ [] +while(high != 0 || cur != 0) { // 当 high 和 cur 同时为 0 时，说明已经越过最高位，因此跳出 + low += cur * digit; // 将 cur 加入 low ，组成下轮 low + cur = high % 10; // 下轮 cur 是本轮 high 的最低位 + high /= 10; // 将本轮 high 最低位删除，得到下轮 high + digit *= 10; // 位因子每轮 × 10 +} +``` + +### 复杂度分析： + +- **时间复杂度 $O(\log n)$ ：** 循环内的计算操作使用 $O(1)$ 时间；循环次数为数字 $n$ 的位数，即 $\log_{10}{n}$ ，因此循环使用 $O(\log n)$ 时间。 +- **空间复杂度 $O(1)$ ：** 几个变量使用常数大小的额外空间。 + + + +### 代码： + +```Python [] +class Solution: + def countDigitOne(self, n: int) -> int: + digit, res = 1, 0 + high, cur, low = n // 10, n % 10, 0 + while high != 0 or cur != 0: + if cur == 0: res += high * digit + elif cur == 1: res += high * digit + low + 1 + else: res += (high + 1) * digit + low += cur * digit + cur = high % 10 + high //= 10 + digit *= 10 + return res +``` + +```Java [] +class Solution { + public int countDigitOne(int n) { + int digit = 1, res = 0; + int high = n / 10, cur = n % 10, low = 0; + while(high != 0 || cur != 0) { + if(cur == 0) res += high * digit; + else if(cur == 1) res += high * digit + low + 1; + else res += (high + 1) * digit; + low += cur * digit; + cur = high % 10; + high /= 10; + digit *= 10; + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int countDigitOne(int n) { + long digit = 1; + int high = n / 10, cur = n % 10, low = 0, res = 0; + while(high != 0 || cur != 0) { + if(cur == 0) res += high * digit; + else if(cur == 1) res += high * digit + low + 1; + else res += (high + 1) * digit; + low += cur * digit; + cur = high % 10; + high /= 10; + digit *= 10; + } + return res; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 44. \346\225\260\345\255\227\345\272\217\345\210\227\344\270\255\346\237\220\344\270\200\344\275\215\347\232\204\346\225\260\345\255\227.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 44. \346\225\260\345\255\227\345\272\217\345\210\227\344\270\255\346\237\220\344\270\200\344\275\215\347\232\204\346\225\260\345\255\227.md" new file mode 100644 index 0000000..692409a --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 44. \346\225\260\345\255\227\345\272\217\345\210\227\344\270\255\346\237\220\344\270\200\344\275\215\347\232\204\346\225\260\345\255\227.md" @@ -0,0 +1,167 @@ +## 解题思路： + +文名词规定如下： + +1. 将 $101112 \cdots$ 中的每一位称为 **数位** ，记为 $n$ ； +2. 将 $10, 11, 12, \cdots$ 称为 **数字** ，记为 $num$ ； +3. 数字 $10$ 是一个两位数，称此数字的 **位数** 为 $2$ ，记为 $digit$ ； +4. 每 $digit$ 位数的起始数字（即：$1, 10, 100, \cdots$），记为 $start$ ； + +![Picture1.png](https://pic.leetcode-cn.com/1599888213-CYhLfm-Picture1.png){:width=400} + +观察上表，可推出各 $digit$ 下的数位数量 $count$ 的计算公式： + +$$ +count = 9 \times start \times digit +$$ + +根据以上分析，可将求解分为三步： + +1. 确定 $n$ 所在 **数字** 的 **位数** ，记为 $digit$ ； +2. 确定 $n$ 所在的 **数字** ，记为 $num$ ； +3. 确定 $n$ 是 $num$ 中的哪一数位，并返回结果； + +### 1. 确定所求数位的所在数字的位数 + +如下图所示，循环执行 $n$ 减去 一位数、两位数、... 的数位数量 $count$ ，直至 $n \leq count$ 时跳出。 + +由于 $n$ 已经减去了一位数、两位数、...、$(digit-1)$ 位数的 **数位数量** $count$ ，因而此时的 $n$ 是从起始数字 $start$ 开始计数的。 + +```Python [] +digit, start, count = 1, 1, 9 +while n > count: + n -= count + start *= 10 # 1, 10, 100, ... + digit += 1 # 1, 2, 3, ... + count = 9 * start * digit # 9, 180, 2700, ... +``` + +```Java [] +int digit = 1; +long start = 1; +long count = 9; +while (n > count) { + n -= count; + start *= 10; // 1, 10, 100, ... + digit += 1; // 1, 2, 3, ... + count = digit * start * 9; // 9, 180, 2700, ... +} +``` + +```C++ [] +int digit = 1; +long start = 1; +long count = 9; +while (n > count) { // 1. + n -= count; + start *= 10; // 1, 10, 100, ... + digit += 1; // 1, 2, 3, ... + count = digit * start * 9; // 9, 180, 2700, ... +} +``` + +**结论：** 所求数位 ① 在某个 $digit$ 位数中； ② 为从数字 $start$ 开始的第 $n$ 个数位。 + +![Picture2.png](https://pic.leetcode-cn.com/1599888496-HivJvS-Picture2.png){:width=500} + +### 2. 确定所求数位所在的数字 + +如下图所示，所求数位 在从数字 $start$ 开始的第 $[(n - 1) / digit]$ 个 **数字** 中（ $start$ 为第 0 个数字）。 + +```Python [] +num = start + (n - 1) // digit +``` + +```Java [] +long num = start + (n - 1) / digit; +``` + +```C++ [] +long num = start + (n - 1) / digit; +``` + +**结论：** 所求数位在数字 $num$ 中。 + +![Picture3.png](https://pic.leetcode-cn.com/1599888213-gCcnEA-Picture3.png){:width=500} + +### 3. 确定所求数位在 $num$ 的哪一数位 + +如下图所示，所求数位为数字 $num$ 的第 $(n - 1) \% digit$ 位（ 数字的首个数位为第 0 位）。 + +```Python [] +s = str(num) # 转化为 string +res = int(s[(n - 1) % digit]) # 获得 num 的 第 (n - 1) % digit 个数位，并转化为 int +``` + +```Java [] +String s = Long.toString(num); // 转化为 string +int res = s.charAt((n - 1) % digit) - '0'; // 获得 num 的 第 (n - 1) % digit 个数位，并转化为 int +``` + +```C++ [] +string s = to_string(num); // 转化为 string +int res = s[(n - 1) % digit] - '0'; // 获得 num 的 第 (n - 1) % digit 个数位，并转化为 int +``` + +**结论：** 所求数位是 $res$ 。 + +![Picture4.png](https://pic.leetcode-cn.com/1599888395-oeWGAH-Picture4.png){:width=500} + +### 复杂度分析： + +- **时间复杂度 $O(\log n)$ ：** 所求数位 $n$ 对应数字 $num$ 的位数 $digit$ 最大为 $O(\log n)$ ；第一步最多循环 $O(\log n)$ 次；第三步中将 $num$ 转化为字符串使用 $O(\log n)$ 时间；因此总体为 $O(\log n)$ 。 +- **空间复杂度 $O(\log n)$ ：** 将数字 $num$ 转化为字符串 `str(num)` ，占用 $O(\log n)$ 的额外空间。 + + + +### 代码： + +```Python [] +class Solution: + def findNthDigit(self, n: int) -> int: + digit, start, count = 1, 1, 9 + while n > count: # 1. + n -= count + start *= 10 + digit += 1 + count = 9 * start * digit + num = start + (n - 1) // digit # 2. + return int(str(num)[(n - 1) % digit]) # 3. +``` + +```Java [] +class Solution { + public int findNthDigit(int n) { + int digit = 1; + long start = 1; + long count = 9; + while (n > count) { // 1. + n -= count; + start *= 10; + digit += 1; + count = digit * start * 9; + } + long num = start + (n - 1) / digit; // 2. + return Long.toString(num).charAt((n - 1) % digit) - '0'; // 3. + } +} +``` + +```C++ [] +class Solution { +public: + int findNthDigit(int n) { + int digit = 1; + long start = 1; + long count = 9; + while (n > count) { // 1. + n -= count; + start *= 10; + digit += 1; + count = digit * start * 9; + } + long num = start + (n - 1) / digit; // 2. + return to_string(num)[(n - 1) % digit] - '0'; // 3. + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 45. \346\212\212\346\225\260\347\273\204\346\216\222\346\210\220\346\234\200\345\260\217\347\232\204\346\225\260.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 45. \346\212\212\346\225\260\347\273\204\346\216\222\346\210\220\346\234\200\345\260\217\347\232\204\346\225\260.md" new file mode 100644 index 0000000..68dd4b4 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 45. \346\212\212\346\225\260\347\273\204\346\216\222\346\210\220\346\234\200\345\260\217\347\232\204\346\225\260.md" @@ -0,0 +1,168 @@ +## 解题思路： + +此题求拼接起来的最小数字，本质上是一个排序问题。设数组 $nums$ 中任意两数字的字符串为 $x$ 和 $y$ ，则规定 **排序判断规则** 为： + +- 若拼接字符串 $x + y > y + x$ ，则 $x$ “大于” $y$ ； +- 反之，若 $x + y < y + x$ ，则 $x$ “小于” $y$ ； + +> $x$ “小于” $y$ 代表：排序完成后，数组中 $x$ 应在 $y$ 左边；“大于” 则反之。 + +根据以上规则，套用任何排序方法对 $nums$ 执行排序即可。 + +![Picture1.png](https://pic.leetcode-cn.com/95e81dbccc44f26292d88c509afd68204a86b37d342f83d109fa7aa0cd4a6049-Picture1.png){:width=450} + +### 算法流程： + +1. **初始化：** 字符串列表 $strs$ ，保存各数字的字符串格式； +2. **列表排序：** 应用以上 “排序判断规则” ，对 $strs$ 执行排序； +3. **返回值：** 拼接 $strs$ 中的所有字符串，并返回。 + +**复杂度分析：** + +- **时间复杂度 $O(N \log N)$ ：** $N$ 为最终返回值的字符数量（ $strs$ 列表的长度 $\leq N$ ）；使用快排或内置函数的平均时间复杂度为 $O(N \log N)$ ，最差为 $O(N^2)$ 。 +- **空间复杂度 $O(N)$ ：** 字符串列表 $strs$ 占用线性大小的额外空间。 + + + +## 代码： + +本文列举 **快速排序** 和 **内置函数** 两种排序方法，其他排序方法也可实现。 + +### 快速排序： + +需修改快速排序函数中的排序判断规则。字符串大小（字典序）对比的实现方法： + +- Python/C++ 中可直接用 `<` , `>`； +- Java 中使用函数 `A.compareTo(B)`； + +```Python [] +class Solution: + def minNumber(self, nums: List[int]) -> str: + def quick_sort(l , r): + if l >= r: return + i, j = l, r + while i < j: + while strs[j] + strs[l] >= strs[l] + strs[j] and i < j: j -= 1 + while strs[i] + strs[l] <= strs[l] + strs[i] and i < j: i += 1 + strs[i], strs[j] = strs[j], strs[i] + strs[i], strs[l] = strs[l], strs[i] + quick_sort(l, i - 1) + quick_sort(i + 1, r) + + strs = [str(num) for num in nums] + quick_sort(0, len(strs) - 1) + return ''.join(strs) +``` + +```Java [] +class Solution { + public String minNumber(int[] nums) { + String[] strs = new String[nums.length]; + for(int i = 0; i < nums.length; i++) + strs[i] = String.valueOf(nums[i]); + quickSort(strs, 0, strs.length - 1); + StringBuilder res = new StringBuilder(); + for(String s : strs) + res.append(s); + return res.toString(); + } + void quickSort(String[] strs, int l, int r) { + if(l >= r) return; + int i = l, j = r; + String tmp = strs[i]; + while(i < j) { + while((strs[j] + strs[l]).compareTo(strs[l] + strs[j]) >= 0 && i < j) j--; + while((strs[i] + strs[l]).compareTo(strs[l] + strs[i]) <= 0 && i < j) i++; + tmp = strs[i]; + strs[i] = strs[j]; + strs[j] = tmp; + } + strs[i] = strs[l]; + strs[l] = tmp; + quickSort(strs, l, i - 1); + quickSort(strs, i + 1, r); + } +} +``` + +```C++ [] +class Solution { +public: + string minNumber(vector& nums) { + vector strs; + for(int i = 0; i < nums.size(); i++) + strs.push_back(to_string(nums[i])); + quickSort(strs, 0, strs.size() - 1); + string res; + for(string s : strs) + res.append(s); + return res; + } +private: + void quickSort(vector& strs, int l, int r) { + if(l >= r) return; + int i = l, j = r; + while(i < j) { + while(strs[j] + strs[l] >= strs[l] + strs[j] && i < j) j--; + while(strs[i] + strs[l] <= strs[l] + strs[i] && i < j) i++; + swap(strs[i], strs[j]); + } + swap(strs[i], strs[l]); + quickSort(strs, l, i - 1); + quickSort(strs, i + 1, r); + } +}; +``` + +### 内置函数： + +需定义排序规则： + +- Python 定义在函数 `sort_rule(x, y)` 中； +- Java 定义为 `(x, y) -> (x + y).compareTo(y + x)` ； +- C++ 定义为 `(string& x, string& y){ return x + y < y + x; }` ； + +```Python [] +class Solution: + def minNumber(self, nums: List[int]) -> str: + def sort_rule(x, y): + a, b = x + y, y + x + if a > b: return 1 + elif a < b: return -1 + else: return 0 + + strs = [str(num) for num in nums] + strs.sort(key = functools.cmp_to_key(sort_rule)) + return ''.join(strs) +``` + +```Java [] +class Solution { + public String minNumber(int[] nums) { + String[] strs = new String[nums.length]; + for(int i = 0; i < nums.length; i++) + strs[i] = String.valueOf(nums[i]); + Arrays.sort(strs, (x, y) -> (x + y).compareTo(y + x)); + StringBuilder res = new StringBuilder(); + for(String s : strs) + res.append(s); + return res.toString(); + } +} +``` + +```C++ [] +class Solution { +public: + string minNumber(vector& nums) { + vector strs; + string res; + for(int i = 0; i < nums.size(); i++) + strs.push_back(to_string(nums[i])); + sort(strs.begin(), strs.end(), [](string& x, string& y){ return x + y < y + x; }); + for(int i = 0; i < strs.size(); i++) + res.append(strs[i]); + return res; + } +}; +``` \ No newline at end of file diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 46. \346\212\212\346\225\260\345\255\227\347\277\273\350\257\221\346\210\220\345\255\227\347\254\246\344\270\262.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 46. \346\212\212\346\225\260\345\255\227\347\277\273\350\257\221\346\210\220\345\255\227\347\254\246\344\270\262.md" new file mode 100644 index 0000000..effe520 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 46. \346\212\212\346\225\260\345\255\227\347\277\273\350\257\221\346\210\220\345\255\227\347\254\246\344\270\262.md" @@ -0,0 +1,237 @@ +## 解题思路： + +根据题意，可按照下图的思路，总结出 “递推公式” （即转移方程）。 + +![Picture1.png](https://pic.leetcode-cn.com/1603462412-iUcKzA-Picture1.png){:width=600} + +因此，此题可用动态规划解决，以下按照流程解题。 + +### 动态规划解析： + +> 记数字 $num$ 第 $i$ 位数字为 $x_i$ ，数字 $num$ 的位数为 $n$ ； +> 例如： $num = 12258$ 的 $n = 5$ , $x_1 = 1$ 。 + +- **状态定义：** 设动态规划列表 $dp$ ，$dp[i]$ 代表以 $x_i$ 为结尾的数字的翻译方案数量。 + +- **转移方程：** 若 $x_i$ 和 $x_{i-1}$ 组成的两位数字可被整体翻译，则 $dp[i] = dp[i - 1] + dp[i - 2]$ ，否则 $dp[i] = dp[i - 1]$ 。 + +$$ +dp[i] = +\begin{cases} +dp[i - 1] + dp[i - 2] & {, (10 x_{i-1} + x_i) \in [10,25]} \\ +dp[i - 1] & {, (10 x_{i-1} + x_i) \in [0, 10) \cup (25, 99]} +\end{cases} +$$ + +> **可被整体翻译的两位数区间分析：** 当 $x_{i-1} = 0$ 时，组成的两位数无法被整体翻译（例如 $00, 01, 02, \cdots$ ），大于 $25$ 的两位数也无法被整体翻译（例如 $26, 27, \cdots$ ），因此区间为 $[10, 25]$ 。 + +- **初始状态：** $dp[0] = dp[1] = 1$ ，即 “无数字” 和 “第 $1$ 位数字” 的翻译方法数量均为 $1$ ； + +- **返回值：** $dp[n]$ ，即此数字的翻译方案数量； + +> **Q：** 无数字情况 $dp[0] = 1$ 从何而来？ +> **A：** 当 $num$ 第 $1, 2$ 位的组成的数字 $\in [10,25]$ 时，显然应有 $2$ 种翻译方法，即 $dp[2] = dp[1] + dp[0] = 2$ ，而显然 $dp[1] = 1$ ，因此推出 $dp[0] = 1$ 。 + +## 方法一：字符串遍历 + +- 为方便获取数字的各位 $x_i$ ，考虑先将数字 $num$ 转化为字符串 $s$ ，通过遍历 $s$ 实现动态规划。 +- 通过字符串切片 $s[i - 2:i]$ 获取数字组合 $10 x_{i-1} + x_i$ ，通过对比字符串 ASCII 码判断字符串对应的数字区间。 +- **空间使用优化：** 由于 $dp[i]$ 只与 $dp[i - 1]$ 有关，因此可使用两个变量 $a, b$ 分别记录 $dp[i]$ , $dp[i - 1]$ ，两变量交替前进即可。此方法可省去 $dp$ 列表使用的 $O(N)$ 的额外空间。 + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为字符串 $s$ 的长度（即数字 $num$ 的位数 $\log(num)$ ），其决定了循环次数。 +- **空间复杂度 $O(N)$ ：** 字符串 $s$ 使用 $O(N)$ 大小的额外空间。 + + + +### 代码： + +由于 Python 语言特性，可不使用临时变量 $tmp$ 和 $c$ ，若阅读不畅可转至第三栏代码。 + +```Java [] +class Solution { + public int translateNum(int num) { + String s = String.valueOf(num); + int a = 1, b = 1; + for(int i = 2; i <= s.length(); i++) { + String tmp = s.substring(i - 2, i); + int c = tmp.compareTo("10") >= 0 && tmp.compareTo("25") <= 0 ? a + b : a; + b = a; + a = c; + } + return a; + } +} +``` + +```C++ [] +class Solution { +public: + int translateNum(int num) { + string s = to_string(num); + int a = 1, b = 1, len = s.size(); + for(int i = 2; i <= len; i++) { + string tmp = s.substr(i - 2, 2); + int c = tmp.compare("10") >= 0 && tmp.compare("25") <= 0 ? a + b : a; + b = a; + a = c; + } + return a; + } +}; +``` + +```Python [] +class Solution: + def translateNum(self, num: int) -> int: + s = str(num) + a = b = 1 + for i in range(2, len(s) + 1): + a, b = (a + b if "10" <= s[i - 2:i] <= "25" else a), a + return a +``` + +```Python [] +class Solution: + def translateNum(self, num: int) -> int: + s = str(num) + a = b = 1 + for i in range(2, len(s) + 1): + tmp = s[i - 2:i] + c = a + b if "10" <= tmp <= "25" else a + b = a + a = c + return a +``` + +此题的动态规划计算是 **对称的** ，即 **从左向右** 遍历（从第 $dp[2]$ 计算至 $dp[n]$ ）和 **从右向左** 遍历（从第 $dp[n - 2]$ 计算至 $dp[0]$ ）所得方案数一致。从右向左遍历的代码如下所示。 + +```Java [] +class Solution { + public int translateNum(int num) { + String s = String.valueOf(num); + int a = 1, b = 1; + for(int i = s.length() - 2; i > -1; i--) { + String tmp = s.substring(i, i + 2); + int c = tmp.compareTo("10") >= 0 && tmp.compareTo("25") <= 0 ? a + b : a; + b = a; + a = c; + } + return a; + } +} +``` + +```C++ [] +class Solution { +public: + int translateNum(int num) { + string s = to_string(num); + int a = 1, b = 1, len = s.size(); + for(int i = len - 2; i > -1; i--) { + string tmp = s.substr(i, 2); + int c = tmp.compare("10") >= 0 && tmp.compare("25") <= 0 ? a + b : a; + b = a; + a = c; + } + return a; + } +}; +``` + +```Python [] +class Solution: + def translateNum(self, num: int) -> int: + s = str(num) + a = b = 1 + for i in range(len(s) - 2, -1, -1): + a, b = (a + b if "10" <= s[i:i + 2] <= "25" else a), a + return a +``` + +## 方法二：数字求余 + +上述方法虽然已经节省了 $dp$ 列表的空间占用，但字符串 $s$ 仍使用了 $O(N)$ 大小的额外空间。 + +### 空间复杂度优化： + +- 利用求余运算 $num \% 10$ 和求整运算 $num // 10$ ，可获取数字 $num$ 的各位数字（获取顺序为个位、十位、百位…）。 +- 运用 **求余** 和 **求整** 运算实现，可实现 **从右向左** 的动态规划计算。而根据上述动态规划 “对称性” ，可知从右向左计算是正确的。 +- 自此，字符串 $s$ 的空间占用也被省去，空间复杂度从 $O(N)$ 降至 $O(1)$ 。 + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为字符串 $s$ 的长度，即数字 $num$ 的位数 $\log(num)$ ，其决定了循环次数。 +- **空间复杂度 $O(1)$ ：** 几个变量使用常数大小的额外空间。 + + + +### 代码： + +由于 Python 语言特性，可不使用临时变量 $tmp$ 和 $c$ ，若阅读不畅可转至第三栏代码。 + +```Java [] +class Solution { + public int translateNum(int num) { + int a = 1, b = 1, x, y = num % 10; + while(num > 9) { + num /= 10; + x = num % 10; + int tmp = 10 * x + y; + int c = (tmp >= 10 && tmp <= 25) ? a + b : a; + b = a; + a = c; + y = x; + } + return a; + } +} +``` + +```C++ [] +class Solution { +public: + int translateNum(int num) { + int a = 1, b = 1, x, y = num % 10; + while(num > 9) { + num /= 10; + x = num % 10; + int tmp = 10 * x + y; + int c = (tmp >= 10 && tmp <= 25) ? a + b : a; + b = a; + a = c; + y = x; + } + return a; + } +}; +``` + +```Python [] +class Solution: + def translateNum(self, num: int) -> int: + a = b = 1 + y = num % 10 + while num > 9: + num //= 10 + x = num % 10 + a, b = (a + b if 10 <= 10 * x + y <= 25 else a), a + y = x + return a +``` + +```Python [] +class Solution: + def translateNum(self, num: int) -> int: + a = b = 1 + y = num % 10 + while num > 9: + num //= 10 + x = num % 10 + tmp = 10 * x + y + c = a + b if 10 <= tmp <= 25 else a + a, b = c, a + y = x + return a +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 47. \347\244\274\347\211\251\347\232\204\346\234\200\345\244\247\344\273\267\345\200\274.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 47. \347\244\274\347\211\251\347\232\204\346\234\200\345\244\247\344\273\267\345\200\274.md" new file mode 100644 index 0000000..a7e7b10 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 47. \347\244\274\347\211\251\347\232\204\346\234\200\345\244\247\344\273\267\345\200\274.md" @@ -0,0 +1,146 @@ +## 解题思路： + +题目说明：从棋盘的左上角开始拿格子里的礼物，并每次 **向右** 或者 **向下** 移动一格、直到到达棋盘的右下角。 +根据题目说明，易得某单元格只可能从上边单元格或左边单元格到达。 + +设 $f(i, j)$ 为从棋盘左上角走至单元格 $(i ,j)$ 的礼物最大累计价值，易得到以下递推关系：$f(i,j)$ 等于 $f(i,j-1)$ 和 $f(i-1,j)$ 中的较大值加上当前单元格礼物价值 $grid(i,j)$ 。 + +$$ +f(i,j) = \max[f(i,j-1), f(i-1,j)] + grid(i,j) +$$ + +因此，可用动态规划解决此问题，以上公式便为转移方程。 + +![Picture1.png](https://pic.leetcode-cn.com/1662cdf7aafd8c9ed6e1eadc41bfc9adf58ea808e11f1a3dd2e8ba4632b9d1ac-Picture1.png){:width=450} + +### 动态规划解析： + +- **状态定义：** 设动态规划矩阵 $dp$ ，$dp(i,j)$ 代表从棋盘的左上角开始，到达单元格 $(i,j)$ 时能拿到礼物的最大累计价值。 +- **转移方程：** + 1. 当 $i = 0$ 且 $j = 0$ 时，为起始元素； + 2. 当 $i = 0$ 且 $j \ne 0$ 时，为矩阵第一行元素，只可从左边到达； + 3. 当 $i \ne 0$ 且 $j = 0$ 时，为矩阵第一列元素，只可从上边到达； + 4. 当 $i \ne 0$ 且 $j \ne 0$ 时，可从左边或上边到达； + +$$ +dp(i,j)= +\begin{cases} +grid(i,j) & {,i=0, j=0}\\ +grid(i,j) + dp(i,j-1) & {,i=0, j \ne 0}\\ +grid(i,j) + dp(i-1,j) & {,i \ne 0, j=0}\\ +grid(i,j) + \max[dp(i-1,j),dp(i,j-1)]& ,{i \ne 0, j \ne 0} +\end{cases} +$$ + +- **初始状态：** $dp[0][0] = grid[0][0]$ ，即到达单元格 $(0,0)$ 时能拿到礼物的最大累计价值为 $grid[0][0]$ ； +- **返回值：** $dp[m-1][n-1]$ ，$m, n$ 分别为矩阵的行高和列宽，即返回 $dp$ 矩阵右下角元素。 + +### 空间复杂度降低： + +- 由于 $dp[i][j]$ 只与 $dp[i-1][j]$ , $dp[i][j-1]$ , $grid[i][j]$ 有关系，因此可以将原矩阵 $grid$ 用作 $dp$ 矩阵，即直接在 $grid$ 上修改即可。 +- 应用此方法可省去 $dp$ 矩阵使用的额外空间，因此空间复杂度从 $O(MN)$ 降至 $O(1)$ 。 + + + +### 复杂度分析： + +- **时间复杂度 $O(MN)$ ：** $M, N$ 分别为矩阵行高、列宽；动态规划需遍历整个 $grid$ 矩阵，使用 $O(MN)$ 时间。 +- **空间复杂度 $O(1)$ ：** 原地修改使用常数大小的额外空间。 + +## 代码： + +```Python [] +class Solution: + def maxValue(self, grid: List[List[int]]) -> int: + for i in range(len(grid)): + for j in range(len(grid[0])): + if i == 0 and j == 0: continue + if i == 0: grid[i][j] += grid[i][j - 1] + elif j == 0: grid[i][j] += grid[i - 1][j] + else: grid[i][j] += max(grid[i][j - 1], grid[i - 1][j]) + return grid[-1][-1] +``` + +```Java [] +class Solution { + public int maxValue(int[][] grid) { + int m = grid.length, n = grid[0].length; + for(int i = 0; i < m; i++) { + for(int j = 0; j < n; j++) { + if(i == 0 && j == 0) continue; + if(i == 0) grid[i][j] += grid[i][j - 1] ; + else if(j == 0) grid[i][j] += grid[i - 1][j]; + else grid[i][j] += Math.max(grid[i][j - 1], grid[i - 1][j]); + } + } + return grid[m - 1][n - 1]; + } +} +``` + +```C++ [] +class Solution { +public: + int maxValue(vector>& grid) { + int m = grid.size(), n = grid[0].size(); + for(int i = 0; i < m; i++) { + for(int j = 0; j < n; j++) { + if(i == 0 && j == 0) continue; + if(i == 0) grid[i][j] += grid[i][j - 1] ; + else if(j == 0) grid[i][j] += grid[i - 1][j]; + else grid[i][j] += max(grid[i][j - 1], grid[i - 1][j]); + } + } + return grid[m - 1][n - 1]; + } +}; +``` + +以上代码逻辑清晰，和转移方程直接对应，但仍可提升效率，这是因为：当 $grid$ 矩阵很大时， $i = 0$ 或 $j = 0$ 的情况仅占极少数，相当循环每轮都冗余了一次判断。因此，可先初始化矩阵第一行和第一列，再开始遍历递推。 + +```Python [] +class Solution: + def maxValue(self, grid: List[List[int]]) -> int: + m, n = len(grid), len(grid[0]) + for j in range(1, n): # 初始化第一行 + grid[0][j] += grid[0][j - 1] + for i in range(1, m): # 初始化第一列 + grid[i][0] += grid[i - 1][0] + for i in range(1, m): + for j in range(1, n): + grid[i][j] += max(grid[i][j - 1], grid[i - 1][j]) + return grid[-1][-1] +``` + +```Java [] +class Solution { + public int maxValue(int[][] grid) { + int m = grid.length, n = grid[0].length; + for(int j = 1; j < n; j++) // 初始化第一行 + grid[0][j] += grid[0][j - 1]; + for(int i = 1; i < m; i++) // 初始化第一列 + grid[i][0] += grid[i - 1][0]; + for(int i = 1; i < m; i++) + for(int j = 1; j < n; j++) + grid[i][j] += Math.max(grid[i][j - 1], grid[i - 1][j]); + return grid[m - 1][n - 1]; + } +} +``` + +```C++ [] +class Solution { +public: + int maxValue(vector>& grid) { + int m = grid.size(), n = grid[0].size(); + for(int j = 1; j < n; j++) // 初始化第一行 + grid[0][j] += grid[0][j - 1]; + for(int i = 1; i < m; i++) // 初始化第一列 + grid[i][0] += grid[i - 1][0]; + for(int i = 1; i < m; i++) + for(int j = 1; j < n; j++) + grid[i][j] += max(grid[i][j - 1], grid[i - 1][j]); + return grid[m - 1][n - 1]; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 48. \346\234\200\351\225\277\344\270\215\345\220\253\351\207\215\345\244\215\345\255\227\347\254\246\347\232\204\345\255\220\345\255\227\347\254\246\344\270\262.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 48. \346\234\200\351\225\277\344\270\215\345\220\253\351\207\215\345\244\215\345\255\227\347\254\246\347\232\204\345\255\220\345\255\227\347\254\246\344\270\262.md" new file mode 100644 index 0000000..ba9563b --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 48. \346\234\200\351\225\277\344\270\215\345\220\253\351\207\215\345\244\215\345\255\227\347\254\246\347\232\204\345\255\220\345\255\227\347\254\246\344\270\262.md" @@ -0,0 +1,222 @@ +## 解题思路： + +长度为 $N$ 的字符串共有 $\frac{(1 + N)N}{2}$ 个子字符串（复杂度为 $O(N^2)$ ），判断长度为 $N$ 的字符串是否有重复字符的复杂度为 $O(N)$ ，因此本题使用暴力法解决的复杂度为 $O(N^3)$ 。 + +考虑使用动态规划降低时间复杂度。 + +### 动态规划解析： + +- **状态定义：** 设动态规划列表 $dp$ ，$dp[j]$ 代表以字符 $s[j]$ 为结尾的 “最长不重复子字符串” 的长度。 +- **转移方程：** 固定右边界 $j$ ，设字符 $s[j]$ 左边距离最近的相同字符为 $s[i]$ ，即 $s[i] = s[j]$ 。 + 1. 当 $i < 0$ ，即 $s[j]$ 左边无相同字符，则 $dp[j] = dp[j-1] + 1$ ； + 2. 当 $dp[j - 1] < j - i$ ，说明字符 $s[i]$ 在子字符串 $dp[j-1]$ **区间之外** ，则 $dp[j] = dp[j - 1] + 1$ ； + 3. 当 $dp[j - 1] \geq j - i$ ，说明字符 $s[i]$ 在子字符串 $dp[j-1]$ **区间之中** ，则 $dp[j]$ 的左边界由 $s[i]$ 决定，即 $dp[j] = j - i$ ； + + > 当 $i < 0$ 时，由于 $dp[j - 1] \leq j$ 恒成立，因而 $dp[j - 1] < j - i$ 恒成立，因此分支 `1.` 和 `2.` 可被合并。 + +$$ +dp[j] = +\begin{cases} +dp[j - 1] + 1 & , dp[j-1] < j - i \\ +j - i & , dp[j-1] \geq j - i +\end{cases} +$$ + +- **返回值：** $\max(dp)$ ，即全局的 “最长不重复子字符串” 的长度。 + +![Picture1.png](https://pic.leetcode-cn.com/1599287290-mTdFye-Picture1.png){:width=500} + +### 空间复杂度降低： + +- 由于返回值是取 $dp$ 列表最大值，因此可借助变量 $tmp$ 存储 $dp[j]$ ，变量 $res$ 每轮更新最大值即可。 +- 此优化可节省 $dp$ 列表使用的 $O(N)$ 大小的额外空间。 + +> 观察转移方程，可知本质问题为：每轮遍历字符 $s[j]$ 时，如何计算索引 $i$ ？ + +## 方法一：动态规划 + 哈希表 + +- **哈希表统计：** 遍历字符串 $s$ 时，使用哈希表（记为 $dic$ ）统计 **各字符最后一次出现的索引位置** 。 +- **左边界 $i$ 获取方式：** 遍历到 $s[j]$ 时，可通过访问哈希表 $dic[s[j]]$ 获取最近的相同字符的索引 $i$ 。 + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为字符串长度，动态规划需遍历计算 $dp$ 列表。 +- **空间复杂度 $O(1)$ ：** 字符的 ASCII 码范围为 $0$ ~ $127$ ，哈希表 $dic$ 最多使用 $O(128) = O(1)$ 大小的额外空间。 + + + +### 代码： + +Python 的 `get(key, default)` 方法和 Java 的 `getOrDefault(key, default)` ，代表当哈希表包含键 `key` 时返回对应 `value` ，不包含时返回默认值 `default` 。 + +```Python [] +class Solution: + def lengthOfLongestSubstring(self, s: str) -> int: + dic = {} + res = tmp = 0 + for j in range(len(s)): + i = dic.get(s[j], -1) # 获取索引 i + dic[s[j]] = j # 更新哈希表 + tmp = tmp + 1 if tmp < j - i else j - i # dp[j - 1] -> dp[j] + res = max(res, tmp) # max(dp[j - 1], dp[j]) + return res +``` + +```Java [] +class Solution { + public int lengthOfLongestSubstring(String s) { + Map dic = new HashMap<>(); + int res = 0, tmp = 0, len = s.length(); + for(int j = 0; j < len; j++) { + int i = dic.getOrDefault(s.charAt(j), -1); // 获取索引 i + dic.put(s.charAt(j), j); // 更新哈希表 + tmp = tmp < j - i ? tmp + 1 : j - i; // dp[j - 1] -> dp[j] + res = Math.max(res, tmp); // max(dp[j - 1], dp[j]) + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int lengthOfLongestSubstring(string s) { + unordered_map dic; + int res = 0, tmp = 0, len = s.size(), i; + for(int j = 0; j < len; j++) { + if(dic.find(s[j]) == dic.end()) i = - 1; + else i = dic.find(s[j])->second; // 获取索引 i + dic[s[j]] = j; // 更新哈希表 + tmp = tmp < j - i ? tmp + 1 : j - i; // dp[j - 1] -> dp[j] + res = max(res, tmp); // max(dp[j - 1], dp[j]) + } + return res; + } +}; +``` + +## 方法二： 动态规划 + 线性遍历 + +- **左边界 $i$ 获取方式：** 遍历到 $s[j]$ 时，初始化索引 $i = j - 1$ ，向左遍历搜索第一个满足 $s[i] = s[j]$ 的字符即可 。 + +### 复杂度分析： + +- **时间复杂度 $O(N^2)$ ：** 其中 $N$ 为字符串长度，动态规划需遍历计算 $dp$ 列表，占用 $O(N)$ ；每轮计算 $dp[j]$ 时搜索 $i$ 需要遍历 $j$ 个字符，占用 $O(N)$ 。 +- **空间复杂度 $O(1)$ ：** 几个变量使用常数大小的额外空间。 + +### 代码： + +```Python [] +class Solution: + def lengthOfLongestSubstring(self, s: str) -> int: + res = tmp = i = 0 + for j in range(len(s)): + i = j - 1 + while i >= 0 and s[i] != s[j]: i -= 1 # 线性查找 i + tmp = tmp + 1 if tmp < j - i else j - i # dp[j - 1] -> dp[j] + res = max(res, tmp) # max(dp[j - 1], dp[j]) + return res +``` + +```Java [] +class Solution { + public int lengthOfLongestSubstring(String s) { + int res = 0, tmp = 0, len = s.length(); + for(int j = 0; j < len; j++) { + int i = j - 1; + while(i >= 0 && s.charAt(i) != s.charAt(j)) i--; // 线性查找 i + tmp = tmp < j - i ? tmp + 1 : j - i; // dp[j - 1] -> dp[j] + res = Math.max(res, tmp); // max(dp[j - 1], dp[j]) + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int lengthOfLongestSubstring(string s) { + int res = 0, tmp = 0, len = s.size(); + for(int j = 0; j < len; j++) { + int i = j - 1; + while(i >= 0 && s[i] != s[j]) i--; // 线性查找 i + tmp = tmp < j - i ? tmp + 1 : j - i; // dp[j - 1] -> dp[j] + res = max(res, tmp); // max(dp[j - 1], dp[j]) + } + return res; + } +}; +``` + +## 方法三：双指针 + 哈希表 + +> 与方法一本质等价，不同点在于左边界 $i$ 的定义不同。 + +- **哈希表 $dic$ 统计：** 指针 $j$ 遍历字符 $s$ ，哈希表统计字符 $s[j]$ **最后一次出现的索引** 。 +- **更新左指针 $i$ ：** 根据上轮左指针 $i$ 和 $dic[s[j]]$ ，每轮更新左边界 $i$ ，保证区间 $[i + 1, j]$ 内无重复字符且最大。 + +$$ +i = \max(dic[s[j]], i) +$$ + +- **更新结果 $res$ ：** 取上轮 $res$ 和本轮双指针区间 $[i + 1,j]$ 的宽度（即 $j - i$ ）中的最大值。 + +$$ +res = \max(res, j - i) +$$ + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为字符串长度，动态规划需遍历计算 $dp$ 列表。 +- **空间复杂度 $O(1)$ ：** 字符的 ASCII 码范围为 $0$ ~ $127$ ，哈希表 $dic$ 最多使用 $O(128) = O(1)$ 大小的额外空间。 + + + +### 代码： + +```Python [] +class Solution: + def lengthOfLongestSubstring(self, s: str) -> int: + dic, res, i = {}, 0, -1 + for j in range(len(s)): + if s[j] in dic: + i = max(dic[s[j]], i) # 更新左指针 i + dic[s[j]] = j # 哈希表记录 + res = max(res, j - i) # 更新结果 + return res +``` + +```Java [] +class Solution { + public int lengthOfLongestSubstring(String s) { + Map dic = new HashMap<>(); + int i = -1, res = 0, len = s.length(); + for(int j = 0; j < len; j++) { + if(dic.containsKey(s.charAt(j))) + i = Math.max(i, dic.get(s.charAt(j))); // 更新左指针 i + dic.put(s.charAt(j), j); // 哈希表记录 + res = Math.max(res, j - i); // 更新结果 + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int lengthOfLongestSubstring(string s) { + unordered_map dic; + int i = -1, res = 0, len = s.size(); + for(int j = 0; j < len; j++) { + if(dic.find(s[j]) != dic.end()) + i = max(i, dic.find(s[j])->second); // 更新左指针 + dic[s[j]] = j; // 哈希表记录 + res = max(res, j - i); // 更新结果 + } + return res; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 49. \344\270\221\346\225\260.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 49. \344\270\221\346\225\260.md" new file mode 100644 index 0000000..6d6a5f9 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 49. \344\270\221\346\225\260.md" @@ -0,0 +1,115 @@ +## 解题思路： + +> **丑数的递推性质：** 丑数只包含因子 $2, 3, 5$ ，因此有 “丑数 $=$ 某较小丑数 $\times$ 某因子” （例如：$10 = 5 \times 2$）。 + +设已知长度为 $n$ 的丑数序列 $x_1, x_2, \cdots , x_n$ ，求第 $n+1$ 个丑数 $x_{n+1}$ 。根根据递推性质，丑数 $x_{n+1}$ 只可能是以下三种情况其中之一（索引 $a, b, c$ 为未知数）： + +$$ +x_{n+1} = +\begin{cases} +x_{a} \times 2 & ,a \in [1, n] \\ +x_{b} \times 3 & ,b \in [1, n] \\ +x_{c} \times 5 & ,c \in [1, n] +\end{cases} +$$ + +**丑数递推公式：** 若索引 $a,b,c$ 满足以上条件，则下个丑数 $x_{n+1}$ 为以下三种情况中的 **最小值** ； + +$$ +x_{n+1} = \min(x_{a} \times 2, x_{b} \times 3, x_{c} \times 5) +$$ + +由于 $x_{n+1}$ 是 **最接近** $x_n$ 的丑数，因此索引 $a, b, c$ 需满足以下条件： + +$$ +\begin{cases} +x_{a} \times 2 > x_n \geq x_{a-1} \times 2 & ，即 x_a 为首个乘以 2 后大于 x_n 的丑数 \\ +x_{b} \times 3 > x_n \geq x_{b-1} \times 3 & ，即 x_b 为首个乘以 3 后大于 x_n 的丑数 \\ +x_{c} \times 5 > x_n \geq x_{c-1} \times 5 & ，即 x_c 为首个乘以 5 后大于 x_n 的丑数 \\ +\end{cases} +$$ + +![Picture1.png](https://pic.leetcode-cn.com/1613651468-icTuYo-Picture1.png){:width=550} + +因此，可设置指针 $a,b,c$ 指向首个丑数（即 $1$ ），循环根据递推公式得到下个丑数，并每轮将对应指针执行 $+1$ 即可。 + +### 动态规划解析： + +- **状态定义：** 设动态规划列表 $dp$ ，$dp[i]$ 代表第 $i + 1$ 个丑数； +- **转移方程：** + 1. 当索引 $a, b, c$ 满足以下条件时， $dp[i]$ 为三种情况的最小值； + 2. 每轮计算 $dp[i]$ 后，需要更新索引 $a, b, c$ 的值，使其始终满足方程条件。实现方法：**分别独立判断** $dp[i]$ 和 $dp[a] \times 2$ , $dp[b] \times 3$ , $dp[c] \times 5$ 的大小关系，若相等则将对应索引 $a$ , $b$ , $c$ 加 $1$ ； + +$$ +\begin{cases} +dp[a] \times 2 > dp[i-1] \geq dp[a-1] \times 2 \\ +dp[b] \times 3 > dp[i-1] \geq dp[b-1] \times 3 \\ +dp[c] \times 5 > dp[i-1] \geq dp[c-1] \times 5 \\ +\end{cases} +$$ + +$$ +dp[i] = \min(dp[a] \times 2, dp[b] \times 3, dp[c] \times 5) +$$ + +- **初始状态：** $dp[0] = 1$ ，即第一个丑数为 $1$ ； +- **返回值：** $dp[n-1]$ ，即返回第 $n$ 个丑数； + + + +**复杂度分析：** + +- **时间复杂度 $O(N)$ ：** 其中 $N = n$ ，动态规划需遍历计算 $dp$ 列表。 +- **空间复杂度 $O(N)$ ：** 长度为 $N$ 的 $dp$ 列表使用 $O(N)$ 的额外空间。 + +## 代码： + +```Python [] +class Solution: + def nthUglyNumber(self, n: int) -> int: + dp, a, b, c = [1] * n, 0, 0, 0 + for i in range(1, n): + n2, n3, n5 = dp[a] * 2, dp[b] * 3, dp[c] * 5 + dp[i] = min(n2, n3, n5) + if dp[i] == n2: a += 1 + if dp[i] == n3: b += 1 + if dp[i] == n5: c += 1 + return dp[-1] +``` + +```Java [] +class Solution { + public int nthUglyNumber(int n) { + int a = 0, b = 0, c = 0; + int[] dp = new int[n]; + dp[0] = 1; + for(int i = 1; i < n; i++) { + int n2 = dp[a] * 2, n3 = dp[b] * 3, n5 = dp[c] * 5; + dp[i] = Math.min(Math.min(n2, n3), n5); + if(dp[i] == n2) a++; + if(dp[i] == n3) b++; + if(dp[i] == n5) c++; + } + return dp[n - 1]; + } +} +``` + +```C++ [] +class Solution { +public: + int nthUglyNumber(int n) { + int a = 0, b = 0, c = 0; + int dp[n]; + dp[0] = 1; + for(int i = 1; i < n; i++) { + int n2 = dp[a] * 2, n3 = dp[b] * 3, n5 = dp[c] * 5; + dp[i] = min(min(n2, n3), n5); + if(dp[i] == n2) a++; + if(dp[i] == n3) b++; + if(dp[i] == n5) c++; + } + return dp[n - 1]; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 50. \347\254\254\344\270\200\344\270\252\345\217\252\345\207\272\347\216\260\344\270\200\346\254\241\347\232\204\345\255\227\347\254\246.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 50. \347\254\254\344\270\200\344\270\252\345\217\252\345\207\272\347\216\260\344\270\200\346\254\241\347\232\204\345\255\227\347\254\246.md" new file mode 100644 index 0000000..c5df04e --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 50. \347\254\254\344\270\200\344\270\252\345\217\252\345\207\272\347\216\260\344\270\200\346\254\241\347\232\204\345\255\227\347\254\246.md" @@ -0,0 +1,148 @@ +本题考察 **哈希表** 的使用，本文介绍 **哈希表** 和 **有序哈希表** 两种解法。其中，在字符串长度较大、重复字符很多时，“有序哈希表” 解法理论上效率更高。 + +## 方法一：哈希表 + +1. 遍历字符串 `s` ，使用哈希表统计 “各字符数量是否 $> 1$ ”。 +2. 再遍历字符串 `s` ，在哈希表中找到首个 “数量为 $1$ 的字符”，并返回。 + +![Picture1.png](https://pic.leetcode-cn.com/ed093aabc9195caff6d088454eaebe3cad875e8ca4a643c004ef25e4e5e9e174-Picture1.png){:width=450} + +### 算法流程： + +1. **初始化：** 字典 (Python)、HashMap(Java)、map(C++)，记为 `dic` ； +2. **字符统计：** 遍历字符串 `s` 中的每个字符 `c` ； + 1. 若 `dic` 中 **不包含** 键(key) `c` ：则向 `dic` 中添加键值对 `(c, True)` ，代表字符 `c` 的数量为 $1$ ； + 2. 若 `dic` 中 **包含** 键(key) `c` ：则修改键 `c` 的键值对为 `(c, False)` ，代表字符 `c` 的数量 $> 1$ 。 +3. **查找数量为 $1$ 的字符：** 遍历字符串 `s` 中的每个字符 `c` ； + 1. 若 `dic`中键 `c` 对应的值为 `True` ：，则返回 `c` 。 +4. 返回 `' '` ，代表字符串无数量为 $1$ 的字符。 + + + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为字符串 `s` 的长度；需遍历 `s` 两轮，使用 $O(N)$ ；HashMap 查找操作的复杂度为 $O(1)$ ； +- **空间复杂度 $O(1)$ ：** 由于题目指出 `s` 只包含小写字母，因此最多有 26 个不同字符，HashMap 存储需占用 $O(26) = O(1)$ 的额外空间。 + +### 代码： + +Python 代码中的 `not c in dic` 整体为一个布尔值； `c in dic` 为判断字典中是否含有键 `c` 。 + +```Python [] +class Solution: + def firstUniqChar(self, s: str) -> str: + dic = {} + for c in s: + dic[c] = not c in dic + for c in s: + if dic[c]: return c + return ' ' +``` + +```Java [] +class Solution { + public char firstUniqChar(String s) { + HashMap dic = new HashMap<>(); + char[] sc = s.toCharArray(); + for(char c : sc) + dic.put(c, !dic.containsKey(c)); + for(char c : sc) + if(dic.get(c)) return c; + return ' '; + } +} +``` + +```C++ [] +class Solution { +public: + char firstUniqChar(string s) { + unordered_map dic; + for(char c : s) + dic[c] = dic.find(c) == dic.end(); + for(char c : s) + if(dic[c]) return c; + return ' '; + } +}; +``` + +## 方法二：有序哈希表 + +在哈希表的基础上，有序哈希表中的键值对是 **按照插入顺序排序** 的。基于此，可通过遍历有序哈希表，实现搜索首个 “数量为 $1$ 的字符”。 + +哈希表是 **去重** 的，即哈希表中键值对数量 $\leq$ 字符串 `s` 的长度。因此，相比于方法一，方法二减少了第二轮遍历的循环次数。当字符串很长（重复字符很多）时，方法二则效率更高。 + +### 复杂度分析： + +时间和空间复杂度均与 “方法一” 相同，而具体分析：方法一 需遍历 `s` 两轮；方法二 遍历 `s` 一轮，遍历 `dic` 一轮（ `dic` 的长度不大于 26 ）。 + +### 代码： + +Python 3.6 后，默认字典就是有序的，因此无需使用 `OrderedDict()` ，详情可见：[为什么Python 3.6以后字典有序并且效率更高？](https://www.cnblogs.com/xieqiankun/p/python_dict.html) + +Java 使用 `LinkedHashMap` 实现有序哈希表。 + +由于 C++ 未提供自带的链式哈希表，因此借助一个 vector 按序存储哈希表 dic 中的 key ，第二轮遍历此 vector 即可。 + +```Python [] +class Solution: + def firstUniqChar(self, s: str) -> str: + dic = collections.OrderedDict() + for c in s: + dic[c] = not c in dic + for k, v in dic.items(): + if v: return k + return ' ' +``` + +```Python [] +class Solution: + def firstUniqChar(self, s: str) -> str: + dic = {} + for c in s: + dic[c] = not c in dic + for k, v in dic.items(): + if v: return k + return ' ' +``` + +```Java [] +class Solution { + public char firstUniqChar(String s) { + Map dic = new LinkedHashMap<>(); + char[] sc = s.toCharArray(); + for(char c : sc) + dic.put(c, !dic.containsKey(c)); + for(Map.Entry d : dic.entrySet()){ + if(d.getValue()) return d.getKey(); + } + return ' '; + } +} +``` + +```C++ [] +class Solution { +public: + char firstUniqChar(string s) { + vector keys; + unordered_map dic; + for(char c : s) { + if(dic.find(c) == dic.end()) + keys.push_back(c); + dic[c] = dic.find(c) == dic.end(); + } + for(char c : keys) { + if(dic[c]) return c; + } + return ' '; + } +}; +``` + +\ +{:style="text-align: center;"} + +欢迎您对「图解算法数据结构」提出评论、意见或期待 +{:style="text-align: center;"} \ No newline at end of file diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 51. \346\225\260\347\273\204\344\270\255\347\232\204\351\200\206\345\272\217\345\257\271.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 51. \346\225\260\347\273\204\344\270\255\347\232\204\351\200\206\345\272\217\345\257\271.md" new file mode 100644 index 0000000..d09468f --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 51. \346\225\260\347\273\204\344\270\255\347\232\204\351\200\206\345\272\217\345\257\271.md" @@ -0,0 +1,148 @@ +## 解题思路： + +> 直观来看，使用暴力统计法即可，即遍历数组的所有数字对并统计逆序对数量。此方法时间复杂度为 $O(N^2)$ ，观察题目给定的数组长度范围 $0 \leq N \leq 50000$ ，可知此复杂度是不能接受的。 + +「归并排序」与「逆序对」是息息相关的。归并排序体现了 “分而治之” 的算法思想，具体为： + +- **分：** 不断将数组从中点位置划分开（即二分法），将整个数组的排序问题转化为子数组的排序问题； +- **治：** 划分到子数组长度为 1 时，开始向上合并，不断将 **较短排序数组** 合并为 **较长排序数组**，直至合并至原数组时完成排序； + +> 如下图所示，为数组 $[7, 3, 2, 6, 0, 1, 5, 4]$ 的归并排序过程。 + +![Picture1.png](https://pic.leetcode-cn.com/1614274007-nBQbZZ-Picture1.png){:width=500} + +**合并阶段** 本质上是 **合并两个排序数组** 的过程，而每当遇到 左子数组当前元素 > 右子数组当前元素 时，意味着 「左子数组当前元素 至 末尾元素」 与 「右子数组当前元素」 构成了若干 「逆序对」 。 + +> 如下图所示，为左子数组 $[2, 3, 6, 7]$ 与 右子数组 $[0, 1, 4, 5]$ 的合并与逆序对统计过程。 + + + +因此，考虑在归并排序的合并阶段统计「逆序对」数量，完成归并排序时，也随之完成所有逆序对的统计。 + +### 算法流程： + +**`merge_sort()` 归并排序与逆序对统计：** + +1. **终止条件：** 当 $l \geq r$ 时，代表子数组长度为 1 ，此时终止划分； +2. **递归划分：** 计算数组中点 $m$ ，递归划分左子数组 `merge_sort(l, m)` 和右子数组 `merge_sort(m + 1, r)` ； +3. **合并与逆序对统计：** + 1. 暂存数组 $nums$ 闭区间 $[l, r]$ 内的元素至辅助数组 $tmp$ ； + 2. **循环合并：** 设置双指针 $i$ , $j$ 分别指向左 / 右子数组的首元素； + - **当 $i = m + 1$ 时：** 代表左子数组已合并完，因此添加右子数组当前元素 $tmp[j]$ ，并执行 $j = j + 1$ ； + - **否则，当 $j = r + 1$ 时：** 代表右子数组已合并完，因此添加左子数组当前元素 $tmp[i]$ ，并执行 $i = i + 1$ ； + - **否则，当 $tmp[i] \leq tmp[j]$ 时：** 添加左子数组当前元素 $tmp[i]$ ，并执行 $i = i + 1$； + - **否则（即 $tmp[i] > tmp[j]$）时：** 添加右子数组当前元素 $tmp[j]$ ，并执行 $j = j + 1$ ；此时构成 $m - i + 1$ 个「逆序对」，统计添加至 $res$ ； +4. **返回值：** 返回直至目前的逆序对总数 $res$ ； + +**`reversePairs()` 主函数：** + +1. **初始化：** 辅助数组 $tmp$ ，用于合并阶段暂存元素； +2. **返回值：** 执行归并排序 `merge_sort()` ，并返回逆序对总数即可； + +> 如下图所示，为数组 $[7, 3, 2, 6, 0, 1, 5, 4]$ 的归并排序与逆序对统计过程。 + +![Picture2.png](https://pic.leetcode-cn.com/1614274007-rtFHbG-Picture2.png){:width=500} + +### 复杂度分析： + +- **时间复杂度 $O(N \log N)$ ：** 其中 $N$ 为数组长度；归并排序使用 $O(N \log N)$ 时间； +- **空间复杂度 $O(N)$ ：** 辅助数组 $tmp$ 占用 $O(N)$ 大小的额外空间； + +## 代码： + +为简化代码，「当 $j = r + 1$ 时」 与 「当 $tmp[i] \leq tmp[j]$ 时」 两判断项可合并。 + +```Python [] +class Solution: + def reversePairs(self, nums: List[int]) -> int: + def merge_sort(l, r): + # 终止条件 + if l >= r: return 0 + # 递归划分 + m = (l + r) // 2 + res = merge_sort(l, m) + merge_sort(m + 1, r) + # 合并阶段 + i, j = l, m + 1 + tmp[l:r + 1] = nums[l:r + 1] + for k in range(l, r + 1): + if i == m + 1: + nums[k] = tmp[j] + j += 1 + elif j == r + 1 or tmp[i] <= tmp[j]: + nums[k] = tmp[i] + i += 1 + else: + nums[k] = tmp[j] + j += 1 + res += m - i + 1 # 统计逆序对 + return res + + tmp = [0] * len(nums) + return merge_sort(0, len(nums) - 1) +``` + +```Java [] +class Solution { + int[] nums, tmp; + public int reversePairs(int[] nums) { + this.nums = nums; + tmp = new int[nums.length]; + return mergeSort(0, nums.length - 1); + } + private int mergeSort(int l, int r) { + // 终止条件 + if (l >= r) return 0; + // 递归划分 + int m = (l + r) / 2; + int res = mergeSort(l, m) + mergeSort(m + 1, r); + // 合并阶段 + int i = l, j = m + 1; + for (int k = l; k <= r; k++) + tmp[k] = nums[k]; + for (int k = l; k <= r; k++) { + if (i == m + 1) + nums[k] = tmp[j++]; + else if (j == r + 1 || tmp[i] <= tmp[j]) + nums[k] = tmp[i++]; + else { + nums[k] = tmp[j++]; + res += m - i + 1; // 统计逆序对 + } + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int reversePairs(vector& nums) { + vector tmp(nums.size()); + return mergeSort(0, nums.size() - 1, nums, tmp); + } +private: + int mergeSort(int l, int r, vector& nums, vector& tmp) { + // 终止条件 + if (l >= r) return 0; + // 递归划分 + int m = (l + r) / 2; + int res = mergeSort(l, m, nums, tmp) + mergeSort(m + 1, r, nums, tmp); + // 合并阶段 + int i = l, j = m + 1; + for (int k = l; k <= r; k++) + tmp[k] = nums[k]; + for (int k = l; k <= r; k++) { + if (i == m + 1) + nums[k] = tmp[j++]; + else if (j == r + 1 || tmp[i] <= tmp[j]) + nums[k] = tmp[i++]; + else { + nums[k] = tmp[j++]; + res += m - i + 1; // 统计逆序对 + } + } + return res; + } +}; +``` \ No newline at end of file diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 52. \344\270\244\344\270\252\351\223\276\350\241\250\347\232\204\347\254\254\344\270\200\344\270\252\345\205\254\345\205\261\350\212\202\347\202\271.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 52. \344\270\244\344\270\252\351\223\276\350\241\250\347\232\204\347\254\254\344\270\200\344\270\252\345\205\254\345\205\261\350\212\202\347\202\271.md" new file mode 100644 index 0000000..04b8d07 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 52. \344\270\244\344\270\252\351\223\276\350\241\250\347\232\204\347\254\254\344\270\200\344\270\252\345\205\254\345\205\261\350\212\202\347\202\271.md" @@ -0,0 +1,81 @@ +## 解题思路： + +设「第一个公共节点」为 `node` ，「链表 `headA`」的节点数量为 $a$ ，「链表 `headB`」的节点数量为 $b$ ，「两链表的公共尾部」的节点数量为 $c$ ，则有： + +- 头节点 `headA` 到 `node` 前，共有 $a - c$ 个节点； +- 头节点 `headB` 到 `node` 前，共有 $b - c$ 个节点； + +![Picture1.png](https://pic.leetcode-cn.com/1615224578-EBRtwv-Picture1.png){:width=500} + +考虑构建两个节点指针 `A​` , `B` 分别指向两链表头节点 `headA` , `headB` ，做如下操作： + +- 指针 `A` 先遍历完链表 `headA` ，再开始遍历链表 `headB` ，当走到 `node` 时，共走步数为： + +$$ +a + (b - c) +$$ + +- 指针 `B` 先遍历完链表 `headB` ，再开始遍历链表 `headA` ，当走到 `node` 时，共走步数为： + +$$ +b + (a - c) +$$ + +如下式所示，此时指针 `A` , `B` 重合，并有两种情况： + +$$ +a + (b - c) = b + (a - c) +$$ + +1. 若两链表 **有** 公共尾部 (即 $c > 0$ ) ：指针 `A` , `B` 同时指向「第一个公共节点」`node` 。 +2. 若两链表 **无** 公共尾部 (即 $c = 0$ ) ：指针 `A` , `B` 同时指向 $null$ 。 + +因此返回 `A` 即可。 + +> 如下图所示，为 $a = 5$ , $b = 3$ , $c = 2$ 示例的算法执行过程。 + + + +### 复杂度分析： + +- **时间复杂度 $O(a + b)$ ：** 最差情况下（即 $|a - b| = 1$ , $c = 0$ ），此时需遍历 $a + b$ 个节点。 +- **空间复杂度 $O(1)$ ：** 节点指针 `A` , `B` 使用常数大小的额外空间。 + +## 代码： + +```Python [] +class Solution: + def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode: + A, B = headA, headB + while A != B: + A = A.next if A else headB + B = B.next if B else headA + return A +``` + +```Java [] +public class Solution { + public ListNode getIntersectionNode(ListNode headA, ListNode headB) { + ListNode A = headA, B = headB; + while (A != B) { + A = A != null ? A.next : headB; + B = B != null ? B.next : headA; + } + return A; + } +} +``` + +```C++ [] +class Solution { +public: + ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { + ListNode *A = headA, *B = headB; + while (A != B) { + A = A != nullptr ? A->next : headB; + B = B != nullptr ? B->next : headA; + } + return A; + } +}; +``` \ No newline at end of file diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 53 - I. \345\234\250\346\216\222\345\272\217\346\225\260\347\273\204\344\270\255\346\237\245\346\211\276\346\225\260\345\255\227 I.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 53 - I. \345\234\250\346\216\222\345\272\217\346\225\260\347\273\204\344\270\255\346\237\245\346\211\276\346\225\260\345\255\227 I.md" new file mode 100644 index 0000000..cb94e06 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 53 - I. \345\234\250\346\216\222\345\272\217\346\225\260\347\273\204\344\270\255\346\237\245\346\211\276\346\225\260\345\255\227 I.md" @@ -0,0 +1,170 @@ +## 解题思路： + +> 排序数组中的搜索问题，首先想到 **二分法** 解决。 + +排序数组 $nums$ 中的所有数字 $target$ 形成一个窗口，记窗口的 **左 / 右边界** 索引分别为 $left$ 和 $right$ ，分别对应窗口左边 / 右边的首个元素。 + +本题要求统计数字 $target$ 的出现次数，可转化为：使用二分法分别找到 **左边界 $left$** 和 **右边界 $right$** ，易得数字 $target$ 的数量为 $right - left - 1$ 。 + +![Picture1.png](https://pic.leetcode-cn.com/1600793982-pWqZGf-Picture1.png){:width=500} + +### 算法解析： + +1. **初始化：** 左边界 $i = 0$ ，右边界 $j = len(nums) - 1$ 。 +2. **循环二分：** 当闭区间 $[i, j]$ 无元素时跳出； + 1. 计算中点 $m = (i + j) / 2$ （向下取整）； + 2. 若 $nums[m] < target$ ，则 $target$ 在闭区间 $[m + 1, j]$ 中，因此执行 $i = m + 1$； + 3. 若 $nums[m] > target$ ，则 $target$ 在闭区间 $[i, m - 1]$ 中，因此执行 $j = m - 1$； + 4. 若 $nums[m] = target$ ，则右边界 $right$ 在闭区间 $[m+1, j]$ 中；左边界 $left$ 在闭区间 $[i, m-1]$ 中。因此分为以下两种情况： + 1. 若查找 **右边界 $right$** ，则执行 $i = m + 1$ ；（跳出时 $i$ 指向右边界） + 2. 若查找 **左边界 $left$** ，则执行 $j = m - 1$ ；（跳出时 $j$ 指向左边界） +3. **返回值：** 应用两次二分，分别查找 $right$ 和 $left$ ，最终返回 $right - left - 1$ 即可。 + +### 效率优化： + +> 以下优化基于：查找完右边界 $right = i$ 后，则 $nums[j]$ 指向最右边的 $target$ （若存在）。 + +1. 查找完右边界后，可用 $nums[j] = target$ 判断数组中是否包含 $target$ ，若不包含则直接提前返回 $0$ ，无需后续查找左边界。 +2. 查找完右边界后，左边界 $left$ 一定在闭区间 $[0, j]$ 中，因此直接从此区间开始二分查找即可。 + + + +### 复杂度分析： + +- **时间复杂度 $O(\log N)$ ：** 二分法为对数级别复杂度。 +- **空间复杂度 $O(1)$ ：** 几个变量使用常数大小的额外空间。 + +### 代码： + +可将 $nums[m] = target$ 情况合并至其他两种情况中。 + +```Python [] +class Solution: + def search(self, nums: [int], target: int) -> int: + # 搜索右边界 right + i, j = 0, len(nums) - 1 + while i <= j: + m = (i + j) // 2 + if nums[m] <= target: i = m + 1 + else: j = m - 1 + right = i + # 若数组中无 target ，则提前返回 + if j >= 0 and nums[j] != target: return 0 + # 搜索左边界 left + i = 0 + while i <= j: + m = (i + j) // 2 + if nums[m] < target: i = m + 1 + else: j = m - 1 + left = j + return right - left - 1 +``` + +```Java [] +class Solution { + public int search(int[] nums, int target) { + // 搜索右边界 right + int i = 0, j = nums.length - 1; + while(i <= j) { + int m = (i + j) / 2; + if(nums[m] <= target) i = m + 1; + else j = m - 1; + } + int right = i; + // 若数组中无 target ，则提前返回 + if(j >= 0 && nums[j] != target) return 0; + // 搜索左边界 right + i = 0; j = nums.length - 1; + while(i <= j) { + int m = (i + j) / 2; + if(nums[m] < target) i = m + 1; + else j = m - 1; + } + int left = j; + return right - left - 1; + } +} +``` + +```C++ [] +class Solution { +public: + int search(vector& nums, int target) { + // 搜索右边界 right + int i = 0, j = nums.size() - 1; + while(i <= j) { + int m = (i + j) / 2; + if(nums[m] <= target) i = m + 1; + else j = m - 1; + } + int right = i; + // 若数组中无 target ，则提前返回 + if(j >= 0 && nums[j] != target) return 0; + // 搜索左边界 right + i = 0; j = nums.size() - 1; + while(i <= j) { + int m = (i + j) / 2; + if(nums[m] < target) i = m + 1; + else j = m - 1; + } + int left = j; + return right - left - 1; + } +}; +``` + +以上代码显得比较臃肿（两轮二分查找代码冗余）。为简化代码，可将**二分查找右边界 $right$ 的代码** 封装至函数 `helper()` 。 +如下图所示，由于数组 $nums$ 中元素都为整数，因此可以分别二分查找 $target$ 和 $target - 1$ 的右边界，将两结果相减并返回即可。 + +![Picture2.png](https://pic.leetcode-cn.com/1600793982-ikEYZs-Picture2.png){:width=450} + +本质上看， `helper()` 函数旨在查找数字 $tar$ 在数组 $nums$ 中的 **插入点** ，且若数组中存在值相同的元素，则插入到这些元素的右边。 + +```Python [] +class Solution: + def search(self, nums: [int], target: int) -> int: + def helper(tar): + i, j = 0, len(nums) - 1 + while i <= j: + m = (i + j) // 2 + if nums[m] <= tar: i = m + 1 + else: j = m - 1 + return i + return helper(target) - helper(target - 1) +``` + +```Java [] +class Solution { + public int search(int[] nums, int target) { + return helper(nums, target) - helper(nums, target - 1); + } + int helper(int[] nums, int tar) { + int i = 0, j = nums.length - 1; + while(i <= j) { + int m = (i + j) / 2; + if(nums[m] <= tar) i = m + 1; + else j = m - 1; + } + return i; + } +} +``` + +```C++ [] +class Solution { +public: + int search(vector& nums, int target) { + return helper(nums, target) - helper(nums, target - 1); + } +private: + int helper(vector& nums, int tar) { + int i = 0, j = nums.size() - 1; + while(i <= j) { + int m = (i + j) / 2; + if(nums[m] <= tar) i = m + 1; + else j = m - 1; + } + return i; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 53 - II. 0\357\275\236n-1 \344\270\255\347\274\272\345\244\261\347\232\204\346\225\260\345\255\227.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 53 - II. 0\357\275\236n-1 \344\270\255\347\274\272\345\244\261\347\232\204\346\225\260\345\255\227.md" new file mode 100644 index 0000000..8dcda52 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 53 - II. 0\357\275\236n-1 \344\270\255\347\274\272\345\244\261\347\232\204\346\225\260\345\255\227.md" @@ -0,0 +1,68 @@ +## 解题思路： + +排序数组中的搜索问题，首先想到 **二分法** 解决。根据题意，数组可以按照以下规则划分为两部分。 + +- **左子数组：** $nums[i] = i$ ； +- **右子数组：** $nums[i] \ne i$ ； + +缺失的数字等于 **“右子数组的首位元素”** 对应的索引；因此考虑使用二分法查找 “右子数组的首位元素” 。 + +![Picture1.png](https://pic.leetcode-cn.com/1600794300-lLZwAz-Picture1.png){:width=500} + +### 算法解析： + +1. **初始化：** 左边界 $i = 0$ ，右边界 $j = len(nums) - 1$ ；代表闭区间 $[i, j]$ 。 +2. **循环二分：** 当 $i \leq j$ 时循环 *（即当闭区间 $[i, j]$ 为空时跳出）* ； + 1. 计算中点 $m = (i + j) // 2$ ，其中 "$//$" 为向下取整除法； + 2. 若 $nums[m] = m$ ，则 “右子数组的首位元素” 一定在闭区间 $[m + 1, j]$ 中，因此执行 $i = m + 1$； + 3. 若 $nums[m] \ne m$ ，则 “左子数组的末位元素” 一定在闭区间 $[i, m - 1]$ 中，因此执行 $j = m - 1$； +3. **返回值：** 跳出时，变量 $i$ 和 $j$ 分别指向 “右子数组的首位元素” 和 “左子数组的末位元素” 。因此返回 $i$ 即可。 + + + +### 复杂度分析： + +- **时间复杂度 $O(\log N)$：** 二分法为对数级别复杂度。 +- **空间复杂度 $O(1)$：** 几个变量使用常数大小的额外空间。 + +## 代码： + +```Python [] +class Solution: + def missingNumber(self, nums: List[int]) -> int: + i, j = 0, len(nums) - 1 + while i <= j: + m = (i + j) // 2 + if nums[m] == m: i = m + 1 + else: j = m - 1 + return i +``` + +```Java [] +class Solution { + public int missingNumber(int[] nums) { + int i = 0, j = nums.length - 1; + while(i <= j) { + int m = (i + j) / 2; + if(nums[m] == m) i = m + 1; + else j = m - 1; + } + return i; + } +} +``` + +```C++ [] +class Solution { +public: + int missingNumber(vector& nums) { + int i = 0, j = nums.size() - 1; + while(i <= j) { + int m = (i + j) / 2; + if(nums[m] == m) i = m + 1; + else j = m - 1; + } + return i; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 54. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\347\232\204\347\254\254 k \345\244\247\350\212\202\347\202\271.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 54. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\347\232\204\347\254\254 k \345\244\247\350\212\202\347\202\271.md" new file mode 100644 index 0000000..eb18100 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 54. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\347\232\204\347\254\254 k \345\244\247\350\212\202\347\202\271.md" @@ -0,0 +1,148 @@ +## 解题思路： + +本文解法基于性质：二叉搜索树的中序遍历为递增序列。根据此性质，易得二叉搜索树的 **中序遍历倒序** 为 **递减序列** 。 +因此，求 “二叉搜索树第 $k$ 大的节点” 可转化为求 “此树的中序遍历倒序的第 $k$ 个节点”。 + +![Picture1.png](https://pic.leetcode-cn.com/1600793852-IaPwtP-Picture1.png){:width=450} + +**中序遍历** 为 “左、根、右” 顺序，递归法代码如下： + +```Python [] +# 打印中序遍历 +def dfs(root): + if not root: return + dfs(root.left) # 左 + print(root.val) # 根 + dfs(root.right) # 右 +``` + +```Java [] +// 打印中序遍历 +void dfs(TreeNode root) { + if(root == null) return; + dfs(root.left); // 左 + System.out.println(root.val); // 根 + dfs(root.right); // 右 +} +``` + +```C++ [] +void dfs(TreeNode* root) { + if(root == nullptr) return; + dfs(root->left); + cout << root->val; + dfs(root->right); +} +``` + +**中序遍历的倒序** 为 “右、根、左” 顺序，递归法代码如下： + +```Python [] +# 打印中序遍历倒序 +def dfs(root): + if not root: return + dfs(root.right) # 右 + print(root.val) # 根 + dfs(root.left) # 左 +``` + +```Java [] +// 打印中序遍历倒序 +void dfs(TreeNode root) { + if(root == null) return; + dfs(root.right); // 右 + System.out.println(root.val); // 根 + dfs(root.left); // 左 +} +``` + +```C++ [] +void dfs(TreeNode* root) { + if(root == nullptr) return; + dfs(root->right); + cout << root->val; + dfs(root->left); +} +``` + +为求第 $k$ 个节点，需要实现以下三项工作： + +1. 递归遍历时计数，统计当前节点的序号； +2. 递归到第 $k$ 个节点时，应记录结果 $res$ ； +3. 记录结果后，后续的遍历即失去意义，应提前终止（即返回）； + +### 递归解析： + +1. **终止条件：** 当节点 $root$ 为空（越过叶节点），则直接返回； +2. **递归右子树：** 即 $dfs(root.right)$ ； +3. **递推工作：** + 1. 提前返回： 若 $k = 0$ ，代表已找到目标节点，无需继续遍历，因此直接返回； + 2. 统计序号： 执行 $k = k - 1$ （即从 $k$ 减至 $0$ ）； + 3. 记录结果： 若 $k = 0$ ，代表当前节点为第 $k$ 大的节点，因此记录 $res = root.val$ ； +4. **递归左子树：** 即 $dfs(root.left)$ ； + + + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 当树退化为链表时（全部为右子节点），无论 $k$ 的值大小，递归深度都为 $N$ ，占用 $O(N)$ 时间。 +- **空间复杂度 $O(N)$ ：** 当树退化为链表时（全部为右子节点），系统使用 $O(N)$ 大小的栈空间。 + +## 代码： + +题目指出：$1 \leq k \leq N$ （二叉搜索树节点个数）；因此无需考虑 $k > N$ 的情况。 +若考虑，可以在中序遍历完成后判断 $k > 0$ 是否成立，若成立则说明 $k > N$ 。 + +```Python [] +class Solution: + def kthLargest(self, root: TreeNode, k: int) -> int: + def dfs(root): + if not root: return + dfs(root.right) + if self.k == 0: return + self.k -= 1 + if self.k == 0: self.res = root.val + dfs(root.left) + + self.k = k + dfs(root) + return self.res +``` + +```Java [] +class Solution { + int res, k; + public int kthLargest(TreeNode root, int k) { + this.k = k; + dfs(root); + return res; + } + void dfs(TreeNode root) { + if(root == null) return; + dfs(root.right); + if(k == 0) return; + if(--k == 0) res = root.val; + dfs(root.left); + } +} +``` + +```C++ [] +class Solution { +public: + int kthLargest(TreeNode* root, int k) { + this->k = k; + dfs(root); + return res; + } +private: + int res, k; + void dfs(TreeNode* root) { + if(root == nullptr) return; + dfs(root->right); + if(k == 0) return; + if(--k == 0) res = root->val; + dfs(root->left); + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 55 - I. \344\272\214\345\217\211\346\240\221\347\232\204\346\267\261\345\272\246.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 55 - I. \344\272\214\345\217\211\346\240\221\347\232\204\346\267\261\345\272\246.md" new file mode 100644 index 0000000..dcb7d03 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 55 - I. \344\272\214\345\217\211\346\240\221\347\232\204\346\267\261\345\272\246.md" @@ -0,0 +1,134 @@ +树的遍历方式总体分为两类：深度优先搜索（DFS）、广度优先搜索（BFS）； +- **常见 DFS ：** 先序遍历、中序遍历、后序遍历； +- **常见 BFS ：** 层序遍历（即按层遍历）； + +求树的深度需要遍历树的所有节点，本文将介绍基于 **后序遍历（DFS）** 和 **层序遍历（BFS）** 的两种解法。 + +## 方法一：后序遍历（DFS） + +- 树的后序遍历 / 深度优先搜索往往利用 **递归** 或 **栈** 实现，本文使用递归实现。 + +- **关键点：** 此树的深度和其左（右）子树的深度之间的关系。显然，**此树的深度** 等于 **左子树的深度** 与 **右子树的深度** 中的 **最大值** $+1$ 。 + +![Picture1.png](https://pic.leetcode-cn.com/1603024336-lXVRDd-Picture1.png){:width=450} + +### 算法解析： + +1. **终止条件：** 当 `root​` 为空，说明已越过叶节点，因此返回 深度 $0$ 。 +2. **递推工作：** 本质上是对树做后序遍历。 + 1. 计算节点 `root​` 的 **左子树的深度** ，即调用 `maxDepth(root.left)`； + 2. 计算节点 `root​` 的 **右子树的深度** ，即调用 `maxDepth(root.right)`； +3. **返回值：** 返回 **此树的深度** ，即 `max(maxDepth(root.left), maxDepth(root.right)) + 1`。 + + + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为树的节点数量，计算树的深度需要遍历所有节点。 +- **空间复杂度 $O(N)$ ：** 最差情况下（当树退化为链表时），递归深度可达到 $N$ 。 + +```Python [] +class Solution: + def maxDepth(self, root: TreeNode) -> int: + if not root: return 0 + return max(self.maxDepth(root.left), self.maxDepth(root.right)) + 1 +``` + +```Java [] +class Solution { + public int maxDepth(TreeNode root) { + if(root == null) return 0; + return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1; + } +} +``` + +```C++ [] +class Solution { +public: + int maxDepth(TreeNode* root) { + if(root == nullptr) return 0; + return max(maxDepth(root->left), maxDepth(root->right)) + 1; + } +}; +``` + +## 方法二：层序遍历（BFS） + +- 树的层序遍历 / 广度优先搜索往往利用 **队列** 实现。 +- **关键点：** 每遍历一层，则计数器 $+1$ ，直到遍历完成，则可得到树的深度。 + +### 算法解析： + +1. **特例处理：** 当 `root​` 为空，直接返回 深度 $0$ 。 +2. **初始化：** 队列 `queue` （加入根节点 `root` ），计数器 `res = 0`。 +3. **循环遍历：** 当 `queue` 为空时跳出。 + 1. 初始化一个空列表 `tmp` ，用于临时存储下一层节点； + 2. 遍历队列： 遍历 `queue` 中的各节点 `node` ，并将其左子节点和右子节点加入 `tmp`； + 3. 更新队列： 执行 `queue = tmp` ，将下一层节点赋值给 `queue`； + 4. 统计层数： 执行 `res += 1` ，代表层数加 $1$； +4. **返回值：** 返回 `res` 即可。 + + + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为树的节点数量，计算树的深度需要遍历所有节点。 +- **空间复杂度 $O(N)$ ：** 最差情况下（当树平衡时），队列 `queue` 同时存储 $N/2$ 个节点。 + +```Python [] +class Solution: + def maxDepth(self, root: TreeNode) -> int: + if not root: return 0 + queue, res = [root], 0 + while queue: + tmp = [] + for node in queue: + if node.left: tmp.append(node.left) + if node.right: tmp.append(node.right) + queue = tmp + res += 1 + return res +``` + +```Java [] +class Solution { + public int maxDepth(TreeNode root) { + if(root == null) return 0; + List queue = new LinkedList<>() {{ add(root); }}, tmp; + int res = 0; + while(!queue.isEmpty()) { + tmp = new LinkedList<>(); + for(TreeNode node : queue) { + if(node.left != null) tmp.add(node.left); + if(node.right != null) tmp.add(node.right); + } + queue = tmp; + res++; + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int maxDepth(TreeNode* root) { + if(root == nullptr) return 0; + vector que; + que.push_back(root); + int res = 0; + while(!que.empty()) { + vector tmp; + for(TreeNode* node : que) { + if(node->left != nullptr) tmp.push_back(node->left); + if(node->right != nullptr) tmp.push_back(node->right); + } + que = tmp; + res++; + } + return res; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 55 - II. \345\271\263\350\241\241\344\272\214\345\217\211\346\240\221.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 55 - II. \345\271\263\350\241\241\344\272\214\345\217\211\346\240\221.md" new file mode 100644 index 0000000..8ded873 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 55 - II. \345\271\263\350\241\241\344\272\214\345\217\211\346\240\221.md" @@ -0,0 +1,162 @@ +> 此题为 [剑指Offer 55 - I. 二叉树的深度](https://leetcode-cn.com/problems/er-cha-shu-de-shen-du-lcof/solution/mian-shi-ti-55-i-er-cha-shu-de-shen-du-xian-xu-bia/) 的拓展，建议先做上一题。 + +以下两种方法均基于以下性质推出： **此树的深度** 等于 **左子树的深度** 与 **右子树的深度** 中的 **最大值** $+1$ 。 + +![Picture1.png](https://pic.leetcode-cn.com/1603024695-GYNvjf-Picture1.png){:width=450} + +## 方法一：后序遍历 + 剪枝 （从底至顶） + +> 此方法为本题的最优解法，但剪枝的方法不易第一时间想到。 + +思路是对二叉树做后序遍历，从底至顶返回子树深度，若判定某子树不是平衡树则 “剪枝” ，直接向上返回。 + +### 算法流程： + +**`recur(root)` 函数：** + +- **返回值：** + 1. 当节点`root` 左 / 右子树的深度差 $\leq 1$ ：则返回当前子树的深度，即节点 `root` 的左 / 右子树的深度最大值 $+1$ （ `max(left, right) + 1` ）； + 2. 当节点`root` 左 / 右子树的深度差 $> 1$ ：则返回 $-1$ ，代表 **此子树不是平衡树** 。 +- **终止条件：** + 1. 当 `root` 为空：说明越过叶节点，因此返回高度 $0$ ； + 2. 当左（右）子树深度为 $-1$ ：代表此树的 **左（右）子树** 不是平衡树，因此剪枝，直接返回 $-1$ ； + +**`isBalanced(root)` 函数：** + +- **返回值：** 若 `recur(root) != -1` ，则说明此树平衡，返回 $true$ ； 否则返回 $false$ 。 + + + +### 复杂度分析： + +- **时间复杂度 $O(N)$：** $N$ 为树的节点数；最差情况下，需要递归遍历树的所有节点。 +- **空间复杂度 $O(N)$：** 最差情况下（树退化为链表时），系统递归需要使用 $O(N)$ 的栈空间。 + +### 代码： + +```Python [] +class Solution: + def isBalanced(self, root: TreeNode) -> bool: + def recur(root): + if not root: return 0 + left = recur(root.left) + if left == -1: return -1 + right = recur(root.right) + if right == -1: return -1 + return max(left, right) + 1 if abs(left - right) <= 1 else -1 + + return recur(root) != -1 +``` + +```Java [] +class Solution { + public boolean isBalanced(TreeNode root) { + return recur(root) != -1; + } + + private int recur(TreeNode root) { + if (root == null) return 0; + int left = recur(root.left); + if(left == -1) return -1; + int right = recur(root.right); + if(right == -1) return -1; + return Math.abs(left - right) < 2 ? Math.max(left, right) + 1 : -1; + } +} +``` + +```C++ [] +class Solution { +public: + bool isBalanced(TreeNode* root) { + return recur(root) != -1; + } +private: + int recur(TreeNode* root) { + if (root == nullptr) return 0; + int left = recur(root->left); + if(left == -1) return -1; + int right = recur(root->right); + if(right == -1) return -1; + return abs(left - right) < 2 ? max(left, right) + 1 : -1; + } +}; +``` + +## 方法二：先序遍历 + 判断深度 （从顶至底） + +> 此方法容易想到，但会产生大量重复计算，时间复杂度较高。 + +思路是构造一个获取当前子树的深度的函数 `depth(root)` （即 [面试题55 - I. 二叉树的深度](https://leetcode-cn.com/problems/er-cha-shu-de-shen-du-lcof/solution/mian-shi-ti-55-i-er-cha-shu-de-shen-du-xian-xu-bia/) ），通过比较某子树的左右子树的深度差 `abs(depth(root.left) - depth(root.right)) <= 1` 是否成立，来判断某子树是否是二叉平衡树。若所有子树都平衡，则此树平衡。 + +### 算法流程： + +**`isBalanced(root)` 函数：** 判断树 `root` 是否平衡 + +- **特例处理：** 若树根节点 `root` 为空，则直接返回 $true$ ； +- **返回值：** 所有子树都需要满足平衡树性质，因此以下三者使用与逻辑 $\&\&$ 连接； + 1. `abs(self.depth(root.left) - self.depth(root.right)) <= 1` ：判断 **当前子树** 是否是平衡树； + 2. `self.isBalanced(root.left)` ： 先序遍历递归，判断 **当前子树的左子树** 是否是平衡树； + 3. `self.isBalanced(root.right)` ： 先序遍历递归，判断 **当前子树的右子树** 是否是平衡树； + +**`depth(root)` 函数：** 计算树 `root` 的深度 + +- **终止条件：** 当 `root` 为空，即越过叶子节点，则返回高度 $0$ ； +- **返回值：** 返回左 / 右子树的深度的最大值 $+1$ 。 + + + +### 复杂度分析： + +- **时间复杂度 $O(N \log N)$：** 最差情况下（为 “满二叉树” 时）， `isBalanced(root)` 遍历树所有节点，判断每个节点的深度 `depth(root)` 需要遍历 **各子树的所有节点** 。 + - 满二叉树高度的复杂度 $O(log N)$ ，将满二叉树按层分为 $log (N+1)$ 层； + - 通过调用 `depth(root)` ，判断二叉树各层的节点的对应子树的深度，需遍历节点数量为 $N \times 1, \frac{N-1}{2} \times 2, \frac{N-3}{4} \times 4, \frac{N-7}{8} \times 8, ..., 1 \times \frac{N+1}{2}$ 。因此各层执行 `depth(root)` 的时间复杂度为 $O(N)$ （每层开始，最多遍历 $N$ 个节点，最少遍历 $\frac{N+1}{2}$ 个节点）。 + > 其中，$\frac{N-3}{4} \times 4$ 代表从此层开始总共需遍历 $N-3$ 个节点，此层共有 $4$ 个节点，因此每个子树需遍历 $\frac{N-3}{4}$ 个节点。 + - 因此，总体时间复杂度 $=$ 每层执行复杂度 $\times$ 层数复杂度 = $O(N \times \log N)$ 。 + +![Picture2.png](https://pic.leetcode-cn.com/1603024695-yyFsRH-Picture2.png){:width=550} + +- **空间复杂度 $O(N)$：** 最差情况下（树退化为链表时），系统递归需要使用 $O(N)$ 的栈空间。 + +### 代码： + +```Python [] +class Solution: + def isBalanced(self, root: TreeNode) -> bool: + if not root: return True + return abs(self.depth(root.left) - self.depth(root.right)) <= 1 and \ + self.isBalanced(root.left) and self.isBalanced(root.right) + + def depth(self, root): + if not root: return 0 + return max(self.depth(root.left), self.depth(root.right)) + 1 +``` + +```Java [] +class Solution { + public boolean isBalanced(TreeNode root) { + if (root == null) return true; + return Math.abs(depth(root.left) - depth(root.right)) <= 1 && isBalanced(root.left) && isBalanced(root.right); + } + + private int depth(TreeNode root) { + if (root == null) return 0; + return Math.max(depth(root.left), depth(root.right)) + 1; + } +} +``` + +```C++ [] +class Solution { +public: + bool isBalanced(TreeNode* root) { + if (root == nullptr) return true; + return abs(depth(root->left) - depth(root->right)) <= 1 && isBalanced(root->left) && isBalanced(root->right); + } +private: + int depth(TreeNode* root) { + if (root == nullptr) return 0; + return max(depth(root->left), depth(root->right)) + 1; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 56 - I. \346\225\260\347\273\204\344\270\255\346\225\260\345\255\227\345\207\272\347\216\260\347\232\204\346\254\241\346\225\260.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 56 - I. \346\225\260\347\273\204\344\270\255\346\225\260\345\255\227\345\207\272\347\216\260\347\232\204\346\254\241\346\225\260.md" new file mode 100644 index 0000000..21b3695 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 56 - I. \346\225\260\347\273\204\344\270\255\346\225\260\345\255\227\345\207\272\347\216\260\347\232\204\346\254\241\346\225\260.md" @@ -0,0 +1,190 @@ +## 解题思路： + +题目要求时间复杂度 $O(N)$ ，空间复杂度 $O(1)$ ，因此首先排除 **暴力法** 和 **哈希表统计法** 。 + +> **简化问题：** 一个整型数组 `nums` 里除 **一个** 数字之外，其他数字都出现了两次。 + +设整型数组 $nums$ 中出现一次的数字为 $x$ ，出现两次的数字为 $a, a, b, b, ...$ ，即： + +$$ +nums = [a, a, b, b, ..., x] +$$ + +异或运算有个重要的性质，两个相同数字异或为 $0$ ，即对于任意整数 $a$ 有 $a \oplus a = 0$ 。因此，若将 $nums$ 中所有数字执行异或运算，留下的结果则为 **出现一次的数字 $x$** ，即： + +$$ +\begin{aligned} +& \ \ a \oplus a \oplus b \oplus b \oplus ... \oplus x \\ += & \ \ 0 \oplus 0 \oplus ... \oplus x \\ += & \ \ x +\end{aligned} +$$ + +异或运算满足交换律 $a \oplus b = b \oplus a$ ，即以上运算结果与 $nums$ 的元素顺序无关。代码如下： + +```Python [] +def singleNumber(self, nums: List[int]) -> List[int]: + x = 0 + for num in nums: # 1. 遍历 nums 执行异或运算 + x ^= num + return x; # 2. 返回出现一次的数字 x +``` + +```Java [] +public int[] singleNumber(int[] nums) { + int x = 0; + for(int num : nums) // 1. 遍历 nums 执行异或运算 + x ^= num; + return x; // 2. 返回出现一次的数字 x +} +``` + +```C++ [] +vector singleNumber(vector& nums) { + int x = 0; + for(int num : nums) // 1. 遍历 nums 执行异或运算 + x ^= num; + return x; // 2. 返回出现一次的数字 x +} +``` + +> 设 $nums = [3, 3, 4, 4, 1]$ ，以上计算流程如下图所示。 + +![Picture1.png](https://pic.leetcode-cn.com/1611393960-EnUIaQ-Picture1.png){:width=500} + +> **本题难点：** 数组 $nums$ 有 **两个** 只出现一次的数字，因此无法通过异或直接得到这两个数字。 + +设两个只出现一次的数字为 $x$ , $y$ ，由于 $x \ne y$ ，则 $x$ 和 $y$ 二进制至少有一位不同（即分别为 $0$ 和 $1$ ），根据此位可以将 $nums$ 拆分为分别包含 $x$ 和 $y$ 的两个子数组。 + +易知两子数组都满足 「除一个数字之外，其他数字都出现了两次」。因此，仿照以上简化问题的思路，分别对两子数组遍历执行异或操作，即可得到两个只出现一次的数字 $x$, $y$ 。 + +### 算法流程： + +1. **遍历 $nums$ 执行异或：** + +- 设整型数组 $nums = [a, a, b, b, ..., x, y]$ ，对 $nums$ 中所有数字执行异或，得到的结果为 $x \oplus y$ ，即： + +$$ +\begin{aligned} +& \ \ a \oplus a \oplus b \oplus b \oplus ... \oplus x \oplus y \\ += & \ \ 0 \oplus 0 \oplus ... \oplus x \oplus y \\ += & \ \ x \oplus y +\end{aligned} +$$ + +2. **循环左移计算 $m$ ：** + +- 根据异或运算定义，若整数 $x \oplus y$ 某二进制位为 $1$ ，则 $x$ 和 $y$ 的此二进制位一定不同。换言之，找到 $x \oplus y$ 某为 $1$ 的二进制位，即可将数组 $nums$ 拆分为上述的两个子数组。根据与运算特点，可知对于任意整数 $a$ 有： + + - 若 $a \& 0001 \ne 0$ ，则 $a$ 的第一位为 $1$ ； + - 若 $a \& 0010 \ne 0$ ，则 $a$ 的第二位为 $1$ ； + - 以此类推…… + +- 因此，初始化一个辅助变量 $m = 1$ ，通过与运算从右向左循环判断，可 **获取整数 $x \oplus y$ 首位 $1$** ，记录于 $m$ 中，代码如下： + +```Python [] +while z & m == 0: # m 循环左移一位，直到 z & m ！= 0 + m <<= 1 +``` + +```Java [] +while(z & m == 0) // m 循环左移一位，直到 z & m ！= 0 + m <<= 1 +``` + +```C++ [] +while(z & m == 0) // m 循环左移一位，直到 z & m ！= 0 + m <<= 1 +``` + +3. **拆分 $nums$ 为两个子数组：** +4. **分别遍历两个子数组执行异或：** + +- 通过遍历判断 $nums$ 中各数字和 $m$ 做与运算的结果，可将数组拆分为两个子数组，并分别对两个子数组遍历求异或，则可得到两个只出现一次的数字，代码如下： + +```Python [] +for num in nums: + if num & m: x ^= num # 若 num & m != 0 , 划分至子数组 1 ，执行遍历异或 + else: y ^= num # 若 num & m == 0 , 划分至子数组 2 ，执行遍历异或 +return x, y # 遍历异或完毕，返回只出现一次的数字 x 和 y +``` + +```Java [] +for(int num: nums) { + if((num & m) != 0) x ^= num; // 若 num & m != 0 , 划分至子数组 1 ，执行遍历异或 + else y ^= num; // 若 num & m == 0 , 划分至子数组 2 ，执行遍历异或 +} +return new int[] {x, y}; // 遍历异或完毕，返回只出现一次的数字 x 和 y +``` + +```C++ [] +for(int num : nums) { + if(num & m) x ^= num; // 若 num & m != 0 , 划分至子数组 1 ，执行遍历异或 + else y ^= num; // 若 num & m == 0 , 划分至子数组 2 ，执行遍历异或 +} +return vector {x, y}; // 遍历异或完毕，返回只出现一次的数字 x 和 y +``` + +5. **返回值**： + +- 返回只出现一次的数字 x, y 即可。 + +> 设 $nums = [3, 3, 4, 4, 1, 6]$ ，以上计算流程如下图所示。 + +![Picture2.png](https://pic.leetcode-cn.com/1614836837-oygHyk-Picture2.png) + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 线性遍历 $nums$ 使用 $O(N)$ 时间，遍历 $x \oplus y$ 二进制位使用 $O(32) = O(1)$ 时间。 +- **空间复杂度 $O(1)$ ：** 辅助变量 $a$ , $b$ , $x$ , $y$ 使用常数大小额外空间。 + +## 代码： + +```Python [] +class Solution: + def singleNumbers(self, nums: List[int]) -> List[int]: + x, y, n, m = 0, 0, 0, 1 + for num in nums: # 1. 遍历异或 + n ^= num + while n & m == 0: # 2. 循环左移，计算 m + m <<= 1 + for num in nums: # 3. 遍历 nums 分组 + if num & m: x ^= num # 4. 当 num & m != 0 + else: y ^= num # 4. 当 num & m == 0 + return x, y # 5. 返回出现一次的数字 +``` + +```Java [] +class Solution { + public int[] singleNumbers(int[] nums) { + int x = 0, y = 0, n = 0, m = 1; + for(int num : nums) // 1. 遍历异或 + n ^= num; + while((n & m) == 0) // 2. 循环左移，计算 m + m <<= 1; + for(int num: nums) { // 3. 遍历 nums 分组 + if((num & m) != 0) x ^= num; // 4. 当 num & m != 0 + else y ^= num; // 4. 当 num & m == 0 + } + return new int[] {x, y}; // 5. 返回出现一次的数字 + } +} +``` + +```C++ [] +class Solution { +public: + vector singleNumbers(vector& nums) { + int x = 0, y = 0, n = 0, m = 1; + for(int num : nums) // 1. 遍历异或 + n ^= num; + while((n & m) == 0) // 2. 循环左移，计算 m + m <<= 1; + for(int num : nums) { // 3. 遍历 nums 分组 + if(num & m) x ^= num; // 4. 当 num & m != 0 + else y ^= num; // 4. 当 num & m == 0 + } + return vector {x, y}; // 5. 返回出现一次的数字 + } +}; +``` \ No newline at end of file diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 56 - II. \346\225\260\347\273\204\344\270\255\346\225\260\345\255\227\345\207\272\347\216\260\347\232\204\346\254\241\346\225\260 II.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 56 - II. \346\225\260\347\273\204\344\270\255\346\225\260\345\255\227\345\207\272\347\216\260\347\232\204\346\254\241\346\225\260 II.md" new file mode 100644 index 0000000..2722426 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 56 - II. \346\225\260\347\273\204\344\270\255\346\225\260\345\255\227\345\207\272\347\216\260\347\232\204\346\254\241\346\225\260 II.md" @@ -0,0 +1,285 @@ +## 解题思路： + +如下图所示，考虑数字的二进制形式，对于出现三次的数字，各 **二进制位** 出现的次数都是 $3$ 的倍数。 +因此，统计所有数字的各二进制位中 $1$ 的出现次数，并对 $3$ 求余，结果则为只出现一次的数字。 + +![Picture1.png](https://pic.leetcode-cn.com/1603022900-quEtJr-Picture1.png){:width=450} + +## 方法一：有限状态自动机 + +各二进制位的 **位运算规则相同** ，因此只需考虑一位即可。如下图所示，对于所有数字中的某二进制位 $1$ 的个数，存在 3 种状态，即对 3 余数为 $0, 1, 2$ 。 + +- 若输入二进制位 $1$ ，则状态按照以下顺序转换； +- 若输入二进制位 $0$ ，则状态不变。 + +$$ +0 \rightarrow 1 \rightarrow 2 \rightarrow 0 \rightarrow \cdots +$$ + +![Picture2.png](https://pic.leetcode-cn.com/1603022900-GNKGMP-Picture2.png){:width=400} + +如下图所示，由于二进制只能表示 $0, 1$ ，因此需要使用两个二进制位来表示 $3$ 个状态。设此两位分别为 $two$ , $one$ ，则状态转换变为： + +$$ +00 \rightarrow 01 \rightarrow 10 \rightarrow 00 \rightarrow \cdots +$$ + +![Picture3.png](https://pic.leetcode-cn.com/1603022900-nPXQLk-Picture3.png){:width=400} + +接下来，需要通过 **状态转换表** 导出 **状态转换的计算公式** 。首先回忆一下位运算特点，对于任意二进制位 $x$ ，有： + +- 异或运算：`x ^ 0 = x`​ ， `x ^ 1 = ~x` +- 与运算：`x & 0 = 0` ， `x & 1 = x` + +**计算 $one$ 方法：** + +设当前状态为 $two$ $one$ ，此时输入二进制位 $n$ 。如下图所示，通过对状态表的情况拆分，可推出 $one$ 的计算方法为： + +```Python +if two == 0: + if n == 0: + one = one + if n == 1: + one = ~one +if two == 1: + one = 0 +``` + +引入 **异或运算** ，可将以上拆分简化为： + +```Python +if two == 0: + one = one ^ n +if two == 1: + one = 0 +``` + +引入 **与运算** ，可继续简化为： + +```Python +one = one ^ n & ~two +``` + +![Picture4.png](https://pic.leetcode-cn.com/1603022900-qIFpAR-Picture4.png){:width=550} + +**计算 $two$ 方法：** + +由于是先计算 $one$ ，因此应在新 $one$ 的基础上计算 $two$ 。 +如下图所示，修改为新 $one$ 后，得到了新的状态图。观察发现，可以使用同样的方法计算 $two$ ，即： + +```Python +two = two ^ n & ~one +``` + +![Picture5.png](https://pic.leetcode-cn.com/1603022900-hnUxBz-Picture5.png){:width=450} + +**返回值：** + +以上是对数字的二进制中 “一位” 的分析，而 `int` 类型的其他 31 位具有相同的运算规则，因此可将以上公式直接套用在 32 位数上。 + +遍历完所有数字后，各二进制位都处于状态 $00$ 和状态 $01$ （取决于 “只出现一次的数字” 的各二进制位是 $1$ 还是 $0$ ），而此两状态是由 $one$ 来记录的（此两状态下 $twos$ 恒为 $0$ ），因此返回 $ones$ 即可。 + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 位数组 $nums$ 的长度；遍历数组占用 $O(N)$ ，每轮中的常数个位运算操作占用 $O(32 \times3 \times 2) = O(1)$ 。 +- **空间复杂度 $O(1)$ ：** 变量 $ones$ , $twos$ 使用常数大小的额外空间。 + + + +### 代码： + +```Python [] +class Solution: + def singleNumber(self, nums: List[int]) -> int: + ones, twos = 0, 0 + for num in nums: + ones = ones ^ num & ~twos + twos = twos ^ num & ~ones + return ones +``` + +```Java [] +class Solution { + public int singleNumber(int[] nums) { + int ones = 0, twos = 0; + for(int num : nums){ + ones = ones ^ num & ~twos; + twos = twos ^ num & ~ones; + } + return ones; + } +} +``` + +```C++ [] +class Solution { +public: + int singleNumber(vector& nums) { + int ones = 0, twos = 0; + for(int num : nums){ + ones = ones ^ num & ~twos; + twos = twos ^ num & ~ones; + } + return ones; + } +}; +``` + +## 方法二：遍历统计 + +> 此方法相对容易理解，但效率较低，总体推荐方法一。 + +使用 **与运算** ，可获取二进制数字 $num$ 的最右一位 $n_1$ ： + +$$ +n_1 = num \& i +$$ + +配合 **右移操作** ，可从低位至高位，获取 $num$ 所有位的值（设 int 类型从低至高的位数为第 0 位 至第 31 位，即 $n_0$ ~ $n_{31}$ ）： + +$$ +num = num >> 1 +$$ + +建立一个长度为 32 的数组 $counts$ ，通过以上方法可记录所有数字的各二进制位的 $1$ 的出现次数之和。 + +```Python [] +counts = [0] * 32 +for num in nums: + for i in range(32): + counts[i] += num & 1 # 更新第 i 位 1 的个数之和 + num >>= 1 # 第 i 位 --> 第 i + 1 位 +``` + +```Java [] +int[] counts = new int[32]; +for(int num : nums) { + for(int i = 0; i < 32; i++) { + counts[i] += num & 1; // 更新第 i 位 1 的个数之和 + num >>= 1; // 第 i 位 --> 第 i + 1 位 + } +} +``` + +```C++ [] +int counts[32] = {0}; // C++ 初始化数组需要写明初始值 0 +for(int num : nums) { + for(int i = 0; i < 32; i++) { + counts[i] += num & 1; // 更新第 i 位 1 的个数之和 + num >>= 1; // 第 i 位 --> 第 i + 1 位 + } +} +``` + +将 $counts$ 各元素对 $3$ 求余，则结果为 “只出现一次的数字” 的各二进制位。 + +```Python [] +for i in range(31, -1, -1): + x = counts[i] %= 3 # 得到 “只出现一次的数字” 的第 i 位 +``` + +```Java [] +for(int i = 31; i >= 0; i--) { + int x = counts[i] %= 3; // 得到 “只出现一次的数字” 的第 i 位 +} +``` + +```C++ [] +for(int i = 31; i >= 0; i--) { + int x = counts[i] % 3; // 得到 “只出现一次的数字” 的第 i 位 +} +``` + +利用 **左移操作** 和 **或运算** ，可将 $counts$ 数组中各二进位的值恢复到数字 $res$ 上。 + +```Python [] +for i in range(31, -1, -1): + res <<= 1 + res |= counts[i] % 3 # 恢复第 i 位 +``` + +```Java [] +for(int i = 31; i >= 0; i--) { + res <<= 1; + res |= counts[i] % 3; // 恢复第 i 位 +} +``` + +```C++ [] +for(int i = 31; i >= 0; i--) { + res <<= 1; + res |= counts[i] % 3; // 恢复第 i 位 +} +``` + +最终返回 $res$ 即可。 + +> 由于 Python 的存储负数的特殊性，需要先将 $0$ - $31$ 位取反（即 `res ^ 0xffffffff` ），再将所有位取反（即 `~` ）。 +> **此组合操作含义：** 将数字 $31$ 以上位取反， $0$ - $31$ 位不变。 + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 位数组 $nums$ 的长度；遍历数组占用 $O(N)$ ，每轮中的常数个位运算操作占用 $O(1)$ 。 +- **空间复杂度 $O(1)$ ：** 数组 $counts$ 长度恒为 $32$ ，占用常数大小的额外空间。 + +### 代码： + +实际上，只需要修改求余数值 $m$ ，即可实现解决 **除了一个数字以外，其余数字都出现 $m$ 次** 的通用问题。 + +> 设 int 类型从低至高的位数为第 0 位 至第 31 位。 + +```Python [] +class Solution: + def singleNumber(self, nums: List[int]) -> int: + counts = [0] * 32 + for num in nums: + for i in range(32): + counts[i] += num & 1 # 更新第 i 位 1 的个数之和 + num >>= 1 # 第 i 位 --> 第 i 位 + res, m = 0, 3 + for i in range(31, -1, -1): + res <<= 1 + res |= counts[i] % m # 恢复第 i 位 + return res if counts[31] % m == 0 else ~(res ^ 0xffffffff) +``` + +```Java [] +class Solution { + public int singleNumber(int[] nums) { + int[] counts = new int[32]; + for(int num : nums) { + for(int i = 0; i < 32; i++) { + counts[i] += num & 1; // 更新第 i 位 1 的个数之和 + num >>= 1; // 第 i 位 --> 第 i 位 + } + } + int res = 0, m = 3; + for(int i = 31; i >= 0; i--) { + res <<= 1; + res |= counts[i] % m; // 恢复第 i 位 + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int singleNumber(vector& nums) { + int counts[32] = {0}; // C++ 初始化数组需要写明初始值 0 + for(int num : nums) { + for(int i = 0; i < 32; i++) { + counts[i] += num & 1; // 更新第 i 位 1 的个数之和 + num >>= 1; // 第 i 位 --> 第 i 位 + } + } + int res = 0, m = 3; + for(int i = 31; i >= 0; i--) { + res <<= 1; + res |= counts[i] % m; // 恢复第 i 位 + } + return res; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 57 - II. \345\222\214\344\270\272 s \347\232\204\350\277\236\347\273\255\346\255\243\346\225\260\345\272\217\345\210\227.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 57 - II. \345\222\214\344\270\272 s \347\232\204\350\277\236\347\273\255\346\255\243\346\225\260\345\272\217\345\210\227.md" new file mode 100644 index 0000000..e86f8ff --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 57 - II. \345\222\214\344\270\272 s \347\232\204\350\277\236\347\273\255\346\255\243\346\225\260\345\272\217\345\210\227.md" @@ -0,0 +1,205 @@ +## 方法一：求和公式 + +设连续正整数序列的左边界 $i$ 和右边界 $j$ ，则此序列的 **元素和** $target$ 等于 **元素平均值 $\frac{i + j}{2}$** 乘以 **元素数量 $(j - i + 1)$** ，即： + +$$ +target = \frac{(i + j) \times (j - i + 1)}{2} +$$ + +观察发现，当确定 元素和 $target$ 与 左边界 $i$ 时，可通过 **解一元二次方程** ，直接计算出右边界 $j$ ，公式推导如下： + +$$ +\begin{aligned} +target & = \frac{(i + j) \times (j - i + 1)}{2} \\ +& = \frac{j^2 + j - i^2 + i}{2} \\ +\end{aligned} +$$ + +整理上式得： + +$$ +0 = j^2 + j - (2 \times target + i^2 - i) +$$ + +根据一元二次方程求根公式得： + +$$ +j = \frac{-1 \pm \sqrt{1 + 4(2 \times target + i^2 - i)}}{2} +$$ + +由于 $j > i$ 恒成立，因此直接 **舍去必为负数的解** ，即 $j$ 的唯一解求取公式为： + +$$ +\begin{aligned} +j & = \frac{-1 + \sqrt{1 + 4(2 \times target + i^2 - i)}}{2} +\end{aligned} +$$ + +因此，通过从小到大遍历左边界 $i$ 来计算 **以 $i$ 为起始数字的连续正整数序列** 。每轮中，由以上公式计算得到右边界 $j$ ，当 $j$ 满足以下两个条件时记录结果： + +1. $j$ 为 **整数** ：符合题目所求「连续正整数序列」； +2. $i < j$ ：满足题目要求「至少含有两个数」； + +> 当 $target = 9$ 时，以上求解流程如下图所示： + +![Picture1.png](https://pic.leetcode-cn.com/1611494538-VUzxtS-Picture1.png){:width=550xl} + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N = target$ ；连续整数序列至少有两个数字，而 $i < j$ 恒成立，因此至多循环 $\frac{target}{2}$ 次，使用 $O(N)$ 时间；循环内，计算 $j$ 使用 $O(1)$ 时间；当 $i = 1$ 时，达到最大序列长度 $\frac{-1 + \sqrt{1 + 8s}}{2}$ ，考虑到解的稀疏性，将列表构建时间简化考虑为 $O(1)$ ； +- **空间复杂度 $O(1)$ ：** 变量 $i$ , $j$ 使用常数大小的额外空间。 + +### 代码： + +计算公式中 $i^2$ 项可能超过 int 类型取值范围，因此在 Java, C++ 中需要转化成 long 类型。 + +```Python [] +class Solution: + def findContinuousSequence(self, target: int): + i, j, res = 1, 2, [] + while i < j: + j = (-1 + (1 + 4 * (2 * target + i * i - i)) ** 0.5) / 2 + if i < j and j == int(j): + res.append(list(range(i, int(j) + 1))) + i += 1 + return res +``` + +```Java [] +class Solution { + public int[][] findContinuousSequence(int target) { + int i = 1; + double j = 2.0; + List res = new ArrayList<>(); + while(i < j) { + j = (-1 + Math.sqrt(1 + 4 * (2 * target + (long) i * i - i))) / 2; + if(i < j && j == (int)j) { + int[] ans = new int[(int)j - i + 1]; + for(int k = i; k <= (int)j; k++) + ans[k - i] = k; + res.add(ans); + } + i++; + } + return res.toArray(new int[0][]); + } +} +``` + +```C++ [] +class Solution { +public: + vector> findContinuousSequence(int target) { + int i = 1; + double j = 2.0; + vector> res; + while(i < j) { + j = (-1 + sqrt(1 + 4 * (2 * target + (long) i * i - i))) / 2; + if(i < j && j == (int)j) { + vector ans; + for(int k = i; k <= (int)j; k++) + ans.push_back(k); + res.push_back(ans); + } + i++; + } + return res; + } +}; +``` + +## 方法二：滑动窗口（双指针） + +设连续正整数序列的左边界 $i$ 和右边界 $j$ ，则可构建滑动窗口从左向右滑动。循环中，每轮判断滑动窗口内元素和与目标值 $target$ 的大小关系，若相等则记录结果，若大于 $target$ 则移动左边界 $i$ （以减小窗口内的元素和），若小于 $target$ 则移动右边界 $j$ （以增大窗口内的元素和）。 + +### 算法流程： + +1. **初始化：** 左边界 $i = 1$ ，右边界 $j = 2$ ，元素和 $s = 3$ ，结果列表 $res$ ； + +2. **循环：** 当 $i \geq j$ 时跳出； + + - 当 $s > target$ 时： 向右移动左边界 $i = i + 1$ ，并更新元素和 $s$ ； + - 当 $s < target$ 时： 向右移动右边界 $j = j + 1$ ，并更新元素和 $s$ ； + - 当 $s = target$ 时： 记录连续整数序列，并向右移动左边界 $i = i + 1$ ； + +3. **返回值：** 返回结果列表 $res$ ； + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N = target$ ；连续整数序列至少有两个数字，而 $i < j$ 恒成立，因此至多循环 $target$ 次（ $i$ , $j$ 都移动到 $\frac{target}{2}$ ），使用 $O(N)$ 时间；当 $i = 1$ 时，达到最大序列长度 $\frac{-1 + \sqrt{1 + 8s}}{2}$ ，考虑到解的稀疏性，将列表构建时间简化考虑为 $O(1)$ ； +- **空间复杂度 $O(1)$ ：** 变量 $i$ , $j$ , $s$ 使用常数大小的额外空间。 + +> 当 $target = 9$ 时，以上求解流程如下图所示： + +![Picture2.png](https://pic.leetcode-cn.com/1611495306-LsrxgS-Picture2.png){:width=600} + +### 代码： + +观察本文的算法流程发现，当 $s = target$ 和 $s > target$ 的移动边界操作相同，因此可以合并，代码如下所示。 + +```Python [] +class Solution: + def findContinuousSequence(self, target: int) -> List[List[int]]: + i, j, s, res = 1, 2, 3, [] + while i < j: + if s == target: + res.append(list(range(i, j + 1))) + if s >= target: + s -= i + i += 1 + else: + j += 1 + s += j + return res +``` + +```Java [] +class Solution { + public int[][] findContinuousSequence(int target) { + int i = 1, j = 2, s = 3; + List res = new ArrayList<>(); + while(i < j) { + if(s == target) { + int[] ans = new int[j - i + 1]; + for(int k = i; k <= j; k++) + ans[k - i] = k; + res.add(ans); + } + if(s >= target) { + s -= i; + i++; + } else { + j++; + s += j; + } + } + return res.toArray(new int[0][]); + } +} +``` + +```C++ [] +class Solution { +public: + vector> findContinuousSequence(int target) { + int i = 1, j = 2, s = 3; + vector> res; + while(i < j) { + if(s == target) { + vector ans; + for(int k = i; k <= j; k++) + ans.push_back(k); + res.push_back(ans); + } + if(s >= target) { + s -= i; + i++; + } else { + j++; + s += j; + } + } + return res; + } +}; +``` \ No newline at end of file diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 57. \345\222\214\344\270\272 s \347\232\204\344\270\244\344\270\252\346\225\260\345\255\227.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 57. \345\222\214\344\270\272 s \347\232\204\344\270\244\344\270\252\346\225\260\345\255\227.md" new file mode 100644 index 0000000..e60456c --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 57. \345\222\214\344\270\272 s \347\232\204\344\270\244\344\270\252\346\225\260\345\255\227.md" @@ -0,0 +1,78 @@ +## 解题思路： + +利用 HashMap 可以通过遍历数组找到数字组合，时间和空间复杂度均为 $O(N)$ ； +注意本题的 $nums$ 是 **排序数组** ，因此可使用 **双指针法** 将空间复杂度降低至 $O(1)$ 。 + +### 算法流程： + +1. **初始化：** 双指针 $i$ , $j$ 分别指向数组 $nums$ 的左右两端 *（俗称对撞双指针）*。 +2. **循环搜索：** 当双指针相遇时跳出； + 1. 计算和 $s = nums[i] + nums[j]$ ； + 2. 若 $s > target$ ，则指针 $j$ 向左移动，即执行 $j = j - 1$ ； + 3. 若 $s < target$ ，则指针 $i$ 向右移动，即执行 $i = i + 1$ ； + 4. 若 $s = target$ ，立即返回数组 $[nums[i], nums[j]]$ ； +3. 若循环结束，则返回空数组，代表无和为 $target$ 的数字组合。 + + + +### 正确性证明： + +> 记每个状态为 $S(i, j)$ ，即 $S(i, j) = nums[i] + nums[j]$ 。假设 $S(i, j) < target$ ，则执行 $i = i + 1$ ，即状态切换至 $S(i + 1, j)$ 。 + +状态 $S(i, j)$ 切换至 $S(i + 1, j)$ ，则会消去一行元素，相当于 **消去了状态集合** {$S(i, i + 1), S(i, i + 2), ..., S(i, j - 2), S(i, j - 1), S(i, j)$ } 。（由于双指针都是向中间收缩，因此这些状态之后不可能再遇到）。 + +由于 $nums$ 是排序数组，因此这些 **消去的状态** 都一定满足 $S(i, j) < target$ ，即这些状态都 **不是解** 。 + +**结论：** 以上分析已证明 “每次指针 $i$ 的移动操作，都不会导致解的丢失” ，即指针 $i$ 的移动操作是安全的；同理，对于指针 $j$ 可得出同样推论；因此，此双指针法是正确的。 + +![Picture1.png](https://pic.leetcode-cn.com/1600794717-VSmNyQ-Picture1.png){:width=550} + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为数组 $nums$ 的长度；双指针共同线性遍历整个数组。 +- **空间复杂度 $O(1)$ ：** 变量 $i$, $j$ 使用常数大小的额外空间。 + +## 代码： + +```Python [] +class Solution: + def twoSum(self, nums: List[int], target: int) -> List[int]: + i, j = 0, len(nums) - 1 + while i < j: + s = nums[i] + nums[j] + if s > target: j -= 1 + elif s < target: i += 1 + else: return nums[i], nums[j] + return [] +``` + +```Java [] +class Solution { + public int[] twoSum(int[] nums, int target) { + int i = 0, j = nums.length - 1; + while(i < j) { + int s = nums[i] + nums[j]; + if(s < target) i++; + else if(s > target) j--; + else return new int[] { nums[i], nums[j] }; + } + return new int[0]; + } +} +``` + +```C++ [] +class Solution { +public: + vector twoSum(vector& nums, int target) { + int i = 0, j = nums.size() - 1; + while(i < j) { + int s = nums[i] + nums[j]; + if(s < target) i++; + else if(s > target) j--; + else return { nums[i], nums[j] }; + } + return {}; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 58 - I. \347\277\273\350\275\254\345\215\225\350\257\215\351\241\272\345\272\217.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 58 - I. \347\277\273\350\275\254\345\215\225\350\257\215\351\241\272\345\272\217.md" new file mode 100644 index 0000000..02b3027 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 58 - I. \347\277\273\350\275\254\345\215\225\350\257\215\351\241\272\345\272\217.md" @@ -0,0 +1,103 @@ +## 方法一：双指针 + +### 算法解析： + +- 倒序遍历字符串 $s$ ，记录单词左右索引边界 $i$ , $j$ ； +- 每确定一个单词的边界，则将其添加至单词列表 $res$ ； +- 最终，将单词列表拼接为字符串，并返回即可。 + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为字符串 $s$ 的长度，线性遍历字符串。 +- **空间复杂度 $O(N)$ ：** 新建的 list(Python) 或 StringBuilder(Java) 中的字符串总长度 $\leq N$ ，占用 $O(N)$ 大小的额外空间。 + + + +### 代码： + +```Python [] +class Solution: + def reverseWords(self, s: str) -> str: + s = s.strip() # 删除首尾空格 + i = j = len(s) - 1 + res = [] + while i >= 0: + while i >= 0 and s[i] != ' ': i -= 1 # 搜索首个空格 + res.append(s[i + 1: j + 1]) # 添加单词 + while i >= 0 and s[i] == ' ': i -= 1 # 跳过单词间空格 + j = i # j 指向下个单词的尾字符 + return ' '.join(res) # 拼接并返回 +``` + +```Java [] +class Solution { + public String reverseWords(String s) { + s = s.trim(); // 删除首尾空格 + int j = s.length() - 1, i = j; + StringBuilder res = new StringBuilder(); + while (i >= 0) { + while (i >= 0 && s.charAt(i) != ' ') i--; // 搜索首个空格 + res.append(s.substring(i + 1, j + 1) + " "); // 添加单词 + while (i >= 0 && s.charAt(i) == ' ') i--; // 跳过单词间空格 + j = i; // j 指向下个单词的尾字符 + } + return res.toString().trim(); // 转化为字符串并返回 + } +} +``` + +## 方法二：分割 + 倒序 + +利用 “字符串分割”、“列表倒序” 的内置函数 *（面试时不建议使用）* ，可简便地实现本题的字符串翻转要求。 + +### 算法解析： + +- **Python ：** 由于 $split()$ 方法将单词间的 “多个空格看作一个空格” （参考自 [split()和split(' ')的区别](https://www.cnblogs.com/python-coder/p/10073329.html) ），因此不会出现多余的 “空单词” 。因此，直接利用 $reverse()$ 方法翻转单词列表 $strs$ ，拼接为字符串并返回即可。 + +![Picture1.png](https://pic.leetcode-cn.com/1600795186-pzCvCm-Picture1.png){:width=500} + +- **Java ：** 以空格为分割符完成字符串分割后，若两单词间有 $x > 1$ 个空格，则在单词列表 $strs$ 中，此两单词间会多出 $x - 1$ 个 “空单词” （即 `""` ）。解决方法：倒序遍历单词列表，并将单词逐个添加至 StringBuilder ，遇到空单词时跳过。 + +![Picture2.png](https://pic.leetcode-cn.com/1600795186-RmKJXL-Picture2.png){:width=500} + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 总体为线性时间复杂度，各函数时间复杂度和参考资料链接如下。 + - [`split()` 方法：](https://softwareengineering.stackexchange.com/questions/331909/whats-the-complexity-of-javas-string-split-function) 为 $O(N)$ ； + - [`trim()` 和 `strip()` 方法：](https://stackoverflow.com/questions/51110114/is-string-trim-faster-than-string-replace) 最差情况下（当字符串全为空格时），为 $O(N)$ ； + - [`join()` 方法：](https://stackoverflow.com/questions/37133547/time-complexity-of-string-concatenation-in-python) 为 $O(N)$ ； + - [`reverse()` 方法：](https://stackoverflow.com/questions/37606159/what-is-the-time-complexity-of-python-list-reverse) 为 $O(N)$ ； +- **空间复杂度 $O(N)$ ：** 单词列表 $strs$ 占用线性大小的额外空间。 + +### 代码： + +```Python [] +class Solution: + def reverseWords(self, s: str) -> str: + s = s.strip() # 删除首尾空格 + strs = s.split() # 分割字符串 + strs.reverse() # 翻转单词列表 + return ' '.join(strs) # 拼接为字符串并返回 +``` + +```Java [] +class Solution { + public String reverseWords(String s) { + String[] strs = s.trim().split(" "); // 删除首尾空格，分割字符串 + StringBuilder res = new StringBuilder(); + for (int i = strs.length - 1; i >= 0; i--) { // 倒序遍历单词列表 + if(strs[i].equals("")) continue; // 遇到空单词则跳过 + res.append(strs[i] + " "); // 将单词拼接至 StringBuilder + } + return res.toString().trim(); // 转化为字符串，删除尾部空格，并返回 + } +} +``` + +Python 可一行实现： + +```Python [] +class Solution: + def reverseWords(self, s: str) -> str: + return ' '.join(s.strip().split()[::-1]) +``` \ No newline at end of file diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 58 - II. \345\267\246\346\227\213\350\275\254\345\255\227\347\254\246\344\270\262.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 58 - II. \345\267\246\346\227\213\350\275\254\345\255\227\347\254\246\344\270\262.md" new file mode 100644 index 0000000..64e0ab8 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 58 - II. \345\267\246\346\227\213\350\275\254\345\255\227\347\254\246\344\270\262.md" @@ -0,0 +1,278 @@ +## 解题思路： + +本题解法较多，本文主要介绍 **字符串切片** ， **列表遍历拼接** ， **字符串遍历拼接** 三种方法，适用于 Python 和 Java 语言。同时，介绍了 **三次翻转法** ，适用于 C++ 语言。 + +## 方法一：字符串切片 + +> 应用字符串切片函数，可方便实现左旋转字符串。 + +获取字符串 `s[n:]` 切片和 `s[:n]` 切片，使用 "$+$" 运算符拼接并返回即可。 + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为字符串 `s` 的长度，字符串切片函数为线性时间复杂度（[参考资料](https://stackoverflow.com/questions/4679746/time-complexity-of-javas-substring)）。 +- **空间复杂度 $O(N)$ ：** 两个字符串切片的总长度为 $N$ 。 + +![Picture1.png](https://pic.leetcode-cn.com/1600793170-eyvDTJ-Picture1.png){:width=500} + +### 代码： + +```Python [] +class Solution: + def reverseLeftWords(self, s: str, n: int) -> str: + return s[n:] + s[:n] +``` + +```Java [] +class Solution { + public String reverseLeftWords(String s, int n) { + return s.substring(n, s.length()) + s.substring(0, n); + } +} +``` + +```C++ [] +class Solution { +public: + string reverseLeftWords(string s, int n) { + return s.substr(n, s.size()) + s.substr(0, n); + } +}; +``` + +## 方法二：列表遍历拼接 + +> 若面试规定不允许使用 **切片函数** ，则使用此方法。 + +## 算法流程： + +1. 新建一个 list (Python) 、StringBuilder (Java) ，记为 `res` ； +2. 先向 `res` 添加 “第 $n + 1$ 位至末位的字符” ； +3. 再向 `res` 添加 “首位至第 $n$ 位的字符” ； +4. 将 `res` 转化为字符串并返回； + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 线性遍历 `s` 并添加，使用线性时间。 +- **空间复杂度 $O(N)$ ：** 新建的辅助 `res` 使用 $O(N)$ 大小的额外空间。 + +![Picture2.png](https://pic.leetcode-cn.com/1600793170-ViWBNV-Picture2.png){:width=550} + +### 代码： + +```Python [] +class Solution: + def reverseLeftWords(self, s: str, n: int) -> str: + res = [] + for i in range(n, len(s)): + res.append(s[i]) + for i in range(n): + res.append(s[i]) + return ''.join(res) +``` + +```Java [] +class Solution { + public String reverseLeftWords(String s, int n) { + StringBuilder res = new StringBuilder(); + for(int i = n; i < s.length(); i++) + res.append(s.charAt(i)); + for(int i = 0; i < n; i++) + res.append(s.charAt(i)); + return res.toString(); + } +} +``` + +利用求余运算，可以简化代码。 + +```Python [] +class Solution: + def reverseLeftWords(self, s: str, n: int) -> str: + res = [] + for i in range(n, n + len(s)): + res.append(s[i % len(s)]) + return ''.join(res) +``` + +```Java [] +class Solution { + public String reverseLeftWords(String s, int n) { + StringBuilder res = new StringBuilder(); + for(int i = n; i < n + s.length(); i++) + res.append(s.charAt(i % s.length())); + return res.toString(); + } +} +``` + +## 方法三：字符串遍历拼接 + +> 若规定 Python 不能使用 `join()` 函数，或规定 Java 只能用 String ，则使用此方法。 + +此方法与 **方法二** 思路一致，区别是使用字符串代替列表。 + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 线性遍历 `s` 并添加，使用线性时间。 +- **空间复杂度 $O(N)$ ：** 假设循环过程中内存会被及时回收，内存中至少同时存在长度为 $N$ 和 $N-1$ 的两个字符串（新建长度为 $N$ 的 `res` 需要使用前一个长度 $N-1$ 的 `res` ），因此至少使用 $O(N)$ 的额外空间。 + +![Picture3.png](https://pic.leetcode-cn.com/1600793170-uasqXO-Picture3.png){:width=550} + +```Python [] +class Solution: + def reverseLeftWords(self, s: str, n: int) -> str: + res = "" + for i in range(n, len(s)): + res += s[i] + for i in range(n): + res += s[i] + return res +``` + +```Java [] +class Solution { + public String reverseLeftWords(String s, int n) { + String res = ""; + for(int i = n; i < s.length(); i++) + res += s.charAt(i); + for(int i = 0; i < n; i++) + res += s.charAt(i); + return res; + } +} +``` + +同理，利用求余运算，可以简化代码。 + +```Python [] +class Solution: + def reverseLeftWords(self, s: str, n: int) -> str: + res = "" + for i in range(n, n + len(s)): + res += s[i % len(s)] + return res +``` + +```Java [] +class Solution { + public String reverseLeftWords(String s, int n) { + String res = ""; + for(int i = n; i < n + s.length(); i++) + res += s.charAt(i % s.length()); + return res; + } +} +``` + +## 效率对比： + +由于本题的多解法涉及到了 **字符串为不可变对象** 的相关概念，导致效率区别较大。以上三种方法的空间使用如下图所示。 + +> 详细分析请参考 [Efficient String Concatenation in Python](https://waymoot.org/home/python_string/) 。 + +以 Python 为例开展三种方法的效率测试，结论同样适用于 Java 语言。 + +![Picture4.png](https://pic.leetcode-cn.com/1600793170-xExsDE-Picture4.png){:width=650} + +### 测试数据： + +长度为 $10000000$ 的全为 `'1'` 的字符串。 + +```Python +s = "1" * 10000000 +``` + +**方法一测试：** + +新建两切片字符串，并将两切片拼接为结果字符串，无冗余操作，效率最高。 + +```Python [] +# 运行时间: 0.01 秒 +def func1(s): + cut = len(s) // 3 + return s[:cut] + s[cut:] +``` + +**方法二测试：** + +列表(Python) 和 StringBuilder(Java) 都是可变对象，每轮遍历拼接字符时，只是向列表尾部添加一个新的字符元素。最终拼接转化为字符串时，系统 **仅申请一次内存** 。 + +```Python [] +# 运行时间: 1.86 秒 +def func2(s): + res = [] + for i in range(len(s)): + res.append(s[i]) # 仅需在列表尾部添加元素 + return ''.join(res) +``` + +**方法三测试：** + +在 Python 和 Java 中，字符串是 “不可变对象” 。因此，每轮遍历拼接字符时，都需要新建一个字符串；因此，系统 **需申请 $N$ 次内存** ，数据量较大时效率低下。 + +```Python [] +# 运行时间: 6.31 秒 +def func3(s): + res = "" + for i in range(len(s)): + res += s[i] # 每次拼接都需要新建一个字符串 + return res +``` + +## 方法四：三次翻转（C++） + +由于 C++ 中的字符串是 **可变类型** ，因此可在原字符串上直接操作实现字符串旋转，实现 $O(1)$ 的空间复杂度。 + +设字符串 $s = s_1 s_2$ ，字符串 $s$ 的反转字符串为 $\hat s$ ，则左旋转字符串 $s_2 s_1$ 计算方法为： + +$$ +s_2 s_1 = \hat{\hat{s_1} \hat{s_2}} +$$ + +> 例如，$s = "abcdefg"$ , $s_1 = "ab"$ , $s_2 = "cdefg"$ ，则有： +> $$ +> \hat{s_1} = "ba" \\ +> \hat{s_2} = "gfedc" \\ +> \hat{\hat{s_1} \hat{s_2}} = \hat{"bagfedc"} = "cdefgba" +> $$ +> 即 $"cdefgba"$ 为所求字符串 $s$ 的左旋转结果。 + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 共线性遍历两轮 `s` 。 +- **空间复杂度 $O(1)$ ：** C++ 原地字符串操作，使用常数大小额外空间。 + +### 代码： + +自行实现字符串翻转函数 `reverseString()` ，代码如下： + +```C++ [] +class Solution { +public: + string reverseLeftWords(string s, int n) { + reverseString(s, 0, n - 1); + reverseString(s, n, s.size() - 1); + reverseString(s, 0, s.size() - 1); + return s; + } +private: + void reverseString(string& s, int i, int j) { + while(i < j) swap(s[i++], s[j--]); + } +}; +``` + +也可使用库函数实现，代码如下： + +```C++ [] +class Solution { +public: + string reverseLeftWords(string s, int n) { + reverse(s.begin(), s.begin() + n); + reverse(s.begin() + n, s.end()); + reverse(s.begin(), s.end()); + return s; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 59 - I. \346\273\221\345\212\250\347\252\227\345\217\243\347\232\204\346\234\200\345\244\247\345\200\274.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 59 - I. \346\273\221\345\212\250\347\252\227\345\217\243\347\232\204\346\234\200\345\244\247\345\200\274.md" new file mode 100644 index 0000000..a7636a0 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 59 - I. \346\273\221\345\212\250\347\252\227\345\217\243\347\232\204\346\234\200\345\244\247\345\200\274.md" @@ -0,0 +1,145 @@ +## 解题思路： + +设窗口区间为 $[i, j]$ ，最大值为 $x_j$ 。当窗口向前移动一格，则区间变为 $[i+1,j+1]$ ，即添加了 $nums[j + 1]$ ，删除了 $nums[i]$ 。 + +若只向窗口 $[i, j]$ 右边添加数字 $nums[j + 1]$ ，则新窗口最大值可以 **通过一次对比** 使用 $O(1)$ 时间得到，即： + +$$ +x_{j+1} = \max(x_{j}, nums[j + 1]) +$$ + +而由于删除的 $nums[i]$ 可能恰好是窗口内唯一的最大值 $x_j$ ，因此不能通过以上方法计算 $x_{j+1}$ ，而必须使用 $O(j-i)$ 时间， **遍历整个窗口区间** 获取最大值，即： + +$$ +x_{j+1} = \max(nums(i+1), \cdots , num(j+1)) +$$ + +根据以上分析，可得 **暴力法** 的时间复杂度为 $O((n-k+1)k) \approx O(nk)$ 。 + +- 设数组 $nums$ 的长度为 $n$ ，则共有 $(n-k+1)$ 个窗口； +- 获取每个窗口最大值需线性遍历，时间复杂度为 $O(k)$ 。 + +![Picture1.png](https://pic.leetcode-cn.com/1600878237-pBiBdf-Picture1.png){:width=650} + +> **本题难点：** 如何在每次窗口滑动后，将 “获取窗口内最大值” 的时间复杂度从 $O(k)$ 降低至 $O(1)$ 。 + +回忆 [剑指Offer 30. 包含 min 函数的栈](https://leetcode-cn.com/problems/bao-han-minhan-shu-de-zhan-lcof/solution/mian-shi-ti-30-bao-han-minhan-shu-de-zhan-fu-zhu-z/) ，其使用 **单调栈** 实现了随意入栈、出栈情况下的 $O(1)$ 时间获取 “栈内最小值” 。本题同理，不同点在于 “出栈操作” 删除的是 “列表尾部元素” ，而 “窗口滑动” 删除的是 “列表首部元素” 。 + +窗口对应的数据结构为 **双端队列** ，本题使用 **单调队列** 即可解决以上问题。遍历数组时，每轮保证单调队列 $deque$ ： + +1. $deque$ 内 **仅包含窗口内的元素** $\Rightarrow$ 每轮窗口滑动移除了元素 $nums[i - 1]$ ，需将 $deque$ 内的对应元素一起删除。 +2. $deque$ 内的元素 **非严格递减** $\Rightarrow$ 每轮窗口滑动添加了元素 $nums[j + 1]$ ，需将 $deque$ 内所有 $< nums[j + 1]$ 的元素删除。 + +### 算法流程： + +1. **初始化：** 双端队列 $deque$ ，结果列表 $res$ ，数组长度 $n$ ； +2. **滑动窗口：** 左边界范围 $i \in [1 - k, n - k]$ ，右边界范围 $j \in [0, n - 1]$ ； + 1. 若 $i > 0$ 且 队首元素 $deque[0]$ $=$ 被删除元素 $nums[i - 1]$ ：则队首元素出队； + 2. 删除 $deque$ 内所有 $< nums[j]$ 的元素，以保持 $deque$ 递减； + 3. 将 $nums[j]$ 添加至 $deque$ 尾部； + 4. 若已形成窗口（即 $i \geq 0$ ）：将窗口最大值（即队首元素 $deque[0]$ ）添加至列表 $res$ ； +3. **返回值：** 返回结果列表 $res$ ； + + + +### 复杂度分析： + +- **时间复杂度 $O(n)$ ：** 其中 $n$ 为数组 $nums$ 长度；线性遍历 $nums$ 占用 $O(n)$ ；每个元素最多仅入队和出队一次，因此单调队列 $deque$ 占用 $O(2n)$ 。 +- **空间复杂度 $O(k)$ ：** 双端队列 $deque$ 中最多同时存储 $k$ 个元素（即窗口大小）。 + +## 代码： + +Python 通过 `zip(range(), range())` 可实现滑动窗口的左右边界 `i, j` 同时遍历。 + +```Python [] +class Solution: + def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]: + deque = collections.deque() + res, n = [], len(nums) + for i, j in zip(range(1 - k, n + 1 - k), range(n)): + # 删除 deque 中对应的 nums[i-1] + if i > 0 and deque[0] == nums[i - 1]: + deque.popleft() + # 保持 deque 递减 + while deque and deque[-1] < nums[j]: + deque.pop() + deque.append(nums[j]) + # 记录窗口最大值 + if i >= 0: + res.append(deque[0]) + return res +``` + +```Java [] +class Solution { + public int[] maxSlidingWindow(int[] nums, int k) { + if(nums.length == 0 || k == 0) return new int[0]; + Deque deque = new LinkedList<>(); + int[] res = new int[nums.length - k + 1]; + for(int j = 0, i = 1 - k; j < nums.length; i++, j++) { + // 删除 deque 中对应的 nums[i-1] + if(i > 0 && deque.peekFirst() == nums[i - 1]) + deque.removeFirst(); + // 保持 deque 递减 + while(!deque.isEmpty() && deque.peekLast() < nums[j]) + deque.removeLast(); + deque.addLast(nums[j]); + // 记录窗口最大值 + if(i >= 0) + res[i] = deque.peekFirst(); + } + return res; + } +} +``` + +可以将 “未形成窗口” 和 “形成窗口后” 两个阶段拆分到两个循环里实现。代码虽变长，但减少了冗余的判断操作。 + +```Python [] +class Solution: + def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]: + if not nums or k == 0: return [] + deque = collections.deque() + # 未形成窗口 + for i in range(k): + while deque and deque[-1] < nums[i]: + deque.pop() + deque.append(nums[i]) + res = [deque[0]] + # 形成窗口后 + for i in range(k, len(nums)): + if deque[0] == nums[i - k]: + deque.popleft() + while deque and deque[-1] < nums[i]: + deque.pop() + deque.append(nums[i]) + res.append(deque[0]) + return res +``` + +```Java [] +class Solution { + public int[] maxSlidingWindow(int[] nums, int k) { + if(nums.length == 0 || k == 0) return new int[0]; + Deque deque = new LinkedList<>(); + int[] res = new int[nums.length - k + 1]; + // 未形成窗口 + for(int i = 0; i < k; i++) { + while(!deque.isEmpty() && deque.peekLast() < nums[i]) + deque.removeLast(); + deque.addLast(nums[i]); + } + res[0] = deque.peekFirst(); + // 形成窗口后 + for(int i = k; i < nums.length; i++) { + if(deque.peekFirst() == nums[i - k]) + deque.removeFirst(); + while(!deque.isEmpty() && deque.peekLast() < nums[i]) + deque.removeLast(); + deque.addLast(nums[i]); + res[i - k + 1] = deque.peekFirst(); + } + return res; + } +} +``` \ No newline at end of file diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 59 - II. \351\230\237\345\210\227\347\232\204\346\234\200\345\244\247\345\200\274.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 59 - II. \351\230\237\345\210\227\347\232\204\346\234\200\345\244\247\345\200\274.md" new file mode 100644 index 0000000..00aa082 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 59 - II. \351\230\237\345\210\227\347\232\204\346\234\200\345\244\247\345\200\274.md" @@ -0,0 +1,127 @@ +## 解题思路： + +> 对于普通队列，入队 `push_back()` 和出队 `pop_front()` 的时间复杂度均为 $O(1)$ ；本题难点为实现查找最大值 `max_value()` 的 $O(1)$ 时间复杂度。 +> 假设队列中存储 $N$ 个元素，从中获取最大值需要遍历队列，时间复杂度为 $O(N)$ ，单从算法上无优化空间。 + +如下图所示，最直观的想法是 **维护一个最大值变量** ，在元素入队时更新此变量即可；但当最大值出队后，并无法确定下一个 **次最大值** ，因此不可行。 + +![Picture1.png](https://pic.leetcode-cn.com/1609261470-WanZuG-Picture1.png){:width=500} + +考虑利用 **数据结构** 来实现，即经常使用的 “空间换时间” 。如下图所示，考虑构建一个递减列表来保存队列 **所有递减的元素** ，递减链表随着入队和出队操作实时更新，这样队列最大元素就始终对应递减列表的首元素，实现了获取最大值 $O(1)$ 时间复杂度。 + +![Picture2.png](https://pic.leetcode-cn.com/1609261470-gMTEAf-Picture2.png){:width=500} + +为了实现此递减列表，需要使用 **双向队列** ，假设队列已经有若干元素： + +1. 当执行入队 `push_back()` 时： 若入队一个比队列某些元素更大的数字 $x$ ，则为了保持此列表递减，需要将双向队列 **尾部所有小于 $x$ 的元素** 弹出。 +2. 当执行出队 `pop_front()` 时： 若出队的元素是最大元素，则 双向队列 需要同时 **将首元素出队** ，以保持队列和双向队列的元素一致性。 + +> 使用双向队列原因：维护递减列表需要元素队首弹出、队尾插入、队尾弹出操作皆为 $O(1)$ 时间复杂度。 + +### 函数设计： + +初始化队列 `queue` ，双向队列 `deque` ； + +**最大值 `max_value()` ：** + +- 当双向队列 `deque` 为空，则返回 $-1$ ； +- 否则，返回 `deque` 首元素； + +**入队 `push_back()` ：** + +1. 将元素 `value` 入队 `queue` ； +2. 将双向队列中队尾 **所有** 小于 `value` 的元素弹出（以保持 `deque` 非单调递减），并将元素 `value` 入队 `deque` ； + +**出队 `pop_front()` ：** + +1. 若队列 `queue` 为空，则直接返回 $-1$ ； +2. 否则，将 `queue` 首元素出队； +3. 若 `deque` 首元素和 `queue` 首元素 **相等** ，则将 `deque` 首元素出队（以保持两队列 **元素一致** ） ； + +> 设计双向队列为 **单调不增** 的原因：若队列 `queue` 中存在两个 **值相同的最大元素** ，此时 `queue` 和 `deque` 同时弹出一个最大元素，而 `queue` 中还有一个此最大元素；即采用单调递减将导致两队列中的元素不一致。 + + + +### 复杂度分析： + +- **时间复杂度 $O(1)$ ：** `max_value()`, `push_back()`, `pop_front()` 方法的均摊时间复杂度均为 $O(1)$ ； +- **空间复杂度 $O(N)$ ：** 当元素个数为 $N$ 时，最差情况下`deque` 中保存 $N$ 个元素，使用 $O(N)$ 的额外空间； + +## 代码： + +```Python [] +import queue + +class MaxQueue: + def __init__(self): + self.queue = queue.Queue() + self.deque = queue.deque() + + def max_value(self) -> int: + return self.deque[0] if self.deque else -1 + + def push_back(self, value: int) -> None: + self.queue.put(value) + while self.deque and self.deque[-1] < value: + self.deque.pop() + self.deque.append(value) + + def pop_front(self) -> int: + if self.queue.empty(): return -1 + val = self.queue.get() + if val == self.deque[0]: + self.deque.popleft() + return val +``` + +```Java [] +class MaxQueue { + Queue queue; + Deque deque; + public MaxQueue() { + queue = new LinkedList<>(); + deque = new LinkedList<>(); + } + public int max_value() { + return deque.isEmpty() ? -1 : deque.peekFirst(); + } + public void push_back(int value) { + queue.offer(value); + while(!deque.isEmpty() && deque.peekLast() < value) + deque.pollLast(); + deque.offerLast(value); + } + public int pop_front() { + if(queue.isEmpty()) return -1; + if(queue.peek().equals(deque.peekFirst())) + deque.pollFirst(); + return queue.poll(); + } +} +``` + +```C++ [] +class MaxQueue { + queue que; + deque deq; +public: + MaxQueue() { } + int max_value() { + return deq.empty() ? -1 : deq.front(); + } + void push_back(int value) { + que.push(value); + while(!deq.empty() && deq.back() < value) + deq.pop_back(); + deq.push_back(value); + } + int pop_front() { + if(que.empty()) return -1; + int val = que.front(); + if(val == deq.front()) + deq.pop_front(); + que.pop(); + return val; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 60. n \344\270\252\351\252\260\345\255\220\347\232\204\347\202\271\346\225\260.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 60. n \344\270\252\351\252\260\345\255\220\347\232\204\347\202\271\346\225\260.md" new file mode 100644 index 0000000..2f8ca18 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 60. n \344\270\252\351\252\260\345\255\220\347\232\204\347\202\271\346\225\260.md" @@ -0,0 +1,112 @@ +## 方法一：暴力法 + +> 此方法超时，但为便于理解「方法二」，建议先理解此方法。 + +给定 $n$ 个骰子，可得： + +- 每个骰子摇到 $1$ 至 $6$ 的概率相等，都为 $\frac{1}{6}$ 。 + +- 将每个骰子的点数看作独立情况，共有 $6^n$ 种「**点数组合**」。例如 $n = 2$ 时的点数组合为： + +$$ +(1,1), (1,2), \cdots, (2, 1), (2, 2), \cdots, (6,1), \cdots, (6, 6) +$$ + +- $n$ 个骰子「**点数和**」的范围为 $[n, 6n]$ ，数量为 $6n - n + 1 = 5n + 1$ 种。 + +**暴力统计：** 每个「点数组合」都对应一个「点数和」，考虑遍历所有点数组合，统计每个点数和的出现次数，最后除以点数组合的总数（即除以 $6^n$ ），即可得到每个点数和的出现概率。 + +> 如下图所示，为输入 $n = 2$ 时，点数组合、点数和、各点数概率的计算过程。 + +![Picture1.png](https://pic.leetcode-cn.com/1615223242-EMOnIR-Picture1.png){:width=550} + +暴力法需要遍历所有点数组合，因此时间复杂度为 $O(6^n)$ ，观察本题输入取值范围 $1 \leq n \leq 11$ ，可知此复杂度是无法接受的。 + +## 方法二：动态规划 + +> 设输入 $n$ 个骰子的解（即概率列表）为 $f(n)$ ，其中「点数和」 $x$ 的概率为 $f(n, x)$ 。 + +假设已知 $n - 1$ 个骰子的解 $f(n - 1)$ ，此时**添加**一枚骰子，求 $n$ 个骰子的点数和为 $x$ 的概率 $f(n, x)$ 。 + +当添加骰子的点数为 $1$ 时，前 $n - 1$ 个骰子的点数和应为 $x - 1$ ，方可组成点数和 $x$ ；同理，当此骰子为 $2$ 时，前 $n - 1$ 个骰子应为 $x - 2$ ；以此类推，直至此骰子点数为 $6$ 。将这 $6$ 种情况的概率相加，即可得到概率 $f(n, x)$ 。递推公式如下所示： +$$ +f(n, x) = \sum_{i=1}^6 f(n - 1, x - i) \times \frac{1}{6} +$$ + +根据以上分析，得知通过子问题的解 $f(n - 1)$ 可递推计算出 $f(n)$ ，而输入一个骰子的解 $f(1)$ 已知，因此可通过解 $f(1)$ 依次递推出任意解 $f(n)$ 。 + +> 如下图所示，为 $n = 2$ , $x = 7$ 的递推计算示例。 + +![Picture2.png](https://pic.leetcode-cn.com/1614960989-tpJNRQ-Picture2.png){:width=550} + +观察发现，以上递推公式虽然可行，但 $f(n - 1, x - i)$ 中的 $x - i$ 会有越界问题。例如，若希望递推计算 $f(2, 2)$ ，由于一个骰子的点数和范围为 $[1, 6]$ ，因此只应求和 $f(1, 1)$ ，即 $f(1, 0)$ , $f(1, -1)$ , ... , $f(1, -4)$ 皆无意义。此越界问题导致代码编写的难度提升。 + +> 如下图所示，以上递推公式是 “逆向” 的，即为了计算 $f(n, x)$ ，将所有与之有关的情况求和；而倘若改换为 “正向” 的递推公式，便可解决越界问题。 + +![Picture3.png](https://pic.leetcode-cn.com/1614960989-mMonMs-Picture3.png){:width=550} + +具体来看，由于新增骰子的点数只可能为 $1$ 至 $6$ ，因此概率 $f(n - 1, x)$ 仅与 $f(n, x + 1)$ , $f(n, x + 2)$, ... , $f(n, x + 6)$ 相关。因而，遍历 $f(n - 1)$ 中各点数和的概率，并将其相加至 $f(n)$ 中所有相关项，即可完成 $f(n - 1)$ 至 $f(n)$ 的递推。 + +> 将 $f(i)$ 记为动态规划列表形式 $dp[i]$ ，则 $i = 1, 2, ..., n$ 的状态转移过程如下图所示。 + + + +### 复杂度分析： + +- **时间复杂度 $O(n ^ 2)$ ：** 状态转移循环 $n - 1$ 轮；每轮中，当 $i = 2, 3, ..., n$ 时，对应循环数量分别为 $6 \times 6, 11 \times 6, ..., [5(n - 1) + 1] \times 6$ ；因此总体复杂度为 $O((n - 1) \times \frac{6 + [5(n - 1) + 1]}{2} \times 6)$ ，即等价于 $O(n^2)$ 。 +- **空间复杂度 $O(n)$ ：** 状态转移过程中，辅助数组 `tmp` 最大长度为 $6(n-1) - [(n-1) - 1] = 5n - 4$ ，因此使用 $O(5n - 4) = O(n)$ 大小的额外空间。 + +## 代码： + +通常做法是声明一个二维数组 $dp$ ，$dp[i][j]$ 代表前 $i$ 个骰子的点数和 $j$ 的概率，并执行状态转移。而由于 $dp[i]$ 仅由 $dp[i-1]$ 递推得出，为降低空间复杂度，只建立两个一维数组 $dp$ , $tmp$ 交替前进即可。 + +```Python [] +class Solution: + def dicesProbability(self, n: int) -> List[float]: + dp = [1 / 6] * 6 + for i in range(2, n + 1): + tmp = [0] * (5 * i + 1) + for j in range(len(dp)): + for k in range(6): + tmp[j + k] += dp[j] / 6 + dp = tmp + return dp +``` + +```Java [] +class Solution { + public double[] dicesProbability(int n) { + double[] dp = new double[6]; + Arrays.fill(dp, 1.0 / 6.0); + for (int i = 2; i <= n; i++) { + double[] tmp = new double[5 * i + 1]; + for (int j = 0; j < dp.length; j++) { + for (int k = 0; k < 6; k++) { + tmp[j + k] += dp[j] / 6.0; + } + } + dp = tmp; + } + return dp; + } +} +``` + +```C++ [] +class Solution { +public: + vector dicesProbability(int n) { + vector dp(6, 1.0 / 6.0); + for (int i = 2; i <= n; i++) { + vector tmp(5 * i + 1, 0); + for (int j = 0; j < dp.size(); j++) { + for (int k = 0; k < 6; k++) { + tmp[j + k] += dp[j] / 6.0; + } + } + dp = tmp; + } + return dp; + } +}; +``` \ No newline at end of file diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 61. \346\211\221\345\205\213\347\211\214\344\270\255\347\232\204\351\241\272\345\255\220.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 61. \346\211\221\345\205\213\347\211\214\344\270\255\347\232\204\351\241\272\345\255\220.md" new file mode 100644 index 0000000..49fabdf --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 61. \346\211\221\345\205\213\347\211\214\344\270\255\347\232\204\351\241\272\345\255\220.md" @@ -0,0 +1,133 @@ +## 解题思路： + +根据题意，此 $5$ 张牌是顺子的 **充分条件** 如下： + +1. 除大小王外，所有牌 **无重复** ； +2. 设此 $5$ 张牌中最大的牌为 $max$ ，最小的牌为 $min$ （大小王除外），则需满足： + +$$ +max - min < 5 +$$ + +因此可将问题转化为：此 $5$ 张牌是否满足以上两个条件？ + +![Picture1.png](https://pic.leetcode-cn.com/1599885716-MGMODX-Picture1.png){:width=650} + +## 方法一： 集合 Set + 遍历 + +- 遍历五张牌，遇到大小王（即 $0$ ）直接跳过。 +- **判别重复：** 利用 Set 实现遍历判重， Set 的查找方法的时间复杂度为 $O(1)$ ； +- **获取最大 / 最小的牌：** 借助辅助变量 $ma$ 和 $mi$ ，遍历统计即可。 + +### 复杂度分析： + +- **时间复杂度 $O(1)$ ：** 本题中给定牌数量 $N \equiv 5$ ；遍历数组使用 $O(N) = O(5) = O(1)$ 时间。 +- **空间复杂度 $O(1)$ ：** 用于判重的辅助 Set 使用 $O(N) = O(1)$ 额外空间。 + + + +### 代码： + +```Python [] +class Solution: + def isStraight(self, nums: List[int]) -> bool: + repeat = set() + ma, mi = 0, 14 + for num in nums: + if num == 0: continue # 跳过大小王 + ma = max(ma, num) # 最大牌 + mi = min(mi, num) # 最小牌 + if num in repeat: return False # 若有重复，提前返回 false + repeat.add(num) # 添加牌至 Set + return ma - mi < 5 # 最大牌 - 最小牌 < 5 则可构成顺子 +``` + +```Java [] +class Solution { + public boolean isStraight(int[] nums) { + Set repeat = new HashSet<>(); + int max = 0, min = 14; + for(int num : nums) { + if(num == 0) continue; // 跳过大小王 + max = Math.max(max, num); // 最大牌 + min = Math.min(min, num); // 最小牌 + if(repeat.contains(num)) return false; // 若有重复，提前返回 false + repeat.add(num); // 添加此牌至 Set + } + return max - min < 5; // 最大牌 - 最小牌 < 5 则可构成顺子 + } +} +``` + +```C++ [] +class Solution { +public: + bool isStraight(vector& nums) { + unordered_set repeat; + int ma = 0, mi = 14; + for(int num : nums) { + if(num == 0) continue; // 跳过大小王 + ma = max(ma, num); // 最大牌 + mi = min(mi, num); // 最小牌 + if(repeat.find(num) != repeat.end()) return false; // 若有重复，提前返回 false + repeat.insert(num); // 添加此牌至 Set + } + return ma - mi < 5; // 最大牌 - 最小牌 < 5 则可构成顺子 + } +}; +``` + +## 方法二：排序 + 遍历 + +- 先对数组执行排序。 +- **判别重复：** 排序数组中的相同元素位置相邻，因此可通过遍历数组，判断 $nums[i] = nums[i + 1]$ 是否成立来判重。 +- **获取最大 / 最小的牌：** 排序后，数组末位元素 $nums[4]$ 为最大牌；元素 $nums[joker]$ 为最小牌，其中 $joker$ 为大小王的数量。 + +### 复杂度分析： + +- **时间复杂度 $O(1)$ ：** 本题中给定牌数量 $N \equiv 5$ ；数组排序使用 $O(N \log N) = O(5 \log 5) = O(1)$ 时间。 +- **空间复杂度 $O(1)$ ：** 变量 $joker$ 使用 $O(1)$ 大小的额外空间。 + + + +### 代码： + +```Python [] +class Solution: + def isStraight(self, nums: List[int]) -> bool: + joker = 0 + nums.sort() # 数组排序 + for i in range(4): + if nums[i] == 0: joker += 1 # 统计大小王数量 + elif nums[i] == nums[i + 1]: return False # 若有重复，提前返回 false + return nums[4] - nums[joker] < 5 # 最大牌 - 最小牌 < 5 则可构成顺子 +``` + +```Java [] +class Solution { + public boolean isStraight(int[] nums) { + int joker = 0; + Arrays.sort(nums); // 数组排序 + for(int i = 0; i < 4; i++) { + if(nums[i] == 0) joker++; // 统计大小王数量 + else if(nums[i] == nums[i + 1]) return false; // 若有重复，提前返回 false + } + return nums[4] - nums[joker] < 5; // 最大牌 - 最小牌 < 5 则可构成顺子 + } +} +``` + +```C++ [] +class Solution { +public: + bool isStraight(vector& nums) { + int joker = 0; + sort(nums.begin(), nums.end()); // 数组排序 + for(int i = 0; i < 4; i++) { + if(nums[i] == 0) joker++; // 统计大小王数量 + else if(nums[i] == nums[i + 1]) return false; // 若有重复，提前返回 false + } + return nums[4] - nums[joker] < 5; // 最大牌 - 最小牌 < 5 则可构成顺子 + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 62. \345\234\206\345\234\210\344\270\255\346\234\200\345\220\216\345\211\251\344\270\213\347\232\204\346\225\260\345\255\227.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 62. \345\234\206\345\234\210\344\270\255\346\234\200\345\220\216\345\211\251\344\270\213\347\232\204\346\225\260\345\255\227.md" new file mode 100644 index 0000000..c90dcb4 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 62. \345\234\206\345\234\210\344\270\255\346\234\200\345\220\216\345\211\251\344\270\213\347\232\204\346\225\260\345\255\227.md" @@ -0,0 +1,121 @@ +## 解题思路： + +模拟整个删除过程最直观，即构建一个长度为 $n$ 的链表，各节点值为对应的顺序索引；每轮删除第 $m$ 个节点，直至链表长度为 1 时结束，返回最后剩余节点的值即可。 + +模拟法需要循环删除 $n - 1$ 轮，每轮在链表中寻找删除节点需要 $m$ 次访问操作（链表线性遍历），因此总体时间复杂度为 $O(nm)$ 。题目给定的 $m, n$ 取值范围如下所示，观察可知此时间复杂度是不可接受的。 + +$$ +1 \leq n \leq 10^5 \\ +1 \leq m \leq 10^6 +$$ + +> 实际上，本题是著名的 “约瑟夫环” 问题，可使用 **动态规划** 解决。 + +输入 $n, m$ ，记此约瑟夫环问题为 「$n, m$ 问题」 ，设解（即最后留下的数字）为 $f(n)$ ，则有： + +- 「$n, m$ 问题」：数字环为 $0, 1, 2, ..., n - 1$ ，解为 $f(n)$ ； +- 「$n-1, m$ 问题」：数字环为 $0, 1, 2, ..., n - 2$ ，解为 $f(n-1)$ ； +- 以此类推…… + +> 请注意，数字环是 **首尾相接** 的，为方便行文，本文使用列表形式表示。 + +对于「$n, m$ 问题」，首轮删除环中第 $m$ 个数字后，得到一个长度为 $n - 1$ 的数字环。由于有可能 $m > n$ ，因此删除的数字为 $(m - 1) \% n$ ，删除后的数字环从下个数字（即 $m \% n$ ）开始，设 $t = m \% n$ ，可得数字环： + +$$ +t, t + 1, t + 2, ..., 0, 1, ..., t - 3, t - 2 +$$ + +删除一轮后的数字环也变为一个「$n-1, m$ 问题」，观察以下数字编号对应关系： + +$$ +\begin{aligned} +「n-1, m 问题」 && \rightarrow && 「n, m 问题」删除后 \\ +0 && \rightarrow && t + 0 \\ +1 && \rightarrow && t + 1 \\ +... && \rightarrow && ... \\ +n - 2 && \rightarrow && t - 2 \\ +\end{aligned} +$$ + +设「$n-1, m$ 问题」某数字为 $x$ ，则可得递推关系： + +$$ +x \rightarrow (x + t) \% n \\ +$$ + +换而言之，若已知「$n-1, m$ 问题」的解 $f(n - 1)$ ，则可通过以上公式计算得到「$n, m$ 问题」的解 $f(n)$ ，即： + +$$ +\begin{aligned} +f(n) & = (f(n - 1) + t) \% n \\ +& = (f(n - 1) + m \% n) \% n \\ +& = (f(n - 1) + m) \% n +\end{aligned} +$$ + +> 以 $n = 5$ , $m = 3$ 的示例如下图所示。 + +![Picture1.png](https://pic.leetcode-cn.com/1615096532-kUoKUe-Picture1.png){:width=550} + +$f(n)$ 可由 $f(n - 1)$ 得到，$f(n - 1)$ 可由 $f(n - 2)$ 得到，……，$f(2)$ 可由 $f(1)$ 得到；因此，若给定 $f(1)$ 的值，就可以递推至任意 $f(n)$ 。而「$1, m$ 问题」的解 $f(1) = 0$ 恒成立，即无论 $m$ 为何值，长度为 1 的数字环留下的是一定是数字 $0$ 。 + +> 以上数学推导本质是得出动态规划的 转移方程 和 初始状态 。 + +### 动态规划解析： + +1. **状态定义：** 设「$i, m$ 问题」的解为 $dp[i]$ ； +2. **转移方程：** 通过以下公式可从 $dp[i - 1]$ 递推得到 $dp[i]$ ； + +$$ +dp[i] = (dp[i - 1] + m) \% i +$$ + +3. **初始状态：**「$1, m$ 问题」的解恒为 $0$ ，即 $dp[1] = 0$ ； +4. **返回值：** 返回「$n, m$ 问题」的解 $dp[n]$ ； + +> 如下图所示，为 $n = 5$ , $m = 3$ 时的状态转移和对应的模拟删除过程。 + +![Picture2.png](https://pic.leetcode-cn.com/1613584667-AQpTlK-Picture2.png) + +### 复杂度分析： + +- **时间复杂度 $O(n)$ ：** 状态转移循环 $n - 1$ 次使用 $O(n)$ 时间，状态转移方程计算使用 $O(1)$ 时间； +- **空间复杂度 $O(1)$ ：** 使用常数大小的额外空间； + +## 代码： + +根据状态转移方程的递推特性，无需建立状态列表 $dp$ ，而使用一个变量 $x$ 执行状态转移即可。 + +```Python [] +class Solution: + def lastRemaining(self, n: int, m: int) -> int: + x = 0 + for i in range(2, n + 1): + x = (x + m) % i + return x +``` + +```Java [] +class Solution { + public int lastRemaining(int n, int m) { + int x = 0; + for (int i = 2; i <= n; i++) { + x = (x + m) % i; + } + return x; + } +} +``` + +```C++ [] +class Solution { +public: + int lastRemaining(int n, int m) { + int x = 0; + for (int i = 2; i <= n; i++) { + x = (x + m) % i; + } + return x; + } +}; +``` \ No newline at end of file diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 63. \350\202\241\347\245\250\347\232\204\346\234\200\345\244\247\345\210\251\346\266\246.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 63. \350\202\241\347\245\250\347\232\204\346\234\200\345\244\247\345\210\251\346\266\246.md" new file mode 100644 index 0000000..09d3c2f --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 63. \350\202\241\347\245\250\347\232\204\346\234\200\345\244\247\345\210\251\346\266\246.md" @@ -0,0 +1,89 @@ +## 解题思路： + +设共有 $n$ 天，第 $a$ 天买，第 $b$ 天卖，则需保证 $a < b$ ；可推出交易方案数共有： + +$$ +(n - 1) + (n - 2) + \cdots + 2 + 1 = n(n - 1) / 2 +$$ + +因此，暴力法的时间复杂度为 $O(n^2)$ 。考虑使用动态规划降低时间复杂度。 + +### 动态规划解析： + +- **状态定义：** 设动态规划列表 $dp$ ，$dp[i]$ 代表以 $prices[i]$ 为结尾的子数组的最大利润（以下简称为 **前 $i$ 日的最大利润** ）。 +- **转移方程：** 由于题目限定 “买卖该股票一次” ，因此前 $i$ 日最大利润 $dp[i]$ 等于前 $i - 1$ 日最大利润 $dp[i-1]$ 和第 $i$ 日卖出的最大利润中的最大值。 + +$$ +dp[i] = \max(dp[i - 1], prices[i] - \min(prices[0:i])) \\ + +↑ \\ + +前 i 日最大利润 = \max(前 (i-1) 日最大利润, 第 i 日价格 - 前 i 日最低价格) +$$ + +- **初始状态：** $dp[0] = 0$ ，即首日利润为 $0$ ； +- **返回值：** $dp[n - 1]$ ，其中 $n$ 为 $dp$ 列表长度。 + +![Picture1.png](https://pic.leetcode-cn.com/1600880605-QGgqZW-Picture1.png){:width=550} + +### **时间复杂度降低：** + +前 $i$ 日的最低价格 $\min(prices[0:i])$ 时间复杂度为 $O(i)$ 。而在遍历 $prices$ 时，可以借助一个变量（记为成本 $cost$ ）每日更新最低价格。优化后的转移方程为： + +$$ +dp[i] = \max(dp[i - 1], prices[i] - \min(cost, prices[i]) +$$ + +### **空间复杂度降低：** + +由于 $dp[i]$ 只与 $dp[i - 1]$ , $prices[i]$ , $cost$ 相关，因此可使用一个变量（记为利润 $profit$ ）代替 $dp$ 列表。优化后的转移方程为： + +$$ +profit = \max(profit, prices[i] - \min(cost, prices[i]) +$$ + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为 $prices$ 列表长度，动态规划需遍历 $prices$ 。 +- **空间复杂度 $O(1)$ ：** 变量 $cost$ 和 $profit$ 使用常数大小的额外空间。 + + + +## 代码： + +```Python [] +class Solution: + def maxProfit(self, prices: List[int]) -> int: + cost, profit = float("+inf"), 0 + for price in prices: + cost = min(cost, price) + profit = max(profit, price - cost) + return profit +``` + +```Java [] +class Solution { + public int maxProfit(int[] prices) { + int cost = Integer.MAX_VALUE, profit = 0; + for(int price : prices) { + cost = Math.min(cost, price); + profit = Math.max(profit, price - cost); + } + return profit; + } +} +``` + +```C++ [] +class Solution { +public: + int maxProfit(vector& prices) { + int cost = INT_MAX, profit = 0; + for(int price : prices) { + cost = min(cost, price); + profit = max(profit, price - cost); + } + return profit; + } +}; +``` \ No newline at end of file diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 64. \346\261\202 1 + 2 + \342\200\246 + n.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 64. \346\261\202 1 + 2 + \342\200\246 + n.md" new file mode 100644 index 0000000..3be51ca --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 64. \346\261\202 1 + 2 + \342\200\246 + n.md" @@ -0,0 +1,151 @@ +## 解题思路： + +本题在简单问题上做了许多限制，需要使用排除法一步步导向答案。 +$1+2+...+(n-1)+n$ 的计算方法主要有三种：平均计算、迭代、递归。 + +**方法一：** 平均计算 +**问题：** 此计算必须使用 **乘除法** ，因此本方法不可取，直接排除。 + +```Java [] +public int sumNums(int n) { + return (1 + n) * n / 2; +} +``` + +```Python [] +def sumNums(n): + return (1 + n) * n // 2 +``` + +```C++ [] +int sumNums(int n) { + return (1 + n) * n / 2; +} +``` + +**方法二：** 迭代 +**问题：** 循环必须使用 $while$ 或 $for$ ，因此本方法不可取，直接排除。 + +```Java [] +public int sumNums(int n) { + int res = 0; + for(int i = 1; i <= n; i++) + res += i; + return res; +} +``` + +```Python [] +def sumNums(n): + res = 0 + for i in range(1, n + 1): + res += i + return res +``` + +```C++ [] +int sumNums(int n) { + int res = 0; + for(int i = 1; i <= n; i++) + res += i; + return res; +} +``` + +**方法三：** 递归 +**问题：** 终止条件需要使用 $if$ ，因此本方法不可取。 +**思考：** 除了 $if$ 和 $switch$ 等判断语句外，是否有其他方法可用来终止递归？ + +```Java [] +public int sumNums(int n) { + if(n == 1) return 1; + n += sumNums(n - 1); + return n; +} +``` + +```Python [] +def sumNums(n): + if n == 1: return 1 + n += sumNums(n - 1) + return n +``` + +```C++ [] +int sumNums(int n) { + if(n == 1) return 1; + n += sumNums(n - 1); + return n; +} +``` + +![Picture1.png](https://pic.leetcode-cn.com/1603023621-WFZwzS-Picture1.png){:width=500} + +### 逻辑运算符的短路效应： + +常见的逻辑运算符有三种，即 “与 $\&\&$ ”，“或 $||$ ”，“非 $!$ ” ；而其有重要的短路效应，如下所示： + +```Java +if(A && B) // 若 A 为 false ，则 B 的判断不会执行（即短路），直接判定 A && B 为 false + +if(A || B) // 若 A 为 true ，则 B 的判断不会执行（即短路），直接判定 A || B 为 true +``` + +本题需要实现 “当 $n = 1$ 时终止递归” 的需求，可通过短路效应实现。 + +```Java +n > 1 && sumNums(n - 1) // 当 n = 1 时 n > 1 不成立 ，此时 “短路” ，终止后续递归 +``` + +### 复杂度分析： + +- **时间复杂度 $O(n)$ ：** 计算 $n + (n-1) + ... + 2 + 1$ 需要开启 $n$ 个递归函数。 +- **空间复杂度 $O(n)$ ：** 递归深度达到 $n$ ，系统使用 $O(n)$ 大小的额外空间。 + + + +## 代码： + +1. Java 中，为构成语句，需加一个辅助布尔量 $x$ ，否则会报错； +2. Java 中，开启递归函数需改写为 `sumNums(n - 1) > 0` ，此整体作为一个布尔量输出，否则会报错； +3. 初始化变量 $res$ 记录结果。（ Java 可使用第二栏的简洁写法，不用借助变量 $res$ ）。 + +```Java [] +class Solution { + int res = 0; + public int sumNums(int n) { + boolean x = n > 1 && sumNums(n - 1) > 0; + res += n; + return res; + } +} +``` + +```Java [] +class Solution { + public int sumNums(int n) { + boolean x = n > 1 && (n += sumNums(n - 1)) > 0; + return n; + } +} +``` + +```Python [] +class Solution: + def __init__(self): + self.res = 0 + def sumNums(self, n: int) -> int: + n > 1 and self.sumNums(n - 1) + self.res += n + return self.res +``` + +```C++ [] +class Solution { +public: + int sumNums(int n) { + n > 1 && (n += sumNums(n - 1)); + return n; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 65. \344\270\215\347\224\250\345\212\240\345\207\217\344\271\230\351\231\244\345\201\232\345\212\240\346\263\225.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 65. \344\270\215\347\224\250\345\212\240\345\207\217\344\271\230\351\231\244\345\201\232\345\212\240\346\263\225.md" new file mode 100644 index 0000000..df854cb --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 65. \344\270\215\347\224\250\345\212\240\345\207\217\344\271\230\351\231\244\345\201\232\345\212\240\346\263\225.md" @@ -0,0 +1,100 @@ +## 解题思路： + +本题考察对位运算的灵活使用，即使用位运算实现加法。 +设两数字的二进制形式 $a, b$ ，其求和 $s = a + b$ ，$a(i)$ 代表 $a$ 的二进制第 $i$ 位，则分为以下四种情况： + +| $a(i)$ | $b(i)$ | 无进位和 $n(i)$ | 进位 $c(i+1)$ | +| :----: | :----: | :-------------: | :-----------: | +| $0$ | $0$ | $0$ | $0$ | +| $0$ | $1$ | $1$ | $0$ | +| $1$ | $0$ | $1$ | $0$ | +| $1$ | $1$ | $0$ | $1$ | + +观察发现，**无进位和** 与 **异或运算** 规律相同，**进位** 和 **与运算** 规律相同（并需左移一位）。因此，无进位和 $n$ 与进位 $c$ 的计算公式如下； + +$$ +\begin{cases} +n = a \oplus b & 非进位和：异或运算 \\ +c = a \& b << 1 & 进位：与运算 + 左移一位 +\end{cases} +$$ + +（和 $s$ ）$=$（非进位和 $n$ ）$+$（进位 $c$ ）。即可将 $s = a + b$ 转化为： + +$$ +s = a + b \Rightarrow s = n + c +$$ + +循环求 $n$ 和 $c$ ，直至进位 $c = 0$ ；此时 $s = n$ ，返回 $n$ 即可。 + +![Picture1.png](https://pic.leetcode-cn.com/9716b1a1ead21824b8216c7d54910bee4d838c011581f4e3d82a14f71cb392a1-Picture1.png){:width=500} + +> **Q ：** 若数字 $a$ 和 $b$ 中有负数，则变成了减法，如何处理？ +> **A ：** 在计算机系统中，数值一律用 **补码** 来表示和存储。**补码的优势：** 加法、减法可以统一处理（CPU只有加法器）。因此，以上方法 **同时适用于正数和负数的加法** 。 + +### 复杂度分析： + +- **时间复杂度 $O(1)$ ：** 最差情况下（例如 $a =$ 0x7fffffff , $b = 1$ 时），需循环 32 次，使用 $O(1)$ 时间；每轮中的常数次位操作使用 $O(1)$ 时间。 +- **空间复杂度 $O(1)$ ：** 使用常数大小的额外空间。 + + + +## 代码： + +```Java [] +class Solution { + public int add(int a, int b) { + while(b != 0) { // 当进位为 0 时跳出 + int c = (a & b) << 1; // c = 进位 + a ^= b; // a = 非进位和 + b = c; // b = 进位 + } + return a; + } +} +``` + +```C++ [] +class Solution { +public: + int add(int a, int b) { + while(b != 0) + { + int c = (unsigned int)(a & b) << 1; + a ^= b; + b = c; + } + return a; + } +}; +``` + +```Python [] +class Solution: + def add(self, a: int, b: int) -> int: + x = 0xffffffff + a, b = a & x, b & x + while b != 0: + a, b = (a ^ b), (a & b) << 1 & x + return a if a <= 0x7fffffff else ~(a ^ x) +``` + +> 由于 Python 的数字存储特点，需要做特殊考虑，以下详细介绍。 + +### Python 负数的存储： + +Python，Java, C++ 等语言中的数字都是以 **补码** 形式存储的。但 Python 没有 `int` , `long` 等不同长度变量，即在编程时无变量位数的概念。 + +**获取负数的补码：** 需要将数字与十六进制数 `0xffffffff` 相与。可理解为舍去此数字 32 位以上的数字（将 32 位以上都变为 $0$ ），从无限长度变为一个 32 位整数。 + +**返回前数字还原：** 若补码 $a$ 为负数（ `0x7fffffff` 是最大的正数的补码 ），需执行 `~(a ^ x)` 操作，将补码还原至 Python 的存储格式。 `a ^ x` 运算将 1 至 32 位按位取反； `~` 运算是将整个数字取反；因此， `~(a ^ x)` 是将 32 位以上的位取反，1 至 32 位不变。 + +```Python +print(hex(1)) # = 0x1 补码 +print(hex(-1)) # = -0x1 负号 + 原码 （ Python 特色，Java 会直接输出补码） + +print(hex(1 & 0xffffffff)) # = 0x1 正数补码 +print(hex(-1 & 0xffffffff)) # = 0xffffffff 负数补码 + +print(-1 & 0xffffffff) # = 4294967295 （ Python 将其认为正数） +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 66. \346\236\204\345\273\272\344\271\230\347\247\257\346\225\260\347\273\204.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 66. \346\236\204\345\273\272\344\271\230\347\247\257\346\225\260\347\273\204.md" new file mode 100644 index 0000000..13abf18 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 66. \346\236\204\345\273\272\344\271\230\347\247\257\346\225\260\347\273\204.md" @@ -0,0 +1,76 @@ +## 解题思路： + +本题的难点在于 **不能使用除法** ，即需要 **只用乘法** 生成数组 $B$ 。根据题目对 $B[i]$ 的定义，可列表格，如下图所示。 + +根据表格的主对角线（全为 $1$ ），可将表格分为 **上三角** 和 **下三角** 两部分。分别迭代计算下三角和上三角两部分的乘积，即可 **不使用除法** 就获得结果。 + +![Picture1.png](https://pic.leetcode-cn.com/1624619180-vpyyqh-Picture1.png){:width=500} + +### 算法流程： + +1. 初始化：数组 $B$ ，其中 $B[0] = 1$ ；辅助变量 $tmp = 1$ ； +2. 计算 $B[i]$ 的 **下三角** 各元素的乘积，直接乘入 $B[i]$ ； +3. 计算 $B[i]$ 的 **上三角** 各元素的乘积，记为 $tmp$ ，并乘入 $B[i]$ ； +4. 返回 $B$ 。 + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为数组长度，两轮遍历数组 $a$ ，使用 $O(N)$ 时间。 +- **空间复杂度 $O(1)$ ：** 变量 $tmp$ 使用常数大小额外空间（数组 $b$ 作为返回值，不计入复杂度考虑）。 + + + +## 代码： + +```Python [] +class Solution: + def constructArr(self, a: List[int]) -> List[int]: + b, tmp = [1] * len(a), 1 + for i in range(1, len(a)): + b[i] = b[i - 1] * a[i - 1] # 下三角 + for i in range(len(a) - 2, -1, -1): + tmp *= a[i + 1] # 上三角 + b[i] *= tmp # 下三角 * 上三角 + return b +``` + +```Java [] +class Solution { + public int[] constructArr(int[] a) { + int len = a.length; + if(len == 0) return new int[0]; + int[] b = new int[len]; + b[0] = 1; + int tmp = 1; + for(int i = 1; i < len; i++) { + b[i] = b[i - 1] * a[i - 1]; + } + for(int i = len - 2; i >= 0; i--) { + tmp *= a[i + 1]; + b[i] *= tmp; + } + return b; + } +} +``` + +```C++ [] +class Solution { +public: + vector constructArr(vector& a) { + int len = a.size(); + if(len == 0) return {}; + vector b(len, 1); + b[0] = 1; + int tmp = 1; + for(int i = 1; i < len; i++) { + b[i] = b[i - 1] * a[i - 1]; + } + for(int i = len - 2; i >= 0; i--) { + tmp *= a[i + 1]; + b[i] *= tmp; + } + return b; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 67. \346\212\212\345\255\227\347\254\246\344\270\262\350\275\254\346\215\242\346\210\220\346\225\264\346\225\260.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 67. \346\212\212\345\255\227\347\254\246\344\270\262\350\275\254\346\215\242\346\210\220\346\225\264\346\225\260.md" new file mode 100644 index 0000000..c31f54f --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 67. \346\212\212\345\255\227\347\254\246\344\270\262\350\275\254\346\215\242\346\210\220\346\225\264\346\225\260.md" @@ -0,0 +1,141 @@ +## 解题思路： + +根据题意，有以下四种字符需要考虑： + +1. **首部空格：** 删除之即可； +2. **符号位：** 三种情况，即 ''$+$'' , ''$-$'' , ''无符号" ；新建一个变量保存符号位，返回前判断正负即可； +3. **非数字字符：** 遇到首个非数字的字符时，应立即返回； +4. **数字字符：** + 1. **字符转数字：** “此数字的 ASCII 码” 与 “ $0$ 的 ASCII 码” 相减即可； + 2. **数字拼接：** 若从左向右遍历数字，设当前位字符为 $c$ ，当前位数字为 $x$ ，数字结果为 $res$ ，则数字拼接公式为： + +$$ +res = 10 \times res + x \\ +x = ascii(c) - ascii('0') +$$ + +![Picture1.png](https://pic.leetcode-cn.com/1600793383-jCgsGU-Picture1.png){:width=450} + +**数字越界处理：** + +> 题目要求返回的数值范围应在 $[-2^{31}, 2^{31} - 1]$ ，因此需要考虑数字越界问题。而由于题目指出 `环境只能存储 32 位大小的有符号整数` ，因此判断数字越界时，要始终保持 $res$ 在 int 类型的取值范围内。 + +在每轮数字拼接前，判断 $res$ **在此轮拼接后是否超过 $2147483647$** ，若超过则加上符号位直接返回。 +设数字拼接边界 $bndry = 2147483647 // 10 = 214748364$ ，则以下两种情况越界： + +$$ +\begin{cases} + res > bndry & 情况一：执行拼接 10 \times res \geq 2147483650 越界 \\ + res = bndry, x > 7 & 情况二：拼接后是 2147483648 或 2147483649 越界 \\ +\end{cases} +$$ + +![Picture2.png](https://pic.leetcode-cn.com/1600793383-JZRYip-Picture2.png){:width=450} + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为字符串长度，线性遍历字符串占用 $O(N)$ 时间。 +- **空间复杂度 $O(N)$ ：** 删除首尾空格后需建立新字符串，最差情况下占用 $O(N)$ 额外空间。 + + + +## 代码： + +```Python [] +class Solution: + def strToInt(self, str: str) -> int: + str = str.strip() # 删除首尾空格 + if not str: return 0 # 字符串为空则直接返回 + res, i, sign = 0, 1, 1 + int_max, int_min, bndry = 2 ** 31 - 1, -2 ** 31, 2 ** 31 // 10 + if str[0] == '-': sign = -1 # 保存负号 + elif str[0] != '+': i = 0 # 若无符号位，则需从 i = 0 开始数字拼接 + for c in str[i:]: + if not '0' <= c <= '9' : break # 遇到非数字的字符则跳出 + if res > bndry or res == bndry and c > '7': return int_max if sign == 1 else int_min # 数字越界处理 + res = 10 * res + ord(c) - ord('0') # 数字拼接 + return sign * res +``` + +```Java [] +class Solution { + public int strToInt(String str) { + char[] c = str.trim().toCharArray(); + if(c.length == 0) return 0; + int res = 0, bndry = Integer.MAX_VALUE / 10; + int i = 1, sign = 1; + if(c[0] == '-') sign = -1; + else if(c[0] != '+') i = 0; + for(int j = i; j < c.length; j++) { + if(c[j] < '0' || c[j] > '9') break; + if(res > bndry || res == bndry && c[j] > '7') return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE; + res = res * 10 + (c[j] - '0'); + } + return sign * res; + } +} +``` + +若不使用 `trim() / strip()` 删除首部空格，而采取遍历跳过空格的方式，则可以将空间复杂度降低至 $O(1)$ ，代码如下： + +```Python [] +class Solution: + def strToInt(self, str: str) -> int: + res, i, sign, length = 0, 0, 1, len(str) + int_max, int_min, bndry = 2 ** 31 - 1, -2 ** 31, 2 ** 31 // 10 + if not str: return 0 # 空字符串，提前返回 + while str[i] == ' ': + i += 1 + if i == length: return 0 # 字符串全为空格，提前返回 + if str[i] == '-': sign = -1 + if str[i] in '+-': i += 1 + for j in range(i, length): + if not '0' <= str[j] <= '9' : break + if res > bndry or res == bndry and str[j] > '7': + return int_max if sign == 1 else int_min + res = 10 * res + ord(str[j]) - ord('0') + return sign * res +``` + +```Java [] +class Solution { + public int strToInt(String str) { + int res = 0, bndry = Integer.MAX_VALUE / 10; + int i = 0, sign = 1, length = str.length(); + if(length == 0) return 0; + while(str.charAt(i) == ' ') + if(++i == length) return 0; + if(str.charAt(i) == '-') sign = -1; + if(str.charAt(i) == '-' || str.charAt(i) == '+') i++; + for(int j = i; j < length; j++) { + if(str.charAt(j) < '0' || str.charAt(j) > '9') break; + if(res > bndry || res == bndry && str.charAt(j) > '7') + return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE; + res = res * 10 + (str.charAt(j) - '0'); + } + return sign * res; + } +} +``` + +```C++ [] +class Solution { +public: + int strToInt(string str) { + int res = 0, bndry = INT_MAX / 10; + int i = 0, sign = 1, length = str.size(); + if(length == 0) return 0; + while(str[i] == ' ') + if(++i == length) return 0; + if(str[i] == '-') sign = -1; + if(str[i] == '-' || str[i] == '+') i++; + for(int j = i; j < length; j++) { + if(str[j] < '0' || str[j] > '9') break; + if(res > bndry || res == bndry && str[j] > '7') + return sign == 1 ? INT_MAX : INT_MIN; + res = res * 10 + (str[j] - '0'); + } + return sign * res; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 68 - I. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\347\232\204\346\234\200\350\277\221\345\205\254\345\205\261\347\245\226\345\205\210.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 68 - I. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\347\232\204\346\234\200\350\277\221\345\205\254\345\205\261\347\245\226\345\205\210.md" new file mode 100644 index 0000000..2dcb415 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 68 - I. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\347\232\204\346\234\200\350\277\221\345\205\254\345\205\261\347\245\226\345\205\210.md" @@ -0,0 +1,183 @@ +## 解题思路： + +**祖先的定义：** 若节点 $p$ 在节点 $root$ 的左（右）子树中，或 $p = root$，则称 $root$ 是 $p$ 的祖先。 + +**最近公共祖先的定义：** 设节点 $root$ 为节点 $p,q$ 的某公共祖先，若其左子节点 $root.left$ 和右子节点 $root.right$ 都不是 $p,q$ 的公共祖先，则称 $root$ 是 “最近的公共祖先” 。 + +![Picture1.png](https://pic.leetcode-cn.com/1599885085-LbAmPd-Picture1.png){:width=400} + +根据以上定义，若 $root$ 是 $p,q$ 的 **最近公共祖先** ，则只可能为以下三种情况之一： + +1. $p$ 和 $q$ 在 $root$ 的子树中，且分列 $root$ 的 **异侧**（即分别在左、右子树中）； +2. $p = root$ 且 $q$ 在 $root$ 的左或右子树中； +3. $q = root$ 且 $p$ 在 $root$ 的左或右子树中； + +![Picture2.png](https://pic.leetcode-cn.com/1599885085-mTpblH-Picture2.png){:width=400} + +本题给定了两个重要条件：① 树为 **二叉搜索树** ，② 树的所有节点的值都是 **唯一** 的。根据以上条件，可方便地判断 $p,q$ 与 $root$ 的子树关系，即： + +- 若 $root.val < p.val$ ，则 $p$ 在 $root$ **右子树** 中； +- 若 $root.val > p.val$ ，则 $p$ 在 $root$ **左子树** 中； +- 若 $root.val = p.val$ ，则 $p$ 和 $root$ 指向 **同一节点** ； + +## 方法一：迭代 + +1. **循环搜索：** 当节点 $root$ 为空时跳出； + 1. 当 $p, q$ 都在 $root$ 的 **右子树** 中，则遍历至 $root.right$ ； + 2. 否则，当 $p, q$ 都在 $root$ 的 **左子树** 中，则遍历至 $root.left$ ； + 3. 否则，说明找到了 **最近公共祖先** ，跳出； +2. **返回值：** 最近公共祖先 $root$ ； + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为二叉树节点数；每循环一轮排除一层，二叉搜索树的层数最小为 $\log N$ （满二叉树），最大为 $N$ （退化为链表）。 +- **空间复杂度 $O(1)$ ：** 使用常数大小的额外空间。 + + + +### 代码： + +```Python [] +class Solution: + def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode': + while root: + if root.val < p.val and root.val < q.val: # p,q 都在 root 的右子树中 + root = root.right # 遍历至右子节点 + elif root.val > p.val and root.val > q.val: # p,q 都在 root 的左子树中 + root = root.left # 遍历至左子节点 + else: break + return root +``` + +```Java [] +class Solution { + public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { + while(root != null) { + if(root.val < p.val && root.val < q.val) // p,q 都在 root 的右子树中 + root = root.right; // 遍历至右子节点 + else if(root.val > p.val && root.val > q.val) // p,q 都在 root 的左子树中 + root = root.left; // 遍历至左子节点 + else break; + } + return root; + } +} +``` + +```C++ [] +class Solution { +public: + TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { + while(root != nullptr) { + if(root->val < p->val && root->val < q->val) // p,q 都在 root 的右子树中 + root = root->right; // 遍历至右子节点 + else if(root->val > p->val && root->val > q->val) // p,q 都在 root 的左子树中 + root = root->left; // 遍历至左子节点 + else break; + } + return root; + } +}; +``` + +代码优化：若可保证 $p.val < q.val$ ，则在循环中可减少判断条件，提升计算效率。 + +```Python [] +class Solution: + def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode': + if p.val > q.val: p, q = q, p # 保证 p.val < q.val + while root: + if root.val < p.val: # p,q 都在 root 的右子树中 + root = root.right # 遍历至右子节点 + elif root.val > q.val: # p,q 都在 root 的左子树中 + root = root.left # 遍历至左子节点 + else: break + return root +``` + +```Java [] +class Solution { + public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { + if(p.val > q.val) { // 保证 p.val < q.val + TreeNode tmp = p; + p = q; + q = tmp; + } + while(root != null) { + if(root.val < p.val) // p,q 都在 root 的右子树中 + root = root.right; // 遍历至右子节点 + else if(root.val > q.val) // p,q 都在 root 的左子树中 + root = root.left; // 遍历至左子节点 + else break; + } + return root; + } +} +``` + +```C++ [] +class Solution { +public: + TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { + if(p->val > q->val) + swap(p, q); + while(root != nullptr) { + if(root->val < p->val) // p,q 都在 root 的右子树中 + root = root->right; // 遍历至右子节点 + else if(root->val > q->val) // p,q 都在 root 的左子树中 + root = root->left; // 遍历至左子节点 + else break; + } + return root; + } +}; +``` + +## 方法二：递归 + +1. **递推工作：** + 1. 当 $p, q$ 都在 $root$ 的 **右子树** 中，则开启递归 $root.right$ 并返回； + 2. 否则，当 $p, q$ 都在 $root$ 的 **左子树** 中，则开启递归 $root.left$ 并返回； +2. **返回值：** 最近公共祖先 $root$ ； + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为二叉树节点数；每循环一轮排除一层，二叉搜索树的层数最小为 $\log N$ （满二叉树），最大为 $N$ （退化为链表）。 +- **空间复杂度 $O(N)$ ：** 最差情况下，即树退化为链表时，递归深度达到树的层数 $N$ 。 + +### 代码： + +```Python [] +class Solution: + def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode': + if root.val < p.val and root.val < q.val: + return self.lowestCommonAncestor(root.right, p, q) + if root.val > p.val and root.val > q.val: + return self.lowestCommonAncestor(root.left, p, q) + return root +``` + +```Java [] +class Solution { + public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { + if(root.val < p.val && root.val < q.val) + return lowestCommonAncestor(root.right, p, q); + if(root.val > p.val && root.val > q.val) + return lowestCommonAncestor(root.left, p, q); + return root; + } +} +``` + +```C++ [] +class Solution { +public: + TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { + if(root->val < p->val && root->val < q->val) + return lowestCommonAncestor(root->right, p, q); + if(root->val > p->val && root->val > q->val) + return lowestCommonAncestor(root->left, p, q); + return root; + } +}; +``` diff --git "a/sword_for_offer/docs/\345\211\221\346\214\207 Offer 68 - II. \344\272\214\345\217\211\346\240\221\347\232\204\346\234\200\350\277\221\345\205\254\345\205\261\347\245\226\345\205\210.md" "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 68 - II. \344\272\214\345\217\211\346\240\221\347\232\204\346\234\200\350\277\221\345\205\254\345\205\261\347\245\226\345\205\210.md" new file mode 100644 index 0000000..e68c5d6 --- /dev/null +++ "b/sword_for_offer/docs/\345\211\221\346\214\207 Offer 68 - II. \344\272\214\345\217\211\346\240\221\347\232\204\346\234\200\350\277\221\345\205\254\345\205\261\347\245\226\345\205\210.md" @@ -0,0 +1,125 @@ +## 解题思路： + +**祖先的定义：** 若节点 $p$ 在节点 $root$ 的左（右）子树中，或 $p = root$ ，则称 $root$ 是 $p$ 的祖先。 + +**最近公共祖先的定义：** 设节点 $root$ 为节点 $p, q$ 的某公共祖先，若其左子节点 $root.left$ 和右子节点 $root.right$ 都不是 $p,q$ 的公共祖先，则称 $root$ 是 “最近的公共祖先” 。 + +![Picture1.png](https://pic.leetcode-cn.com/1599885247-rxcHcZ-Picture1.png){:width=450} + +根据以上定义，若 $root$ 是 $p, q$ 的 **最近公共祖先** ，则只可能为以下情况之一： + +1. $p$ 和 $q$ 在 $root$ 的子树中，且分列 $root$ 的 **异侧**（即分别在左、右子树中）； +2. $p = root$ ，且 $q$ 在 $root$ 的左或右子树中； +3. $q = root$ ，且 $p$ 在 $root$ 的左或右子树中； + +![Picture2.png](https://pic.leetcode-cn.com/1599885247-mgYjRv-Picture2.png){:width=450} + +考虑通过递归对二叉树进行先序遍历，当遇到节点 $p$ 或 $q$ 时返回。从底至顶回溯，当节点 $p, q$ 在节点 $root$ 的异侧时，节点 $root$ 即为最近公共祖先，则向上返回 $root$ 。 + +### 递归解析： + +1. **终止条件：** + 1. 当越过叶节点，则直接返回 $null$ ； + 2. 当 $root$ 等于 $p, q$ ，则直接返回 $root$ ； +2. **递推工作：** + 1. 开启递归左子节点，返回值记为 $left$ ； + 2. 开启递归右子节点，返回值记为 $right$ ； +3. **返回值：** 根据 $left$ 和 $right$ ，可展开为四种情况； + 1. 当 $left$ 和 $right$ **同时为空** ：说明 $root$ 的左 / 右子树中都不包含 $p,q$ ，返回 $null$ ； + 2. 当 $left$ 和 $right$ **同时不为空** ：说明 $p, q$ 分列在 $root$ 的 **异侧** （分别在 左 / 右子树），因此 $root$ 为最近公共祖先，返回 $root$ ； + 3. 当 $left$ **为空** ，$right$ **不为空** ：$p,q$ 都不在 $root$ 的左子树中，直接返回 $right$ 。具体可分为两种情况： + 1. $p,q$ 其中一个在 $root$ 的 **右子树** 中，此时 $right$ 指向 $p$（假设为 $p$ ）； + 2. $p,q$ 两节点都在 $root$ 的 **右子树** 中，此时的 $right$ 指向 **最近公共祖先节点** ； + 4. 当 $left$ **不为空** ， $right$ **为空** ：与情况 `3.` 同理； + +> 观察发现， 情况 `1.` 可合并至 `3.` 和 `4.` 内，详见文章末尾代码。 + + + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为二叉树节点数；最差情况下，需要递归遍历树的所有节点。 +- **空间复杂度 $O(N)$ ：** 最差情况下，递归深度达到 $N$ ，系统使用 $O(N)$ 大小的额外空间。 + +### 代码： + +```Python [] +class Solution: + def lowestCommonAncestor(self, root: TreeNode, p: TreeNode, q: TreeNode) -> TreeNode: + if not root or root == p or root == q: return root + left = self.lowestCommonAncestor(root.left, p, q) + right = self.lowestCommonAncestor(root.right, p, q) + if not left: return right + if not right: return left + return root +``` + +```Java [] +class Solution { + public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { + if(root == null || root == p || root == q) return root; + TreeNode left = lowestCommonAncestor(root.left, p, q); + TreeNode right = lowestCommonAncestor(root.right, p, q); + if(left == null) return right; + if(right == null) return left; + return root; + } +} +``` + +```C++ [] +class Solution { +public: + TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { + if(root == nullptr || root == p || root == q) return root; + TreeNode *left = lowestCommonAncestor(root->left, p, q); + TreeNode *right = lowestCommonAncestor(root->right, p, q); + if(left == nullptr) return right; + if(right == nullptr) return left; + return root; + } +}; +``` + +情况 `1.` , `2.` , `3.` , `4.` 的展开写法如下。 + +```Python [] +class Solution: + def lowestCommonAncestor(self, root: TreeNode, p: TreeNode, q: TreeNode) -> TreeNode: + if not root or root == p or root == q: return root + left = self.lowestCommonAncestor(root.left, p, q) + right = self.lowestCommonAncestor(root.right, p, q) + if not left and not right: return # 1. + if not left: return right # 3. + if not right: return left # 4. + return root # 2. if left and right: +``` + +```Java [] +class Solution { + public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { + if(root == null || root == p || root == q) return root; + TreeNode left = lowestCommonAncestor(root.left, p, q); + TreeNode right = lowestCommonAncestor(root.right, p, q); + if(left == null && right == null) return null; // 1. + if(left == null) return right; // 3. + if(right == null) return left; // 4. + return root; // 2. if(left != null and right != null) + } +} +``` + +```C++ [] +class Solution { +public: + TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { + if(root == nullptr || root == p || root == q) return root; + TreeNode *left = lowestCommonAncestor(root->left, p, q); + TreeNode *right = lowestCommonAncestor(root->right, p, q); + if(left == nullptr && right == nullptr) return nullptr; // 1. + if(left == nullptr) return right; // 3. + if(right == nullptr) return left; // 4. + return root; // 2. if(left != null and right != null) + } +}; +``` From 406decbc81fd9156d0f62ab21d262887ddb3cc35 Mon Sep 17 00:00:00 2001 From: krahets Date: Tue, 10 Oct 2023 19:50:04 +0800 Subject: [PATCH 10/23] Update LICENSE --- LICENSE.md => LICENSE | 62 +++++++++++-------- LICENSE-SAMPLECODE | 21 +++++++ README.md | 134 ++++-------------------------------------- 3 files changed, 70 insertions(+), 147 deletions(-) rename LICENSE.md => LICENSE (90%) create mode 100644 LICENSE-SAMPLECODE diff --git a/LICENSE.md b/LICENSE similarity index 90% rename from LICENSE.md rename to LICENSE index 3b7b82d..cbe5ad1 100644 --- a/LICENSE.md +++ b/LICENSE @@ -1,4 +1,4 @@ -Attribution-ShareAlike 4.0 International +Attribution-NonCommercial-ShareAlike 4.0 International ======================================================================= @@ -33,7 +33,7 @@ exhaustive, and do not form part of our licenses. material not subject to the license. This includes other CC- licensed material, or material used under an exception or limitation to copyright. More considerations for licensors: - wiki.creativecommons.org/Considerations_for_licensors + wiki.creativecommons.org/Considerations_for_licensors Considerations for the public: By using one of our public licenses, a licensor grants the public permission to use the @@ -48,24 +48,24 @@ exhaustive, and do not form part of our licenses. rights in the material. A licensor may make special requests, such as asking that all changes be marked or described. Although not required by our licenses, you are encouraged to - respect those requests where reasonable. More_considerations + respect those requests where reasonable. 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Sui Generis Database Rights means rights other than copyright resulting from Directive 96/9/EC of the European Parliament and of the Council of 11 March 1996 on the legal protection of databases, as amended and/or succeeded, as well as other essentially equivalent rights anywhere in the world. - m. You means the individual or entity exercising the Licensed Rights + n. You means the individual or entity exercising the Licensed Rights under this Public License. Your has a corresponding meaning. @@ -150,9 +158,10 @@ Section 2 -- Scope. exercise the Licensed Rights in the Licensed Material to: a. reproduce and Share the Licensed Material, in whole or - in part; and + in part, for NonCommercial purposes only; and - b. produce, reproduce, and Share Adapted Material. + b. produce, reproduce, and Share Adapted Material for + NonCommercial purposes only. 2. Exceptions and Limitations. For the avoidance of doubt, where Exceptions and Limitations apply to Your use, this Public @@ -220,7 +229,9 @@ Section 2 -- Scope. Rights, whether directly or through a collecting society under any voluntary or waivable statutory or compulsory licensing scheme. In all other cases the Licensor expressly - reserves any right to collect such royalties. + reserves any right to collect such royalties, including when + the Licensed Material is used other than for NonCommercial + purposes. Section 3 -- License Conditions. @@ -265,7 +276,6 @@ following conditions. reasonable to satisfy the conditions by providing a URI or hyperlink to a resource that includes the required information. - 3. If requested by the Licensor, You must remove any of the information required by Section 3(a)(1)(A) to the extent reasonably practicable. @@ -277,7 +287,7 @@ following conditions. 1. The Adapter's License You apply must be a Creative Commons license with the same License Elements, this version or - later, or a BY-SA Compatible License. + later, or a BY-NC-SA Compatible License. 2. You must include the text of, or the URI or hyperlink to, the Adapter's License You apply. You may satisfy this condition @@ -297,14 +307,15 @@ apply to Your use of the Licensed Material: a. for the avoidance of doubt, Section 2(a)(1) grants You the right to extract, reuse, reproduce, and Share all or a substantial - portion of the contents of the database; + portion of the contents of the database for NonCommercial purposes + only; b. if You include all or a substantial portion of the database contents in a database in which You have Sui Generis Database Rights, then the database in which You have Sui Generis Database Rights (but not its individual contents) is Adapted Material, - including for purposes of Section 3(b); and + c. You must comply with the conditions in Section 3(a) if You Share all or a substantial portion of the contents of the database. @@ -404,7 +415,6 @@ Section 8 -- Interpretation. that apply to the Licensor or You, including from the legal processes of any jurisdiction or authority. - ======================================================================= Creative Commons is not a party to its public diff --git a/LICENSE-SAMPLECODE b/LICENSE-SAMPLECODE new file mode 100644 index 0000000..12c4642 --- /dev/null +++ b/LICENSE-SAMPLECODE @@ -0,0 +1,21 @@ +MIT License + +Copyright (c) 2023 Krahets + +Permission is hereby granted, free of charge, to any person obtaining a copy +of this software and associated documentation files (the "Software"), to deal +in the Software without restriction, including without limitation the rights +to use, copy, modify, merge, publish, distribute, sublicense, and/or sell +copies of the Software, and to permit persons to whom the Software is +furnished to do so, subject to the following conditions: + +The above copyright notice and this permission notice shall be included in all +copies or substantial portions of the Software. + +THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR +IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, +FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE +AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER +LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, +OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE +SOFTWARE. diff --git a/README.md b/README.md index 8b48835..a6eab34 100755 --- a/README.md +++ b/README.md @@ -13,13 +13,15 @@ [《图解算法数据结构》](https://leetcode-cn.com/leetbook/detail/illustration-of-algorithm/)是一本面向算法初学者和互联网求职者编写的 LeetBook 手册。 -- 图文并茂地讲解《剑指 Offer》75 道题目，题库活跃于各大互联网公司笔面试。 -- 重点解析部分算法知识，包括复杂度分析、数据结构、排序算法和动态规划等。 -- 提供简洁的 Python, Java, C++ 解题代码和一个月刷题计划，让刷题有迹可循。 +- 图文详解 75 道题目，覆盖主要算法知识点，非常适合作为算法学习的第一份题库。 +- 题目活跃于各大互联网公司招聘中，可使笔面试准备事半功倍。 +- 致力于行文深入浅出、图文搭配，提供简洁的 Python, Java, C++ 解题代码。 > 若本 LeetBook 对您有所帮助，请在页面右上角点个 Star :star: 支持一下，谢谢！ ---- +## 如何入门算法？ + +### 第一步：看入门书

@@ -27,7 +29,7 @@

- 推荐算法初学者阅读《Hello 算法》，夯实基础后可大幅提升刷题效率 + 《Hello 算法》：动画图解、一键运行的数据结构与算法教程

@@ -40,123 +42,13 @@

---- - -本《剑指 Offer》刷题计划已被[力扣（LeetCode）官方题单](https://leetcode.cn/study-plan/lcof/)采用。 - -行百里者半九十，坚持一个月刷完，一起加油！ +### 第二步：刷算法题 -| 日程 | 题目 | -| :--------: | :------------------------------------------------ | -| **Day 1** | **栈与队列（简单）** | -| | [剑指 Offer 09. 用两个栈实现队列](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5d3i87/) | -| | [剑指 Offer 30. 包含 min 函数的栈](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/50bp33/) | -| **Day 2** | **链表（简单）** | -| | [剑指 Offer 06. 从尾到头打印链表](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5dt66m/) | -| | [剑指 Offer 24. 反转链表](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/9pdjbm/) | -| | [剑指 Offer 35. 复杂链表的复制](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/9p0yy1/) | -| **Day 3** | **字符串（简单）** | -| | [剑指 Offer 05. 替换空格](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/50ywkd/) | -| | [剑指 Offer 58 - II. 左旋转字符串](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/589fz2/) | -| **Day 4** | **查找算法（简单）** | -| | [剑指 Offer 03. 数组中重复的数字](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/59bjss/) | -| | [剑指 Offer 53 - I. 在排序数组中查找数字 I](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5874p1/) | -| | [剑指 Offer 53 - II. 0～n-1 中缺失的数字](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/58iqo5/) | -| **Day 5** | **查找算法（中等）** | -| | [剑指 Offer 04. 二维数组中的查找](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5v76yi/) | -| | [剑指 Offer 11. 旋转数组的最小数字](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/50xofm/) | -| | [剑指 Offer 50. 第一个只出现一次的字符](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5viisg/) | -| **Day 6** | **搜索与回溯算法（简单）** | -| | [剑指 Offer 32 - I. 从上到下打印二叉树](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/9ackoe/) | -| | [剑指 Offer 32 - II. 从上到下打印二叉树 II](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5vawr3/) | -| | [剑指 Offer 32 - III. 从上到下打印二叉树 III](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5vnp91/) | -| **Day 7** | **搜索与回溯算法（简单）** | -| | [剑指 Offer 26. 树的子结构](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5dshwe/) | -| | [剑指 Offer 27. 二叉树的镜像](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/59zt5i/) | -| | [剑指 Offer 28. 对称的二叉树](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5d412v/) | -| **Day 8** | **动态规划（简单）** | -| | [剑指 Offer 10- I. 斐波那契数列](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/50fxu1/) | -| | [剑指 Offer 10- II. 青蛙跳台阶问题](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/57hyl5/) | -| | [剑指 Offer 63. 股票的最大利润](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/58nn7r/) | -| **Day 9** | **动态规划（中等）** | -| | [剑指 Offer 42. 连续子数组的最大和](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/59gq9c/) | -| | [剑指 Offer 47. 礼物的最大价值](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5vokvr/) | -| **Day 10** | **动态规划（中等）** | -| | [剑指 Offer 46. 把数字翻译成字符串](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/99wd55/) | -| | [剑指 Offer 48. 最长不含重复字符的子字符串](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5dgr0c/) | -| **Day 11** | **双指针（简单）** | -| | [剑指 Offer 18. 删除链表的节点](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/505fc7/) | -| | [剑指 Offer 22. 链表中倒数第 k 个节点](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/58tl52/) | -| **Day 12** | **双指针（简单）** | -| | [剑指 Offer 25. 合并两个排序的链表](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5vq98s/) | -| | [剑指 Offer 52. 两个链表的第一个公共节点](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/oe5os3/) | -| **Day 13** | **双指针（简单）** | -| | [剑指 Offer 21. 调整数组顺序使奇数位于偶数前面](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5v8a6t/) | -| | [剑指 Offer 57. 和为 s 的两个数字](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5832fi/) | -| | [剑指 Offer 58 - I. 翻转单词顺序](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/586ecg/) | -| **Day 14** | **搜索与回溯算法（中等）** | -| | [剑指 Offer 12. 矩阵中的路径](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/58wowd/) | -| | [剑指 Offer 13. 机器人的运动范围](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/9h6vo2/) | -| **Day 15** | **搜索与回溯算法（中等）** | -| | [剑指 Offer 34. 二叉树中和为某一值的路径](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5dy6pt/) | -| | [剑指 Offer 36. 二叉搜索树与双向链表](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5dbies/) | -| | [剑指 Offer 54. 二叉搜索树的第 k 大节点](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/58df23/) | -| **Day 16** | **排序（简单）** | -| | [剑指 Offer 45. 把数组排成最小的数](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/59ypcj/) | -| | [剑指 Offer 61. 扑克牌中的顺子](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/57mpoj/) | -| **Day 17** | **排序（中等）** | -| | [剑指 Offer 40. 最小的 k 个数](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/ohvl0d/) | -| | [剑指 Offer 41. 数据流中的中位数](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5vd1j2/) | -| **Day 18** | **搜索与回溯算法（中等）** | -| | [剑指 Offer 55 - I. 二叉树的深度](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/9hgr5i/) | -| | [剑指 Offer 55 - II. 平衡二叉树](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/9hzffg/) | -| **Day 19** | **搜索与回溯算法（中等）** | -| | [剑指 Offer 64. 求 1 + 2 + … + n](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/9h44cj/) | -| | [剑指 Offer 68 - I. 二叉搜索树的最近公共祖先](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/575kd2/) | -| | [剑指 Offer 68 - II. 二叉树的最近公共祖先](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/57euni/) | -| **Day 20** | **分治算法（中等）** | -| | [剑指 Offer 07. 重建二叉树](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/99lxci/) | -| | [剑指 Offer 16. 数值的整数次方](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/57rwmg/) | -| | [剑指 Offer 33. 二叉搜索树的后序遍历序列](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5vwxx5/) | -| **Day 21** | **位运算（简单）** | -| | [剑指 Offer 15. 二进制中 1 的个数](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5vk1l3/) | -| | [剑指 Offer 65. 不用加减乘除做加法](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5vz6d1/) | -| **Day 22** | **位运算（中等）** | -| | [剑指 Offer 56 - I. 数组中数字出现的次数](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/eubbnm/) | -| | [剑指 Offer 56 - II. 数组中数字出现的次数 II](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/9hyq1r/) | -| **Day 23** | **数学（简单）** | -| | [剑指 Offer 39. 数组中出现次数超过一半的数字](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/99iy4g/) | -| | [剑指 Offer 66. 构建乘积数组](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/57d8cm/) | -| **Day 24** | **数学（中等）** | -| | [剑指 Offer 14- I. 剪绳子](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5v1026/) | -| | [剑指 Offer 57 - II. 和为 s 的连续正数序列](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/eufzm7/) | -| | [剑指 Offer 62. 圆圈中最后剩下的数字](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/oxrkot/) | -| **Day 25** | **模拟（中等）** | -| | [剑指 Offer 29. 顺时针打印矩阵](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5vfh9g/) | -| | [剑指 Offer 31. 栈的压入、弹出序列](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5wh1hj/) | -| **Day 26** | **字符串（中等）** | -| | [剑指 Offer 20. 表示数值的字符串](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5d6vi6/) | -| | [剑指 Offer 67. 把字符串转换成整数](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/58pq8g/) | -| **Day 27** | **栈与队列（困难）** | -| | [剑指 Offer 59 - I. 滑动窗口的最大值](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/58o46i/) | -| | [剑指 Offer 59 - II. 队列的最大值](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/e2t5ug/) | -| **Day 28** | **搜索与回溯算法（困难）** | -| | [剑指 Offer 37. 序列化二叉树](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/990pf2/) | -| | [剑指 Offer 38. 字符串的排列](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5dfv5h/) | -| **Day 29** | **动态规划（困难）** | -| | [剑指 Offer 19. 正则表达式匹配](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/9a1ypc/) | -| | [剑指 Offer 49. 丑数](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/9h3im5/) | -| | [剑指 Offer 60. n 个骰子的点数](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/ozzl1r/) | -| **Day 30** | **分治算法（困难）** | -| | [剑指 Offer 17. 打印从 1 到最大的 n 位数](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/594wfg/) | -| | [剑指 Offer 51. 数组中的逆序对](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/o58jfs/) | -| **Day 31** | **数学（困难）** | -| | [剑指 Offer 14- II. 剪绳子 II](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5vcapc/) | -| | [剑指 Offer 43. 1～n 整数中 1 出现的次数](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/572jxs/) | -| | [剑指 Offer 44. 数字序列中某一位的数字](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/57vzfh/) | - ---- +1. [Krahets 笔面试精选 88 题](https://leetcode.cn/studyplan/selected-coding-interview/)：由“剑指 Offer”和“Hot 100”组成的高频面试题单。 +2. [图解算法数据结构](https://leetcode.cn/studyplan/coding-interviews/)：题目更贴近实际应用，相对“剑指 Offer”难度有所增加。 ## License -This work is licensed under CC-BY-SA-4.0 license. +The texts and figures in this repository are licensed under [CC BY-NC-SA-4.0 license](https://creativecommons.org/licenses/by-nc-sa/4.0/). + +The sample code within this repository is made available under [MIT license](LICENSE-SAMPLECODE). From 6def47ca3f80efa15b77bd316d93f9b48abdde9f Mon Sep 17 00:00:00 2001 From: krahets Date: Tue, 10 Oct 2023 19:56:11 +0800 Subject: [PATCH 11/23] Update README --- LICENSE-SAMPLECODE | 21 --------------------- README.md | 8 +++----- 2 files changed, 3 insertions(+), 26 deletions(-) delete mode 100644 LICENSE-SAMPLECODE diff --git a/LICENSE-SAMPLECODE b/LICENSE-SAMPLECODE deleted file mode 100644 index 12c4642..0000000 --- a/LICENSE-SAMPLECODE +++ /dev/null @@ -1,21 +0,0 @@ -MIT License - -Copyright (c) 2023 Krahets - -Permission is hereby granted, free of charge, to any person obtaining a copy -of this software and associated documentation files (the "Software"), to deal -in the Software without restriction, including without limitation the rights -to use, copy, modify, merge, publish, distribute, sublicense, and/or sell -copies of the Software, and to permit persons to whom the Software is -furnished to do so, subject to the following conditions: - -The above copyright notice and this permission notice shall be included in all -copies or substantial portions of the Software. - -THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR -IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY, -FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE -AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER -LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM, -OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE -SOFTWARE. diff --git a/README.md b/README.md index a6eab34..de7f97e 100755 --- a/README.md +++ b/README.md @@ -40,15 +40,13 @@ 在线阅读 > -

+

### 第二步：刷算法题 1. [Krahets 笔面试精选 88 题](https://leetcode.cn/studyplan/selected-coding-interview/)：由“剑指 Offer”和“Hot 100”组成的高频面试题单。 -2. [图解算法数据结构](https://leetcode.cn/studyplan/coding-interviews/)：题目更贴近实际应用，相对“剑指 Offer”难度有所增加。 +2. [图解算法数据结构](https://leetcode-cn.com/leetbook/detail/illustration-of-algorithm/)或[Interview-75](https://leetcode.cn/studyplan/coding-interviews/)：题目更贴近实际应用，相对“剑指 Offer”难度有所增加。 ## License -The texts and figures in this repository are licensed under [CC BY-NC-SA-4.0 license](https://creativecommons.org/licenses/by-nc-sa/4.0/). - -The sample code within this repository is made available under [MIT license](LICENSE-SAMPLECODE). +The texts, codes and images in this repository are licensed under [CC BY-NC-SA-4.0](https://creativecommons.org/licenses/by-nc-sa/4.0/). From de1a505d63c04f8f84cede673a96824a25daacbb Mon Sep 17 00:00:00 2001 From: krahets Date: Tue, 10 Oct 2023 20:22:09 +0800 Subject: [PATCH 12/23] Add documents of leetbook IOA and selected coding interview. --- .../docs/# 0 \345\274\225\350\250\200.md" | 26 + ...67\351\242\230\345\273\272\350\256\256.md" | 14 + ...30\347\233\256\345\210\206\347\261\273.md" | 81 +++ ...23\346\236\204\347\256\200\344\273\213.md" | 629 ++++++++++++++++++ ...25\345\244\215\346\235\202\345\272\246.md" | 15 + ...64\345\244\215\346\235\202\345\272\246.md" | 514 ++++++++++++++ ...64\345\244\215\346\235\202\345\272\246.md" | 547 +++++++++++++++ ...43\351\242\230\346\241\206\346\236\266.md" | 482 ++++++++++++++ ...27\346\263\225\347\256\200\344\273\213.md" | 136 ++++ ...22\346\263\241\346\216\222\345\272\217.md" | 125 ++++ ...53\351\200\237\346\216\222\345\272\217.md" | 244 +++++++ ...22\345\271\266\346\216\222\345\272\217.md" | 130 ++++ ...07\344\273\266\345\211\257\346\234\254.md" | 140 ++++ ...14\347\273\264\346\225\260\347\273\204.md" | 72 ++ ...57\345\276\204\345\212\240\345\257\206.md" | 46 ++ ...\344\271\246\346\225\264\347\220\206 I.md" | 126 ++++ ...06\344\272\214\345\217\211\346\240\221.md" | 117 ++++ ...344\271\246\346\225\264\347\220\206 II.md" | 97 +++ ...42\351\202\243\345\245\221\346\225\260.md" | 97 +++ ...63\350\267\203\350\256\255\347\273\203.md" | 93 +++ ...\345\255\230\347\256\241\347\220\206 I.md" | 161 +++++ ...27\346\257\215\350\277\267\345\256\253.md" | 103 +++ ...43\346\251\261\346\225\264\347\220\206.md" | 242 +++++++ ...\347\240\215\347\253\271\345\255\220 I.md" | 164 +++++ ...347\240\215\347\253\271\345\255\220 II.md" | 166 +++++ ...1 \347\232\204\344\270\252\346\225\260.md" | 126 ++++ leetbook_ioa/docs/LCR 134. 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\346\227\213\350\275\254\345\255\227\347\254\246\344\270\262.md" create mode 100644 "selected_coding_interview/docs/8. \345\255\227\347\254\246\344\270\262\350\275\254\346\215\242\346\225\264\346\225\260 (atoi).md" create mode 100644 "selected_coding_interview/docs/86. \345\210\206\351\232\224\351\223\276\350\241\250.md" create mode 100644 "selected_coding_interview/docs/876. \351\223\276\350\241\250\347\232\204\344\270\255\351\227\264\347\273\223\347\202\271.md" create mode 100644 "selected_coding_interview/docs/89. \346\240\274\351\233\267\347\274\226\347\240\201.md" create mode 100644 "selected_coding_interview/docs/946. \351\252\214\350\257\201\346\240\210\345\272\217\345\210\227.md" diff --git "a/leetbook_ioa/docs/# 0 \345\274\225\350\250\200.md" "b/leetbook_ioa/docs/# 0 \345\274\225\350\250\200.md" new file mode 100755 index 0000000..b3f6ee9 --- /dev/null +++ "b/leetbook_ioa/docs/# 0 \345\274\225\350\250\200.md" @@ -0,0 +1,26 @@ +# 引言 + +《图解算法数据结构》面向算法初学者、互联网求职者设计，主要内容包括： + +### 精选面试题图文解析 + +- 图文详解 75 道题目，覆盖主要算法知识点，非常适合作为算法学习的 **第一份题库**。 +- 题库活跃于各大互联网公司招聘中，可使笔面试准备事半功倍。 +- 致力于行文深入浅出、图文搭配，提供简洁的 **Python, Java, C++** 解题代码。 +- 笔者整理了 **题目分类** 和 **刷题计划** ，让刷题有迹可循。 + +### 数据结构与算法专栏 + +- **基础知识：** 时间复杂度、空间复杂度等算法知识。 +- **数据结构：** 数组、链表、字符串、栈、队列、哈希表、树、图、堆。 +- **算法专题：** 搜索与回溯、分治、动态规划、贪心、排序、位运算、模拟、数学。 + +## 配套代码 + +为方便各位 Debug 算法题目，笔者整理了本 LeetBook 的配套代码，包括： + +- 「题解代码」提供 Python, Java, C++ 语言。 +- 「测试样例」与运行调用代码。 +- 「数据结构」封装，提升 LeetCode 刷题效率。 + +与本 LeetBook 配合食用更佳，仓库链接：https://github.com/krahets/LeetCode-Book diff --git "a/leetbook_ioa/docs/# 0.1 \345\210\267\351\242\230\345\273\272\350\256\256.md" "b/leetbook_ioa/docs/# 0.1 \345\210\267\351\242\230\345\273\272\350\256\256.md" new file mode 100755 index 0000000..a2e4c90 --- /dev/null +++ "b/leetbook_ioa/docs/# 0.1 \345\210\267\351\242\230\345\273\272\350\256\256.md" @@ -0,0 +1,14 @@ +# 刷题建议 + +本书专为算法初学者设计，特别针对有意进入互联网行业的求职者。内容覆盖如下主题： + +- **算法**：搜索、查找、排序、双指针、回溯、分治、动态规划、贪心、位运算、数学等。 +- **数据结构**：数组、栈、队列、字符串、链表、树、图、堆、哈希表等。 + +所有题目已经进行分类，并按照难易程度排序。对于初学者，这里提供几条刷题建议： + +1. 建议每日刷 2~3 题。若能轻松完成，可以尝试增加至 5～8 题，但请记住：刷题的质量远重要于数量。务必确保你真正理解了每个题目的解法及背后的算法原理。 +2. 建议你按照目录顺序逐题解答。如果碰到某些难以解决的题目，可以先跳过，稍后回顾时再挑战。 +3. 很多题目都有不止一种解法，请你注意比较和探讨各种方法的特点和适用情况。 +4. 如果你发现自己忘记了某个题目的解法，不必灰心。艾宾浩斯遗忘曲线指出，为了真正掌握一个知识点，通常需要复习至少3次。 +5. 行百里者半九十。坚持至关重要，加油，相信你可以做到！ diff --git "a/leetbook_ioa/docs/# 0.2 \351\242\230\347\233\256\345\210\206\347\261\273.md" "b/leetbook_ioa/docs/# 0.2 \351\242\230\347\233\256\345\210\206\347\261\273.md" new file mode 100755 index 0000000..3bfb3e9 --- /dev/null +++ "b/leetbook_ioa/docs/# 0.2 \351\242\230\347\233\256\345\210\206\347\261\273.md" @@ -0,0 +1,81 @@ +# 题目分类 + +题目可能存在多种解法，下表仅列举最优解法（时间与空间复杂度最低）的算法和数据结构分类。 + +| 题目 | 算法分类 | 数据结构分类 | +| -------------------------------- | ---------------- | -------------- | +| 寻找文件副本 | 查找 | 数组 | +| 寻找目标值 - 二维数组 | 查找 | 数组 | +| 路径加密 | | 字符串 | +| 图书整理 I | | 栈与队列，链表 | +| 推理二叉树 | 分治 | 树，哈希表 | +| 图书整理 II | | 栈与队列 | +| 斐波那契数 | 动态规划 | 数组 | +| 跳跃训练 | 动态规划 | 数组 | +| 库存管理 I | 查找 | 数组 | +| 字母迷宫 | 回溯，搜索 | 数组，图 | +| 衣橱整理 | 回溯，搜索 | 数组，图 | +| 砍竹子 I | 贪心，数学 | | +| 砍竹子 II | 贪心，分治，数学 | | +| 位 1 的个数 | 位运算 | | +| Pow(x, n) | 分治，位运算 | | +| 报数 | | 数组 | +| 删除链表节点 | 双指针 | 链表 | +| 模糊搜索验证 | 动态规划 | 字符串 | +| 有效数字 | | 字符串 | +| 训练计划 I | 双指针 | 数组 | +| 训练计划 II | 双指针 | 链表 | +| 训练计划 III | 双指针 | 链表 | +| 训练计划 IV | 双指针 | 链表 | +| 子结构判断 | 搜索 | 树 | +| 翻转二叉树 | 搜索 | 栈与队列，树 | +| 判断对称二叉树 | 搜索 | 树 | +| 螺旋遍历二维数组 | 模拟 | 数组 | +| 最小栈 | 排序 | 栈与队列 | +| 验证图书取出顺序 | 模拟 | 栈与队列 | +| 彩灯装饰记录 I | 搜索 | 栈与队列，树 | +| 彩灯装饰记录 II | 搜索 | 栈与队列，树 | +| 彩灯装饰记录 III | 搜索 | 栈与队列，树 | +| 验证二叉搜索树的后序遍历序列 | 分治 | 栈与队列，树 | +| 二叉树中和为目标值的路径 | 回溯，搜索 | 树 | +| 随机链表的复制 | | 链表 | +| 将二叉搜索树转化为排序的双向链表 | 搜索，双指针 | 树 | +| 序列化与反序列化二叉树 | 搜索 | 树 | +| 套餐内商品的排列顺序 | 回溯 | 字符串，哈希表 | +| 库存管理 II | | 数组 | +| 库存管理 III | 排序 | 数组，堆 | +| 数据流中的中位数 | 排序 | 堆 | +| 连续天数的最高销售额 | 动态规划 | 数组 | +| 数字 1 的个数 | 数学 | | +| 找到第 k 位数字 | 数学 | | +| 破解闯关密码 | 排序 | 字符串 | +| 解密数字 | 动态规划 | 字符串 | +| 珠宝的最高价值 | 动态规划 | 数组 | +| 招式拆解 I | 动态规划，双指针 | 哈希表 | +| 丑数 | 动态规划 | | +| 招式拆解 II | | 哈希表 | +| 交易逆序对的总数 | 分治 | 数组 | +| 训练计划 V | 双指针 | 链表 | +| 统计目标成绩的出现次数 | 查找 | 数组 | +| 点名 | 查找 | 数组 | +| 寻找二叉搜索树中的目标节点 | 搜索 | 树 | +| 计算二叉树的深度 | 搜索 | 树 | +| 判断是否为平衡二叉树 | 搜索 | 树 | +| 撞色搭配 | 位运算 | 数组 | +| 训练计划 VI | 位运算 | 数组 | +| 查找总价格为目标值的两个商品 | 双指针 | 数组 | +| 文件组合 | 双指针 | 数组 | +| 字符串中的单词反转 | 双指针 | 字符串 | +| 动态口令 | | 字符串 | +| 望远镜中最高的海拔 | 排序 | 数组，栈与队列 | +| 设计自助结算系统 | 排序 | 数组，栈与队列 | +| 统计结果概率 | 动态规划 | | +| 文物朝代判断 | 排序 | 数组，哈希表 | +| 破冰游戏 | 数学 | | +| 买卖芯片的最佳时机 | 动态规划 | 数组 | +| 设计机械累加器 | | | +| 加密运算 | 位运算 | | +| 按规则计算统计结果 | 数学 | 数组 | +| 不使用库函数的字符串转整数 | | 字符串 | +| 求二叉搜索树的最近公共祖先 | 搜索 | 树 | +| 寻找二叉树的最近公共祖先 | 搜索 | 树 | diff --git "a/leetbook_ioa/docs/# 1.1 \346\225\260\346\215\256\347\273\223\346\236\204\347\256\200\344\273\213.md" "b/leetbook_ioa/docs/# 1.1 \346\225\260\346\215\256\347\273\223\346\236\204\347\256\200\344\273\213.md" new file mode 100755 index 0000000..9b4e077 --- /dev/null +++ "b/leetbook_ioa/docs/# 1.1 \346\225\260\346\215\256\347\273\223\346\236\204\347\256\200\344\273\213.md" @@ -0,0 +1,629 @@ +# 数据结构简介 + +数据结构是为实现对计算机数据有效使用的各种数据组织形式，服务于各类计算机操作。不同的数据结构具有各自对应的适用场景，旨在降低各种算法计算的时间与空间复杂度，达到最佳的任务执行效率。 + +如下图所示，常见的数据结构可分为「线性数据结构」与「非线性数据结构」，具体为：「数组」、「链表」、「栈」、「队列」、「树」、「图」、「散列表」、「堆」。 + +![Picture1.png](https://pic.leetcode-cn.com/1599638810-SZDwfK-Picture1.png){:width=500} + +从零开始学习算法的同学对数据结构的使用方法可能尚不熟悉，本节将初步介绍各数据结构的基本特点，与 Python3 , Java , C++ 语言中各数据结构的初始化与构建方法。 + +> 代码运行可使用本地 IDE 或 [力扣 PlayGround](https://leetcode-cn.com/playground/) 。 + +--- + +## 数组 + +数组是将相同类型的元素存储于连续内存空间的数据结构，其长度不可变。 + +如下图所示，构建此数组需要在初始化时给定长度，并对数组每个索引元素赋值，代码如下： + +```Java [] +// 初始化一个长度为 5 的数组 array +int[] array = new int[5]; +// 元素赋值 +array[0] = 2; +array[1] = 3; +array[2] = 1; +array[3] = 0; +array[4] = 2; +``` + +```C++ [] +// 初始化一个长度为 5 的数组 array +int array[5]; +// 元素赋值 +array[0] = 2; +array[1] = 3; +array[2] = 1; +array[3] = 0; +array[4] = 2; +``` + +或者可以使用直接赋值的初始化方式，代码如下： + +```Java [] +int[] array = {2, 3, 1, 0, 2}; +``` + +```C++ [] +int array[] = {2, 3, 1, 0, 2}; +``` + +![Picture2.png](https://pic.leetcode-cn.com/1599587176-JAxwpf-Picture2.png){:width=500} + +「可变数组」是经常使用的数据结构，其基于数组和扩容机制实现，相比普通数组更加灵活。常用操作有：访问元素、添加元素、删除元素。 + +```Java [] +// 初始化可变数组 +List array = new ArrayList<>(); + +// 向尾部添加元素 +array.add(2); +array.add(3); +array.add(1); +array.add(0); +array.add(2); +``` + +```Python [] +# 初始化可变数组 +array = [] + +# 向尾部添加元素 +array.append(2) +array.append(3) +array.append(1) +array.append(0) +array.append(2) +``` + +```C++ [] +// 初始化可变数组 +vector array; + +// 向尾部添加元素 +array.push_back(2); +array.push_back(3); +array.push_back(1); +array.push_back(0); +array.push_back(2); +``` + +--- + +## 链表 + +链表以节点为单位，每个元素都是一个独立对象，在内存空间的存储是非连续的。链表的节点对象具有两个成员变量：「值 `val`」，「后继节点引用 `next`」 。 + +```Java [] +class ListNode { + int val; // 节点值 + ListNode next; // 后继节点引用 + ListNode(int x) { val = x; } +} +``` + +```Python [] +class ListNode: + def __init__(self, x): + self.val = x # 节点值 + self.next = None # 后继节点引用 +``` + +```C++ [] +struct ListNode { + int val; // 节点值 + ListNode *next; // 后继节点引用 + ListNode(int x) : val(x), next(NULL) {} +}; +``` + +如下图所示，建立此链表需要实例化每个节点，并构建各节点的引用指向。 + +```Java [] +// 实例化节点 +ListNode n1 = new ListNode(4); // 节点 head +ListNode n2 = new ListNode(5); +ListNode n3 = new ListNode(1); + +// 构建引用指向 +n1.next = n2; +n2.next = n3; +``` + +```Python [] +# 实例化节点 +n1 = ListNode(4) # 节点 head +n2 = ListNode(5) +n3 = ListNode(1) + +# 构建引用指向 +n1.next = n2 +n2.next = n3 +``` + +```C++ [] +// 实例化节点 +ListNode *n1 = new ListNode(4); // 节点 head +ListNode *n2 = new ListNode(5); +ListNode *n3 = new ListNode(1); + +// 构建引用指向 +n1->next = n2; +n2->next = n3; +``` + +![Picture3.png](https://pic.leetcode-cn.com/1599578767-zgLjYw-Picture3.png){:width=500} + +--- + +## 栈 + +栈是一种具有 「先入后出」 特点的抽象数据结构，可使用数组或链表实现。 + +```Java [] +Stack stack = new Stack<>(); +``` + +```Python [] +stack = [] # Python 可将列表作为栈使用 +``` + +```C++ [] +stack stk; +``` + +如下图所示，通过常用操作「入栈 `push()`」,「出栈 `pop()`」，展示了栈的先入后出特性。 + +```Java [] +stack.push(1); // 元素 1 入栈 +stack.push(2); // 元素 2 入栈 +stack.pop(); // 出栈 -> 元素 2 +stack.pop(); // 出栈 -> 元素 1 +``` + +```Python [] +stack.append(1) # 元素 1 入栈 +stack.append(2) # 元素 2 入栈 +stack.pop() # 出栈 -> 元素 2 +stack.pop() # 出栈 -> 元素 1 +``` + +```C++ [] +stk.push(1); // 元素 1 入栈 +stk.push(2); // 元素 2 入栈 +stk.pop(); // 出栈 -> 元素 2 +stk.pop(); // 出栈 -> 元素 1 +``` + +![Picture4.png](https://pic.leetcode-cn.com/1599578767-ZifMEX-Picture4.png){:width=500} + +> 注意：通常情况下，不推荐使用 Java 的 `Vector` 以及其子类 `Stack` ，而一般将 `LinkedList` 作为栈来使用。详细说明请见：[Stack，ArrayDeque，LinkedList 的区别](https://blog.csdn.net/cartoon_/article/details/87992743) 。 + +```Java [] +LinkedList stack = new LinkedList<>(); + +stack.addLast(1); // 元素 1 入栈 +stack.addLast(2); // 元素 2 入栈 +stack.removeLast(); // 出栈 -> 元素 2 +stack.removeLast(); // 出栈 -> 元素 1 +``` + +--- + +## 队列 + +队列是一种具有 「先入先出」 特点的抽象数据结构，可使用链表实现。 + +```Java [] +Queue queue = new LinkedList<>(); +``` + +```Python [] +# Python 通常使用双端队列 collections.deque +from collections import deque + +queue = deque() +``` + +```C++ [] +queue que; +``` + +如下图所示，通过常用操作「入队 `push()`」,「出队 `pop()`」，展示了队列的先入先出特性。 + +```Java [] +queue.offer(1); // 元素 1 入队 +queue.offer(2); // 元素 2 入队 +queue.poll(); // 出队 -> 元素 1 +queue.poll(); // 出队 -> 元素 2 +``` + +```Python [] +queue.append(1) # 元素 1 入队 +queue.append(2) # 元素 2 入队 +queue.popleft() # 出队 -> 元素 1 +queue.popleft() # 出队 -> 元素 2 +``` + +```C++ [] +que.push(1); // 元素 1 入队 +que.push(2); // 元素 2 入队 +que.pop(); // 出队 -> 元素 1 +que.pop(); // 出队 -> 元素 2 +``` + +![Picture5.png](https://pic.leetcode-cn.com/1599588416-Majmwh-Picture5.png){:width=500} + +--- + +## 树 + +树是一种非线性数据结构，根据子节点数量可分为 「二叉树」 和 「多叉树」，最顶层的节点称为「根节点 `root`」。以二叉树为例，每个节点包含三个成员变量：「值 `val`」、「左子节点 `left`」、「右子节点 `right`」 。 + +```Java [] +class TreeNode { + int val; // 节点值 + TreeNode left; // 左子节点 + TreeNode right; // 右子节点 + TreeNode(int x) { val = x; } +} +``` + +```Python [] +class TreeNode: + def __init__(self, x): + self.val = x # 节点值 + self.left = None # 左子节点 + self.right = None # 右子节点 +``` + +```C++ [] +struct TreeNode { + int val; // 节点值 + TreeNode *left; // 左子节点 + TreeNode *right; // 右子节点 + TreeNode(int x) : val(x), left(NULL), right(NULL) {} +}; +``` + +如下图所示，建立此二叉树需要实例化每个节点，并构建各节点的引用指向。 + +```Java [] +// 初始化节点 +TreeNode n1 = new TreeNode(3); // 根节点 root +TreeNode n2 = new TreeNode(4); +TreeNode n3 = new TreeNode(5); +TreeNode n4 = new TreeNode(1); +TreeNode n5 = new TreeNode(2); + +// 构建引用指向 +n1.left = n2; +n1.right = n3; +n2.left = n4; +n2.right = n5; +``` + +```Python [] +# 初始化节点 +n1 = TreeNode(3) # 根节点 root +n2 = TreeNode(4) +n3 = TreeNode(5) +n4 = TreeNode(1) +n5 = TreeNode(2) + +# 构建引用指向 +n1.left = n2 +n1.right = n3 +n2.left = n4 +n2.right = n5 +``` + +```C++ [] +// 初始化节点 +TreeNode *n1 = new TreeNode(3); // 根节点 root +TreeNode *n2 = new TreeNode(4); +TreeNode *n3 = new TreeNode(5); +TreeNode *n4 = new TreeNode(1); +TreeNode *n5 = new TreeNode(2); + +// 构建引用指向 +n1->left = n2; +n1->right = n3; +n2->left = n4; +n2->right = n5; +``` + +![Picture6.png](https://pic.leetcode-cn.com/1599579136-bBARpC-Picture6.png){:width=500} + +--- + +## 图 + +图是一种非线性数据结构，由「节点（顶点）`vertex`」和「边 `edge`」组成，每条边连接一对顶点。根据边的方向有无，图可分为「有向图」和「无向图」。本文 **以无向图为例** 开展介绍。 + +如下图所示，此无向图的 **顶点** 和 **边** 集合分别为： + +- 顶点集合： `vertices = {1, 2, 3, 4, 5}` +- 边集合： `edges = {(1, 2), (1, 3), (1, 4), (1, 5), (2, 4), (3, 5), (4, 5)}` + +![Picture7.png](https://pic.leetcode-cn.com/1599579136-Fxseew-Picture7.png){:width=500} + +表示图的方法通常有两种： + +1. **邻接矩阵：** 使用数组 $vertices$ 存储顶点，邻接矩阵 $edges$ 存储边； $edges[i][j]$ 代表节点 $i + 1$ 和 节点 $j + 1$ 之间是否有边。 + +$$ +vertices = [1, 2, 3, 4, 5] \\ + +edges = \left[ \begin{matrix} 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 1 \\ 1 & 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 1 & 0 \\ \end{matrix} \right] +$$ + +```Python [] +vertices = [1, 2, 3, 4, 5] +edges = [[0, 1, 1, 1, 1], + [1, 0, 0, 1, 0], + [1, 0, 0, 0, 1], + [1, 1, 0, 0, 1], + [1, 0, 1, 1, 0]] +``` + +```Java [] +int[] vertices = {1, 2, 3, 4, 5}; +int[][] edges = {{0, 1, 1, 1, 1}, + {1, 0, 0, 1, 0}, + {1, 0, 0, 0, 1}, + {1, 1, 0, 0, 1}, + {1, 0, 1, 1, 0}}; +``` + +```C++ [] +int vertices[5] = {1, 2, 3, 4, 5}; +int edges[5][5] = {{0, 1, 1, 1, 1}, + {1, 0, 0, 1, 0}, + {1, 0, 0, 0, 1}, + {1, 1, 0, 0, 1}, + {1, 0, 1, 1, 0}}; +``` + +2. **邻接表：** 使用数组 $vertices$ 存储顶点，邻接表 $edges$ 存储边。 $edges$ 为一个二维容器，第一维 $i$ 代表顶点索引，第二维 $edges[i]$ 存储此顶点对应的边集和；例如 $edges[0] = [1, 2, 3, 4]$ 代表 $vertices[0]$ 的边集合为 $[1, 2, 3, 4]$ 。 + +$$ +vertices = [1, 2, 3, 4, 5] \\ + +edges = \left[ \begin{matrix} [ & 1 & 2 & 3 & 4 & ] \\ [ & 0 & 3 & ] \\ [ & 0 & 4 & ] \\ [ & 0 & 1 & 4 & ] \\ [ & 0 & 2 & 3 & ] \end{matrix} \right] +$$ + +```Python [] +vertices = [1, 2, 3, 4, 5] +edges = [[1, 2, 3, 4], + [0, 3], + [0, 4], + [0, 1, 4], + [0, 2, 3]] +``` + +```Java [] +int[] vertices = {1, 2, 3, 4, 5}; +List> edges = new ArrayList<>(); + +List edge_1 = new ArrayList<>(Arrays.asList(1, 2, 3, 4)); +List edge_2 = new ArrayList<>(Arrays.asList(0, 3)); +List edge_3 = new ArrayList<>(Arrays.asList(0, 4)); +List edge_4 = new ArrayList<>(Arrays.asList(0, 1, 4)); +List edge_5 = new ArrayList<>(Arrays.asList(0, 2, 3)); +edges.add(edge_1); +edges.add(edge_2); +edges.add(edge_3); +edges.add(edge_4); +edges.add(edge_5); +``` + +```C++ [] +int vertices[5] = {1, 2, 3, 4, 5}; +vector> edges; + +vector edge_1 = {1, 2, 3, 4}; +vector edge_2 = {0, 3}; +vector edge_3 = {0, 4}; +vector edge_4 = {0, 1, 4}; +vector edge_5 = {0, 2, 3}; +edges.push_back(edge_1); +edges.push_back(edge_2); +edges.push_back(edge_3); +edges.push_back(edge_4); +edges.push_back(edge_5); +``` + +> **邻接矩阵 VS 邻接表 ：** +> +> 邻接矩阵的大小只与节点数量有关，即 $N^2$ ，其中 $N$ 为节点数量。因此，当边数量明显少于节点数量时，使用邻接矩阵存储图会造成较大的内存浪费。 +> 因此，**邻接表** 适合存储稀疏图（顶点较多、边较少）； **邻接矩阵** 适合存储稠密图（顶点较少、边较多）。 + +--- + +## 散列表 + +散列表是一种非线性数据结构，通过利用 Hash 函数将指定的「键 `key`」映射至对应的「值 `value`」，以实现高效的元素查找。 + +> 设想一个简单场景：小力、小特、小扣的学号分别为 10001, 10002, 10003 。 +> 现需求从「姓名」查找「学号」。 + +则可通过建立姓名为 `key` ，学号为 `value` 的散列表实现此需求，代码如下： + +```Java [] +// 初始化散列表 +Map dic = new HashMap<>(); + +// 添加 key -> value 键值对 +dic.put("小力", 10001); +dic.put("小特", 10002); +dic.put("小扣", 10003); + +// 从姓名查找学号 +dic.get("小力"); // -> 10001 +dic.get("小特"); // -> 10002 +dic.get("小扣"); // -> 10003 +``` + +```Python [] +# 初始化散列表 +dic = {} + +# 添加 key -> value 键值对 +dic["小力"] = 10001 +dic["小特"] = 10002 +dic["小扣"] = 10003 + +# 从姓名查找学号 +dic["小力"] # -> 10001 +dic["小特"] # -> 10002 +dic["小扣"] # -> 10003 +``` + +```C++ [] +// 初始化散列表 +unordered_map dic; + +// 添加 key -> value 键值对 +dic["小力"] = 10001; +dic["小特"] = 10002; +dic["小扣"] = 10003; + +// 从姓名查找学号 +dic.find("小力")->second; // -> 10001 +dic.find("小特")->second; // -> 10002 +dic.find("小扣")->second; // -> 10003 +``` + +![Picture8.png](https://pic.leetcode-cn.com/1599811794-ruXMOV-Picture8.png){:width=550} + +### Hash 函数设计示例 ： + +> 假设需求：从「学号」查找「姓名」。 + +将三人的姓名存储至以下数组中，则各姓名在数组中的索引分别为 0, 1, 2 。 + +```Java [] +String[] names = { "小力", "小特", "小扣" }; +``` + +```Python [] +names = [ "小力", "小特", "小扣" ] +``` + +```C++ [] +string names[] = { "小力", "小特", "小扣" }; +``` + +此时，我们构造一个简单的 Hash 函数（ $\%$ 为取余符号 ），公式和封装函数如下所示： + +$$ +hash(key) = (key - 1) \% 10000 +$$ + +```Java [] +int hash(int id) { + int index = (id - 1) % 10000; + return index; +} +``` + +```Python [] +def hash(id): + index = (id - 1) % 10000 + return index +``` + +```C++ [] +int hash(int id) { + int index = (id - 1) % 10000; + return index; +} +``` + +则我们构建了以学号为 `key` 、姓名对应的数组索引为 `value` 的散列表。利用此 Hash 函数，则可在 $O(1)$ 时间复杂度下通过学号查找到对应姓名，即： + +```Java +names[hash(10001)] // 小力 +names[hash(10002)] // 小特 +names[hash(10003)] // 小扣 +``` + +![Picture8-1.png](https://pic.leetcode-cn.com/1599811794-NfbpfW-Picture8-1.png){:width=550} + +以上设计只适用于此示例，实际的 Hash 函数需保证低碰撞率、 高鲁棒性等，以适用于各类数据和场景。 + +--- + +## 堆： + +堆是一种基于「完全二叉树」的数据结构，可使用数组实现。以堆为原理的排序算法称为「堆排序」，基于堆实现的数据结构为「优先队列」。堆分为「大顶堆」和「小顶堆」，大（小）顶堆：任意节点的值不大于（小于）其父节点的值。 + +> **完全二叉树定义：** 设二叉树深度为 $k$ ，若二叉树除第 $k$ 层外的其它各层（第 $1$ 至 $k-1$ 层）的节点达到最大个数，且处于第 $k$ 层的节点都连续集中在最左边，则称此二叉树为完全二叉树。 + +如下图所示，为包含 `1, 4, 2, 6, 8` 元素的小顶堆。将堆（完全二叉树）中的结点按层编号，即可映射到右边的数组存储形式。 + +![Picture9.png](https://pic.leetcode-cn.com/1599584901-xoiGEQ-Picture9.png){:width=550} + +通过使用「优先队列」的「压入 `push()`」和「弹出 `pop()`」操作，即可完成堆排序，实现代码如下： + +```Java [] +// 初始化小顶堆 +Queue heap = new PriorityQueue<>(); + +// 元素入堆 +heap.add(1); +heap.add(4); +heap.add(2); +heap.add(6); +heap.add(8); + +// 元素出堆（从小到大） +heap.poll(); // -> 1 +heap.poll(); // -> 2 +heap.poll(); // -> 4 +heap.poll(); // -> 6 +heap.poll(); // -> 8 +``` + +```Python [] +from heapq import heappush, heappop + +# 初始化小顶堆 +heap = [] + +# 元素入堆 +heappush(heap, 1) +heappush(heap, 4) +heappush(heap, 2) +heappush(heap, 6) +heappush(heap, 8) + +# 元素出堆（从小到大） +heappop(heap) # -> 1 +heappop(heap) # -> 2 +heappop(heap) # -> 4 +heappop(heap) # -> 6 +heappop(heap) # -> 8 +``` + +```C++ [] +// 初始化小顶堆 +priority_queue, greater> heap; + +// 元素入堆 +heap.push(1); +heap.push(4); +heap.push(2); +heap.push(6); +heap.push(8); + +// 元素出堆（从小到大） +heap.pop(); // -> 1 +heap.pop(); // -> 2 +heap.pop(); // -> 4 +heap.pop(); // -> 6 +heap.pop(); // -> 8 +``` diff --git "a/leetbook_ioa/docs/# 1.2 \347\256\227\346\263\225\345\244\215\346\235\202\345\272\246.md" "b/leetbook_ioa/docs/# 1.2 \347\256\227\346\263\225\345\244\215\346\235\202\345\272\246.md" new file mode 100755 index 0000000..8e3bc55 --- /dev/null +++ "b/leetbook_ioa/docs/# 1.2 \347\256\227\346\263\225\345\244\215\346\235\202\345\272\246.md" @@ -0,0 +1,15 @@ +# 算法复杂度 + +算法复杂度旨在计算在输入数据量 $N$ 的情况下，算法的「时间使用」和「空间使用」情况；体现算法运行使用的时间和空间随「数据大小 $N$ 」而增大的速度。 + +算法复杂度主要可从 **时间** 、**空间** 两个角度评价： + +- **时间：** 假设各操作的运行时间为固定常数，统计算法运行的「计算操作的数量」 ，以代表算法运行所需时间； +- **空间：** 统计在最差情况下，算法运行所需使用的「最大空间」； + +「输入数据大小 $N$ 」指算法处理的输入数据量；根据不同算法，具有不同定义，例如： + +- **排序算法：** $N$ 代表需要排序的元素数量； +- **搜索算法：** $N$ 代表搜索范围的元素总数，例如数组大小、矩阵大小、二叉树节点数、图节点和边数等； + +接下来，我们将分别从概念定义、符号表示、常见种类、时空权衡、示例解析、示例题目等角度入手，学习「时间复杂度」和「空间复杂度」。 diff --git "a/leetbook_ioa/docs/# 1.3 \346\227\266\351\227\264\345\244\215\346\235\202\345\272\246.md" "b/leetbook_ioa/docs/# 1.3 \346\227\266\351\227\264\345\244\215\346\235\202\345\272\246.md" new file mode 100755 index 0000000..fd3a46f --- /dev/null +++ "b/leetbook_ioa/docs/# 1.3 \346\227\266\351\227\264\345\244\215\346\235\202\345\272\246.md" @@ -0,0 +1,514 @@ +# 时间复杂度 + +根据定义，时间复杂度指输入数据大小为 $N$ 时，算法运行所需花费的时间。需要注意： + +- 统计的是算法的「计算操作数量」，而不是「运行的绝对时间」。计算操作数量和运行绝对时间呈正相关关系，并不相等。算法运行时间受到「编程语言 、计算机处理器速度、运行环境」等多种因素影响。例如，同样的算法使用 Python 或 C++ 实现、使用 CPU 或 GPU 、使用本地 IDE 或力扣平台提交，运行时间都不同。 +- 体现的是计算操作随数据大小 $N$ 变化时的变化情况。假设算法运行总共需要「 $1$ 次操作」、「 $100$ 次操作」，此两情况的时间复杂度都为常数级 $O(1)$ ；需要「 $N$ 次操作」、「 $100N$ 次操作」的时间复杂度都为 $O(N)$ 。 + +--- + +## 符号表示 + +根据输入数据的特点，时间复杂度具有「最差」、「平均」、「最佳」三种情况，分别使用 $O$ , $\Theta$ , $\Omega$ 三种符号表示。以下借助一个查找算法的示例题目帮助理解。 + +> **题目：** 输入长度为 $N$ 的整数数组 `nums` ，判断此数组中是否有数字 $7$ ，若有则返回 `true` ，否则返回 $\text{false}$ 。 +> +> **解题算法：** 线性查找，即遍历整个数组，遇到 $7$ 则返回 `true` 。 +> +> **代码：** +> +> ```Python [] +> def find_seven(nums): +> for num in nums: +> if num == 7: +> return True +> return False +> ``` +> +> ```Java [] +> boolean findSeven(int[] nums) { +> for (int num : nums) { +> if (num == 7) +> return true; +> } +> return false; +> } +> ``` +> +> ```C++ [] +> bool findSeven(vector& nums) { +> for (int num : nums) { +> if (num == 7) +> return true; +> } +> return false; +> } +> ``` + +- **最佳情况 $\Omega(1)$ ：** `nums = [7, a, b, c, ...]` ，即当数组首个数字为 $7$ 时，无论 `nums` 有多少元素，线性查找的循环次数都为 $1$ 次； +- **最差情况 $O(N)$ ：** `nums = [a, b, c, ...]` 且 `nums` 中所有数字都不为 $7$ ，此时线性查找会遍历整个数组，循环 $N$ 次； +- **平均情况 $\Theta$ ：** 需要考虑输入数据的分布情况，计算所有数据情况下的平均时间复杂度；例如本题目，需要考虑数组长度、数组元素的取值范围等； + +> 大 $O$ 是最常使用的时间复杂度评价渐进符号，下文示例与本 LeetBook 题目解析皆使用 $O$ 。 + +--- + +## 常见种类 + +根据从小到大排列，常见的算法时间复杂度主要有： + +$$ +O(1) < O(\log N) < O(N) < O(N\log N) < O(N^2) < O(2^N) < O(N!) +$$ + +![Picture1.png](https://pic.leetcode-cn.com/1623519242-UTNefQ-Picture1.png) + +对于以下所有示例，设输入数据大小为 $N$ ，计算操作数量为 $count$ 。图中每个「**蓝色方块**」代表一个单元计算操作。 + +### 常数 $O(1)$ ： + +运行次数与 $N$ 大小呈常数关系，即不随输入数据大小 $N$ 的变化而变化。 + +```Python [] +def algorithm(N): + a = 1 + b = 2 + x = a * b + N + return 1 +``` + +```Java [] +int algorithm(int N) { + int a = 1; + int b = 2; + int x = a * b + N; + return 1; +} +``` + +```C++ [] +int algorithm(int N) { + int a = 1; + int b = 2; + int x = a * b + N; + return 1; +} +``` + +对于以下代码，无论 $a$ 取多大，都与输入数据大小 $N$ 无关，因此时间复杂度仍为 $O(1)$ 。 + +```Python [] +def algorithm(N): + count = 0 + a = 10000 + for i in range(a): + count += 1 + return count +``` + +```Java [] +int algorithm(int N) { + int count = 0; + int a = 10000; + for (int i = 0; i < a; i++) { + count++; + } + return count; +} +``` + +```C++ [] +int algorithm(int N) { + int count = 0; + int a = 10000; + for (int i = 0; i < a; i++) { + count++; + } + return count; +} +``` + +![Picture2.png](https://pic.leetcode-cn.com/1623779241-lViysV-Picture2.png){:width=500} + +### 线性 $O(N)$ ： + +循环运行次数与 $N$ 大小呈线性关系，时间复杂度为 $O(N)$ 。 + +```Python [] +def algorithm(N): + count = 0 + for i in range(N): + count += 1 + return count +``` + +```Java [] +int algorithm(int N) { + int count = 0; + for (int i = 0; i < N; i++) + count++; + return count; +} +``` + +```C++ [] +int algorithm(int N) { + int count = 0; + for (int i = 0; i < N; i++) + count++; + return count; +} +``` + +对于以下代码，虽然是两层循环，但第二层与 $N$ 大小无关，因此整体仍与 $N$ 呈线性关系。 + +```Python [] +def algorithm(N): + count = 0 + a = 10000 + for i in range(N): + for j in range(a): + count += 1 + return count +``` + +```Java [] +int algorithm(int N) { + int count = 0; + int a = 10000; + for (int i = 0; i < N; i++) { + for (int j = 0; j < a; j++) { + count++; + } + } + return count; +} +``` + +```C++ [] +int algorithm(int N) { + int count = 0; + int a = 10000; + for (int i = 0; i < N; i++) { + for (int j = 0; j < a; j++) { + count++; + } + } + return count; +} +``` + +![Picture3.png](https://pic.leetcode-cn.com/1623519242-AhnqvJ-Picture3.png){:width=500} + +### 平方 $O(N^2)$ ： + +两层循环相互独立，都与 $N$ 呈线性关系，因此总体与 $N$ 呈平方关系，时间复杂度为 $O(N^2)$ 。 + +```Python [] +def algorithm(N): + count = 0 + for i in range(N): + for j in range(N): + count += 1 + return count +``` + +```Java [] +int algorithm(int N) { + int count = 0; + for (int i = 0; i < N; i++) { + for (int j = 0; j < N; j++) { + count++; + } + } + return count; +} +``` + +```C++ [] +int algorithm(int N) { + int count = 0; + for (int i = 0; i < N; i++) { + for (int j = 0; j < N; j++) { + count++; + } + } + return count; +} +``` + +以「冒泡排序」为例，其包含两层独立循环： + +1. 第一层复杂度为 $O(N)$ ； +2. 第二层平均循环次数为 $\frac{N}{2}$ ，复杂度为 $O(N)$ ，推导过程如下： + +$$ +O(\frac{N}{2}) = O(\frac{1}{2})O(N) = O(1)O(N) = O(N) +$$ + +因此，冒泡排序的总体时间复杂度为 $O(N^2)$ ，代码如下所示。 + +```Python [] +def bubble_sort(nums): + N = len(nums) + for i in range(N - 1): + for j in range(N - 1 - i): + if nums[j] > nums[j + 1]: + nums[j], nums[j + 1] = nums[j + 1], nums[j] + return nums +``` + +```Java [] +int[] bubbleSort(int[] nums) { + int N = nums.length; + for (int i = 0; i < N - 1; i++) { + for (int j = 0; j < N - 1 - i; j++) { + if (nums[j] > nums[j + 1]) { + int tmp = nums[j]; + nums[j] = nums[j + 1]; + nums[j + 1] = tmp; + } + } + } + return nums; +} +``` + +```C++ [] +vector bubbleSort(vector& nums) { + int N = nums.size(); + for (int i = 0; i < N - 1; i++) { + for (int j = 0; j < N - 1 - i; j++) { + if (nums[j] > nums[j + 1]) { + swap(nums[j], nums[j + 1]); + } + } + } + return nums; +} +``` + +![Picture4.png](https://pic.leetcode-cn.com/1623519242-piiPrs-Picture4.png){:width=450} + +### 指数 $O(2^N)$ ： + +生物学科中的 “细胞分裂” 即是指数级增长。初始状态为 $1$ 个细胞，分裂一轮后为 $2$ 个，分裂两轮后为 $4$ 个，……，分裂 $N$ 轮后有 $2^N$ 个细胞。 + +算法中，指数阶常出现于递归，算法原理图与代码如下所示。 + +```Python [] +def algorithm(N): + if N <= 0: return 1 + count_1 = algorithm(N - 1) + count_2 = algorithm(N - 1) + return count_1 + count_2 +``` + +```Java [] +int algorithm(int N) { + if (N <= 0) return 1; + int count_1 = algorithm(N - 1); + int count_2 = algorithm(N - 1); + return count_1 + count_2; +} +``` + +```C++ [] +int algorithm(int N) { + if (N <= 0) return 1; + int count_1 = algorithm(N - 1); + int count_2 = algorithm(N - 1); + return count_1 + count_2; +} +``` + +![Picture5.png](https://pic.leetcode-cn.com/1623519242-XLBkIT-Picture5.png){:width=600} + +### 阶乘 $O(N!)$ ： + +阶乘阶对应数学上常见的 “全排列” 。即给定 $N$ 个互不重复的元素，求其所有可能的排列方案，则方案数量为： + +$$ +N \times (N - 1) \times (N - 2) \times \cdots \times 2 \times 1 = N! +$$ + +如下图与代码所示，阶乘常使用递归实现，算法原理：第一层分裂出 $N$ 个，第二层分裂出 $N - 1$ 个，…… ，直至到第 $N$ 层时终止并回溯。 + +```Python [] +def algorithm(N): + if N <= 0: return 1 + count = 0 + for _ in range(N): + count += algorithm(N - 1) + return count +``` + +```Java [] +int algorithm(int N) { + if (N <= 0) return 1; + int count = 0; + for (int i = 0; i < N; i++) { + count += algorithm(N - 1); + } + return count; +} +``` + +```C++ [] +int algorithm(int N) { + if (N <= 0) return 1; + int count = 0; + for (int i = 0; i < N; i++) { + count += algorithm(N - 1); + } + return count; +} +``` + +![Picture6.png](https://pic.leetcode-cn.com/1623519242-AFSqrK-Picture6.png){:width=600} + +### 对数 $O(\log N)$ ： + +对数阶与指数阶相反，指数阶为 “每轮分裂出两倍的情况” ，而对数阶是 “每轮排除一半的情况” 。对数阶常出现于「二分法」、「分治」等算法中，体现着 “一分为二” 或 “一分为多” 的算法思想。 + +设循环次数为 $m$ ，则输入数据大小 $N$ 与 $2 ^ m$ 呈线性关系，两边同时取 $log_2$ 对数，则得到循环次数 $m$ 与 $\log_2 N$ 呈线性关系，即时间复杂度为 $O(\log N)$ 。 + +```Python [] +def algorithm(N): + count = 0 + i = N + while i > 1: + i = i / 2 + count += 1 + return count +``` + +```Java [] +int algorithm(int N) { + int count = 0; + float i = N; + while (i > 1) { + i = i / 2; + count++; + } + return count; +} +``` + +```C++ [] +int algorithm(int N) { + int count = 0; + float i = N; + while (i > 1) { + i = i / 2; + count++; + } + return count; +} +``` + +如以下代码所示，对于不同 $a$ 的取值，循环次数 $m$ 与 $\log_a N$ 呈线性关系 ，时间复杂度为 $O(\log_a N)$ 。而无论底数 $a$ 取值，时间复杂度都可记作 $O(\log N)$ ，根据对数换底公式的推导如下： + +$$ +O(\log_a N) = \frac{O(\log_2 N)}{O(\log_2 a)} = O(\log N) +$$ + +```Python [] +def algorithm(N): + count = 0 + i = N + a = 3 + while i > 1: + i = i / a + count += 1 + return count +``` + +```Java [] +int algorithm(int N) { + int count = 0; + float i = N; + int a = 3; + while (i > 1) { + i = i / a; + count++; + } + return count; +} +``` + +```C++ [] +int algorithm(int N) { + int count = 0; + float i = N; + int a = 3; + while (i > 1) { + i = i / a; + count++; + } + return count; +} +``` + +> 如下图所示，为二分查找的时间复杂度示意图，每次二分将搜索区间缩小一半。 + +![Picture7.png](https://pic.leetcode-cn.com/1623519242-WthaZa-Picture7.png){:width=600} + +### 线性对数 $O(N \log N)$ ： + +两层循环相互独立，第一层和第二层时间复杂度分别为 $O(\log N)$ 和 $O(N)$ ，则总体时间复杂度为 $O(N \log N)$ ； + +```Python [] +def algorithm(N): + count = 0 + i = N + while i > 1: + i = i / 2 + for j in range(N): + count += 1 +``` + +```Java [] +int algorithm(int N) { + int count = 0; + float i = N; + while (i > 1) { + i = i / 2; + for (int j = 0; j < N; j++) + count++; + } + return count; +} +``` + +```C++ [] +int algorithm(int N) { + int count = 0; + float i = N; + while (i > 1) { + i = i / 2; + for (int j = 0; j < N; j++) + count++; + } + return count; +} +``` + +线性对数阶常出现于排序算法，例如「快速排序」、「归并排序」、「堆排序」等，其时间复杂度原理如下图所示。 + +![Picture8.png](https://pic.leetcode-cn.com/1623519242-rhCOIh-Picture8.png) + +--- + +## 示例题目 + +以下列举本 LeetBook 中各时间复杂度的对应示例题解，以帮助加深理解。 + +| 时间复杂度 | 示例题解 | +| ------------- | ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | +| $O(1)$ | [砍竹子 I](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5vyva2/)、[文物朝代判断](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/572x9r/) | +| $O(\log N)$ | [Pow(x, n)](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/57p2pv/)、[统计目标成绩的出现次数](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/58lgr7/) | +| $O(N)$ | [训练计划 III](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/9p7s17/)、[斐波那契数](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/50fji7/) | +| $O(N \log N)$ | [破解闯关密码](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/59ceyt/)、[交易逆序对的总数](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/o53yjd/) | +| $O(N^2)$ | [验证二叉搜索树的后序遍历序列](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5vwbf6/)、[招式拆解 I](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5dz9di/) | +| $O(N!)$ | [套餐内商品的排列顺序](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/50hah3/) | diff --git "a/leetbook_ioa/docs/# 1.4 \347\251\272\351\227\264\345\244\215\346\235\202\345\272\246.md" "b/leetbook_ioa/docs/# 1.4 \347\251\272\351\227\264\345\244\215\346\235\202\345\272\246.md" new file mode 100755 index 0000000..646ae0a --- /dev/null +++ "b/leetbook_ioa/docs/# 1.4 \347\251\272\351\227\264\345\244\215\346\235\202\345\272\246.md" @@ -0,0 +1,547 @@ +# 空间复杂度 + +空间复杂度涉及的空间类型有： + +- **输入空间：** 存储输入数据所需的空间大小； +- **暂存空间：** 算法运行过程中，存储所有中间变量和对象等数据所需的空间大小； +- **输出空间：** 算法运行返回时，存储输出数据所需的空间大小； + +通常情况下，空间复杂度指在输入数据大小为 $N$ 时，算法运行所使用的「暂存空间」+「输出空间」的总体大小。 + +![Picture1.png](https://pic.leetcode-cn.com/1623769147-NnmTbL-Picture1.png){:width=500} + +而根据不同来源，算法使用的内存空间分为三类： + +**指令空间：** + +编译后，程序指令所使用的内存空间。 + +**数据空间：** + +算法中的各项变量使用的空间，包括：声明的常量、变量、动态数组、动态对象等使用的内存空间。 + +```Python [] +class Node: + def __init__(self, val): + self.val = val + self.next = None + +def algorithm(N): + num = N # 变量 + nums = [0] * N # 动态数组 + node = Node(N) # 动态对象 +``` + +```Java [] +class Node { + int val; + Node next; + Node(int x) { val = x; } +} + +void algorithm(int N) { + int num = N; // 变量 + int[] nums = new int[N]; // 动态数组 + Node node = new Node(N); // 动态对象 +} +``` + +```C++ [] +struct Node { + int val; + Node *next; + Node(int x) : val(x), next(NULL) {} +}; + +void algorithm(int N) { + int num = N; // 变量 + int nums[N]; // 动态数组 + Node* node = new Node(N); // 动态对象 +} +``` + +**栈帧空间：** + +程序调用函数是基于栈实现的，函数在调用期间，占用常量大小的栈帧空间，直至返回后释放。如以下代码所示，在循环中调用函数，每轮调用 `test()` 返回后，栈帧空间已被释放，因此空间复杂度仍为 $O(1)$ 。 + +```Python [] +def test(): + return 0 + +def algorithm(N): + for _ in range(N): + test() +``` + +```Java [] +int test() { + return 0; +} + +void algorithm(int N) { + for (int i = 0; i < N; i++) { + test(); + } +} +``` + +```C++ [] +int test() { + return 0; +} + +void algorithm(int N) { + for (int i = 0; i < N; i++) { + test(); + } +} +``` + +算法中，栈帧空间的累计常出现于递归调用。如以下代码所示，通过递归调用，会同时存在 $N$ 个未返回的函数 `algorithm()` ，此时累计使用 $O(N)$ 大小的栈帧空间。 + +```Python [] +def algorithm(N): + if N <= 1: return 1 + return algorithm(N - 1) + 1 +``` + +```Java [] +int algorithm(int N) { + if (N <= 1) return 1; + return algorithm(N - 1) + 1; +} +``` + +```C++ [] +int algorithm(int N) { + if (N <= 1) return 1; + return algorithm(N - 1) + 1; +} +``` + +--- + +## 符号表示 + +通常情况下，空间复杂度统计算法在 “最差情况” 下使用的空间大小，以体现算法运行所需预留的空间量，使用符号 $O$ 表示。 + +最差情况有两层含义，分别为「最差输入数据」、算法运行中的「最差运行点」。例如以下代码： + +> 输入整数 $N$ ，取值范围 $N \geq 1$ ； + +- **最差输入数据：** 当 $N \leq 10$ 时，数组 `nums` 的长度恒定为 10 ，空间复杂度为 $O(10) = O(1)$ ；当 $N > 10$ 时，数组 $nums$ 长度为 $N$ ，空间复杂度为 $O(N)$ ；因此，空间复杂度应为最差输入数据情况下的 $O(N)$ 。 +- **最差运行点：** 在执行 `nums = [0] * 10` 时，算法仅使用 $O(1)$ 大小的空间；而当执行 `nums = [0] * N` 时，算法使用 $O(N)$ 的空间；因此，空间复杂度应为最差运行点的 $O(N)$ 。 + +```Python [] +def algorithm(N): + num = 5 # O(1) + nums = [0] * 10 # O(1) + if N > 10: + nums = [0] * N # O(N) +``` + +```Java [] +void algorithm(int N) { + int num = 5; // O(1) + int[] nums = new int[10]; // O(1) + if (N > 10) { + nums = new int[N]; // O(N) + } +} +``` + +```C++ [] +void algorithm(int N) { + int num = 5; // O(1) + vector nums(10); // O(1) + if (N > 10) { + nums.resize(N); // O(N) + } +} +``` + +--- + +## 常见种类 + +根据从小到大排列，常见的算法空间复杂度有： + +$$ +O(1) < O(\log N) < O(N) < O(N^2) < O(2^N) +$$ + +![Picture2.png](https://pic.leetcode-cn.com/1623769147-SyWGvV-Picture2.png) + +对于以下所有示例，设输入数据大小为正整数 $N$ ，节点类 `Node` 、函数 `test()` 如以下代码所示。 + +```Python [] +# 节点类 Node +class Node: + def __init__(self, val): + self.val = val + self.next = None + +# 函数 test() +def test(): + return 0 +``` + +```Java [] +// 节点类 Node +class Node { + int val; // 变量 + Node next; // 动态数组 + Node(int x) { val = x; } // 动态对象 +} + +// 函数 test() +int test() { + return 0; +} +``` + +```C++ [] +// 节点类 Node +struct Node { + int val; + Node *next; + Node(int x) : val(x), next(NULL) {} +}; + +// 函数 test() +int test() { + return 0; +} +``` + +### 常数 $O(1)$ ： + +普通常量、变量、对象、元素数量与输入数据大小 $N$ 无关的集合，皆使用常数大小的空间。 + +```Python [] +def algorithm(N): + num = 0 + nums = [0] * 10000 + node = Node(0) + dic = { 0: '0' } +``` + +```Java [] +void algorithm(int N) { + int num = 0; + int[] nums = new int[10000]; + Node node = new Node(0); + Map dic = new HashMap<>() {{ put(0, "0"); }}; +} +``` + +```C++ [] +void algorithm(int N) { + int num = 0; + int nums[10000]; + Node* node = new Node(0); + unordered_map dic; + dic.emplace(0, "0"); +} +``` + +如以下代码所示，虽然函数 `test()` 调用了 $N$ 次，但每轮调用后 `test()` 已返回，无累计栈帧空间使用，因此空间复杂度仍为 $O(1)$ 。 + +```Python [] +def algorithm(N): + for _ in range(N): + test() +``` + +```Java [] +void algorithm(int N) { + for (int i = 0; i < N; i++) { + test(); + } +} +``` + +```C++ [] +void algorithm(int N) { + for (int i = 0; i < N; i++) { + test(); + } +} +``` + +### 线性 $O(N)$ ： + +元素数量与 $N$ 呈线性关系的任意类型集合（常见于一维数组、链表、哈希表等），皆使用线性大小的空间。 + +```Python [] +def algorithm(N): + nums_1 = [0] * N + nums_2 = [0] * (N // 2) + + nodes = [Node(i) for i in range(N)] + + dic = {} + for i in range(N): + dic[i] = str(i) +``` + +```Java [] +void algorithm(int N) { + int[] nums_1 = new int[N]; + int[] nums_2 = new int[N / 2]; + + List nodes = new ArrayList<>(); + for (int i = 0; i < N; i++) { + nodes.add(new Node(i)); + } + + Map dic = new HashMap<>(); + for (int i = 0; i < N; i++) { + dic.put(i, String.valueOf(i)); + } +} +``` + +```C++ [] +void algorithm(int N) { + int nums_1[N]; + int nums_2[N / 2 + 1]; + + vector nodes; + for (int i = 0; i < N; i++) { + nodes.push_back(new Node(i)); + } + + unordered_map dic; + for (int i = 0; i < N; i++) { + dic.emplace(i, to_string(i)); + } +} +``` + +如下图与代码所示，此递归调用期间，会同时存在 $N$ 个未返回的 `algorithm()` 函数，因此使用 $O(N)$ 大小的栈帧空间。 + +```Python [] +def algorithm(N): + if N <= 1: return 1 + return algorithm(N - 1) + 1 +``` + +```Java [] +int algorithm(int N) { + if (N <= 1) return 1; + return algorithm(N - 1) + 1; +} +``` + +```C++ [] +int algorithm(int N) { + if (N <= 1) return 1; + return algorithm(N - 1) + 1; +} +``` + +![Picture3.png](https://pic.leetcode-cn.com/1623771937-urjBWX-Picture3.png) + +### 平方 $O(N^2)$ ： + +元素数量与 $N$ 呈平方关系的任意类型集合（常见于矩阵），皆使用平方大小的空间。 + +```Python [] +def algorithm(N): + num_matrix = [[0 for j in range(N)] for i in range(N)] + node_matrix = [[Node(j) for j in range(N)] for i in range(N)] +``` + +```Java [] +void algorithm(int N) { + int num_matrix[][] = new int[N][N]; + + List> node_matrix = new ArrayList<>(); + for (int i = 0; i < N; i++) { + List nodes = new ArrayList<>(); + for (int j = 0; j < N; j++) { + nodes.add(new Node(j)); + } + node_matrix.add(nodes); + } +} +``` + +```C++ [] +void algorithm(int N) { + vector> num_matrix; + for (int i = 0; i < N; i++) { + vector nums; + for (int j = 0; j < N; j++) { + nums.push_back(0); + } + num_matrix.push_back(nums); + } + + vector> node_matrix; + for (int i = 0; i < N; i++) { + vector nodes; + for (int j = 0; j < N; j++) { + nodes.push_back(new Node(j)); + } + node_matrix.push_back(nodes); + } +} +``` + +如下图与代码所示，递归调用时同时存在 $N$ 个未返回的 `algorithm()` 函数，使用 $O(N)$ 栈帧空间；每层递归函数中声明了数组，平均长度为 $\frac{N}{2}$ ，使用 $O(N)$ 空间；因此总体空间复杂度为 $O(N^2)$ 。 + +```Python [] +def algorithm(N): + if N <= 0: return 0 + nums = [0] * N + return algorithm(N - 1) +``` + +```Java [] +int algorithm(int N) { + if (N <= 0) return 0; + int[] nums = new int[N]; + return algorithm(N - 1); +} +``` + +```C++ [] +int algorithm(int N) { + if (N <= 0) return 0; + int nums[N]; + return algorithm(N - 1); +} +``` + +![Picture4.png](https://pic.leetcode-cn.com/1623769147-PYKjhh-Picture4.png) + +### 指数 $O(2^N)$ ： + +指数阶常见于二叉树、多叉树。例如，高度为 $N$ 的「满二叉树」的节点数量为 $2^N$ ，占用 $O(2^N)$ 大小的空间；同理，高度为 $N$ 的「满 $m$ 叉树」的节点数量为 $m^N$ ，占用 $O(m^N) = O(2^N)$ 大小的空间。 + +![Picture5.png](https://pic.leetcode-cn.com/1623769147-FLxBQi-Picture5.png){:width=600} + +### 对数 $O(\log N)$ ： + +对数阶常出现于分治算法的栈帧空间累计、数据类型转换等，例如： + +- **快速排序** ，平均空间复杂度为 $\Theta(\log N)$ ，最差空间复杂度为 $O(N)$ 。拓展知识：通过应用 [尾递归优化](https://stackoverflow.com/questions/310974/what-is-tail-call-optimization) ，可以将快速排序的最差空间复杂度限定至 $O(N)$ 。 +- **数字转化为字符串** ，设某正整数为 $N$ ，则字符串的空间复杂度为 $O(\log N)$ 。推导如下：正整数 $N$ 的位数为 $log_{10} N$ ，即转化的字符串长度为 $\log_{10} N$ ，因此空间复杂度为 $O(\log N)$ 。 + +--- + +## 时空权衡 + +对于算法的性能，需要从时间和空间的使用情况来综合评价。优良的算法应具备两个特性，即时间和空间复杂度皆较低。而实际上，对于某个算法问题，同时优化时间复杂度和空间复杂度是非常困难的。降低时间复杂度，往往是以提升空间复杂度为代价的，反之亦然。 + +> 由于当代计算机的内存充足，通常情况下，算法设计中一般会采取「空间换时间」的做法，即牺牲部分计算机存储空间，来提升算法的运行速度。 + +以 LeetCode 全站第一题 [两数之和](https://leetcode-cn.com/problems/two-sum/) 为例，「暴力枚举」和「辅助哈希表」分别为「空间最优」和「时间最优」的两种算法。 + +### 方法一：暴力枚举 + +时间复杂度 $O(N^2)$ ，空间复杂度 $O(1)$ ；属于「时间换空间」，虽然仅使用常数大小的额外空间，但运行速度过慢。 + +```Python [] +class Solution: + def twoSum(self, nums: List[int], target: int) -> List[int]: + for i in range(len(nums) - 1): + for j in range(i + 1, len(nums)): + if nums[i] + nums[j] == target: + return i, j + return +``` + +```Java [] +class Solution { + public int[] twoSum(int[] nums, int target) { + int size = nums.length; + for (int i = 0; i < size - 1; i++) { + for (int j = i + 1; j < size; j++) { + if (nums[i] + nums[j] == target) + return new int[] { i, j }; + } + } + return new int[0]; + } +} +``` + +```C++ [] +class Solution { +public: + vector twoSum(vector& nums, int target) { + int size = nums.size(); + for (int i = 0; i < size - 1; i++) { + for (int j = i + 1; j < size; j++) { + if (nums[i] + nums[j] == target) + return { i, j }; + } + } + return {}; + } +}; +``` + +### 方法二：辅助哈希表 + +时间复杂度 $O(N)$ ，空间复杂度 $O(N)$ ；属于「空间换时间」，借助辅助哈希表 `dic` ，通过保存数组元素值与索引的映射来提升算法运行效率，是本题的最佳解法。 + +```Python [] +class Solution: + def twoSum(self, nums: List[int], target: int) -> List[int]: + dic = {} + for i in range(len(nums)): + if target - nums[i] in dic: + return dic[target - nums[i]], i + dic[nums[i]] = i + return [] +``` + +```Java [] +class Solution { + public int[] twoSum(int[] nums, int target) { + int size = nums.length; + Map dic = new HashMap<>(); + for (int i = 0; i < size; i++) { + if (dic.containsKey(target - nums[i])) { + return new int[] { dic.get(target - nums[i]), i }; + } + dic.put(nums[i], i); + } + return new int[0]; + } +} +``` + +```C++ [] +class Solution { +public: + vector twoSum(vector& nums, int target) { + int size = nums.size(); + unordered_map dic; + for (int i = 0; i < size; i++) { + if (dic.find(target - nums[i]) != dic.end()) { + return { dic[target - nums[i]], i }; + } + dic.emplace(nums[i], i); + } + return {}; + } +}; +``` + +--- + +## 示例题目 + +在 LeetCode 题目中，「输入空间」和「输出空间」往往是固定的，是必须使用的内存空间。因希望专注于算法性能对比，本 LeetBook 的题目解析的空间复杂度仅统计「暂存空间」大小。 + +| 空间复杂度 | 示例题解 | +| ----------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ | +| $O(1)$ | [斐波那契数](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/50fji7/)、[训练计划 III](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/9p7s17/) | +| $O(\log N)$ | [库存管理 III](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/ohwddh/)、[找到第 k 位数字](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/57w6b3/) | +| $O(N)$ | [图书整理 I](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5d8831/)、[动态口令](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/58eckc/) | +| $O(N^2)$ | [衣橱整理](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/9hka9c/)、[套餐内商品的排列顺序](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/50hah3/) | diff --git "a/leetbook_ioa/docs/# 11.1 \345\212\250\346\200\201\350\247\204\345\210\222\350\247\243\351\242\230\346\241\206\346\236\266.md" "b/leetbook_ioa/docs/# 11.1 \345\212\250\346\200\201\350\247\204\345\210\222\350\247\243\351\242\230\346\241\206\346\236\266.md" new file mode 100755 index 0000000..af0b0cd --- /dev/null +++ "b/leetbook_ioa/docs/# 11.1 \345\212\250\346\200\201\350\247\204\345\210\222\350\247\243\351\242\230\346\241\206\346\236\266.md" @@ -0,0 +1,482 @@ +# 动态规划解题框架 + +动态规划是算法与数据结构的重难点之一，其包含了「分治思想」、「空间换时间」、「最优解」等多种基石算法思想，常作为笔面试中的中等困难题出现。为帮助读者全面理解动态规划，知晓其来龙去脉，本文将从以下几个角度切入介绍： + +1. 动态规划问题特点，**动态规划**和**分治算法**的联系与区别； +2. 借助例题介绍**重叠子问题**和**最优子结构**分别是什么，以及动态规划是如何解决它们的； +3. 动态规划的**解题框架**总结； +4. 动态规划的**练习例题**，从易到难排序； + +--- + +## 动态规划特点 + +「分治」是算法中的一种基本思想，其通过**将原问题分解为子问题**，不断递归地将子问题分解为更小的子问题，并通过**组合子问题的解**来得到原问题的解。 + +类似于分治算法，「动态规划」也通过组合子问题的解得到原问题的解。不同的是，适合用动态规划解决的问题具有「重叠子问题」和「最优子结构」两大特性。 + +### 重叠子问题 + +动态规划的子问题是有**重叠的**，即各个子问题中包含**重复的更小子问题**。若使用暴力法穷举，求解这些相同子问题会产生大量的重复计算，效率低下。 + +动态规划在第一次求解某子问题时，会将子问题的解保存；后续遇到重叠子问题时，则直接通过查表获取解，保证每个**独立子问题只被计算一次**，从而降低算法的时间复杂度。 + +### 最优子结构 + +如果一个问题的最优解可以由其子问题的最优解组合构成，并且这些子问题可以独立求解，那么称此问题具有最优子结构。 + +动态规划从基础问题的解开始，不断迭代**组合、选择子问题的最优解**，最终得到原问题最优解。 + +--- + +## 重叠子问题示例：斐波那契数列 + +> 斐波那契数形成的数列为 $[0, 1, 1, 2, 3, 5, 8, 13, \cdots]$ ，数学定义如下： +> $$ +> \begin{aligned} +> & F_0 = 0 \\ +> & F_1 = 1 \\ +> & F_n = F_{n-1} + F_{n-2} +> \end{aligned} +> $$ +> **题目：** 求取第 $n$ 个斐波那契数（从第 0 个斐波那契数开始）。 + +以下，本文从「暴力递归」$\rightarrow$「记忆化递归」$\rightarrow$「动态规划」三种解法，介绍**重叠子问题**的概念与解决方案。 + +### 方法一：暴力递归 + +设斐波那契数列第 $n$ 个数字为 $f(n)$ 。根据数列定义，可得 $f(n) = f(n - 1) + f(n - 2)$ ，且第 0 , 1 个斐波那契数分别为 $f(0) = 0$ , $f(1) = 1$ 。 + +我们很容易联想到使用分治思想来求取 $f(n)$ ，即将求原问题 $f(n)$ 分解为求子问题 $f(n-1)$ 和 $f(n-2)$ ，向下递归直至已知的 $f(0)$ 和 $f(1)$ ，最终组合这些子问题求取原问题 $f(n)$ 。 + +```Python [] +# 求第 n 个斐波那契数 +def fibonacci(n): + if n == 0: return 0 # 返回 f(0) + if n == 1: return 1 # 返回 f(1) + return fibonacci(n - 1) + fibonacci(n - 2) # 分解为两个子问题求解 +``` + +```Java [] +// 求第 n 个斐波那契数 +int fibonacci(int n) { + if (n == 0) return 0; // 返回 f(0) + if (n == 1) return 1; // 返回 f(1) + return fibonacci(n - 1) + fibonacci(n - 2); // 分解为两个子问题求解 +} +``` + +```C++ [] +int fibonacci(int n) { + if (n == 0) return 0; // 返回 f(0) + if (n == 1) return 1; // 返回 f(1) + return fibonacci(n - 1) + fibonacci(n - 2); // 分解为两个子问题求解 +} +``` + +![Picture1.png](https://pic.leetcode-cn.com/1635075778-GADtbi-Picture1.png) + +如上图所示，为暴力递归求斐波那契数 $f(5)$ 形成的二叉树，树中的每个节点代表着执行了一次 `fibonacci()` 函数，且有： + +- 执行一次 `fibonacci()` 函数的时间复杂度为 $O(1)$ ； +- 二叉树节点数为指数级 $O(2^n)$ ； + +因此，暴力递归的总体时间复杂度为 $O(2^n)$ 。此方法效率低下，随着 $n$ 的增长产生指数级爆炸。 + +### 方法二：记忆化递归 + +观察发现，暴力递归中的子问题多数都是**重叠子问题**，即： + +$$ +\begin{aligned} +& f(n) = f(n - 1) + f(n - 2) & 包含 f(n - 2) \\ +& f(n - 1) = f(n - 2) + f(n - 3) & 重复 f(n - 2) \\ +& f(n - 2) = f(n - 3) + f(n - 4) & 重复 f(n - 3) \\ +& \cdots &以此类推 +\end{aligned} +$$ + +这些重叠子问题产生了大量的递归树节点，其**不应被重复计算**。实际上，可以在递归中**第一次求解子问题**时，就将它们**保存**；后续递归中再次遇到相同子问题时，直接访问内存赋值即可。记忆化递归的代码如下所示。 + +```Python [] +def fibonacci(n, dp): + if n == 0: return 0 # 返回 f(0) + if n == 1: return 1 # 返回 f(1) + if dp[n] != 0: return dp[n] # 若 f(n) 以前已经计算过，则直接返回记录的解 + dp[n] = fibonacci(n - 1, dp) + fibonacci(n - 2, dp) # 将 f(n) 则记录至 dp + return dp[n] + +# 求第 n 个斐波那契数 +def fibonacci_memorized(n): + dp = [0] * (n + 1) # 用于保存 f(0) 至 f(n) 问题的解 + return fibonacci(n, dp) +``` + +```Java [] +int fibonacci(int n, int[] dp) { + if (n == 0) return 0; // 返回 f(0) + if (n == 1) return 1; // 返回 f(1) + if (dp[n] != 0) return dp[n]; // 若 f(n) 以前已经计算过，则直接返回记录的解 + dp[n] = fibonacci(n - 1, dp) + fibonacci(n - 2, dp); // 将 f(n) 则记录至 dp + return dp[n]; +} + + +// 求第 n 个斐波那契数 +int fibonacciMemorized(int n) { + int[] dp = new int[n + 1]; // 用于保存 f(0) 至 f(n) 问题的解 + return fibonacci(n, dp); +} +``` + +```C++ [] +int fibonacci(int n, vector dp) { + if (n == 0) return 0; // 返回 f(0) + if (n == 1) return 1; // 返回 f(1) + if (dp[n] != 0) return dp[n]; // 若 f(n) 以前已经计算过，则直接返回记录的解 + dp[n] = fibonacci(n - 1, dp) + fibonacci(n - 2, dp); // 将 f(n) 则记录至 dp + return dp[n]; +} + + +// 求第 n 个斐波那契数 +int fibonacciMemorized(int n) { + vector dp(n + 1, 0); // 用于保存 f(0) 至 f(n) 问题的解 + return fibonacci(n, dp); +} +``` + +如下图所示，应用记忆化递归方法后，递归树中绝大部分节点被**剪枝**。此时，`fibonacci()` 函数的调用次数从 $O(2^n)$ 指数级别降低至 $O(n)$ 线性级别，时间复杂度大大降低。 + +![Picture2.png](https://pic.leetcode-cn.com/1635075778-pJXkPc-Picture2.png) + +### 方法三：动态规划 + +递归本质上是基于分治思想的从顶至底的解法。借助记忆化递归思想，可应用动态规划从底至顶求取 $f(n)$ ，代码如下所示。 + +```Python [] +# 求第 n 个斐波那契数 +def fibonacci(n): + if n == 0: return 0 # 若求 f(0) 则直接返回 0 + dp = [0] * (n + 1) # 初始化 dp 列表 + dp[0], dp[1] = 0, 1 # 初始化 f(0), f(1) + for i in range(2, n + 1): # 状态转移求取 f(2), f(3), ..., f(n) + dp[i] = dp[i - 1] + dp[i - 2] + return dp[n] # 返回 f(n) +``` + +```Java [] +// 求第 n 个斐波那契数 +int fibonacci(int n) { + if (n == 0) return 0; // 若求 f(0) 则直接返回 0 + int[] dp = new int[n + 1]; // 初始化 dp 列表 + dp[1] = 1; // 初始化 f(0), f(1) + for (int i = 2; i <= n; i++) { // 状态转移求取 f(2), f(3), ..., f(n) + dp[i] = dp[i - 1] + dp[i - 2]; + } + return dp[n]; // 返回 f(n) +} +``` + +```C++ [] +// 求第 n 个斐波那契数 +int fibonacci(int n) { + if (n == 0) return 0; // 若求 f(0) 则直接返回 0 + vector dp(n + 1, 0); // 初始化 dp 列表 + dp[1] = 1; // 初始化 f(0), f(1) + for (int i = 2; i <= n; i++) { // 状态转移求取 f(2), f(3), ..., f(n) + dp[i] = dp[i - 1] + dp[i - 2]; + } + return dp[n]; // 返回 f(n) +} +``` + +如下图所示，为动态规划求解 $f(5)$ 的迭代流程，其是转移方程 $f(n) = f(n - 1) + f(n - 2)$ 的体现。 + +![Picture3.png](https://pic.leetcode-cn.com/1635075778-EeBEoS-Picture3.png) + +上述动态规划解法借助了一个 `dp` 数组保存子问题的解，其空间复杂度为 $O(N)$ 。而由于 $f(n)$ 只与 $f(n - 1)$ 和 $f(n - 2)$ 有关，因此我们可以仅使用两个变量 $a$ , $b$ 交替前进计算即可。此时动态规划的空间复杂度降低至 $O(1)$ ，代码如下所示。 + +```Python [] +# 求第 n 个斐波那契数 +def fibonacci(n): + if n == 0: return 0 # 若求 f(0) 则直接返回 0 + a, b = 0, 1 # 初始化 f(0), f(1) + for i in range(2, n + 1): # 状态转移求取 f(2), f(3), ..., f(n) + a, b = b, a + b + return b # 返回 f(n) +``` + +```Java [] +// 求第 n 个斐波那契数 +int fibonacci(int n) { + if (n == 0) return 0; // 若求 f(0) 则直接返回 0 + int a = 0, b = 1; // 初始化 f(0), f(1) + for (int i = 2; i <= n; i++) { // 状态转移求取 f(2), f(3), ..., f(n) + int tmp = a; + a = b; + b = tmp + b; + } + return b; // 返回 f(n) +} +``` + +```C++ [] +// 求第 n 个斐波那契数 +int fibonacci(int n) { + if (n == 0) return 0; // 若求 f(0) 则直接返回 0 + int a = 0, b = 1; // 初始化 f(0), f(1) + for (int i = 2; i <= n; i++) { // 状态转移求取 f(2), f(3), ..., f(n) + int tmp = a; + a = b; + b = tmp + b; + } + return b; // 返回 f(n) +} +``` + +### 示例小结 + +记忆化递归和动态规划的本质思想是一致的，是对斐波那契数列定义的不同表现形式： + +- **记忆化递归 — 从顶至低：** 求 $f(n)$ 需要 $f(n - 1)$ 和 $f(n - 2)$ ； $\cdots$ ；求 $f(2)$ 需要 $f(1)$ 和 $f(0)$ ；而 $f(1)$ 和 $f(0)$ 已知； +- **动态规划 — 从底至顶：** 将已知 $f(0)$ 和 $f(1)$ 组合得到 $f(2)$ ；$\cdots$ ；将 $f(n - 2)$ 和 $f(n - 1)$ 组合得到 $f(n)$ ； + +斐波那契数列问题不包含「最优子结构」，只需计算每个子问题的解，避免重复计算即可，并不需要从子问题组合中**选择最优组合**。接下来，本文借助「最高蛋糕售价方案」，介绍动态规划的**最优子结构**概念。 + +--- + +## 最优子结构示例：蛋糕最高售价 + +> 小力开了一家蛋糕店，并针对不同重量的蛋糕设定了不同售价，分别为： +> +> | 蛋糕重量 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | +> | :------: | :--: | :--: | :--: | :--: | :--: | :--: | :--: | +> | 售价 | 0 | 2 | 3 | 6 | 7 | 11 | 15 | +> +> **问题：** 现给定一个重量为 $n$ 的蛋糕，问小力应该如何切分蛋糕，达到最高的蛋糕总售价。 + +设重量为 $n$ 蛋糕的售价为 $p(n)$ ，切分的最高总售价为 $f(n)$ 。 + +- **子问题：** $f(n)$ 的子问题包括 $f(0), f(1), f(2), \cdots, f(n - 1)$ ，分别代表重量为 $0, 1, 2, \cdots, n - 1$ 蛋糕的最高售价。 已知无蛋糕时 $f(0) = 0$ ，蛋糕重量为 1 时不可切分 $f(1) = p(1)$ ； +- **最优子结构：** + - **定义：** 如果一个问题最优解可以由其子问题最优解组合构成，那么称此问题具有最优子结构。 + - **对于本题：** 重量为 $n$ 的蛋糕的总售价可切分为 $n$ 种组合，即重量为 $0, 1, 2, ..., n - 1$ 蛋糕**最高售价**加上 $n, n - 1, n - 2, \cdots, 1$ 剩余重量蛋糕的**售价**；从这些组合中，售价最高的组合便是原问题的解 $f(n)$ ，这便是本题的最优子结构。 + +- **状态转移方程：** 找出最优子结构后，易构建出如下的状态转移方程。 + +$$ +f(n) = \max_{0 \leq i < n} (f(i) + p(n - i)) +$$ + +根据以上推导，本题也能使用「暴力递归」$\rightarrow$「记忆化递归」$\rightarrow$「动态规划」三种方法解决。 + +### 方法一：暴力递归 + +暴力递归解法的代码如下，其时间复杂度为指数级 $O(2^n)$ 。 + +```Python [] +# 输入蛋糕价格列表 price_list ，求重量为 n 蛋糕的最高售价 +def max_cake_price(n, price_list): + if n <= 1: return price_list[n] # 蛋糕重量 <= 1 时直接返回 + f_n = 0 + for i in range(n): # 从 n 种组合种选择最高售价的组合作为 f(n) + f_n = max(f_n, max_cake_price(i, price_list) + price_list[n - i]) + return f_n # 返回 f(n) + +max_cake_price(4, [0, 2, 3, 6, 7, 11, 15]) +``` + +```Java [] +// 输入蛋糕价格列表 priceList ，求重量为 n 蛋糕的最高售价 +int maxCakePrice(int n, int[] priceList) { + if (n <= 1) return priceList[n]; // 蛋糕重量 <= 1 时直接返回 + int f_n = 0; + for (int i = 0; i < n; i++) // 从 n 种组合种选择最高售价的组合作为 f(n) + f_n = Math.max(f_n, maxCakePrice(i, priceList) + priceList[n - i]); + return f_n; // 返回 f(n) +} +``` + +```C++ [] +// 输入蛋糕价格列表 priceList ，求重量为 n 蛋糕的最高售价 +int maxCakePrice(int n, vector priceList) { + if (n <= 1) return priceList[n]; // 蛋糕重量 <= 1 时直接返回 + int f_n = 0; + for (int i = 0; i < n; i++) // 从 n 种组合种选择最高售价的组合作为 f(n) + f_n = max(f_n, maxCakePrice(i, priceList) + priceList[n - i]); + return f_n; // 返回 f(n) +} +``` + +如下图所示，为暴力递归求解 $f(4)$ 形成的多叉树。 + +![Picture4.png](https://pic.leetcode-cn.com/1635075778-AljQEJ-Picture4.png) + +### 方法二：记忆化递归 + +观察发现，递归树中存在大量**重叠子问题**，可通过记忆化处理避免重复计算。记忆化递归的算法的时间复杂度为 $O(n^2)$ ，包括： + +- $f(2)$ 至 $f(n)$ 共 $n - 1$ 个待计算子问题，使用 $O(n)$ 时间； +- 计算某 $f(i)$ 需遍历 $i - 1$ 种子问题组合，使用 $O(n)$ 时间； + +```Python [] +# 输入蛋糕价格列表 price_list ，求重量为 n 蛋糕的最高售价 +def max_cake_price(n, price_list, dp): + if n <= 1: return price_list[n] # 蛋糕重量 <= 1 时直接返回 + f_n = 0 + for i in range(n): # 从 n 种组合种选择最高售价的组合作为 f(n) + # 若 f(i) 以前已经计算过，则调取记录的解；否则，递归计算 f(i) + f_i = dp[i] if dp[i] != 0 else max_cake_price(i, price_list, dp) + f_n = max(f_n, f_i + price_list[n - i]) + dp[n] = f_n # 记录 f(n) 至 dp 数组 + return f_n # 返回 f(n) + +def max_cake_price_memorized(n, price_list): + dp = [0] * (n + 1) + return max_cake_price(n, price_list, dp) +``` + +```Java [] +// 输入蛋糕价格列表 priceList ，求重量为 n 蛋糕的最高售价 +int maxCakePrice(int n, int[] priceList, int[] dp) { + if (n <= 1) return priceList[n]; // 蛋糕重量 <= 1 时直接返回 + int f_n = 0; + for (int i = 0; i < n; i++) { // 从 n 种组合种选择最高售价的组合作为 f(n) + int f_i = dp[i] != 0 ? dp[i] : maxCakePrice(i, priceList, dp); + f_n = Math.max(f_n, f_i + priceList[n - i]); + } + dp[n] = f_n; // 记录 f(n) 至 dp 数组 + return f_n; // 返回 f(n) +} + +int maxCakePriceMemorized(int n, int[] priceList) { + int[] dp = new int[n + 1]; + return maxCakePrice(n, priceList, dp); +} +``` + +```C++ [] +// 输入蛋糕价格列表 priceList ，求重量为 n 蛋糕的最高售价 +int maxCakePrice(int n, vector &priceList, vector dp) { + if (n <= 1) return priceList[n]; // 蛋糕重量 <= 1 时直接返回 + int f_n = 0; + for (int i = 0; i < n; i++) { // 从 n 种组合种选择最高售价的组合作为 f(n) + int f_i = dp[i] != 0 ? dp[i] : maxCakePrice(i, priceList, dp); + f_n = max(f_n, f_i + priceList[n - i]); + } + dp[n] = f_n; // 记录 f(n) 至 dp 数组 + return f_n; // 返回 f(n) +} + +int maxCakePriceMemorized(int n, vector priceList) { + vector dp(n + 1, 0); + return maxCakePrice(n, priceList, dp); +} +``` + +如下图所示，为记忆化递归求解 $f(4)$ 形成的多叉树。观察得知，重叠子问题皆被**剪枝**。 + +![Picture5.png](https://pic.leetcode-cn.com/1635075778-tsDBrs-Picture5.png) + +### 方法三：动态规划 + +相较于记忆化递归的从顶至底方法，易得动态规划的从底至顶方法，代码如下所示。 + +```Python [] +# 输入蛋糕价格列表 price_list ，求重量为 n 蛋糕的最高售价 +def max_cake_price(n, price_list): + if n <= 1: return price_list[n] # 蛋糕重量 <= 1 时直接返回 + dp = [0] * (n + 1) # 初始化 dp 列表 + for j in range(1, n + 1): # 按顺序计算 f(1), f(2), ..., f(n) + for i in range(j): # 从 j 种组合种选择最高售价的组合作为 f(j) + dp[j] = max(dp[j], dp[i] + price_list[j - i]) + return dp[n] +``` + +```Java [] +// 输入蛋糕价格列表 priceList ，求重量为 n 蛋糕的最高售价 +int maxCakePrice(int n, int[] priceList) { + if (n <= 1) return priceList[n]; // 蛋糕重量 <= 1 时直接返回 + int[] dp = new int[n + 1]; // 初始化 dp 列表 + for (int j = 1; j <= n; j++) { // 按顺序计算 f(1), f(2), ..., f(n) + for (int i = 0; i < j; i++) // 从 j 种组合种选择最高售价的组合作为 f(j) + dp[j] = Math.max(dp[j], dp[i] + priceList[j - i]); + } + return dp[n]; +} +``` + +```C++ [] +// 输入蛋糕价格列表 priceList ，求重量为 n 蛋糕的最高售价 +int maxCakePrice(int n, vector priceList) { + if (n <= 1) return priceList[n]; // 蛋糕重量 <= 1 时直接返回 + vector dp(n + 1, 0); // 初始化 dp 列表 + for (int j = 1; j <= n; j++) { // 按顺序计算 f(1), f(2), ..., f(n) + for (int i = 0; i < j; i++) // 从 j 种组合种选择最高售价的组合作为 f(j) + dp[j] = max(dp[j], dp[i] + priceList[j - i]); + } + return dp[n]; +} +``` + +如下图所示，为动态规划求解 $f(4)$ 的迭代流程，其是转移方程 $f(n) = \max_{0 \leq i < n} (f(i) + p(n - i))$ 的体现。 + +![Picture6.png](https://pic.leetcode-cn.com/1635075778-PBtyyB-Picture6.png) + +### 示例小结 + +本题同时包含「重叠子问题」和「最优子结构」，为动态规划的典型问题。动态规划通过填表避免了重复计算问题，并通过状态转移方程、初始状态实现对问题的迭代求解。 + +普遍来看，**求最值** 的问题一般都具有「重叠子问题」和「最优子结构」特点，因此此类问题往往适合用动态规划解决。 + +--- + +## 动态规划解题框架 + +若确定给定问题具有重叠子问题和最优子结构，那么就可以使用动态规划求解。总体上看，求解可分为四步： + +1. **状态定义：** 构建问题最优解模型，包括问题**最优解的定义**、有哪些**计算解的自变量**； +2. **初始状态：** 确定**基础子问题的解**（即已知解），原问题和子问题的解都是以基础子问题的解为起始点，在迭代计算中得到的； +3. **转移方程：** 确定原问题的解与子问题的解之间的关系是什么，以及使用何种**选择规则**从子问题最优解组合中选出原问题最优解； +4. **返回值：** 确定应返回的问题的解是什么，即动态规划**在何处停止迭代**； + +完成以上步骤后，便容易写出对应的解题代码。 + +### 示例：斐波那契数列 + +- 状态定义：一维 $dp$ 列表，设第 $i$ 个斐波那契数为 $dp[i]$ ； +- 初始状态：已知第 $0$ , $1$ 个斐波那契数分别为 $dp[0] = 0$ , $dp[1] = 1$ ； +- 转移方程：后一个数字等于前两个数字之和，即 + +$$ +dp[i] = dp[i - 1] + dp[i - 2] +$$ + +- 返回值：需求取的第 $n$ 个斐波那契数 $dp[n]$ ； + +### 示例：蛋糕最高售价 + +- 状态定义：一维 $dp$ 列表，设重量为 $i$ 蛋糕的售价为 $p(i)$ ，重量为 $i$ 蛋糕切分后的最高售价为 $dp[i]$ ； +- 初始状态：已知重量为 0 蛋糕的最高售价为 0 ，重量为 1 的蛋糕最高售价为 $p(1)$ ； +- 转移方程：$dp[n]$ 为 $n$ 种切分组合中的最高售价组合，即 + +$$ +dp[n] = \max_{0 \leq i < n} (dp[i] + p(n - i)) +$$ + +- 返回值：需求取的重量为 $n$ 的蛋糕最高售价 $dp[n]$ ； + +--- + +## 例题练习 + +动态规划的问题种类多，难度跨度较大，需要充足练习、熟能生巧。以下给出若干典型例题，供读者巩固理解本文内容。 + +| 题目 | 难度 | 描述 | +| ----------------------------------------------------------------------------------------------- | ---- | ----------------------------------------------------------------------- | +| [跳跃训练](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/57hyl5/) | 简单 | 与本文的斐波那契数列例题等价 | +| [连续天数的最高销售额](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/59gq9c/) | 简单 | 求最大值问题，关键点在于状态定义 | +| [珠宝的最高价值](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/5vokvr/) | 简单 | 求最大值问题，特点是其 $dp$ 列表是二维的 | +| [统计结果概率](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/ozzl1r/) | 中等 | 容易想到暴力枚举方法，难点为列出状态转移方程，且正向递推方法比较 tricky | +| [模糊搜索验证](https://leetcode-cn.com/leetbook/read/illustration-of-algorithm/9a1ypc/) | 困难 | 状态定义容易得出，但状态转移方程复杂、选择规则分支多 | diff --git "a/leetbook_ioa/docs/# 7.1 \346\216\222\345\272\217\347\256\227\346\263\225\347\256\200\344\273\213.md" "b/leetbook_ioa/docs/# 7.1 \346\216\222\345\272\217\347\256\227\346\263\225\347\256\200\344\273\213.md" new file mode 100755 index 0000000..47df189 --- /dev/null +++ "b/leetbook_ioa/docs/# 7.1 \346\216\222\345\272\217\347\256\227\346\263\225\347\256\200\344\273\213.md" @@ -0,0 +1,136 @@ +# 排序算法简介 + +排序算法用作实现列表的排序，列表元素可以是整数，也可以是浮点数、字符串等其他数据类型。生活中有许多需要排序算法的场景，例如： + +- **整数排序：** 对于一个整数数组，我们希望将所有数字从小到大排序； +- **字符串排序：** 对于一个姓名列表，我们希望将所有单词按照字符先后排序； +- **自定义排序：** 对于任意一个 **已定义比较规则** 的集合，我们希望将其按规则排序； + +![Picture1.png](https://pic.leetcode-cn.com/1629483616-HHvvqY-Picture1.png) + +同时，某些算法需要在排序算法的基础上使用（即在排序数组上运行），例如： + +- **二分查找：** 根据数组已排序的特性，才能每轮确定排除两部分中的哪一部分； +- **双指针：** 例如合并两个排序链表，根据已排序特性，才能通过双指针移动在线性时间内将其合并为一个排序链表。 + +> 接下来，本文将从「常见排序算法」、「分类方法」、「时间与空间复杂度」三方面入手，简要介绍排序算法。「各排序算法详细介绍」请见后续专栏文章。 + +--- + +## 常见算法 + +常见排序算法包括「冒泡排序」、「插入排序」、「选择排序」、「快速排序」、「归并排序」、「堆排序」、「基数排序」、「桶排ss序」。如下图所示，为各排序算法的核心特性与时空复杂度总结。 + +![Picture2.png](https://pic.leetcode-cn.com/1629483637-tmENTT-Picture2.png) + +如下图所示，为在 「随机乱序」、「接近有序」、「完全倒序」、「少数独特」四类输入数据下，各常见排序算法的排序过程。 + +![krahets-bubble-sort.gif](https://pic.leetcode-cn.com/1629482805-alxVCi-krahets-bubble-sort.gif) + +![krahets-insertion-sort.gif](https://pic.leetcode-cn.com/1629482805-MMIXPp-krahets-insertion-sort.gif) + +![krahets-selection-sort.gif](https://pic.leetcode-cn.com/1629482805-vdWamx-krahets-selection-sort.gif) + +![krahets-quick-sort.gif](https://pic.leetcode-cn.com/1629482805-DYNZPE-krahets-quick-sort.gif) + +![krahets-merge-sort.gif](https://pic.leetcode-cn.com/1629482805-IjWwSz-krahets-merge-sort.gif) + +![krahets-heap-sort.gif](https://pic.leetcode-cn.com/1629482805-apljJY-krahets-heap-sort.gif) + +--- + +## 分类方法 + +排序算法主要可根据 **稳定性** 、**就地性** 、**自适应性** 分类。理想的排序算法具有以下特性： + +- 具有稳定性，即相等元素的相对位置不变化； +- 具有就地性，即不使用额外的辅助空间； +- 具有自适应性，即时间复杂度受元素分布影响； + +特别地，任意排序算法都 **不同时具有以上所有特性** 。因此，排序算法的选型使用取决于具体的列表类型、元素数量、元素分布情况等应用场景特点。 + +### 稳定性： + +根据 **相等元素** 在数组中的 **相对顺序** 是否被改变，排序算法可分为「稳定排序」和「非稳定排序」两类。 + +- 「稳定排序」在完成排序后，**不改变** 相等元素在数组中的相对顺序。例如：冒泡排序、插入排序、归并排序、基数排序、桶排序。 +- 「非稳定排序」在完成排序后，相等素在数组中的相对位置 **可能被改变**。例如：选择排序、快速排序、堆排序。 + +> **何时需考虑排序算法的稳定性？** +> +> 数组排序中，由于元素皆为数字，因此稳定和非稳定排序皆可输出相同结果，此时无需考虑排序算法的稳定性。 +> +> 非稳定排序会改变相等元素的相对次序，这在实际应用场景中可能是不能接受的。如以下代码所示，非稳定排序破坏了输入列表 `people` 按姓名排序的性质。 +> +> ```Python +> # 人 = (姓名, 年龄) ，按姓名排序 +> people = [ +> ('A', 19), +> ('B', 18), +> ('C', 21), +> ('D', 19), +> ('E', 23) +> ] +> +> # 非稳定排序（按年龄） +> sort_by_age(people) +> +> # 人 = (姓名, 年龄) ，按年龄排序 +> people = [ +> ('B', 18), +> ('D', 19), # ('D', 19) 和 ('A', 19) 的相对位置改变，输入时按姓名排序的性质丢失 +> ('A', 19), +> ('C', 21), +> ('E', 23) +> ] +> ``` + +### 就地性： + +根据排序过程中 **是否使用额外内存（辅助数组）**，排序算法可分为「原地排序」和「异地排序」两类。一般地，由于不使用外部内存，原地排序相比非原地排序的执行效率更高。 + +- 「原地排序」不使用额外辅助数组，例如：冒泡排序、插入排序、选择排序、快速排序、堆排序。 +- 「非原地排序」使用额外辅助数组，例如：归并排序、基数排序、桶排序。 + +### 自适应性： + +根据算法 **时间复杂度** 是否 **受待排序数组的元素分布影响** ，排序算法可分为「自适应排序」和「非自适应排序」两类。 + +- 「自适应排序」的时间复杂度受元素分布影响；例如：冒泡排序、插入排序、快速排序、桶排序。 +- 「非自适应排序」的时间复杂度恒定；例如：选择排序、归并排序、堆排序、基数排序。 + +### 是否基于比较： + +比较类排序基于元素之间的 **比较算子**（小于、相等、大于）来决定元素的相对顺序；相对的，非比较排序则不基于比较算子实现。 + +- 「基于比较排序」基于元素之间的比较完成排序，例如：冒泡排序、插入排序、选择排序、快速排序、归并排序、堆排序。 +- 「非基于比较排序」不基于元素之间的比较完成排序，例如：基数排序、桶排序。 + +> 基于比较的排序算法的平均时间复杂度最优为 $O(N \log N)$ ，而非比较排序算法可以达到线性级别的时间复杂度。 + +--- + +## 时空复杂度 + +总体上看，排序算法追求时间与空间复杂度最低。而即使某些排序算法的时间复杂度相等，但实际性能还受 **输入列表性质、元素数量、元素分布等** 等因素影响。 + +> 设输入列表元素数量为 $N$ ，常见排序算法的「时间复杂度」和「空间复杂度」如下图所示。 + +| 算法 | 最佳时间 | 平均时间 | 最差时间 | 最差空间 | +| :------: | :-----------------: | :----------------: | :-----------: | :---------: | +| 冒泡排序 | $\Omega(N)$ | $\Theta(N^2)$ | $O(N^2)$ | $O(1)$ | +| 插入排序 | $\Omega(N)$ | $\Theta(N^2)$ | $O(N^2)$ | $O(1)$ | +| 选择排序 | $\Omega(N^2)$ | $\Theta(N^2)$ | $O(N^2)$ | $O(1)$ | +| 快速排序 | $\Omega(N \log N )$ | $\Theta(N \log N)$ | $O(N^2)$ | $O(\log N)$ | +| 归并排序 | $\Omega(N \log N)$ | $\Theta(N \log N)$ | $O(N \log N)$ | $O(N)$ | +| 堆排序 | $\Omega(N \log N)$ | $\Theta(N \log N)$ | $O(N \log N)$ | $O(1)$ | +| 基数排序 | $\Omega(Nk)$ | $\Theta(Nk)$ | $O(Nk)$ | $O(N + k)$ | +| 桶排序 | $\Omega(N + k)$ | $\Theta(N + k)$ | $O(N^2)$ | $O(N)$ | + +对于上表，需要特别注意： + +- 「基数排序」适用于正整数、字符串、特定格式的浮点数排序，$k$ 为最大数字的位数；「桶排序」中 $k$ 为桶的数量。 +- 普通「冒泡排序」的最佳时间复杂度为 $O(N^2)$ ，通过增加标志位实现 **提前返回** ，可以将最佳时间复杂度降低至 $O(N)$ 。 +- 在输入列表完全倒序下，普通「快速排序」的空间复杂度劣化至 $O(N)$ ，通过代码优化 **尾递归优化** 保持算法递归较短子数组，可以将最差递归深度降低至 $\log N$ 。 +- 普通「快速排序」总以最左或最右元素为基准数，因此在输入列表有序或倒序下，时间复杂度劣化至 $O(N^2)$ ；通过 **随机选择基准数** ，可极大减少此类最差情况发生，尽可能地保持 $O(N \log N)$ 的时间复杂度。 +- 若输入列表是数组，则归并排序的空间复杂度为 $O(N)$ ；而若排序 **链表** ，则「归并排序」不需要借助额外辅助空间，空间复杂度可以降低至 $O(1)$ 。 diff --git "a/leetbook_ioa/docs/# 7.2 \345\206\222\346\263\241\346\216\222\345\272\217.md" "b/leetbook_ioa/docs/# 7.2 \345\206\222\346\263\241\346\216\222\345\272\217.md" new file mode 100755 index 0000000..98e9fe1 --- /dev/null +++ "b/leetbook_ioa/docs/# 7.2 \345\206\222\346\263\241\346\216\222\345\272\217.md" @@ -0,0 +1,125 @@ +# 冒泡排序 + +冒泡排序是最基础的排序算法，由于其直观性，经常作为首个介绍的排序算法。其原理为： + +- **内循环：** 使用相邻双指针 `j` , `j + 1` 从左至右遍历，依次比较相邻元素大小，若左元素大于右元素则将它们交换；遍历完成时，**最大元素会被交换至数组最右边** 。 +- **外循环：** 不断重复「内循环」，每轮将当前最大元素交换至 **剩余未排序数组最右边** ，直至所有元素都被交换至正确位置时结束。 + +如下图所示，首轮「内循环」后，数组最大元素已被交换至数组最右边；接下来，只需要完成数组剩余 $N - 1$ 个元素的排序即可（设数组元素数量为 $N$ ）。同理，对剩余 $N - 1$ 个元素执行「内循环」，可将第二大元素交换至剩余数组最右端，以此类推…… + + + +如下图所示，冒泡排序的「外循环」共 $N - 1$ 轮，每轮「内循环」都将当前最大元素交换至数组最右边，从而完成对整个数组的排序。 + +![Picture1.png](https://pic.leetcode-cn.com/1628616643-PvqWBM-Picture1.png){:width=550} + +如下图所示，为示例数组 `nums = [4, 1, 3, 1, 5, 2]` 的冒泡排序算法运行过程。 + + + +## 代码 + +```Python [] +def bubble_sort(nums): + N = len(nums) + for i in range(N - 1): # 外循环 + for j in range(N - i - 1): # 内循环 + if nums[j] > nums[j + 1]: + # 交换 nums[j], nums[j + 1] + nums[j], nums[j + 1] = nums[j + 1], nums[j] +``` + +```Java [] +void bubbleSort(int[] nums) { + int N = nums.length; + for (int i = 0; i < N - 1; i++) { // 外循环 + for (int j = 0; j < N - i - 1; j++) { // 内循环 + if (nums[j] > nums[j + 1]) { + // 交换 nums[j], nums[j + 1] + int tmp = nums[j]; + nums[j] = nums[j + 1]; + nums[j + 1] = tmp; + } + } + } +} +``` + +```C++ [] +void bubbleSort(vector &nums) { + int N = nums.size(); + for (int i = 0; i < N - 1; i++) { // 外循环 + for (int j = 0; j < N - i - 1; j++) { // 内循环 + if (nums[j] > nums[j + 1]) { + // 交换 nums[j], nums[j + 1] + swap(nums[j], nums[j + 1]); + } + } + } +} +``` + +## 算法特性 + +- **时间复杂度 $O(N^2)$ ：** + - **最佳 $\Omega(N)$ ：** 普通冒泡排序的时间复杂度恒为 $O(N^2)$ ，对于近似排序数组，通过加入标志位可实现提前返回（详情请见下文）。 + - **平均与最差 $O(N^2)$ ：**「外循环」共 $N - 1$ 轮，使用 $O(N)$ 时间；每轮「内循环」分别遍历 $N - 1$ , $N - 2$ , $\cdots$ , $2$ , $1$ 次，平均 $\frac{N}{2}$ 次，使用 $O(\frac{N}{2}) = O(N)$ 时间；因此，总体时间复杂度为 $O(N^2)$ 。 +- **空间复杂度 $O(1)$ ：** 只需原地交换元素，使用常数大小的额外空间。 +- 冒泡排序是通过不断 **交换元素** 实现排序（交换 2 个元素需要 3 次赋值操作），因此速度较慢； +- **原地：** 指针变量仅使用常数大小额外空间，空间复杂度为 $O(1)$ ； +- **稳定：** 元素值相同时不交换，因此不会改变相同元素的相对位置； +- **自适应：** 通过增加一个标志位 `flag` ，若某轮内循环未执行任何交换操作时，说明已经完成排序，因此直接返回。此优化使冒泡排序的最优时间复杂度达到 $O(N)$（当输入数组已排序时）； + +## 标志位优化 + +> 普通冒泡排序的时间复杂度恒为 $O(N^2)$​ ，与输入数组的元素分布无关。 + +通过增加一个标志位 `flag` ，若在某轮「内循环」中未执行任何交换操作，则说明数组已经完成排序，直接返回结果即可。 + +优化后的冒泡排序的最差和平均时间复杂度仍为 $O(N^2)$ ；在输入数组 **已排序** 时，达到 **最佳时间复杂度** $\Omega(N)$ 。 + +```Python [] +def bubble_sort(nums): + N = len(nums) + for i in range(N - 1): + flag = False # 初始化标志位 + for j in range(N - i - 1): + if nums[j] > nums[j + 1]: + nums[j], nums[j + 1] = nums[j + 1], nums[j] + flag = True # 记录交换元素 + if not flag: break # 内循环未交换任何元素，则跳出 +``` + +```Java [] +void bubbleSort(int[] nums) { + int N = nums.length; + for (int i = 0; i < N - 1; i++) { + boolean flag = false; // 初始化标志位 + for (int j = 0; j < N - i - 1; j++) { + if (nums[j] > nums[j + 1]) { + int tmp = nums[j]; + nums[j] = nums[j + 1]; + nums[j + 1] = tmp; + flag = true; // 记录交换元素 + } + } + if (!flag) break; // 内循环未交换任何元素，则跳出 + } +} +``` + +```C++ [] +void bubbleSort(vector &nums) { + int N = nums.size(); + for (int i = 0; i < N - 1; i++) { + bool flag = false; // 初始化标志位 + for (int j = 0; j < N - i - 1; j++) { + if (nums[j] > nums[j + 1]) { + swap(nums[j], nums[j + 1]); + flag = true; // 记录交换元素 + } + } + if (!flag) break; // 内循环未交换任何元素，则跳出 + } +} +``` diff --git "a/leetbook_ioa/docs/# 7.3 \345\277\253\351\200\237\346\216\222\345\272\217.md" "b/leetbook_ioa/docs/# 7.3 \345\277\253\351\200\237\346\216\222\345\272\217.md" new file mode 100755 index 0000000..fad9abb --- /dev/null +++ "b/leetbook_ioa/docs/# 7.3 \345\277\253\351\200\237\346\216\222\345\272\217.md" @@ -0,0 +1,244 @@ +# 快速排序 + +快速排序算法有两个核心点，分别为 **哨兵划分** 和 **递归** 。 + +**哨兵划分**：以数组某个元素（一般选取首元素）为 **基准数** ，将所有小于基准数的元素移动至其左边，大于基准数的元素移动至其右边。 + +> 下图展示了哨兵划分操作流程。经过一轮 **哨兵划分** ，可将数组排序问题拆分为 **两个较短数组的排序问题** （本文称之为左（右）子数组）。 + + + +**递归**：对 **左子数组** 和 **右子数组** 分别递归执行 **哨兵划分**，直至子数组长度为 1 时终止递归，即可完成对整个数组的排序。 + +> 下图展示了数组 `[2,4,1,0,3,5]` 的快速排序流程。观察发现，快速排序和 **二分法** 的原理类似，都是以 $\log$ 时间复杂度实现搜索区间缩小。 + +![Picture1.png](https://pic.leetcode-cn.com/1612615552-rifQwI-Picture1.png){:width=550} + +## 代码 + +```Python [] +def quick_sort(nums, l, r): + # 子数组长度为 1 时终止递归 + if l >= r: return + # 哨兵划分操作 + i = partition(nums, l, r) + # 递归左（右）子数组执行哨兵划分 + quick_sort(nums, l, i - 1) + quick_sort(nums, i + 1, r) + +def partition(nums, l, r): + # 以 nums[l] 作为基准数 + i, j = l, r + while i < j: + while i < j and nums[j] >= nums[l]: j -= 1 + while i < j and nums[i] <= nums[l]: i += 1 + nums[i], nums[j] = nums[j], nums[i] + nums[l], nums[i] = nums[i], nums[l] + return i + +# 调用 +nums = [3, 4, 1, 5, 2] +quick_sort(nums, 0, len(nums) - 1) +``` + +```Java [] +void quickSort(int[] nums, int l, int r) { + // 子数组长度为 1 时终止递归 + if (l >= r) return; + // 哨兵划分操作 + int i = partition(nums, l, r); + // 递归左（右）子数组执行哨兵划分 + quickSort(nums, l, i - 1); + quickSort(nums, i + 1, r); +} + +int partition(int[] nums, int l, int r) { + // 以 nums[l] 作为基准数 + int i = l, j = r; + while (i < j) { + while (i < j && nums[j] >= nums[l]) j--; + while (i < j && nums[i] <= nums[l]) i++; + swap(nums, i, j); + } + swap(nums, i, l); + return i; +} + +void swap(int[] nums, int i, int j) { + // 交换 nums[i] 和 nums[j] + int tmp = nums[i]; + nums[i] = nums[j]; + nums[j] = tmp; +} + +// 调用 +int[] nums = { 4, 1, 3, 2, 5 }; +quickSort(nums, 0, nums.length - 1); +``` + +```C++ [] +int partition(vector& nums, int l, int r) { + // 以 nums[l] 作为基准数 + int i = l, j = r; + while (i < j) { + while (i < j && nums[j] >= nums[l]) j--; + while (i < j && nums[i] <= nums[l]) i++; + swap(nums[i], nums[j]); + } + swap(nums[i], nums[l]); + return i; +} + +void quickSort(vector& nums, int l, int r) { + // 子数组长度为 1 时终止递归 + if (l >= r) return; + // 哨兵划分操作 + int i = partition(nums, l, r); + // 递归左（右）子数组执行哨兵划分 + quickSort(nums, l, i - 1); + quickSort(nums, i + 1, r); +} + +// 调用 +vector nums = { 4, 1, 3, 2, 5, 1 }; +quickSort(nums, 0, nums.size() - 1); +``` + +## 算法特性 + +- **时间复杂度：** + - **最佳 $\Omega(N \log N )$ ：** 最佳情况下， 每轮哨兵划分操作将数组划分为等长度的两个子数组；哨兵划分操作为线性时间复杂度 $O(N)$ ；递归轮数共 $O(\log N)$ 。 + - **平均 $\Theta(N \log N)$ ：** 对于随机输入数组，哨兵划分操作的递归轮数也为 $O(\log N)$ 。 + - **最差 $O(N^2)$ ：** 对于某些特殊输入数组，每轮哨兵划分操作都将长度为 $N$ 的数组划分为长度为 $1$ 和 $N - 1$ 的两个子数组，此时递归轮数达到 $N$ 。 + > 通过 「随机选择基准数」优化，可尽可能避免出现最差情况，详情请见下文。 +- **空间复杂度 $O(N)$ ：** 快速排序的递归深度最好与平均皆为 $\log N$ ；输入数组完全倒序下，达到最差递归深度 $N$ 。 + > 通过「尾递归」优化，可将最差空间复杂度降低至 $O(\log N)$ ，详情请见下文。 +- 虽然平均时间复杂度与「归并排序」和「堆排序」一致，但在实际使用中快速排序 **效率更高** ，这是因为： + - **最差情况稀疏性：** 虽然快速排序的最差时间复杂度为 $O(N^2)$ ，差于归并排序和堆排序，但统计意义上看，这种情况出现的机率很低。大部分情况下，快速排序以 $O(N \log N)$ 复杂度运行。 + - **缓存使用效率高：** 哨兵划分操作时，将整个子数组加载入缓存中，访问元素效率很高；堆排序需要跳跃式访问元素，因此不具有此特性。 + - **常数系数低：** 在提及的三种算法中，快速排序的 **比较**、**赋值**、**交换** 三种操作的综合耗时最低（类似于插入排序快于冒泡排序的原理）。 +- **原地：** 不用借助辅助数组的额外空间，递归仅使用 $O(\log N)$ 大小的栈帧空间。 +- **非稳定：** 哨兵划分操作可能改变相等元素的相对顺序。 +- **自适应：** 对于极少输入数据，每轮哨兵划分操作都将长度为 $N$ 的数组划分为长度 $1$ 和 $N - 1$ 两个子数组，此时时间复杂度劣化至 $O(N^2)$ 。 + +## 算法优化 + +快速排序的常见优化手段有「尾递归」和「随机基准数」两种。 + +### 尾递归： + +由于普通快速排序每轮选取「子数组最左元素」作为「基准数」，因此在输入数组 **完全倒序** 时， `partition()` 的递归深度会达到 $N$ ，即 **最差空间复杂度** 为 $O(N)$ 。 + +每轮递归时，仅对 **较短的子数组** 执行哨兵划分 `partition()` ，就可将最差的递归深度控制在 $O(\log N)$ （每轮递归的子数组长度都 $\leq$ 当前数组长度 $/ 2$ ），即实现最差空间复杂度 $O(\log N)$ 。 + +> 代码仅需修改 `quick_sort()` 方法，其余方法不变，在此省略。 + +```Python [] +def quick_sort(nums, l, r): + # 子数组长度为 1 时终止递归 + while l < r: + # 哨兵划分操作 + i = partition(nums, l, r) + # 仅递归至较短子数组，控制递归深度 + if i - l < r - i: + quick_sort(nums, l, i - 1) + l = i + 1 + else: + quick_sort(nums, i + 1, r) + r = i - 1 +``` + +```Java [] +void quickSort(int[] nums, int l, int r) { + // 子数组长度为 1 时终止递归 + while (l < r) { + // 哨兵划分操作 + int i = partition(nums, l, r); + // 仅递归至较短子数组，控制递归深度 + if (i - l < r - i) { + quickSort(nums, l, i - 1); + l = i + 1; + } else { + quickSort(nums, i + 1, r); + r = i - 1; + } + } +} +``` + +```C++ [] +void quickSort(vector& nums, int l, int r) { + // 子数组长度为 1 时终止递归 + while (l < r) { + // 哨兵划分操作 + int i = partition(nums, l, r); + // 仅递归至较短子数组，控制递归深度 + if (i - l < r - i) { + quickSort(nums, l, i - 1); + l = i + 1; + } else { + quickSort(nums, i + 1, r); + r = i - 1; + } + } +} +``` + +### 随机基准数： + +同样地，由于快速排序每轮选取「子数组最左元素」作为「基准数」，因此在输入数组 **完全有序** 或 **完全倒序** 时， `partition()` 每轮只划分一个元素，达到最差时间复杂度 $O(N^2)$ 。 + +因此，可使用 **随机函数** ，每轮在子数组中随机选择一个元素作为基准数，这样就可以极大概率避免以上劣化情况。 + +值得注意的是，由于仍然可能出现最差情况，因此快速排序的最差时间复杂度仍为 $O(N^2)$ 。 + +> 代码仅需修改 `partition()` 方法，其余方法不变，在此省略。 + +```Python [] +def partition(nums, l, r): + # 在闭区间 [l, r] 随机选取任意索引，并与 nums[l] 交换 + ra = random.randrange(l, r + 1) + nums[l], nums[ra] = nums[ra], nums[l] + # 以 nums[l] 作为基准数 + i, j = l, r + while i < j: + while i < j and nums[j] >= nums[l]: j -= 1 + while i < j and nums[i] <= nums[l]: i += 1 + nums[i], nums[j] = nums[j], nums[i] + nums[l], nums[i] = nums[i], nums[l] + return i +``` + +```Java [] +int partition(int[] nums, int l, int r) { + // 在闭区间 [l, r] 随机选取任意索引，并与 nums[l] 交换 + int ra = (int)(l + Math.random() * (r - l + 1)); + swap(nums, l, ra); + // 以 nums[l] 作为基准数 + int i = l, j = r; + while (i < j) { + while (i < j && nums[j] >= nums[l]) j--; + while (i < j && nums[i] <= nums[l]) i++; + swap(nums, i, j); + } + swap(nums, i, l); + return i; +} +``` + +```C++ [] +int partition(vector& nums, int l, int r) { + // 在闭区间 [l, r] 随机选取任意索引，并与 nums[l] 交换 + int ra = l + rand() % (r - l + 1); + swap(nums[l], nums[ra]); + // 以 nums[l] 作为基准数 + int i = l, j = r; + while (i < j) { + while (i < j && nums[j] >= nums[l]) j--; + while (i < j && nums[i] <= nums[l]) i++; + swap(nums[i], nums[j]); + } + swap(nums[i], nums[l]); + return i; +} +``` diff --git "a/leetbook_ioa/docs/# 7.4 \345\275\222\345\271\266\346\216\222\345\272\217.md" "b/leetbook_ioa/docs/# 7.4 \345\275\222\345\271\266\346\216\222\345\272\217.md" new file mode 100755 index 0000000..5c0ceaf --- /dev/null +++ "b/leetbook_ioa/docs/# 7.4 \345\275\222\345\271\266\346\216\222\345\272\217.md" @@ -0,0 +1,130 @@ +# 归并排序 + +归并排序体现了 “分而治之” 的算法思想，具体为： + +- **「分」：** 不断将数组从 **中点位置** 划分开，将原数组的排序问题转化为子数组的排序问题； +- **「治」：** 划分到子数组长度为 1 时，开始向上合并，不断将 **左右两个较短排序数组** 合并为 **一个较长排序数组**，直至合并至原数组时完成排序； + +> 如下图所示，为数组 `[7,3,2,6,0,1,5,4]` 的归并排序过程。 + +![Picture1.png](https://pic.leetcode-cn.com/1632675739-CNHaOu-Picture1.png){:width=500} + +## 算法流程 + +1. **递归划分：** + 1. 计算数组中点 $m$ ，递归划分左子数组 `merge_sort(l, m)` 和右子数组 `merge_sort(m + 1, r)` ； + 2. 当 $l \geq r$ 时，代表子数组长度为 1 或 0 ，此时 **终止划分** ，开始合并； + +2. **合并子数组：** + 1. 暂存数组 $nums$ 闭区间 $[l, r]$ 内的元素至辅助数组 $tmp$ ； + 2. **循环合并：** 设置双指针 $i$ , $j$ 分别指向 $tmp$ 的左 / 右子数组的首元素； + > **注意：** $nums$ 子数组的左边界、中点、右边界分别为 $l$ , $m$ , $r$ ，而辅助数组 $tmp$ 中的对应索引为 $0$ , $m - l$ , $r - l$ ； + - **当 $i == m - l + 1$ 时：** 代表左子数组已合并完，因此添加右子数组元素 $tmp[j]$ ，并执行 $j = j + 1$ ； + - **否则，当 $j == r - l + 1$ 时：** 代表右子数组已合并完，因此添加左子数组元素 $tmp[i]$ ，并执行 $i = i + 1$ ； + - **否则，当 $tmp[i] \leq tmp[j]$ 时：** 添加左子数组元素 $tmp[i]$ ，并执行 $i = i + 1$ ； + - **否则（即当 $tmp[i] > tmp[j]$ 时）：** 添加右子数组元素 $tmp[j]$ ，并执行 $j = j + 1$ ； + +> 如下动图所示，为数组 `[7,3,2,6]` 的归并排序过程。 + + + +## 代码 + +为简化代码，「当 $j = r + 1$ 时」 与 「当 $tmp[i] \leq tmp[j]$ 时」 两判断项可合并。 + +```Python [] +def merge_sort(nums, l, r): + # 终止条件 + if l >= r: return + # 递归划分数组 + m = (l + r) // 2 + merge_sort(nums, l, m) + merge_sort(nums, m + 1, r) + # 合并子数组 + tmp = nums[l:r + 1] # 暂存需合并区间元素 + i, j = 0, m - l + 1 # 两指针分别指向左/右子数组的首个元素 + for k in range(l, r + 1): # 遍历合并左/右子数组 + if i == m - l + 1: + nums[k] = tmp[j] + j += 1 + elif j == r - l + 1 or tmp[i] <= tmp[j]: + nums[k] = tmp[i] + i += 1 + else: + nums[k] = tmp[j] + j += 1 + +# 调用 +nums = [3, 4, 1, 5, 2, 1] +merge_sort(0, len(nums) - 1) +``` + +```Java [] +void mergeSort(int[] nums, int l, int r) { + // 终止条件 + if (l >= r) return; + // 递归划分 + int m = (l + r) / 2; + mergeSort(nums, l, m); + mergeSort(nums, m + 1, r); + // 合并子数组 + int[] tmp = new int[r - l + 1]; // 暂存需合并区间元素 + for (int k = l; k <= r; k++) + tmp[k - l] = nums[k]; + int i = 0, j = m - l + 1; // 两指针分别指向左/右子数组的首个元素 + for (int k = l; k <= r; k++) { // 遍历合并左/右子数组 + if (i == m - l + 1) + nums[k] = tmp[j++]; + else if (j == r - l + 1 || tmp[i] <= tmp[j]) + nums[k] = tmp[i++]; + else { + nums[k] = tmp[j++]; + } + } +} + +// 调用 +int[] nums = { 3, 4, 1, 5, 2, 1 }; +mergeSort(nums, 0, len(nums) - 1); +``` + +```C++ [] +void mergeSort(vector& nums, int l, int r) { + // 终止条件 + if (l >= r) return; + // 递归划分 + int m = (l + r) / 2; + mergeSort(nums, l, m); + mergeSort(nums, m + 1, r); + // 合并阶段 + int tmp[r - l + 1]; // 暂存需合并区间元素 + for (int k = l; k <= r; k++) + tmp[k - l] = nums[k]; + int i = 0, j = m - l + 1; // 两指针分别指向左/右子数组的首个元素 + for (int k = l; k <= r; k++) { // 遍历合并左/右子数组 + if (i == m - l + 1) + nums[k] = tmp[j++]; + else if (j == r - l + 1 || tmp[i] <= tmp[j]) + nums[k] = tmp[i++]; + else { + nums[k] = tmp[j++]; + } + } +} + +// 调用 +vector nums = { 4, 1, 3, 2, 5, 1 }; +mergeSort(nums, 0, nums.size() - 1); +``` + +## 算法特性 + +- **时间复杂度：** 最佳 $\Omega(N \log N )$ ，平均 $\Theta(N \log N)$ ，最差 $O(N \log N)$ 。 +- **空间复杂度 $O(N)$ ：** 合并过程中需要借助辅助数组 $tmp$ ，使用 $O(N)$ 大小的额外空间；划分的递归深度为 $\log N$ ，使用 $O(\log N)$ 大小的栈帧空间。 +- 若输入数据是 **链表** ，则归并排序的空间复杂度可被优化至 $O(1)$ ，这是因为： + - 通过应用「双指针法」，可在 $O(1)$ 空间下完成两个排序链表的合并，省去辅助数组 $tmp$ 使用的额外空间； + - 通过使用「迭代」代替「递归划分」，可省去递归使用的栈帧空间； + > 详情请参考：[148. 排序链表](https://leetcode-cn.com/problems/sort-list/solution/sort-list-gui-bing-pai-xu-lian-biao-by-jyd/) +- **非原地：** 辅助数组 $tmp$ 需要使用额外空间。 +- **稳定：** 归并排序不改变相等元素的相对顺序。 +- **非自适应：** 对于任意输入数据，归并排序的时间复杂度皆相同。 diff --git "a/leetbook_ioa/docs/LCR 120. \345\257\273\346\211\276\346\226\207\344\273\266\345\211\257\346\234\254.md" "b/leetbook_ioa/docs/LCR 120. \345\257\273\346\211\276\346\226\207\344\273\266\345\211\257\346\234\254.md" new file mode 100755 index 0000000..646d56a --- /dev/null +++ "b/leetbook_ioa/docs/LCR 120. \345\257\273\346\211\276\346\226\207\344\273\266\345\211\257\346\234\254.md" @@ -0,0 +1,140 @@ +## 方法一：哈希表 + +利用数据结构特点，容易想到使用哈希表（Set）记录数组的各个数字，当查找到重复数字则直接返回。 + +### 算法流程： + +1. 初始化： 新建 HashSet ，记为 $hmap$ ； +2. 遍历数组 $documents$ 中的每个数字 $doc$ ： + 1. 当 $doc$ 在 $hmap$ 中，说明重复，直接返回 $doc$ ； + 2. 将 $doc$ 添加至 $hmap$ 中； +3. 返回 $-1$ 。本题中一定有重复数字，因此这里返回多少都可以。 + +> 下图中的 `nums` 对应本题的 `documents` 。 + + + +### 代码： + +```Python [] +class Solution: + def findRepeatDocument(self, documents: List[int]) -> int: + hmap = set() + for doc in documents: + if doc in hmap: return doc + hmap.add(doc) + return -1 +``` + +```Java [] +class Solution { + public int findRepeatDocument(int[] documents) { + Set hmap = new HashSet<>(); + for(int doc : documents) { + if(hmap.contains(doc)) return doc; + hmap.add(doc); + } + return -1; + } +} +``` + +```C++ [] +class Solution { +public: + int findRepeatDocument(vector& documents) { + unordered_map map; + for(int doc : documents) { + if(map[doc]) return doc; + map[doc] = true; + } + return -1; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 遍历数组使用 $O(N)$ ，HashSet 添加与查找元素皆为 $O(1)$ 。 +- **空间复杂度 $O(N)$ ：** HashSet 占用 $O(N)$ 大小的额外空间。 + +## 方法二：原地交换 + +题目说明尚未被充分使用，即 `在一个长度为 n 的数组 documents 里的所有数字都在 0 ~ n-1 的范围内` 。 此说明含义：数组元素的 **索引** 和 **值** 是 **一对多** 的关系。 +因此，可遍历数组并通过交换操作，使元素的 **索引** 与 **值** 一一对应（即 $documents[i] = i$ ）。因而，就能通过索引映射对应的值，起到与字典等价的作用。 + +![Picture0.png](https://pic.leetcode-cn.com/1618146573-bOieFQ-Picture0.png){:align=center width=500} + +遍历中，第一次遇到数字 $x$ 时，将其交换至索引 $x$ 处；而当第二次遇到数字 $x$ 时，一定有 $documents[x] = x$ ，此时即可得到一组重复数字。 + +### 算法流程： + +1. 遍历数组 $documents$ ，设索引初始值为 $i = 0$ : + 1. **若 $documents[i] = i$ ：** 说明此数字已在对应索引位置，无需交换，因此跳过； + 2. **若 $documents[documents[i]] = documents[i]$ ：** 代表索引 $documents[i]$ 处和索引 $i$ 处的元素值都为 $documents[i]$ ，即找到一组重复值，返回此值 $documents[i]$ ； + 3. **否则：** 交换索引为 $i$ 和 $documents[i]$ 的元素值，将此数字交换至对应索引位置。 + +2. 若遍历完毕尚未返回，则返回 $-1$ 。 + + + +### 代码： + +Python 中，$a, b = c, d$ 操作的原理是先暂存元组 $(c, d)$ ，然后 “按左右顺序” 赋值给 a 和 b 。 +因此，若写为 $documents[i], documents[documents[i]] = documents[documents[i]], documents[i]$ ，则 $documents[i]$ 会先被赋值，之后 $documents[documents[i]]$ 指向的元素则会出错。 + +```Python [] +class Solution: + def findRepeatDocument(self, documents: List[int]) -> int: + i = 0 + while i < len(documents): + if documents[i] == i: + i += 1 + continue + if documents[documents[i]] == documents[i]: return documents[i] + documents[documents[i]], documents[i] = documents[i], documents[documents[i]] + return -1 +``` + +```Java [] +class Solution { + public int findRepeatDocument(int[] documents) { + int i = 0; + while(i < documents.length) { + if(documents[i] == i) { + i++; + continue; + } + if(documents[documents[i]] == documents[i]) return documents[i]; + int tmp = documents[i]; + documents[i] = documents[tmp]; + documents[tmp] = tmp; + } + return -1; + } +} +``` + +```C++ [] +class Solution { +public: + int findRepeatDocument(vector& documents) { + int i = 0; + while(i < documents.size()) { + if(documents[i] == i) { + i++; + continue; + } + if(documents[documents[i]] == documents[i]) + return documents[i]; + swap(documents[i],documents[documents[i]]); + } + return -1; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 遍历数组使用 $O(N)$ ，每轮遍历的判断和交换操作使用 $O(1)$ 。 +- **空间复杂度 $O(1)$ ：** 使用常数复杂度的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 121. \345\257\273\346\211\276\347\233\256\346\240\207\345\200\274 - \344\272\214\347\273\264\346\225\260\347\273\204.md" "b/leetbook_ioa/docs/LCR 121. \345\257\273\346\211\276\347\233\256\346\240\207\345\200\274 - \344\272\214\347\273\264\346\225\260\347\273\204.md" new file mode 100755 index 0000000..17898e7 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 121. \345\257\273\346\211\276\347\233\256\346\240\207\345\200\274 - \344\272\214\347\273\264\346\225\260\347\273\204.md" @@ -0,0 +1,72 @@ +## 解题思路： + +> 若使用暴力法遍历矩阵 `plants` ，则时间复杂度为 $O(NM)$ 。暴力法未利用矩阵 **“从上到下递增、从左到右递增”** 的特点，显然不是最优解法。 + +如下图所示，我们将矩阵逆时针旋转 45° ，并将其转化为图形式，发现其类似于 **二叉搜索树** ，即对于每个元素，其左分支元素更小、右分支元素更大。 + +因此，考虑从 “根节点” 开始搜索，遇到比 `target` 大的元素就向左，反之向右，即可找到目标值 `target` 。 + +![Picture1.png](https://pic.leetcode-cn.com/6584ea93812d27112043d203ea90e4b0950117d45e0452d0c630fcb247fbc4af-Picture1.png){:align=center width=450} + +### 算法流程： + +“根节点” 对应的是矩阵的 “左下角” 和 “右上角” 元素。以 `plants` 中的 **左下角元素** 为起始点，则有: + +1. 从矩阵 `plants` 左下角元素（索引设为 `(i, j)` ）开始遍历，并与目标值对比： + - 当 `plants[i][j] > target` 时，执行 `i--` ，即消去第 `i` 行元素； + - 当 `plants[i][j] < target` 时，执行 `j++` ，即消去第 `j` 列元素； + - 当 `plants[i][j] = target` 时，返回 $\text{true}$ ，代表找到目标值。 +2. 若行索引或列索引越界，则代表矩阵中无目标值，返回 $\text{false}$ 。 + +> 每轮 `i` 或 `j` 移动后，相当于生成了“消去一行（列）的新矩阵”， 索引`(i,j)` 指向新矩阵的左下角元素，因此可重复使用以上性质消去行（列）。 + + + +## 代码： + +```Python [] +class Solution: + def findTargetIn2DPlants(self, plants: List[List[int]], target: int) -> bool: + i, j = len(plants) - 1, 0 + while i >= 0 and j < len(plants[0]): + if plants[i][j] > target: i -= 1 + elif plants[i][j] < target: j += 1 + else: return True + return False +``` + +```Java [] +class Solution { + public boolean findTargetIn2DPlants(int[][] plants, int target) { + int i = plants.length - 1, j = 0; + while(i >= 0 && j < plants[0].length) + { + if(plants[i][j] > target) i--; + else if(plants[i][j] < target) j++; + else return true; + } + return false; + } +} +``` + +```C++ [] +class Solution { +public: + bool findTargetIn2DPlants(vector>& plants, int target) { + int i = plants.size() - 1, j = 0; + while(i >= 0 && j < plants[0].size()) + { + if(plants[i][j] > target) i--; + else if(plants[i][j] < target) j++; + else return true; + } + return false; + } +}; +``` + +### 复杂度分析： + +- 时间复杂度 $O(M+N)$ ：其中，$N$ 和 $M$ 分别为矩阵行数和列数，此算法最多循环 $M+N$ 次。 +- 空间复杂度 $O(1)$ : `i`, `j` 指针使用常数大小额外空间。 diff --git "a/leetbook_ioa/docs/LCR 122. \350\267\257\345\276\204\345\212\240\345\257\206.md" "b/leetbook_ioa/docs/LCR 122. \350\267\257\345\276\204\345\212\240\345\257\206.md" new file mode 100755 index 0000000..fcf1f3d --- /dev/null +++ "b/leetbook_ioa/docs/LCR 122. \350\267\257\345\276\204\345\212\240\345\257\206.md" @@ -0,0 +1,46 @@ +## 方法一：遍历添加 + +在 Python 和 Java 等语言中，字符串都被设计成「不可变」的类型，即无法直接修改字符串的某一位字符，需要新建一个字符串实现。 + +### 算法流程： + +1. 初始化一个 `list` (Python) 或 `StringBuilder` (Java) ，记为 `res` ； +2. 遍历列表 `path` 中的每个字符 `c` ： + - 当 `c` 为空格时：向 `res` 后添加空格 " " ； + - 当 `c` 不为空格时：向 `res` 后添加字符 `c` ； +3. 将列表 `res` 转化为字符串并返回。 + +> 下图中的 `s` 对应本题的 `path` 。 + + + +## 代码： + +```Python [] +class Solution: + def pathEncryption(self, path: str) -> str: + res = [] + for c in path: + if c == '.': res.append(' ') + else: res.append(c) + return "".join(res) +``` + +```Java [] +class Solution { + public String pathEncryption(String path) { + StringBuilder res = new StringBuilder(); + for(Character c : path.toCharArray()) + { + if(c == '.') res.append(' '); + else res.append(c); + } + return res.toString(); + } +} +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 遍历使用 $O(N)$ ，每轮添加（修改）字符操作使用 $O(1)$ ； +- **空间复杂度 $O(N)$ ：** Python 新建的 list 和 Java 新建的 StringBuilder 都使用了线性大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 123. \345\233\276\344\271\246\346\225\264\347\220\206 I.md" "b/leetbook_ioa/docs/LCR 123. \345\233\276\344\271\246\346\225\264\347\220\206 I.md" new file mode 100755 index 0000000..197f59e --- /dev/null +++ "b/leetbook_ioa/docs/LCR 123. \345\233\276\344\271\246\346\225\264\347\220\206 I.md" @@ -0,0 +1,126 @@ +## 方法一：递归 + +利用递归，先递推至链表末端；回溯时，依次将节点值加入列表，即可实现链表值的倒序输出。 + +1. **终止条件：** 当 `head == None` 时，代表越过了链表尾节点，则返回空列表； +2. **递推工作：** 访问下一节点 `head.next` ； +3. **回溯阶段：** + - **Python：** 返回 `当前 list + 当前节点值 [head.val]` ； + - **Java / C++：** 将当前节点值 `head.val` 加入列表 `tmp` ； + + + +### 代码： + +```Python [] +class Solution: + def reverseBookList(self, head: Optional[ListNode]) -> List[int]: + return self.reverseBookList(head.next) + [head.val] if head else [] +``` + +```Java [] +class Solution { + ArrayList tmp = new ArrayList(); + public int[] reverseBookList(ListNode head) { + recur(head); + int[] res = new int[tmp.size()]; + for(int i = 0; i < res.length; i++) + res[i] = tmp.get(i); + return res; + } + void recur(ListNode head) { + if(head == null) return; + recur(head.next); + tmp.add(head.val); + } +} +``` + +```C++ [] +class Solution { +public: + vector reverseBookList(ListNode* head) { + recur(head); + return res; + } +private: + vector res; + void recur(ListNode* head) { + if(head == nullptr) return; + recur(head->next); + res.push_back(head->val); + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$：** 遍历链表，递归 $N$ 次。 +- **空间复杂度 $O(N)$：** 系统递归需要使用 $O(N)$ 的栈空间。 + +## 方法二：辅助栈法 + +链表只能 **从前至后** 访问每个节点，而题目要求 **倒序输出** 各节点值，这种 **先入后出** 的需求可以借助 **栈** 来实现。 + +### 算法流程： + +1. **入栈：** 遍历链表，将各节点值 `push` 入栈。 +2. **出栈：** 将各节点值 `pop` 出栈，存储于数组并返回。 + +> 图解以 Java 代码为例，Python 无需将 `stack` 转移至 `res`，而是直接返回倒序数组。 + + + +### 代码： + +Java 数组长度不可变，因此使用 List 先存储，再转为数组并返回。 + +```Python [] +class Solution: + def reverseBookList(self, head: ListNode) -> List[int]: + stack = [] + while head: + stack.append(head.val) + head = head.next + return stack[::-1] +``` + +```Java [] +class Solution { + public int[] reverseBookList(ListNode head) { + LinkedList stack = new LinkedList(); + while(head != null) { + stack.addLast(head.val); + head = head.next; + } + int[] res = new int[stack.size()]; + for(int i = 0; i < res.length; i++) + res[i] = stack.removeLast(); + return res; + } +} +``` + +```C++ [] +class Solution { +public: + vector reverseBookList(ListNode* head) { + stack stk; + while(head != nullptr) { + stk.push(head->val); + head = head->next; + } + vector res; + while(!stk.empty()) { + res.push_back(stk.top()); + stk.pop(); + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$：** 入栈和出栈共使用 $O(N)$ 时间。 +- **空间复杂度 $O(N)$：** 辅助栈 `stack` 和数组 `res` 共使用 $O(N)$ 的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 124. \346\216\250\347\220\206\344\272\214\345\217\211\346\240\221.md" "b/leetbook_ioa/docs/LCR 124. \346\216\250\347\220\206\344\272\214\345\217\211\346\240\221.md" new file mode 100755 index 0000000..d91cbf9 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 124. \346\216\250\347\220\206\344\272\214\345\217\211\346\240\221.md" @@ -0,0 +1,117 @@ +## 解题思路： + +前序遍历性质： 节点按照 `[ 根节点 | 左子树 | 右子树 ]` 排序。 +中序遍历性质： 节点按照 `[ 左子树 | 根节点 | 右子树 ]` 排序。 + +> 以题目示例为例： +> +> - 前序遍历划分 `[ 3 | 9 | 20 15 7 ]` +> - 中序遍历划分 `[ 9 | 3 | 15 20 7 ]` + +根据以上性质，可得出以下推论： + +1. 前序遍历的首元素 为 树的根节点 `node` 的值。 +2. 在中序遍历中搜索根节点 `node` 的索引 ，可将 中序遍历 划分为 `[ 左子树 | 根节点 | 右子树 ]` 。 +3. 根据中序遍历中的左（右）子树的节点数量，可将 前序遍历 划分为 `[ 根节点 | 左子树 | 右子树 ] ` 。 + +![Picture1.png](https://pic.leetcode-cn.com/1629825510-roByLr-Picture1.png){:align=center width=550} + +通过以上三步，可确定 **三个节点** ：1.树的根节点、2.左子树根节点、3.右子树根节点。 + +根据「分治算法」思想，对于树的左、右子树，仍可复用以上方法划分子树的左右子树。 + +### 分治解析： + +**递推参数：** 根节点在前序遍历的索引 `root` 、子树在中序遍历的左边界 `left` 、子树在中序遍历的右边界 `right` ； + +**终止条件：** 当 `left > right` ，代表已经越过叶节点，此时返回 $\text{null}$ ； + +**递推工作：** + +1. **建立根节点 `node` ：** 节点值为 `preorder[root]` ； +2. **划分左右子树：** 查找根节点在中序遍历 `inorder` 中的索引 `i` ； + +> 为了提升效率，本文使用哈希表 `hmap` 存储中序遍历的值与索引的映射，查找操作的时间复杂度为 $O(1)$ ； + +3. **构建左右子树：** 开启左右子树递归； + +| | 根节点索引 | 中序遍历左边界 | 中序遍历右边界 | +| ---------- | --------------------- | -------------- | -------------- | +| **左子树** | `root + 1` | `left` | `i - 1` | +| **右子树** | `i - left + root + 1` | `i + 1` | `right` | + +> **TIPS：** `i - left + root + 1`含义为 `根节点索引 + 左子树长度 + 1` + +**返回值：** 回溯返回 `node` ，作为上一层递归中根节点的左 / 右子节点； + + + +## 代码： + +> 注意：本文方法只适用于 “无重复节点值” 的二叉树。 + +```Python [] +class Solution: + def deduceTree(self, preorder: List[int], inorder: List[int]) -> TreeNode: + def recur(root, left, right): + if left > right: return # 递归终止 + node = TreeNode(preorder[root]) # 建立根节点 + i = hmap[preorder[root]] # 划分根节点、左子树、右子树 + node.left = recur(root + 1, left, i - 1) # 开启左子树递归 + node.right = recur(i - left + root + 1, i + 1, right) # 开启右子树递归 + return node # 回溯返回根节点 + + hmap, preorder = {}, preorder + for i in range(len(inorder)): + hmap[inorder[i]] = i + return recur(0, 0, len(inorder) - 1) +``` + +```Java [] +class Solution { + int[] preorder; + HashMap hmap = new HashMap<>(); + public TreeNode deduceTree(int[] preorder, int[] inorder) { + this.preorder = preorder; + for(int i = 0; i < inorder.length; i++) + hmap.put(inorder[i], i); + return recur(0, 0, inorder.length - 1); + } + TreeNode recur(int root, int left, int right) { + if(left > right) return null; // 递归终止 + TreeNode node = new TreeNode(preorder[root]); // 建立根节点 + int i = hmap.get(preorder[root]); // 划分根节点、左子树、右子树 + node.left = recur(root + 1, left, i - 1); // 开启左子树递归 + node.right = recur(root + i - left + 1, i + 1, right); // 开启右子树递归 + return node; // 回溯返回根节点 + } +} +``` + +```C++ [] +class Solution { +public: + TreeNode* deduceTree(vector& preorder, vector& inorder) { + this->preorder = preorder; + for(int i = 0; i < inorder.size(); i++) + hmap[inorder[i]] = i; + return recur(0, 0, inorder.size() - 1); + } +private: + vector preorder; + unordered_map hmap; + TreeNode* recur(int root, int left, int right) { + if(left > right) return nullptr; // 递归终止 + TreeNode* node = new TreeNode(preorder[root]); // 建立根节点 + int i = hmap[preorder[root]]; // 划分根节点、左子树、右子树 + node->left = recur(root + 1, left, i - 1); // 开启左子树递归 + node->right = recur(root + i - left + 1, i + 1, right); // 开启右子树递归 + return node; // 回溯返回根节点 + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为树的节点数量。初始化 HashMap 需遍历 `inorder` ，占用 $O(N)$ 。递归共建立 $N$ 个节点，每层递归中的节点建立、搜索操作占用 $O(1)$ ，因此使用 $O(N)$ 时间。 +- **空间复杂度 $O(N)$ ：** HashMap 使用 $O(N)$ 额外空间；最差情况下（输入二叉树为链表时），递归深度达到 $N$ ，占用 $O(N)$ 的栈帧空间；因此总共使用 $O(N)$ 空间。 diff --git "a/leetbook_ioa/docs/LCR 125. \345\233\276\344\271\246\346\225\264\347\220\206 II.md" "b/leetbook_ioa/docs/LCR 125. \345\233\276\344\271\246\346\225\264\347\220\206 II.md" new file mode 100755 index 0000000..0d35601 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 125. \345\233\276\344\271\246\346\225\264\347\220\206 II.md" @@ -0,0 +1,97 @@ +## 解题思路： + +> 我们可将两个书车看作两个“栈”，本题可被转化为“用两个栈实现一个队列”。 + +栈实现队列的出队操作效率低下：栈底元素（对应队首元素）无法直接删除，需要将上方所有元素出栈。 + +列表倒序操作可使用双栈实现：设有含三个元素的栈 `A = [1,2,3]` 和空栈 `B = []` 。若循环执行 `A` 元素出栈并添加入栈 `B` ，直到栈 `A` 为空，则 `A = []` , `B = [3,2,1]` ，即栈 `B` 元素为栈 `A` 元素倒序。 + +利用栈 `B` 删除队首元素：倒序后，`B` 执行出栈则相当于删除了 `A` 的栈底元素，即对应队首元素。 + +![Picture1.png](https://pic.leetcode-cn.com/1599286207-HnnMhX-Picture1.png){:align=center width=500} + +题目要求实现 **加入队尾**`appendTail()` 和 **删除队首**`deleteHead()` 两个函数的正常工作。因此，可以设计栈 `A` 用于加入队尾操作，栈 `B` 用于将元素倒序，从而实现删除队首元素。 + +### 函数设计： + +1. **加入队尾 `appendTail()` ：** 将数字 `val` 加入栈 `A` 即可。 +2. **删除队首`deleteHead()` ：** 有以下三种情况。 + 1. **当栈 `B` 不为空：** `B`中仍有已完成倒序的元素，因此直接返回 `B` 的栈顶元素。 + 2. **否则，当 `A` 为空：** 即两个栈都为空，无元素，因此返回 -1 。 + 3. **否则：** 将栈 `A` 元素全部转移至栈 `B` 中，实现元素倒序，并返回栈 `B` 的栈顶元素。 + + + +## 代码： + +Python 和 Java 的栈的 `pop()` 函数返回栈顶元素，而 C++ 不返回；因此对于 C++ ，需要先使用 `top()` 方法暂存栈顶元素，再执行 `pop()` 出栈操作。 + +```Python [] +class CQueue: + def __init__(self): + self.A, self.B = [], [] + + def appendTail(self, value: int) -> None: + self.A.append(value) + + def deleteHead(self) -> int: + if self.B: return self.B.pop() + if not self.A: return -1 + while self.A: + self.B.append(self.A.pop()) + return self.B.pop() +``` + +```Java [] +class CQueue { + LinkedList A, B; + public CQueue() { + A = new LinkedList(); + B = new LinkedList(); + } + public void appendTail(int value) { + A.addLast(value); + } + public int deleteHead() { + if(!B.isEmpty()) return B.removeLast(); + if(A.isEmpty()) return -1; + while(!A.isEmpty()) + B.addLast(A.removeLast()); + return B.removeLast(); + } +} +``` + +```C++ [] +class CQueue { +public: + stack A, B; + CQueue() {} + void appendTail(int value) { + A.push(value); + } + int deleteHead() { + if(!B.empty()) { + int tmp = B.top(); + B.pop(); + return tmp; + } + if(A.empty()) return -1; + while(!A.empty()) { + int tmp = A.top(); + A.pop(); + B.push(tmp); + } + int tmp = B.top(); + B.pop(); + return tmp; + } +}; +``` + +### 复杂度分析： + +> 以下分析仅满足添加 $N$ 个元素并删除 $N$ 个元素，即栈初始和结束状态下都为空的情况。 + +- **时间复杂度：** `appendTail()`函数为 $O(1)$ ；`deleteHead()` 函数在 $N$ 次队首元素删除操作中总共需完成 $N$ 个元素的倒序。 +- **空间复杂度 $O(N)$ ：** 最差情况下，栈 `A` 和 `B` 共保存 $N$ 个元素。 diff --git "a/leetbook_ioa/docs/LCR 126. \346\226\220\346\263\242\351\202\243\345\245\221\346\225\260.md" "b/leetbook_ioa/docs/LCR 126. \346\226\220\346\263\242\351\202\243\345\245\221\346\225\260.md" new file mode 100755 index 0000000..c4e4a05 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 126. \346\226\220\346\263\242\351\202\243\345\245\221\346\225\260.md" @@ -0,0 +1,97 @@ +## 解题思路： + +斐波那契数列的定义是 $f(n + 1) = f(n) + f(n - 1)$ ，生成第 $n$ 项的做法有以下几种： + +1. **递归：** + - **原理：** 把 $f(n)$ 问题的计算拆分成 $f(n-1)$ 和 $f(n-2)$ 两个子问题的计算，并递归，以 $f(0)$ 和 $f(1)$ 为终止条件。 + - **缺点：** 大量重复的递归计算，例如 $f(n)$ 和 $f(n - 1)$ 两者向下递归需要 **各自计算** $f(n - 2)$ 的值。 +2. **记忆化递归：** + - **原理：** 在递归的基础上，新建一个长度为 $n$ 的数组，用于在递归时存储 $f(0)$ 至 $f(n)$ 的数字值，重复遇到某数字则直接从数组取用，避免了重复的递归计算。 + - **缺点：** 记忆化存储需要使用 $O(N)$ 的额外空间。 +3. **动态规划：** + - **原理：** 以斐波那契数列性质 $f(n + 1) = f(n) + f(n - 1)$ 为转移方程。 + - 从计算效率、空间复杂度上看，动态规划是本题的最佳解法。 + +> 下图帮助理解递归的 “重复计算” 概念。 + +![Picture1.png](https://pic.leetcode-cn.com/1599882883-mtYecf-Picture1.png){:align=center width=500} + +### 动态规划解析： + +- **状态定义：** 设 $dp$ 为一维数组，其中 $dp[i]$ 的值代表 斐波那契数列第 $i$ 个数字 。 +- **转移方程：** $dp[i + 1] = dp[i] + dp[i - 1]$ ，即对应数列定义 $f(n + 1) = f(n) + f(n - 1)$ ； +- **初始状态：** $dp[0] = 0$, $dp[1] = 1$ ，即初始化前两个数字； +- **返回值：** $dp[n]$ ，即斐波那契数列的第 $n$ 个数字。 + +### 空间优化： + +> 若新建长度为 $n$ 的 $dp$ 列表，则空间复杂度为 $O(N)$ 。 + +- 由于 $dp$ 列表第 $i$ 项只与第 $i-1$ 和第 $i-2$ 项有关，因此只需要初始化三个整形变量 `sum`, `a`, `b` ，利用辅助变量 $sum$ 使 $a, b$ 两数字交替前进即可 *（具体实现见代码）* 。 +- 节省了 $dp$ 列表空间，因此空间复杂度降至 $O(1)$ 。 + +### 循环求余法： + +> **大数越界：** 随着 $n$ 增大, $f(n)$ 会超过 `Int32` 甚至 `Int64` 的取值范围，导致最终的返回值错误。 + +- **求余运算规则：** 设正整数 $x, y, p$ ，求余符号为 $\odot$ ，则有 $(x + y) \odot p = (x \odot p + y \odot p) \odot p$ 。 +- **解析：** 根据以上规则，可推出 $f(n) \odot p = [f(n-1) \odot p + f(n-2) \odot p] \odot p$ ，从而可以在循环过程中每次计算 $sum = (a + b) \odot 1000000007$ ，此操作与最终返回前取余等价。 + + + +## 代码： + +```Python [] +class Solution: + def fib(self, n: int) -> int: + a, b = 0, 1 + for _ in range(n): + a, b = b, (a + b) % 1000000007 + return a +``` + +```Java [] +class Solution { + public int fib(int n) { + int a = 0, b = 1, sum; + for(int i = 0; i < n; i++){ + sum = (a + b) % 1000000007; + a = b; + b = sum; + } + return a; + } +} +``` + +```C++ [] +class Solution { +public: + int fib(int n) { + int a = 0, b = 1, sum; + for(int i = 0; i < n; i++){ + sum = (a + b) % 1000000007; + a = b; + b = sum; + } + return a; + } +}; +``` + +由于 Python 中整形数字的大小限制取决计算机的内存（可理解为无限大），因此也可不考虑大数越界问题；但当数字很大时，加法运算的效率也会降低，因此不推荐此方法。 + +```Python [] +# 不考虑大数越界问题 +class Solution: + def fib(self, n: int) -> int: + a, b = 0, 1 + for _ in range(n): + a, b = b, a + b + return a % 1000000007 +``` + +### 复杂度分析： + +- **时间复杂度 $O(n)$ ：** 计算 $f(n)$ 需循环 $n$ 次，每轮循环内计算操作使用 $O(1)$ 。 +- **空间复杂度 $O(1)$ ：** 几个标志变量使用常数大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 127. \350\267\263\350\267\203\350\256\255\347\273\203.md" "b/leetbook_ioa/docs/LCR 127. \350\267\263\350\267\203\350\256\255\347\273\203.md" new file mode 100755 index 0000000..adff689 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 127. \350\267\263\350\267\203\350\256\255\347\273\203.md" @@ -0,0 +1,93 @@ +## 解题思路： + +设跳上 $n$ 级平台有 $f(n)$ 种跳法。在所有跳法中，青蛙的最后一步只有两种情况： **跳上 $1$ 级或 $2$ 级平台**。 + +1. **当为 $1$ 级平台：** 剩 $n-1$ 个平台，此情况共有 $f(n-1)$ 种跳法； +2. **当为 $2$ 级平台：** 剩 $n-2$ 个平台，此情况共有 $f(n-2)$ 种跳法。 + +即 $f(n)$ 为以上两种情况之和，即 $f(n)=f(n-1)+f(n-2)$ ，以上递推性质为斐波那契数列。因此，本题可转化为 **求斐波那契数列第 $n$ 项的值** ，唯一的不同在于起始数字不同。 + +- 跳跃训练问题： $f(0)=1$ , $f(1)=1$ , $f(2)=2$ ； +- 斐波那契数列问题： $f(0)=0$ , $f(1)=1$ , $f(2)=1$ 。 + +![Picture1.png](https://pic.leetcode-cn.com/1599883153-UckfTw-Picture1.png){:align=center width=500} + +### 动态规划解析： + +- **状态定义：** 设 $dp$ 为一维数组，其中 $dp[i]$ 的值代表斐波那契数列的第 $i$ 个数字。 +- **转移方程：** $dp[i + 1] = dp[i] + dp[i - 1]$ ，即对应数列定义 $f(n + 1) = f(n) + f(n - 1)$ ； +- **初始状态：** $dp[0] = 1$, $dp[1] = 1$ ，即初始化前两个数字； +- **返回值：** $dp[n]$ ，即斐波那契数列的第 $n$ 个数字。 + +### 空间优化： + +> 若新建长度为 $n$ 的 $dp$ 列表，则空间复杂度为 $O(N)$ 。 + +- 由于 $dp$ 列表第 $i$ 项只与第 $i-1$ 和第 $i-2$ 项有关，因此只需要初始化三个整形变量 `sum`, `a`, `b` ，利用辅助变量 $sum$ 使 $a, b$ 两数字交替前进即可 *（具体实现见代码）* 。 +- 因为节省了 $dp$ 列表空间，因此空间复杂度降至 $O(1)$ 。 + +### 循环求余法： + +> **大数越界：** 随着 $n$ 增大, $f(n)$ 会超过 `Int32` 甚至 `Int64` 的取值范围，导致最终的返回值错误。 + +- **求余运算规则：** 设正整数 $x, y, p$ ，求余符号为 $\odot$ ，则有 $(x + y) \odot p = (x \odot p + y \odot p) \odot p$ 。 +- **解析：** 根据以上规则，可推出 $f(n) \odot p = [f(n-1) \odot p + f(n-2) \odot p] \odot p$ ，从而可以在循环过程中每次计算 $sum = a + b \odot 1000000007$ ，此操作与最终返回前取余等价。 + + + +## 代码： + +```Python [] +class Solution: + def trainWays(self, num: int) -> int: + a, b = 1, 1 + for _ in range(num): + a, b = b, (a + b) % 1000000007 + return a +``` + +```Java [] +class Solution { + public int trainWays(int num) { + int a = 1, b = 1, sum; + for(int i = 0; i < num; i++){ + sum = (a + b) % 1000000007; + a = b; + b = sum; + } + return a; + } +} +``` + +```C++ [] +class Solution { +public: + int trainWays(int num) { + int a = 1, b = 1, sum; + for(int i = 0; i < num; i++){ + sum = (a + b) % 1000000007; + a = b; + b = sum; + } + return a; + } +}; +``` + +由于 Python 中整形数字的大小限制取决计算机的内存（可理解为无限大），因此也可不考虑大数越界问题；但当数字很大时，加法运算的效率也会降低，因此不推荐此方法。 + +```Python [] +# 不考虑大数越界问题 +class Solution: + def trainWays(self, num: int) -> int: + a, b = 1, 1 + for _ in range(num): + a, b = b, a + b + return a % 1000000007 +``` + +### 复杂度分析： + +- **时间复杂度 $O(n)$ ：** 计算 $f(n)$ 需循环 $n$ 次，每轮循环内计算操作使用 $O(1)$ 。 +- **空间复杂度 $O(1)$ ：** 几个标志变量使用常数大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 128. \345\272\223\345\255\230\347\256\241\347\220\206 I.md" "b/leetbook_ioa/docs/LCR 128. \345\272\223\345\255\230\347\256\241\347\220\206 I.md" new file mode 100755 index 0000000..e4c2887 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 128. \345\272\223\345\255\230\347\256\241\347\220\206 I.md" @@ -0,0 +1,161 @@ +## 解题思路： + +如下图所示，寻找旋转数组的最小元素即为寻找 **右排序数组** 的首个元素 $stock[x]$ ，称 $x$ 为 **旋转点** 。 + +> 下图中的 `numbers` 对应本题的 `stock` 。 + +![Picture1.png](https://pic.leetcode-cn.com/1599404042-JMvjtL-Picture1.png){:align=center width=450} + +排序数组的查找问题首先考虑使用 **二分法** 解决，其可将 **遍历法** 的 **线性级别** 时间复杂度降低至 **对数级别** 。 + +### 算法流程： + +1. **初始化：** 声明 $i$, $j$ 双指针分别指向 $stock$ 数组左右两端； +2. **循环二分：** 设 $m = (i + j) / 2$ 为每次二分的中点（ "`/`" 代表向下取整除法，因此恒有 $i \leq m < j$ ），可分为以下三种情况： + 1. **当 $stock[m] > stock[j]$ 时：** $m$ 一定在 左排序数组 中，即旋转点 $x$ 一定在 $[m + 1, j]$ 闭区间内，因此执行 $i = m + 1$； + 2. **当 $stock[m] < stock[j]$ 时：** $m$ 一定在 右排序数组 中，即旋转点 $x$ 一定在$[i, m]$ 闭区间内，因此执行 $j = m$； + 3. **当 $stock[m] = stock[j]$ 时：** 无法判断 $m$ 在哪个排序数组中，即无法判断旋转点 $x$ 在 $[i, m]$ 还是 $[m + 1, j]$ 区间中。**解决方案：** 执行 $j = j - 1$ 缩小判断范围，分析见下文。 +3. **返回值：** 当 $i = j$ 时跳出二分循环，并返回 **旋转点的值** $stock[i]$ 即可。 + +### 正确性证明： + +当 $stock[m] = stock[j]$ 时，无法判定 $m$ 在左（右）排序数组，自然也无法通过二分法安全地缩小区间，因为其会导致旋转点 $x$ 不在区间 $[i, j]$ 内。举例如下： + +> 设以下两个旋转点值为 $0$ 的示例数组，则当 $i = 0$, $j = 4$ 时 $m = 2$ ，两示例结果不同。 +> 示例一 $[1, 0, 1, 1, 1]$ ：旋转点 $x = 1$ ，因此 $m = 2$ 在 **右排序数组** 中。 +> 示例二 $[1, 1, 1, 0, 1]$ ：旋转点 $x = 3$ ，因此 $m = 2$ 在 **左排序数组** 中。 + +而证明 $j = j - 1$ 正确（缩小区间安全性），需分为两种情况： + +1. **当 $x < j$ 时：** 易得执行 $j = j - 1$ 后，旋转点 $x$ 仍在区间 $[i, j]$ 内。 +2. **当 $x = j$ 时：** 执行 $j = j - 1$ 后越过（丢失）了旋转点 $x$ ，但最终返回的元素值 $stock[i]$ 仍等于旋转点值 $stock[x]$ 。 + + 1. 由于 $x = j$ ，因此 $stock[x] = stock[j] = stock[m] \leq number[i]$ ; + 2. 又由于 $i \leq m 综上所述，此方法可以保证返回值 $stock[i]$ 等于旋转点值 $stock[x]$ ，但在少数特例下 $i \ne x$ ；而本题目只要求返回 “旋转点的值” ，因此本方法正确。 + +**补充思考：** 为什么本题二分法不用 $stock[m]$ 和 $stock[i]$ 作比较？ + +二分目的是判断 $m$ 在哪个排序数组中，从而缩小区间。而在 $stock[m] > stock[i]$情况下，无法判断 $m$ 在哪个排序数组中。本质上是由于 $j$ 初始值肯定在右排序数组中；$i$ 初始值无法确定在哪个排序数组中。举例如下： + +> 对于以下两示例，当 $i = 0, j = 4, m = 2$ 时，有 `stock[m] > stock[i]` ，而结果不同。 +> $[1, 2, 3, 4 ,5]$ 旋转点 $x = 0$ ： $m$ 在右排序数组（此示例只有右排序数组）； +> $[3, 4, 5, 1 ,2]$ 旋转点 $x = 3$ ： $m$ 在左排序数组。 + + + +### 复杂度分析： + +- **时间复杂度 $O(\log N)$ ：** 在特例情况下（例如 $[1, 1, 1, 1]$），会退化到 $O(N)$。 +- **空间复杂度 $O(1)$ ：** $i$ , $j$ , $m$ 变量使用常数大小的额外空间。 + +## 代码： + +```Python [] +class Solution: + def stockManagement(self, stock: List[int]) -> int: + i, j = 0, len(stock) - 1 + while i < j: + m = (i + j) // 2 + if stock[m] > stock[j]: i = m + 1 + elif stock[m] < stock[j]: j = m + else: j -= 1 + return stock[i] +``` + +```Java [] +class Solution { + public int stockManagement(int[] stock) { + int i = 0, j = stock.length - 1; + while (i < j) { + int m = (i + j) / 2; + if (stock[m] > stock[j]) i = m + 1; + else if (stock[m] < stock[j]) j = m; + else j--; + } + return stock[i]; + } +} +``` + +```C++ [] +class Solution { +public: + int stockManagement(vector& stock) { + int i = 0, j = stock.size() - 1; + while (i < j) { + int m = (i + j) / 2; + if (stock[m] > stock[j]) i = m + 1; + else if (stock[m] < stock[j]) j = m; + else j--; + } + return stock[i]; + } +}; +``` + +实际上，当出现 $stock[m] = stock[j]$ 时，一定有区间 $[i, m]$ 内所有元素相等 或 区间 $[m, j]$ 内所有元素相等（或两者皆满足）。对于寻找此类数组的最小值问题，可直接放弃二分查找，而使用线性查找替代。 + +```Python [] +class Solution: + def stockManagement(self, stock: List[int]) -> int: + i, j = 0, len(stock) - 1 + while i < j: + m = (i + j) // 2 + if stock[m] > stock[j]: i = m + 1 + elif stock[m] < stock[j]: j = m + else: return min(stock[i:j]) + return stock[i] +``` + +```Java [] +class Solution { + public int stockManagement(int[] stock) { + int i = 0, j = stock.length - 1; + while (i < j) { + int m = (i + j) / 2; + if (stock[m] > stock[j]) i = m + 1; + else if (stock[m] < stock[j]) j = m; + else { + int x = i; + for(int k = i + 1; k < j; k++) { + if(stock[k] < stock[x]) x = k; + } + return stock[x]; + } + } + return stock[i]; + } +} +``` + +```C++ [] +class Solution { +public: + int stockManagement(vector& stock) { + int i = 0, j = stock.size() - 1; + while (i < j) { + int m = (i + j) / 2; + if (stock[m] > stock[j]) i = m + 1; + else if (stock[m] < stock[j]) j = m; + else { + int x = i; + for(int k = i + 1; k < j; k++) { + if(stock[k] < stock[x]) x = k; + } + return stock[x]; + } + } + return stock[i]; + } +}; +``` diff --git "a/leetbook_ioa/docs/LCR 129. \345\255\227\346\257\215\350\277\267\345\256\253.md" "b/leetbook_ioa/docs/LCR 129. \345\255\227\346\257\215\350\277\267\345\256\253.md" new file mode 100755 index 0000000..7d26483 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 129. \345\255\227\346\257\215\350\277\267\345\256\253.md" @@ -0,0 +1,103 @@ +## 解题思路： + +本问题是典型的回溯问题，可使用 **深度优先搜索（DFS）+ 剪枝** 解决。 + +- **深度优先搜索：** 可以理解为暴力法遍历矩阵中所有字符串可能性。DFS 通过递归，先朝一个方向搜到底，再回溯至上个节点，沿另一个方向搜索，以此类推。 +- **剪枝：** 在搜索中，遇到 `这条路不可能和目标字符串匹配成功` 的情况（*例如：此矩阵元素和目标字符不同、此元素已被访问）*，则应立即返回，称之为 `可行性剪枝` 。 + +> 下图中的 `word` 对应本题的 `target` 。 + +![Picture0.png](https://pic.leetcode-cn.com/1604944042-glmqJO-Picture0.png){:align=center width=500} + +### DFS 解析： + +- **递归参数：** 当前元素在矩阵 `grid` 中的行列索引 `i` 和 `j` ，当前目标字符在 `target` 中的索引 `k` 。 +- **终止条件：** + 1. 返回 $\text{false}$ ： (1) 行或列索引越界 **或** (2) 当前矩阵元素与目标字符不同 **或** (3) 当前矩阵元素已访问过 （ (3) 可合并至 (2) ） 。 + 2. 返回 $\text{true}$ ： `k = len(target) - 1` ，即字符串 `target` 已全部匹配。 +- **递推工作：** + 1. 标记当前矩阵元素： 将 `grid[i][j]` 修改为 **空字符** `''` ，代表此元素已访问过，防止之后搜索时重复访问。 + 2. 搜索下一单元格： 朝当前元素的 **上、下、左、右** 四个方向开启下层递归，使用 `或` 连接 （代表只需找到一条可行路径就直接返回，不再做后续 DFS ），并记录结果至 `res` 。 + 3. 还原当前矩阵元素： 将 `grid[i][j]` 元素还原至初始值，即 `target[k]` 。 +- **返回值：** 返回布尔量 `res` ，代表是否搜索到目标字符串。 + +> 使用空字符（Python: `''` , Java/C++: `'\0'` ）做标记是为了防止标记字符与矩阵原有字符重复。当存在重复时，此算法会将矩阵原有字符认作标记字符，从而出现错误。 + + + +## 代码： + +```Python [] +class Solution: + def wordPuzzle(self, grid: List[List[str]], target: str) -> bool: + def dfs(i, j, k): + if not 0 <= i < len(grid) or not 0 <= j < len(grid[0]) or grid[i][j] != target[k]: return False + if k == len(target) - 1: return True + grid[i][j] = '' + res = dfs(i + 1, j, k + 1) or dfs(i - 1, j, k + 1) or dfs(i, j + 1, k + 1) or dfs(i, j - 1, k + 1) + grid[i][j] = target[k] + return res + + for i in range(len(grid)): + for j in range(len(grid[0])): + if dfs(i, j, 0): return True + return False +``` + +```Java [] +class Solution { + public boolean wordPuzzle(char[][] grid, String target) { + char[] words = target.toCharArray(); + for(int i = 0; i < grid.length; i++) { + for(int j = 0; j < grid[0].length; j++) { + if(dfs(grid, words, i, j, 0)) return true; + } + } + return false; + } + boolean dfs(char[][] grid, char[] target, int i, int j, int k) { + if(i >= grid.length || i < 0 || j >= grid[0].length || j < 0 || grid[i][j] != target[k]) return false; + if(k == target.length - 1) return true; + grid[i][j] = '\0'; + boolean res = dfs(grid, target, i + 1, j, k + 1) || dfs(grid, target, i - 1, j, k + 1) || + dfs(grid, target, i, j + 1, k + 1) || dfs(grid, target, i , j - 1, k + 1); + grid[i][j] = target[k]; + return res; + } +} +``` + +```C++ [] +class Solution { +public: + bool wordPuzzle(vector>& grid, string target) { + rows = grid.size(); + cols = grid[0].size(); + for(int i = 0; i < rows; i++) { + for(int j = 0; j < cols; j++) { + if(dfs(grid, target, i, j, 0)) return true; + } + } + return false; + } +private: + int rows, cols; + bool dfs(vector>& grid, string target, int i, int j, int k) { + if(i >= rows || i < 0 || j >= cols || j < 0 || grid[i][j] != target[k]) return false; + if(k == target.size() - 1) return true; + grid[i][j] = '\0'; + bool res = dfs(grid, target, i + 1, j, k + 1) || dfs(grid, target, i - 1, j, k + 1) || + dfs(grid, target, i, j + 1, k + 1) || dfs(grid, target, i , j - 1, k + 1); + grid[i][j] = target[k]; + return res; + } +}; +``` + +### 复杂度分析： + +> $M, N$ 分别为矩阵行列大小，$K$ 为字符串 `target` 长度。 + +- **时间复杂度 $O(3^KMN)$ ：** 最差情况下，需要遍历矩阵中长度为 $K$ 字符串的所有方案，时间复杂度为 $O(3^K)$；矩阵中共有 $MN$ 个起点，时间复杂度为 $O(MN)$ 。 + - **方案数计算：** 设字符串长度为 $K$ ，搜索中每个字符有上、下、左、右四个方向可以选择，舍弃回头（上个字符）的方向，剩下 $3$ 种选择，因此方案数的复杂度为 $O(3^K)$ 。 +- **空间复杂度 $O(K)$ ：** 搜索过程中的递归深度不超过 $K$ ，因此系统因函数调用累计使用的栈空间占用 $O(K)$ （因为函数返回后，系统调用的[栈空间会释放](https://leetcode-cn.com/explore/orignial/card/recursion-i/259/complexity-analysis/1223/)）。最坏情况下 $K = MN$ ，递归深度为 $MN$ ，此时系统栈使用 $O(MN)$ 的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 130. \350\241\243\346\251\261\346\225\264\347\220\206.md" "b/leetbook_ioa/docs/LCR 130. \350\241\243\346\251\261\346\225\264\347\220\206.md" new file mode 100755 index 0000000..7d1e2b5 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 130. \350\241\243\346\251\261\346\225\264\347\220\206.md" @@ -0,0 +1,242 @@ +## 解题思路： + +为提升回溯的计算效率，首先讲述两项前置工作： **数位之和计算** 、 **可达解分析** 。 + +### 数位之和计算： + +设一数字 $x$ ，向下取整除法符号 $//$ ，求余符号 $\odot$ ，则有： + +- $x \odot 10$ ：得到 $x$ 的个位数字； +- $x // 10$ ： 令 $x$ 的十进制数向右移动一位，即删除个位数字。 + +因此，可通过循环求得数位和 $s$ ，数位和计算的封装函数如下所示： + +```Python [] +def sums(x): + s = 0 + while x != 0: + s += x % 10 + x = x // 10 + return s +``` + +```Java [] +int sums(int x) + int s = 0; + while(x != 0) { + s += x % 10; + x = x / 10; + } + return s; +``` + +```C++ [] +int sums(int x) + int s = 0; + while(x != 0) { + s += x % 10; + x = x / 10; + } + return s; +``` + +由于机器人每次只能移动一格（即只能从 $x$ 运动至 $x \pm 1$），因此每次只需计算 $x$ 到 $x \pm 1$ 的**数位和增量**。本题说明 $1 \leq n,m \leq 100$ ，以下公式仅在此范围适用。 + +**数位和增量公式：** 设 $x$ 的数位和为 $s_x$ ，$x+1$ 的数位和为 $s_{x+1}$ ； + +1. **当 $(x + 1) \odot 10 = 0$ 时：** $s_{x+1} = s_x - 8$ ，例如 $19, 20$ 的数位和分别为 $10, 2$ ； +2. **当 $(x + 1) \odot 10 \neq 0$ 时：** $s_{x+1} = s_x + 1$ ，例如 $1, 2$ 的数位和分别为 $1, 2$ 。 + +> 以下代码为增量公式的三元表达式写法，将整合入最终代码中。 + +```Python [] +s_x + 1 if (x + 1) % 10 else s_x - 8 +``` + +```Java [] +(x + 1) % 10 != 0 ? s_x + 1 : s_x - 8; +``` + +```C++ [] +(x + 1) % 10 != 0 ? s_x + 1 : s_x - 8; +``` + +### 可达解分析： + +根据数位和增量公式得知，数位和每逢 **进位** 突变一次。根据此特点，矩阵中 **满足数位和的解** 构成的几何形状形如多个 **等腰直角三角形** ，每个三角形的直角顶点位于 $0, 10, 20, ...$ 等数位和突变的矩阵索引处 。 + +三角形内的解虽然都满足数位和要求，但由于机器人每步只能走一个单元格，而三角形间不一定是连通的，因此机器人不一定能到达，称之为 **不可达解** ；同理，可到达的解称为 **可达解** *（本题求此解）* 。 + +> 下图展示了 $n,m = 20$ ，$cnt \in [6, 19]$ 的可达解、不可达解、非解，以及连通性的变化。其中 $k$ 对应本题的 $cnt$ 。 + + + +根据可达解的结构和连通性，易推出机器人可 **仅通过向右和向下移动，访问所有可达解** 。 + +- **三角形内部：** 全部连通，易证； +- **两三角形连通处：** 若某三角形内的解为可达解，则必与其左边或上边的三角形连通（即相交），即机器人必可从左边或上边走进此三角形。 + +![Picture9.png](https://pic.leetcode-cn.com/1603024999-XMpudY-Picture9.png){:align=center width=500} + +## 方法一：深度优先遍历（DFS） + +**深度优先搜索：** 可以理解为暴力法模拟机器人在矩阵中的所有路径。DFS 通过递归，先朝一个方向搜到底，再回溯至上个节点，沿另一个方向搜索，以此类推。 + +**剪枝：** 在搜索中，遇到数位和超出目标值、此元素已访问，则应立即返回，称之为 `可行性剪枝` 。 + +### 算法解析： + +- **递归参数：** 当前元素在矩阵中的行列索引 `i` 和 `j` ，两者的数位和 `si`, `sj` 。 +- **终止条件：** 当 (1) 行列索引越界 **或** (2) 数位和超出目标值 `cnt` **或** (3) 当前元素已访问过 时，返回 $0$ ，代表不计入可达解。 +- **递推工作：** + 1. **标记当前单元格** ：将索引 `(i, j)` 存入 Set `visited` 中，代表此单元格已被访问过。 + 2. **搜索下一单元格：** 计算当前元素的 **下、右** 两个方向元素的数位和，并开启下层递归 。 +- **回溯返回值：** 返回 `1 + 右方搜索的可达解总数 + 下方搜索的可达解总数`，代表从本单元格递归搜索的可达解总数。 + + + +### 代码： + +> Java/C++ 代码中 `visited` 为辅助矩阵，Python 中为 Set 。 + +```Python [] +class Solution: + def wardrobeFinishing(self, m: int, n: int, cnt: int) -> int: + def dfs(i, j, si, sj): + if i >= m or j >= n or cnt < si + sj or (i, j) in visited: return 0 + visited.add((i,j)) + return 1 + dfs(i + 1, j, si + 1 if (i + 1) % 10 else si - 8, sj) + dfs(i, j + 1, si, sj + 1 if (j + 1) % 10 else sj - 8) + + visited = set() + return dfs(0, 0, 0, 0) +``` + +```Java [] +class Solution { + int m, n, cnt; + boolean[][] visited; + public int wardrobeFinishing(int m, int n, int cnt) { + this.m = m; this.n = n; this.cnt = cnt; + this.visited = new boolean[m][n]; + return dfs(0, 0, 0, 0); + } + public int dfs(int i, int j, int si, int sj) { + if(i >= m || j >= n || cnt < si + sj || visited[i][j]) return 0; + visited[i][j] = true; + return 1 + dfs(i + 1, j, (i + 1) % 10 != 0 ? si + 1 : si - 8, sj) + dfs(i, j + 1, si, (j + 1) % 10 != 0 ? sj + 1 : sj - 8); + } +} +``` + +```C++ [] +class Solution { +public: + int wardrobeFinishing(int m, int n, int cnt) { + vector> visited(m, vector(n, 0)); + return dfs(0, 0, 0, 0, visited, m, n, cnt); + } +private: + int dfs(int i, int j, int si, int sj, vector> &visited, int m, int n, int cnt) { + if(i >= m || j >= n || cnt < si + sj || visited[i][j]) return 0; + visited[i][j] = true; + return 1 + dfs(i + 1, j, (i + 1) % 10 != 0 ? si + 1 : si - 8, sj, visited, m, n, cnt) + + dfs(i, j + 1, si, (j + 1) % 10 != 0 ? sj + 1 : sj - 8, visited, m, n, cnt); + } +}; +``` + +### 复杂度分析： + +> 设矩阵行列数分别为 $M, N$ 。 + +- **时间复杂度 $O(MN)$ ：** 最差情况下，机器人遍历矩阵所有单元格，此时时间复杂度为 $O(MN)$ 。 +- **空间复杂度 $O(MN)$ ：** 最差情况下，Set `visited` 内存储矩阵所有单元格的索引，使用 $O(MN)$ 的额外空间。 + +## 方法二：广度优先遍历（BFS） + +BFS 和 DFS 的目标都是遍历整个矩阵，不同点在于搜索顺序不同。DFS 是朝一个方向走到底，再回退，以此类推；BFS 则是按照“平推”的方式向前搜索。 + +**BFS 实现：** 通常利用队列实现广度优先遍历。 + +### 算法解析： + +- **初始化：** 将机器人初始点 $(0, 0)$ 加入队列 `queue` ； +- **迭代终止条件：** `queue` 为空。代表已遍历完所有可达解。 +- **迭代工作：** + 1. **单元格出队：** 将队首单元格的 索引、数位和 弹出，作为当前搜索单元格。 + 2. **判断是否跳过：** 若 (1) 行列索引越界 **或** (2) 数位和超出目标值 `cnt` **或** (3) 当前元素已访问过 时，执行 `continue` 。 + 3. **标记当前单元格** ：将单元格索引 `(i, j)` 存入 Set `visited` 中，代表此单元格 **已被访问过** 。 + 4. **单元格入队：** 将当前元素的 **下方、右方** 单元格的 **索引、数位和** 加入 `queue` 。 +- **返回值：** Set `visited` 的长度 `len(visited)` ，即可达解的数量。 + +> Java/C++ 使用了辅助变量 `res` 统计可达解数量； Python 直接返回 Set 的元素数 `len(visited)` 即可。 + + + +### 代码： + +> Java/C++ 代码中 `visited` 为辅助矩阵，Python 中为 Set 。 + +```Python [] +class Solution: + def wardrobeFinishing(self, m: int, n: int, cnt: int) -> int: + queue, visited = [(0, 0, 0, 0)], set() + while queue: + i, j, si, sj = queue.pop(0) + if i >= m or j >= n or cnt < si + sj or (i, j) in visited: continue + visited.add((i,j)) + queue.append((i + 1, j, si + 1 if (i + 1) % 10 else si - 8, sj)) + queue.append((i, j + 1, si, sj + 1 if (j + 1) % 10 else sj - 8)) + return len(visited) +``` + +```Java [] +class Solution { + public int wardrobeFinishing(int m, int n, int cnt) { + boolean[][] visited = new boolean[m][n]; + int res = 0; + Queue queue= new LinkedList(); + queue.add(new int[] { 0, 0, 0, 0 }); + while(queue.size() > 0) { + int[] x = queue.poll(); + int i = x[0], j = x[1], si = x[2], sj = x[3]; + if(i >= m || j >= n || cnt < si + sj || visited[i][j]) continue; + visited[i][j] = true; + res ++; + queue.add(new int[] { i + 1, j, (i + 1) % 10 != 0 ? si + 1 : si - 8, sj }); + queue.add(new int[] { i, j + 1, si, (j + 1) % 10 != 0 ? sj + 1 : sj - 8 }); + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int wardrobeFinishing(int m, int n, int cnt) { + vector> visited(m, vector(n, 0)); + int res = 0; + queue> que; + que.push({ 0, 0, 0, 0 }); + while(que.size() > 0) { + vector x = que.front(); + que.pop(); + int i = x[0], j = x[1], si = x[2], sj = x[3]; + if(i >= m || j >= n || cnt < si + sj || visited[i][j]) continue; + visited[i][j] = true; + res++; + que.push({ i + 1, j, (i + 1) % 10 != 0 ? si + 1 : si - 8, sj }); + que.push({ i, j + 1, si, (j + 1) % 10 != 0 ? sj + 1 : sj - 8 }); + } + return res; + } +}; +``` + +### 复杂度分析： + +> 设矩阵行列数分别为 $M, N$ 。 + +- **时间复杂度 $O(MN)$ ：** 最差情况下，机器人遍历矩阵所有单元格，此时时间复杂度为 $O(MN)$ 。 +- **空间复杂度 $O(MN)$ ：** 最差情况下，Set `visited` 内存储矩阵所有单元格的索引，使用 $O(MN)$ 的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 131. \347\240\215\347\253\271\345\255\220 I.md" "b/leetbook_ioa/docs/LCR 131. \347\240\215\347\253\271\345\255\220 I.md" new file mode 100755 index 0000000..e00337c --- /dev/null +++ "b/leetbook_ioa/docs/LCR 131. \347\240\215\347\253\271\345\255\220 I.md" @@ -0,0 +1,164 @@ +## 解题思路： + +设将长度为 $n$ 的竹子切为 $a$ 段： + +$$ +n = n_1 + n_2 + ... + n_a +$$ + +本题等价于求解： + +$$ +\max(n_1 \times n_2 \times ... \times n_a) +$$ + +> 以下数学推导总体分为两步：(1) 当所有绳段长度相等时，乘积最大。(2) 最优的绳段长度为 $3$ 。 + +### 数学推导： + +以下公式为“算术几何均值不等式” ，等号当且仅当 $n_1 = n_2 = ... = n_a$ 时成立。 + +$$ +\frac{n_1 + n_2 + ... + n_a}{a} \geq \sqrt[a]{n_1 n_2 ... n_a} +$$ + +> **推论一：** 将竹子 **以相等的长度等分为多段** ，得到的乘积最大。 + +设将竹子按照 $x$ 长度等分为 $a$ 段，即 $n = ax$ ，则乘积为 $x^a$ 。观察以下公式，由于 $n$ 为常数，因此当 $x^{\frac{1}{x}}$ 取最大值时， 乘积达到最大值。 + +$$ +x^a = x^{\frac{n}{x}} = (x^{\frac{1}{x}})^n +$$ + +根据分析，可将问题转化为求 $y = x^{\frac{1}{x}}$ 的极大值，因此对 $x$ 求导数。 + +$$ +\begin{aligned} + \ln y & = \frac{1}{x} \ln x & \text{取对数} \\ + \frac{1}{y} \dot {y} & = \frac{1}{x^2} - \frac{1}{x^2} \ln x & \text{对 $x$ 求导} \\ + & = \frac{1 - \ln x}{x^2} \\ + \dot {y} & = \frac{1 - \ln x}{x^2} x^{\frac{1}{x}} & \text{整理得} +\end{aligned} +$$ + +令 $\dot {y} = 0$ ，则 $1 - \ln x = 0$ ，易得驻点为 $x_0 = e \approx 2.7$ ；根据以下公式，可知 $x_0$ 为极大值点。 + +$$ +\dot {y} +\begin{cases} + > 0 & , x \in [- \infty, e) \\ + < 0 & , x \in (e, \infty] \\ +\end{cases} +$$ + +由于切分长度 $x$ 必须为整数，最接近 $e$ 的整数为 $2$ 或 $3$ 。如下式所示，代入 $x = 2$ 和 $x = 3$ ，得出 $x = 3$ 时，乘积达到最大。 + +$$ +y(3) = 3^{1/3} \approx 1.44 \\ +y(2) = 2^{1/2} \approx 1.41 +$$ + +口算对比方法：给两数字同时取 $6$ 次方，再对比。 + +$$ +y(3)^6 = (3^{1/3})^6 = 9 \\ +y(2)^6 = (2^{1/2})^6 = 8 +$$ + +> **推论二：** 尽可能将竹子以长度 $3$ 等分为多段时，乘积最大。 + +### 切分规则： + +1. **最优：** $3$ 。把竹子尽可能切为多个长度为 $3$ 的片段，留下的最后一段竹子的长度可能为 $0,1,2$ 三种情况。 +2. **次优：** $2$ 。若最后一段竹子长度为 $2$ ；则保留，不再拆为 $1+1$ 。 +3. **最差：** $1$ 。若最后一段竹子长度为 $1$ ；则应把一份 $3 + 1$ 替换为 $2 + 2$，因为 $2 \times 2 > 3 \times 1$。 + +### 算法流程： + +1. 当 $n \leq 3$ 时，按照规则应不切分，但由于题目要求必须剪成 $m>1$ 段，因此必须剪出一段长度为 $1$ 的竹子，即返回 $n - 1$ 。 +2. 当 $n>3$ 时，求 $n$ 除以 $3$ 的 整数部分 $a$ 和 余数部分 $b$ （即 $n = 3a + b$ ），并分为以下三种情况： + - 当 $b = 0$ 时，直接返回 $3^a$； + - 当 $b = 1$ 时，要将一个 $1 + 3$ 转换为 $2+2$，因此返回 $3^{a-1} \times 4$； + - 当 $b = 2$ 时，返回 $3^a \times 2$。 + +![Picture1.png](https://pic.leetcode-cn.com/1f9adeaa7b9fff0ab19c9d29e3a8f98749011d22dc162d67bdbe223f1a38119f-Picture1.png){:align=center width=600} + +## 代码： + +> Python 中常见有三种幂计算函数： **`*`** 和 **`pow()`** 的时间复杂度均为 $O(\log a)$ ；而 **`math.pow()`** 始终调用 C 库的 `pow()` 函数，其执行浮点取幂，时间复杂度为 $O(1)$ 。 + +```Python [] +class Solution: + def cuttingBamboo(self, bamboo_len: int) -> int: + if bamboo_len <= 3: return bamboo_len - 1 + a, b = bamboo_len // 3, bamboo_len % 3 + if b == 0: return int(math.pow(3, a)) + if b == 1: return int(math.pow(3, a - 1) * 4) + return int(math.pow(3, a) * 2) +``` + +```Java [] +class Solution { + public int cuttingBamboo(int bamboo_len) { + if(bamboo_len <= 3) return bamboo_len - 1; + int a = bamboo_len / 3, b = bamboo_len % 3; + if(b == 0) return (int)Math.pow(3, a); + if(b == 1) return (int)Math.pow(3, a - 1) * 4; + return (int)Math.pow(3, a) * 2; + } +} +``` + +```C++ [] +class Solution { +public: + int cuttingBamboo(int bamboo_len) { + if(bamboo_len <= 3) return bamboo_len - 1; + int a = bamboo_len / 3, b = bamboo_len % 3; + if(b == 0) return pow(3, a); + if(b == 1) return pow(3, a - 1) * 4; + return pow(3, a) * 2; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(1)$ ：** 仅有求整、求余、次方运算。 + - [求整和求余运算](https://stackoverflow.com/questions/35189851/time-complexity-of-modulo-operator-in-python)：资料提到不超过机器数的整数可以看作是 $O(1)$ ； + - [幂运算](https://stackoverflow.com/questions/32418731/java-math-powa-b-time-complexity)：查阅资料，提到浮点取幂为 $O(1)$ 。 +- **空间复杂度 $O(1)$ ：** 变量 `a` 和 `b` 使用常数大小额外空间。 + +## 贪心思路： + +数学推导需要一定的知识基础，贪心算法的思路更加适合快速解题。 + +> 设一竹子长度为 $n$ ( $n>1$ )，则其必可被切分为两段 $n=n_1+n_2$ 。 +> 根据经验推测，切分的两数字乘积往往原数字更大，即往往有 $n_1 \times n_2 > n_1 + n_2 = n$ 。 +> +> - **例如竹子长度为 $6$ ：** $6 = 3 + 3 < 3 \times 3 = 9$ ； +> - **也有少数反例，例如 $2$ ：** $2 = 1 + 1 > 1 \times 1 = 1$ 。 + +- **推论一：** 合理的切分方案可以带来更大的乘积。 + +> 设一竹子长度为 $n$ ( $n>1$ )，**切分为两段** $n=n_1+n_2$ ，**切分为三段** $n=n_1+n_2+n_3$ 。 +> 根据经验推测，**三段** 的乘积往往更大，即往往有 $n_1 n_2 n_3 > n_1 n_2$ 。 +> +> - **例如竹子长度为 $9$ ：** 两段 $9=4+5$ 和 三段 $9=3+3+3$，则有 $4 \times 5 < 3 \times 3 \times 3$ 。 +> - **也有少数反例，例如 $6$ ：** 两段 $6=3+3$ 和 三段 $6=2+2+2$，则有 $3 \times 3 > 2 \times 2 \times 2$ 。 + +- **推论二：** 若切分方案合理，竹子段切分的越多，乘积越大。 + +> 总体上看，貌似长竹子切分为越多段乘积越大，但其实到某个长度分界点后，乘积到达最大值，就不应再切分了。 +> **问题转化：** 是否有**优先级最高的长度** $x$ 存在？若有，则应该尽可能把竹子以 $x$ 长度切为多段，以获取最大乘积。 + +- **推论三：** 为使乘积最大，只有长度为 $2$ 和 $3$ 的竹子不应再切分，且 $3$ 比 $2$ 更优 *（详情见下表）* 。 + +| 竹子切分方案 | 乘积 | 结论 | +| ------------- | ------------------------------------------ | ----------------------------------------------------------------- | +| $2 = 1 + 1$ | $1 \times 1 = 1$ | $2$ 不应切分 | +| $3=1+2$ | $1 \times 2 = 2$ | $3$ 不应切分 | +| $4=2+2=1+3$ | $2 \times 2 = 4 > 1 \times 3 = 3$ | $4$ 和 $2$ 等价，且 $2+2$ 比 $1+3$ 更优 | +| $5=2+3=1+4$ | $2 \times 3 = 6 > 1 \times 4 = 4$ | $5$ 应切分为 $2+3$ | +| $6=3+3=2+2+2$ | $3 \times 3 = 9 > 2 \times 2 \times 2 = 8$ | $6$ 应切分为 $3+3$ ，进而**推出 $3$ 比 $2$ 更优** | +| $>7$ | ... | **长绳**（长度>7）可转化为多个**短绳**（长度1~6），因此肯定应切分 | diff --git "a/leetbook_ioa/docs/LCR 132. \347\240\215\347\253\271\345\255\220 II.md" "b/leetbook_ioa/docs/LCR 132. \347\240\215\347\253\271\345\255\220 II.md" new file mode 100755 index 0000000..e31e4b0 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 132. \347\240\215\347\253\271\345\255\220 II.md" @@ -0,0 +1,166 @@ +## 解题思路： + +> 切分规则的推导流程请见上一题「砍竹子 I」。 + +### 切分规则： + +1. **最优：** $3$ 。把竹子尽可能切为多个长度为 $3$ 的片段，留下的最后一段竹子的长度可能为 $0,1,2$ 三种情况。 +2. **次优：** $2$ 。若最后一段竹子长度为 $2$ ；则保留，不再拆为 $1+1$ 。 +3. **最差：** $1$ 。若最后一段竹子长度为 $1$ ；则应把一份 $3 + 1$ 替换为 $2 + 2$，因为 $2 \times 2 > 3 \times 1$。 + +### 算法流程： + +1. 当 $n \leq 3$ 时，按照规则应不切分，但由于题目要求必须剪成 $m>1$ 段，因此必须剪出一段长度为 $1$ 的竹子，即返回 $n - 1$ 。 +2. 当 $n>3$ 时，求 $n$ 除以 $3$ 的 整数部分 $a$ 和 余数部分 $b$ （即 $n = 3a + b$ ），并分为以下三种情况（设求余操作符号为 "$\odot$" ）： + - 当 $b = 0$ 时，直接返回 $3^a \odot 1000000007$； + - 当 $b = 1$ 时，要将一个 $1 + 3$ 转换为 $2+2$，因此返回 $(3^{a-1} \times 4)\odot 1000000007$； + - 当 $b = 2$ 时，返回 $(3^a \times 2) \odot 1000000007$。 + +![Picture1.png](https://pic.leetcode-cn.com/1f9adeaa7b9fff0ab19c9d29e3a8f98749011d22dc162d67bdbe223f1a38119f-Picture1.png){:align=center width=600} + +### 大数求余解法： + +**大数越界：** 当 $a$ 增大时，最后返回的 $3^a$ 大小以指数级别增长，可能超出 `int32` 甚至 `int64` 的取值范围，导致返回值错误。 +**大数求余问题：** 在仅使用 `int32` 类型存储的前提下，正确计算 $x^a$ 对 $p$ 求余（即 $x^a \odot p$ ）的值。 +**解决方案：** *循环求余* 、 *快速幂求余* ，其中后者的时间复杂度更低，两种方法均基于以下求余运算规则推出： + +$$ +(xy) \odot p = [(x \odot p)(y \odot p)] \odot p +$$ + +### 1. 循环求余： + +根据求余运算性质推出（∵ 本题中 $x 0: + if a % 2: rem = (rem * x) % p + x = x ** 2 % p + a //= 2 + return rem +``` + +**帮助理解：** 根据下表， 初始状态 $rem=1$, $x=3$, $a=19$, $p=1000000007$ ，最后会将 $rem \times (x^a \odot p)$ 化为 $rem \times (x^0 \odot p) = rem \times 1$ 的形式，即 $rem$ 为余数答案。 + +| $n$ | $rem \times (x^a \odot p)$ | $rem_n=rem_{n-1} \times x_{n-1} \odot p$ | $x_n=x_{n-1}^2 \odot p$ | $a_n=a_{n-1}//2$ | +| --- | -----------------------------------------: | ---------------------------------------: | -----------------------------: | :--------------: | +| $1$ | $1 \times (3^{19} \odot p)$ | $1$ | $3$ | $19$ | +| $2$ | $3 \times (9^{9} \odot p)$ | $3=1\times3\odot p$ | $9=3^2 \odot p$ | $9=19//2$ | +| $3$ | $27 \times (81^{4} \odot p)$ | $27 = 3 \times 9 \odot p$ | $81=9^2\odot p$ | $4=9//2$ | +| $4$ | $27 \times (6561^{2} \odot p)$ | $27$ | $6561=81^2 \odot p$ | $2=4//2$ | +| $5$ | $27 \times (43046721^{1} \odot p)$ | $27$ | $43046721=6561^2 \odot p$ | $1=2//2$ | +| $6$ | $162261460 \times (175880701^{0} \odot p)$ | $162261460=27 \times 43046721 \odot p$ | $175880701=43046721^2 \odot p$ | $0=1//2$ | + +## 代码： + +**Python 代码：** 由于语言特性，理论上 Python 中的变量取值范围由系统内存大小决定（无限大），因此在 Python 中其实不用考虑大数越界问题。 +**Java/C++ 代码：** 根据二分法计算原理，至少要保证变量 `x` 和 `rem` 可以正确存储 $1000000007^2$ ，而 $2^{64} > 1000000007^2 > 2^{32}$ ，因此我们选取 `long` 类型。 + +```Python [] +class Solution: + def cuttingBamboo(self, bamboo_len: int) -> int: + if bamboo_len <= 3: return bamboo_len - 1 + a, b, p, x, rem = bamboo_len // 3 - 1, bamboo_len % 3, 1000000007, 3 , 1 + while a > 0: + if a % 2: rem = (rem * x) % p + x = x ** 2 % p + a //= 2 + if b == 0: return (rem * 3) % p # = 3^(a+1) % p + if b == 1: return (rem * 4) % p # = 3^a * 4 % p + return (rem * 6) % p # = 3^(a+1) * 2 % p +``` + +```Java [] +class Solution { + public int cuttingBamboo(int bamboo_len) { + if(bamboo_len <= 3) return bamboo_len - 1; + int b = bamboo_len % 3, p = 1000000007; + long rem = 1, x = 3; + for(int a = bamboo_len / 3 - 1; a > 0; a /= 2) { + if(a % 2 == 1) rem = (rem * x) % p; + x = (x * x) % p; + } + if(b == 0) return (int)(rem * 3 % p); + if(b == 1) return (int)(rem * 4 % p); + return (int)(rem * 6 % p); + } +} +``` + +```C++ [] +class Solution { +public: + int cuttingBamboo(int bamboo_len) { + if(bamboo_len <= 3) return bamboo_len - 1; + int b = bamboo_len % 3, p = 1000000007; + long rem = 1, x = 3; + for(int a = bamboo_len / 3 - 1; a > 0; a /= 2) { + if(a % 2 == 1) rem = (rem * x) % p; + x = (x * x) % p; + } + if(b == 0) return (int)(rem * 3 % p); + if(b == 1) return (int)(rem * 4 % p); + return (int)(rem * 6 % p); + } +}; +``` + +```Python [] +# 由于语言特性，Python 可以不考虑大数越界问题 +class Solution: + def cuttingBamboo(self, bamboo_len: int) -> int: + if bamboo_len <= 3: return bamboo_len - 1 + a, b, p = bamboo_len // 3, bamboo_len % 3, 1000000007 + if b == 0: return 3 ** a % p + if b == 1: return 3 ** (a - 1) * 4 % p + return 3 ** a * 2 % p +``` + +### 复杂度分析： + +> 以下为 **二分求余法** 的复杂度。 + +- **时间复杂度 $O(\log N)$ ：** 其中 $N=a$ ，二分法为对数级别复杂度，每轮仅有求整、求余、次方运算。 + - [求整和求余运算](https://stackoverflow.com/questions/35189851/time-complexity-of-modulo-operator-in-python)：资料提到不超过机器数的整数可以看作是 $O(1)$ ； + - [幂运算](https://stackoverflow.com/questions/32418731/java-math-powa-b-time-complexity)：查阅资料，提到浮点取幂为 $O(1)$ 。 +- **空间复杂度 $O(1)$ ：** 变量 `a, b, p, x, rem` 使用常数大小额外空间。 diff --git "a/leetbook_ioa/docs/LCR 133. \344\275\215 1 \347\232\204\344\270\252\346\225\260.md" "b/leetbook_ioa/docs/LCR 133. \344\275\215 1 \347\232\204\344\270\252\346\225\260.md" new file mode 100755 index 0000000..9cb4025 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 133. \344\275\215 1 \347\232\204\344\270\252\346\225\260.md" @@ -0,0 +1,126 @@ +## 方法一：逐位判断 + +根据 **与运算** 定义，设二进制数字 $n$ ，则有： + +- 若 $n \& 1 = 0$ ，则 $n$ 二进制 **最右一位** 为 $0$ ； +- 若 $n \& 1 = 1$ ，则 $n$ 二进制 **最右一位** 为 $1$ 。 + +根据以上特点，考虑以下 **循环判断** ： + +1. 判断 $n$ 最右一位是否为 $1$ ，根据结果计数。 +2. 将 $n$ 右移一位（本题要求把数字 $n$ 看作无符号数，因此使用 **无符号右移** 操作）。 + +### 算法流程： + +1. 初始化数量统计变量 $res = 0$ 。 +2. 循环逐位判断： 当 $n = 0$ 时跳出。 + 1. **`res += n & 1` ：** 若 $n \& 1 = 1$ ，则统计数 $res$ 加一。 + 2. **`n >>= 1` ：** 将二进制数字 $n$ 无符号右移一位（ Java 中无符号右移为 "$>>>$" ） 。 +3. 返回统计数量 $res$ 。 + + + +### 代码： + +```Python [] +class Solution: + def hammingWeight(self, n: int) -> int: + res = 0 + while n: + res += n & 1 + n >>= 1 + return res +``` + +```Java [] +public class Solution { + public int hammingWeight(int n) { + int res = 0; + while(n != 0) { + res += n & 1; + n >>>= 1; + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int hammingWeight(uint32_t n) { + unsigned int res = 0; // c++ 使用无符号数 + while(n != 0) { + res += n & 1; + n >>= 1; + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(\log_2 n)$ ：** 此算法循环内部仅有 **移位、与、加** 等基本运算，占用 $O(1)$ ；逐位判断需循环 $log_2 n$ 次，其中 $\log_2 n$ 代表数字 $n$ 最高位 $1$ 的所在位数（例如 $\log_2 4 = 2$, $\log_2 16 = 4$）。 +- **空间复杂度 $O(1)$ ：** 变量 $res$ 使用常数大小额外空间。 + +## 方法二：巧用 $n \& (n - 1)$ + +- **$(n - 1)$ 解析：** 二进制数字 $n$ 最右边的 $1$ 变成 $0$ ，此 $1$ 右边的 $0$ 都变成 $1$ 。 +- **$n \& (n - 1)$ 解析：** 二进制数字 $n$ 最右边的 $1$ 变成 $0$ ，其余不变。 + +![Picture1.png](https://pic.leetcode-cn.com/f23d9ef4fcfd65d7fbe29e477cbf36110b2f34558020e8cff09a1e13c0275c43-Picture1.png){:align=center width=450} + +### 算法流程： + +1. 初始化数量统计变量 $res$ 。 +2. 循环消去最右边的 $1$ ：当 $n = 0$ 时跳出。 + 1. **`res += 1` ：** 统计变量加 $1$ ； + 2. **`n &= n - 1` ：** 消去数字 $n$ 最右边的 $1$ 。 +3. 返回统计数量 $res$ 。 + + + +### 代码： + +```Python [] +class Solution: + def hammingWeight(self, n: int) -> int: + res = 0 + while n: + res += 1 + n &= n - 1 + return res +``` + +```Java [] +public class Solution { + public int hammingWeight(int n) { + int res = 0; + while(n != 0) { + res++; + n &= n - 1; + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int hammingWeight(uint32_t n) { + int res = 0; + while(n != 0) { + res++; + n &= n - 1; + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(M)$ ：** $n \& (n - 1)$ 操作仅有减法和与运算，占用 $O(1)$ ；设 $M$ 为二进制数字 $n$ 中 $1$ 的个数，则需循环 $M$ 次（每轮消去一个 $1$ ），占用 $O(M)$ 。 +- **空间复杂度 $O(1)$ ：** 变量 $res$ 使用常数大小额外空间。 diff --git a/leetbook_ioa/docs/LCR 134. Pow(x, n).md b/leetbook_ioa/docs/LCR 134. Pow(x, n).md new file mode 100755 index 0000000..fef1472 --- /dev/null +++ b/leetbook_ioa/docs/LCR 134. Pow(x, n).md @@ -0,0 +1,134 @@ +## 解题思路： + +求 $x^n$ 最简单的方法是通过循环将 $n$ 个 $x$ 乘起来，依次求 $x^1, x^2, ..., x^{n-1}, x^n$ ，时间复杂度为 $O(n)$ 。 +**快速幂法** 可将时间复杂度降低至 $O(\log n)$ ，以下从 「分治法」 和 「二进制」 两个角度解析快速幂法。 + +### 快速幂解析（分治法角度）： + +> 快速幂实际上是分治思想的一种应用。 + +**二分推导：** $x^n = x^{n/2} \times x^{n/2} = (x^2)^{n/2}$ ，令 $n/2$ 为整数，则需要分为奇偶两种情况（设向下取整除法符号为 "$//$" ）： + +$$ +x^n = +\begin{cases} + (x^2)^{n//2} & , n 为偶数 \\ + x(x^2)^{n//2} & , n 为奇数 \\ +\end{cases} +$$ + +> 观察发现，当 $n$ 为奇数时，二分后会多出一项 $x$ 。 + +**幂结果获取：** + +- 根据推导，可通过循环 $x = x^2$ 操作，每次把幂从 $n$ 降至 $n//2$ ，直至将幂降为 $0$ ； +- 设 $res=1$ ，则初始状态 $x^n = x^n \times res$ 。在循环二分时，每当 $n$ 为奇数时，将多出的一项 $x$ 乘入 $res$ ，则最终可化至 $x^n = x^0 \times res = res$ ，返回 $res$ 即可。 + +![Picture2.png](https://pic.leetcode-cn.com/1599885604-YzlkAN-Picture2.png){:align=center width=500} + +**转化为位运算：** + +- 向下整除 $n // 2$ **等价于** 右移一位 $n >> 1$ ； +- 取余数 $n \mod 2$ **等价于** 判断二进制最右位 $n \& 1$ ； + +### 快速幂解析（二进制角度）： + +> 利用十进制数字 $n$ 的二进制表示，可对快速幂进行数学化解释。 + +对于任何十进制正整数 $n$ ，设其二进制为 "$b_m...b_3b_2b_1$"（ $b_i$ 为二进制某位值，$i \in [1,m]$ ），则有： + +- **二进制转十进制：** $n = 1b_1 + 2b_2 + 4b_3 + ... + 2^{m-1}b_m$ *（即二进制转十进制公式）* ； +- **幂的二进制展开：** $x^n = x^{1b_1 + 2b_2 + 4b_3 + ... + 2^{m-1}b_m} = x^{1b_1}x^{2b_2}x^{4b_3}...x^{2^{m-1}b_m}$ ； + +根据以上推导，可把计算 $x^n$ 转化为解决以下两个问题： + +- **计算 $x^1, x^2, x^4, ..., x^{2^{m-1}}$ 的值：** 循环赋值操作 $x = x^2$ 即可； +- **获取二进制各位 $b_1, b_2, b_3, ..., b_m$ 的值：** 循环执行以下操作即可。 + 1. **$n \& 1$ （与操作）：** 判断 $n$ 二进制最右一位是否为 $1$ ； + 2. **$n>>1$ （移位操作）：** $n$ 右移一位（可理解为删除最后一位）。 + +因此，应用以上操作，可在循环中依次计算 $x^{2^{0}b_1}, x^{2^{1}b_2}, ..., x^{2^{m-1}b_m}$ 的值，并将所有 $x^{2^{i-1}b_i}$ 累计相乘即可，其中： + +$$ +x^{2^{i-1}b_i}= +\begin{cases} + 1 & , b_i = 0 \\ + x^{2^{i-1}} & , b_i = 1 \\ +\end{cases} +$$ + +![Picture1.png](https://pic.leetcode-cn.com/1599885604-yDzVYK-Picture1.png){:align=center width=500} + +### 算法流程： + +1. 当 $x = 0.0$ 时：直接返回 $0.0$ ，以避免后续 $1$ 除以 $0$ 操作报错。**分析：** 数字 $0$ 的正数次幂恒为 $0$ ；$0$ 的 $0$ 次幂和负数次幂没有意义，因此直接返回 $0.0$ 即可。 +2. 初始化 $res = 1$ 。 +3. 当 $n < 0$ 时：把问题转化至 $n \geq 0$ 的范围内，即执行 $x = 1/x$ ，$n = - n$ 。 +4. 循环计算：当 $n = 0$ 时跳出。 + 1. 当 $n \& 1 = 1$ 时：将当前 $x$ 乘入 $res$ （即 $res *= x$ ）。 + 2. 执行 $x = x^2$ （即 $x *= x$ ）。 + 3. 执行 $n$ 右移一位（即 $n >>= 1$）。 +5. 返回 $res$ 。 + +## 代码： + +Java 中 int32 变量区间 $n \in [-2147483648, 2147483647]$ ，因此当 $n = -2147483648$ 时执行 $n = -n$ 会因越界而赋值出错。解决方法是先将 $n$ 存入 long 变量 $b$ ，后面用 $b$ 操作即可。 + +```Python [] +class Solution: + def myPow(self, x: float, n: int) -> float: + if x == 0.0: return 0.0 + res = 1 + if n < 0: x, n = 1 / x, -n + while n: + if n & 1: res *= x + x *= x + n >>= 1 + return res +``` + +```Java [] +class Solution { + public double myPow(double x, int n) { + if(x == 0.0f) return 0.0d; + long b = n; + double res = 1.0; + if(b < 0) { + x = 1 / x; + b = -b; + } + while(b > 0) { + if((b & 1) == 1) res *= x; + x *= x; + b >>= 1; + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + double myPow(double x, int n) { + if(x == 0.0f) return 0.0; + long b = n; + double res = 1.0; + if(b < 0) { + x = 1 / x; + b = -b; + } + while(b > 0) { + if((b & 1) == 1) res *= x; + x *= x; + b >>= 1; + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(\log n)$ ：** 二分的时间复杂度为对数级别。 +- **空间复杂度 $O(1)$ ：** $res$, $b$ 等变量占用常数大小额外空间。 diff --git "a/leetbook_ioa/docs/LCR 135. \346\212\245\346\225\260.md" "b/leetbook_ioa/docs/LCR 135. \346\212\245\346\225\260.md" new file mode 100755 index 0000000..7ec06f2 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 135. \346\212\245\346\225\260.md" @@ -0,0 +1,269 @@ +## 解题思路： + +题目要求打印 “从 $1$ 至 $cnt$ 的数字” ，因此需考虑以下两个问题： + +1. **最大的 $cnt$ 位数（记为 $end$ ）和位数 $cnt$ 的关系：** 例如最大的 $1$ 位数是 $9$ ，最大的 $2$ 位数是 $99$ ，最大的 $3$ 位数是 $999$ 。则可推出公式： + +$$ +end = 10^{cnt} - 1 +$$ + +2. **大数越界问题：** 当 $cnt$ 较大时，$end$ 会超出 $int32$ 整型的取值范围，超出取值范围的数字无法正常存储。但由于本题要求返回 int 类型数组，相当于默认所有数字都在 int32 整型取值范围内，因此不考虑大数越界问题。 + +因此，只需定义区间 $[1, 10^{cnt} - 1]$ 和步长 $1$ ，通过 $for$ 循环生成结果列表 $res$ 并返回即可。 + +### 代码： + +```Python [] +class Solution: + def countNumbers(self, cnt: int) -> List[int]: + res = [] + for i in range(1, 10 ** cnt): + res.append(i) + return res +``` + +```Java [] +class Solution { + public int[] countNumbers(int cnt) { + int end = (int)Math.pow(10, cnt) - 1; + int[] res = new int[end]; + for(int i = 0; i < end; i++) + res[i] = i + 1; + return res; + } +} +``` + +利用 Python 的语言特性，可以简化代码：先使用 `range()` 方法生成可迭代对象，再使用 `list()` 方法转化为列表并返回即可。 + +```Python +class Solution: + def countNumbers(self, cnt: int) -> List[int]: + return list(range(1, 10 ** cnt)) +``` + +### 复杂度分析： + +- **时间复杂度 $O(10^{cnt})$ ：** 生成长度为 $10^{cnt}$ 的列表需使用 $O(10^{cnt})$ 时间。 +- **空间复杂度 $O(1)$ ：** 建立列表需使用 $O(1)$ 大小的额外空间（ 列表作为返回结果，不计入额外空间 ）。 + +## 大数打印拓展： + +实际上，本题的主要考点是大数越界情况下的打印。需要解决以下三个问题： + +**1. 表示大数的变量类型：** + +- 无论是 short / int / long ... 任意变量类型，数字的取值范围都是有限的。因此，大数的表示应用字符串 String 类型。 + +**2. 生成数字的字符串集：** + +- 使用 int 类型时，每轮可通过 $+1$ 生成下个数字，而此方法无法应用至 String 类型。并且， String 类型的数字的进位操作效率较低，例如 `"9999"` 至 `"10000"` 需要从个位到千位循环判断，进位 4 次。 +- 观察可知，生成的列表实际上是 $cnt$ 位 $0$ - $9$ 的 **全排列** ，因此可避开进位操作，通过递归生成数字的 String 列表。 + +**3. 递归生成全排列：** + +- 基于分治算法的思想，先固定高位，向低位递归，当个位已被固定时，添加数字的字符串。例如当 $cnt = 2$ 时（数字范围 $1 - 99$ ），固定十位为 $0$ - $9$ ，按顺序依次开启递归，固定个位 $0$ - $9$ ，终止递归并添加数字字符串。 + +> 下图中的 `n` 对应本题中的 `cnt` 。 + +![Picture1.png](https://pic.leetcode-cn.com/83f4b5930ddc1d42b05c724ea2950ee7f00427b11150c86b45bd88405f8c7c87-Picture1.png){:align=center width=500} + +根据以上方法，可初步编写全排列代码： + +```Python [] +class Solution: + def countNumbers(self, cnt: int) -> [int]: + def dfs(x): + if x == cnt: # 终止条件：已固定完所有位 + res.append(''.join(num)) # 拼接 num 并添加至 res 尾部 + return + for i in range(10): # 遍历 0 - 9 + num[x] = str(i) # 固定第 x 位为 i + dfs(x + 1) # 开启固定第 x + 1 位 + + num = ['0'] * cnt # 起始数字定义为 cnt 个 0 组成的字符列表 + res = [] # 数字字符串列表 + dfs(0) # 开启全排列递归 + return ','.join(res) # 拼接所有数字字符串，使用逗号隔开，并返回 +``` + +```Java [] +class Solution { + StringBuilder res; + int count = 0, cnt; + char[] num, loop = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9'}; + public String countNumbers(int cnt) { + this.cnt = cnt; + res = new StringBuilder(); // 数字字符串集 + num = new char[cnt]; // 定义长度为 cnt 的字符列表 + dfs(0); // 开启全排列递归 + res.deleteCharAt(res.length() - 1); // 删除最后多余的逗号 + return res.toString(); // 转化为字符串并返回 + } + void dfs(int x) { + if(x == cnt) { // 终止条件：已固定完所有位 + res.append(String.valueOf(num) + ","); // 拼接 num 并添加至 res 尾部，使用逗号隔开 + return; + } + for(char i : loop) { // 遍历 ‘0‘ - ’9‘ + num[x] = i; // 固定第 x 位为 i + dfs(x + 1); // 开启固定第 x + 1 位 + } + } +} +``` + +在此方法下，各数字字符串被逗号隔开，共同组成长字符串。返回的数字集字符串如下所示： + +```yaml +输入：n = 1 +输出："0,1,2,3,4,5,6,7,8,9" + +输入：n = 2 +输出："00,01,02,...,10,11,12,...,97,98,99" + +输入：n = 3 +输出："000,001,002,...,100,101,102,...,997,998,999" +``` + +观察可知，当前的生成方法仍有以下问题： + +1. 诸如 $00, 01, 02, \cdots$ 应显示为 $0, 1, 2, \cdots$ ，即应 **删除高位多余的 $0$** ; +2. 此方法从 $0$ 开始生成，而题目要求 **列表从 $1$ 开始** ； + +以上两个问题的解决方法如下： + +**1. 删除高位多余的 $0$ ：** + +- **字符串左边界定义：** 声明变量 $start$ 规定字符串的左边界，以保证添加的数字字符串 `num[start:]` 中无高位多余的 $0$ 。例如当 $cnt = 2$ 时，$1 - 9$ 时 $start = 1$ ，$10 - 99$ 时 $start = 0$ 。 + +- **左边界 $start$ 变化规律：** 观察可知，当输出数字的所有位都是 $9$ 时，则下个数字需要向更高位进 $1$ ，此时左边界 $start$ 需要减 $1$ （即高位多余的 $0$ 减少一个）。例如当 $cnt = 3$ （数字范围 $1 - 999$ ）时，左边界 $start$ 需要减 $1$ 的情况有： "009" 进位至 "010" ， "099" 进位至 "100" 。设数字各位中 $9$ 的数量为 $nine$ ，所有位都为 $9$ 的判断条件可用以下公式表示： + +$$ +cnt - start = nine +$$ + +- **统计 $nine$ 的方法：** 固定第 $x$ 位时，当 $i = 9$ 则执行 $nine = nine + 1$ ，并在回溯前恢复 $nine = nine - 1$ 。 + +**2. 列表从 $1$ 开始：** + +- 在以上方法的基础上，添加数字字符串前判断其是否为 `"0"` ，若为 `"0"` 则直接跳过。 + + + +### 代码： + +为 **正确表示大数** ，以下代码的返回值为数字字符串集拼接而成的长字符串。 + +```Python [] +class Solution: + def countNumbers(self, cnt: int) -> [int]: + def dfs(x): + if x == cnt: + s = ''.join(num[self.start:]) + if s != '0': res.append(s) + if cnt - self.start == self.nine: self.start -= 1 + return + for i in range(10): + if i == 9: self.nine += 1 + num[x] = str(i) + dfs(x + 1) + self.nine -= 1 + + num, res = ['0'] * cnt, [] + self.nine = 0 + self.start = cnt - 1 + dfs(0) + return ','.join(res) +``` + +```Java [] +class Solution { + StringBuilder res; + int nine = 0, count = 0, start, cnt; + char[] num, loop = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9'}; + public String countNumbers(int cnt) { + this.cnt = cnt; + res = new StringBuilder(); + num = new char[cnt]; + start = cnt - 1; + dfs(0); + res.deleteCharAt(res.length() - 1); + return res.toString(); + } + void dfs(int x) { + if(x == cnt) { + String s = String.valueOf(num).substring(start); + if(!s.equals("0")) res.append(s + ","); + if(cnt - start == nine) start--; + return; + } + for(char i : loop) { + if(i == '9') nine++; + num[x] = i; + dfs(x + 1); + } + nine--; + } +} +``` + +本题要求输出 int 类型数组。为 **运行通过** ，可在添加数字字符串 $s$ 前，将其转化为 int 类型。代码如下所示： + +```Python [] +class Solution: + def countNumbers(self, cnt: int) -> [int]: + def dfs(x): + if x == cnt: + s = ''.join(num[self.start:]) + if s != '0': res.append(int(s)) + if cnt - self.start == self.nine: self.start -= 1 + return + for i in range(10): + if i == 9: self.nine += 1 + num[x] = str(i) + dfs(x + 1) + self.nine -= 1 + + num, res = ['0'] * cnt, [] + self.nine = 0 + self.start = cnt - 1 + dfs(0) + return res +``` + +```Java [] +class Solution { + int[] res; + int nine = 0, count = 0, start, cnt; + char[] num, loop = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9'}; + public int[] countNumbers(int cnt) { + this.cnt = cnt; + res = new int[(int)Math.pow(10, cnt) - 1]; + num = new char[cnt]; + start = cnt - 1; + dfs(0); + return res; + } + void dfs(int x) { + if(x == cnt) { + String s = String.valueOf(num).substring(start); + if(!s.equals("0")) res[count++] = Integer.parseInt(s); + if(cnt - start == nine) start--; + return; + } + for(char i : loop) { + if(i == '9') nine++; + num[x] = i; + dfs(x + 1); + } + nine--; + } +} +``` + +### 复杂度分析： + +- **时间复杂度 $O(10^{cnt})$ ：** 递归的生成的排列的数量为 $10^{cnt}$ 。 +- **空间复杂度 $O(10^{cnt})$ ：** 结果列表 $res$ 的长度为 $10^{cnt} - 1$ ，各数字字符串的长度区间为 $1, 2, ..., cnt$ ，因此占用 $O(10^{cnt})$ 大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 136. \345\210\240\351\231\244\351\223\276\350\241\250\350\212\202\347\202\271.md" "b/leetbook_ioa/docs/LCR 136. \345\210\240\351\231\244\351\223\276\350\241\250\350\212\202\347\202\271.md" new file mode 100755 index 0000000..d271763 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 136. \345\210\240\351\231\244\351\223\276\350\241\250\350\212\202\347\202\271.md" @@ -0,0 +1,69 @@ +## 解题思路： + +本题删除值为 `val` 的节点分需为两步：定位节点、修改引用。 + +1. **定位节点：** 遍历链表，直到 `head.val == val` 时跳出，即可定位目标节点。 +2. **修改引用：** 设节点 `cur` 的前驱节点为 `pre` ，后继节点为 `cur.next` ；则执行 `pre.next = cur.next` ，即可实现删除 `cur` 节点。 + +![Picture1.png](https://pic.leetcode-cn.com/1613757478-NBOvjn-Picture1.png){:align=center width=450} + +### 算法流程： + +1. **特例处理：** 当应删除头节点 `head` 时，直接返回 `head.next` 即可。 +2. **初始化：** `pre = head` , `cur = head.next` 。 +3. **定位节点：** 当 `cur` 为空 **或** `cur` 节点值等于 `val` 时跳出。 + 1. 保存当前节点索引，即 `pre = cur` 。 + 2. 遍历下一节点，即 `cur = cur.next` 。 +4. **删除节点：** 若 `cur` 指向某节点，则执行 `pre.next = cur.next` ；若 `cur` 指向 $\text{null}$ ，代表链表中不包含值为 `val` 的节点。 +5. **返回值：** 返回链表头部节点 `head` 即可。 + + + +## 代码： + +```Python [] +class Solution: + def deleteNode(self, head: ListNode, val: int) -> ListNode: + if head.val == val: return head.next + pre, cur = head, head.next + while cur and cur.val != val: + pre, cur = cur, cur.next + if cur: pre.next = cur.next + return head +``` + +```Java [] +class Solution { + public ListNode deleteNode(ListNode head, int val) { + if(head.val == val) return head.next; + ListNode pre = head, cur = head.next; + while(cur != null && cur.val != val) { + pre = cur; + cur = cur.next; + } + if(cur != null) pre.next = cur.next; + return head; + } +} +``` + +```C++ [] +class Solution { +public: + ListNode* deleteNode(ListNode* head, int val) { + if(head->val == val) return head->next; + ListNode *pre = head, *cur = head->next; + while(cur != nullptr && cur->val != val) { + pre = cur; + cur = cur->next; + } + if(cur != nullptr) pre->next = cur->next; + return head; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为链表长度，删除操作平均需循环 $N/2$ 次，最差 $N$ 次。 +- **空间复杂度 $O(1)$ ：** `cur`, `pre` 占用常数大小额外空间。 diff --git "a/leetbook_ioa/docs/LCR 137. \346\250\241\347\263\212\346\220\234\347\264\242\351\252\214\350\257\201.md" "b/leetbook_ioa/docs/LCR 137. \346\250\241\347\263\212\346\220\234\347\264\242\351\252\214\350\257\201.md" new file mode 100755 index 0000000..134cad1 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 137. \346\250\241\347\263\212\346\220\234\347\264\242\351\252\214\350\257\201.md" @@ -0,0 +1,185 @@ +## 解题思路： + +> 设 $s$ 的长度为 $n$ ，$p$ 的长度为 $m$ ；将 $s$ 的第 $i$ 个字符记为 $s_i$ ，$p$ 的第 $j$ 个字符记为 $p_j$ ，将 $s$ 的前 $i$ 个字符组成的子字符串记为 $s[:i]$ , 同理将 $p$ 的前 $j$ 个字符组成的子字符串记为 $p[:j]$ 。 +> +> 因此，本题可转化为求 $s[:n]$ 是否能和 $p[:m]$ 匹配。 + +总体思路是从 $s[:1]$ 和 $p[:1]$ 是否能匹配开始判断，每轮添加一个字符并判断是否能匹配，直至添加完整个字符串 $s$ 和 $p$ 。展开来看，假设 $s[:i]$ 与 $p[:j]$ 可以匹配，那么下一状态有两种： + +1. 添加一个字符 $s_{i+1}$ 后是否能匹配？ +2. 添加字符 $p_{j+1}$ 后是否能匹配？ + +![Picture1.png](https://pic.leetcode-cn.com/1614516402-HyzAil-Picture1.png){:align=center width=500} + +因此，本题的状态共有 $m \times n$ 种，应定义状态矩阵 $dp$ ，$dp[i][j]$ 代表 $s[:i]$ 与 $p[:j]$ 是否可以匹配。 + +做好状态定义，接下来就是根据 「`普通字符`」 , 「`.`」 , 「`*`」三种字符的功能定义，分析出动态规划的转移方程。 + +### 动态规划解析： + +**状态定义：** 设动态规划矩阵 `dp` ，`dp[i][j]` 代表字符串 `s` 的前 `i` 个字符和 `p` 的前 `j` 个字符能否匹配。 + +**转移方程：** 需要注意，由于 `dp[0][0]` 代表的是空字符的状态， 因此 `dp[i][j]` 对应的添加字符是 `s[i - 1]` 和 `p[j - 1]` 。 + +- 当 `p[j - 1] = '*'` 时，`dp[i][j]` 在当以下任一情况为 $\text{true}$ 时等于 $\text{true}$ ： + + 1. **`dp[i][j - 2]`：** 即将字符组合 `p[j - 2] *` 看作出现 0 次时，能否匹配； + 2. **`dp[i - 1][j]` 且 `s[i - 1] = p[j - 2]`:** 即让字符 `p[j - 2]` 多出现 1 次时，能否匹配； + 3. **`dp[i - 1][j]` 且 `p[j - 2] = '.'`:** 即让字符 `'.'` 多出现 1 次时，能否匹配； + +- 当 `p[j - 1] != '*'` 时，`dp[i][j]` 在当以下任一情况为 $\text{true}$ 时等于 $\text{true}$ ： + + 1. **`dp[i - 1][j - 1]` 且 `s[i - 1] = p[j - 1]`：** 即让字符 `p[j - 1]` 多出现一次时，能否匹配； + 2. **`dp[i - 1][j - 1]` 且 `p[j - 1] = '.'`：** 即将字符 `.` 看作字符 `s[i - 1]` 时，能否匹配； + +**初始化：** 需要先初始化 `dp` 矩阵首行，以避免状态转移时索引越界。 + +- **`dp[0][0] = true`：** 代表两个空字符串能够匹配。 +- **`dp[0][j] = dp[0][j - 2]` 且 `p[j - 1] = '*'`：** 首行 `s` 为空字符串，因此当 `p` 的偶数位为 `*` 时才能够匹配（即让 `p` 的奇数位出现 0 次，保持 `p` 是空字符串）。因此，循环遍历字符串 `p` ，步长为 2（即只看偶数位）。 + +**返回值：** `dp` 矩阵右下角字符，代表字符串 `s` 和 `p` 能否匹配。 + + + +## 代码： + +```Python [] +class Solution: + def articleMatch(self, s: str, p: str) -> bool: + m, n = len(s) + 1, len(p) + 1 + dp = [[False] * n for _ in range(m)] + dp[0][0] = True + for j in range(2, n, 2): + dp[0][j] = dp[0][j - 2] and p[j - 1] == '*' + for i in range(1, m): + for j in range(1, n): + dp[i][j] = dp[i][j - 2] or dp[i - 1][j] and (s[i - 1] == p[j - 2] or p[j - 2] == '.') \ + if p[j - 1] == '*' else \ + dp[i - 1][j - 1] and (p[j - 1] == '.' or s[i - 1] == p[j - 1]) + return dp[-1][-1] +``` + +```Java [] +class Solution { + public boolean articleMatch(String s, String p) { + int m = s.length() + 1, n = p.length() + 1; + boolean[][] dp = new boolean[m][n]; + dp[0][0] = true; + for(int j = 2; j < n; j += 2) + dp[0][j] = dp[0][j - 2] && p.charAt(j - 1) == '*'; + for(int i = 1; i < m; i++) { + for(int j = 1; j < n; j++) { + dp[i][j] = p.charAt(j - 1) == '*' ? + dp[i][j - 2] || dp[i - 1][j] && (s.charAt(i - 1) == p.charAt(j - 2) || p.charAt(j - 2) == '.') : + dp[i - 1][j - 1] && (p.charAt(j - 1) == '.' || s.charAt(i - 1) == p.charAt(j - 1)); + } + } + return dp[m - 1][n - 1]; + } +} +``` + +```C++ [] +class Solution { +public: + bool articleMatch(string s, string p) { + int m = s.size() + 1, n = p.size() + 1; + vector> dp(m, vector(n, false)); + dp[0][0] = true; + for(int j = 2; j < n; j += 2) + dp[0][j] = dp[0][j - 2] && p[j - 1] == '*'; + for(int i = 1; i < m; i++) { + for(int j = 1; j < n; j++) { + dp[i][j] = p[j - 1] == '*' ? + dp[i][j - 2] || dp[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == '.'): + dp[i - 1][j - 1] && (p[j - 1] == '.' || s[i - 1] == p[j - 1]); + } + } + return dp[m - 1][n - 1]; + } +}; +``` + +以上代码利用布尔运算实现简短长度，若阅读不畅，可先理解以下代码，与文中内容一一对应： + +```Python [] +class Solution: + def articleMatch(self, s: str, p: str) -> bool: + m, n = len(s) + 1, len(p) + 1 + dp = [[False] * n for _ in range(m)] + dp[0][0] = True + # 初始化首行 + for j in range(2, n, 2): + dp[0][j] = dp[0][j - 2] and p[j - 1] == '*' + # 状态转移 + for i in range(1, m): + for j in range(1, n): + if p[j - 1] == '*': + if dp[i][j - 2]: dp[i][j] = True # 1. + elif dp[i - 1][j] and s[i - 1] == p[j - 2]: dp[i][j] = True # 2. + elif dp[i - 1][j] and p[j - 2] == '.': dp[i][j] = True # 3. + else: + if dp[i - 1][j - 1] and s[i - 1] == p[j - 1]: dp[i][j] = True # 1. + elif dp[i - 1][j - 1] and p[j - 1] == '.': dp[i][j] = True # 2. + return dp[-1][-1] +``` + +```Java [] +class Solution { + public boolean articleMatch(String s, String p) { + int m = s.length() + 1, n = p.length() + 1; + boolean[][] dp = new boolean[m][n]; + dp[0][0] = true; + // 初始化首行 + for(int j = 2; j < n; j += 2) + dp[0][j] = dp[0][j - 2] && p.charAt(j - 1) == '*'; + // 状态转移 + for(int i = 1; i < m; i++) { + for(int j = 1; j < n; j++) { + if(p.charAt(j - 1) == '*') { + if(dp[i][j - 2]) dp[i][j] = true; // 1. + else if(dp[i - 1][j] && s.charAt(i - 1) == p.charAt(j - 2)) dp[i][j] = true; // 2. + else if(dp[i - 1][j] && p.charAt(j - 2) == '.') dp[i][j] = true; // 3. + } else { + if(dp[i - 1][j - 1] && s.charAt(i - 1) == p.charAt(j - 1)) dp[i][j] = true; // 1. + else if(dp[i - 1][j - 1] && p.charAt(j - 1) == '.') dp[i][j] = true; // 2. + } + } + } + return dp[m - 1][n - 1]; + } +} +``` + +```C++ [] +class Solution { +public: + bool articleMatch(string s, string p) { + int m = s.size() + 1, n = p.size() + 1; + vector> dp(m, vector(n, false)); + dp[0][0] = true; + // 初始化首行 + for(int j = 2; j < n; j += 2) + dp[0][j] = dp[0][j - 2] && p[j - 1] == '*'; + // 状态转移 + for(int i = 1; i < m; i++) { + for(int j = 1; j < n; j++) { + if(p[j - 1] == '*') { + if(dp[i][j - 2]) dp[i][j] = true; // 1. + else if(dp[i - 1][j] && s[i - 1] == p[j - 2]) dp[i][j] = true; // 2. + else if(dp[i - 1][j] && p[j - 2] == '.') dp[i][j] = true; // 3. + } else { + if(dp[i - 1][j - 1] && s[i - 1] == p[j - 1]) dp[i][j] = true; // 1. + else if(dp[i - 1][j - 1] && p[j - 1] == '.') dp[i][j] = true; // 2. + } + } + } + return dp[m - 1][n - 1]; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(MN)$ ：** 其中 $M, N$ 分别为 `s` 和 `p` 的长度，状态转移需遍历整个 `dp` 矩阵。 +- **空间复杂度 $O(MN)$ ：** 状态矩阵 `dp` 使用 $O(MN)$ 的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 138. \346\234\211\346\225\210\346\225\260\345\255\227.md" "b/leetbook_ioa/docs/LCR 138. \346\234\211\346\225\210\346\225\260\345\255\227.md" new file mode 100755 index 0000000..7f68180 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 138. \346\234\211\346\225\210\346\225\260\345\255\227.md" @@ -0,0 +1,112 @@ +## 解题思路： + +本题使用有限状态自动机。根据字符类型和合法数值的特点，先定义状态，再画出状态转移图，最后编写代码即可。 + +**字符类型：** + +空格 「 」、数字「 $0—9$ 」 、正负号 「 $+$, $-$ 」 、小数点 「 $.$ 」 、幂符号 「 $e$, $E$ 」 。 + +**状态定义：** + +按照字符串从左到右的顺序，定义以下 9 种状态。 + +0. 开始的空格 +1. 幂符号前的正负号 +2. 小数点前的数字 +3. 小数点、小数点后的数字 +4. 当小数点前为空格时，小数点、小数点后的数字 +5. 幂符号 +6. 幂符号后的正负号 +7. 幂符号后的数字 +8. 结尾的空格 + +**结束状态：** + +合法的结束状态有 2, 3, 7, 8 。 + +![Picture1.png](https://pic.leetcode-cn.com/1599283151-YmPMis-Picture1.png){:align=center width=650} + +### 算法流程： + +1. **初始化：** + 1. **状态转移表 `states` ：** 设 `states[i]` ，其中 `i` 为所处状态，`states[i]` 使用哈希表存储可转移至的状态。键值对 `(key, value)` 含义：输入字符 `key` ，则从状态 `i` 转移至状态 `value` 。 + 2. **当前状态 `p` ：** 起始状态初始化为 `p = 0` 。 + +2. **状态转移循环：** 遍历字符串 `s` 的每个字符 `c` 。 + 1. **记录字符类型 `t` ：** 分为四种情况。 + - 当 `c` 为正负号时，执行 `t = 's'` ; + - 当 `c` 为数字时，执行 `t = 'd'` ; + - 当 `c` 为 `e` 或 `E` 时，执行 `t = 'e'` ; + - 当 `c` 为 `.` 或 `空格` 时，执行 `t = c` （即用字符本身表示字符类型）; + - 否则，执行 `t = '?'` ，代表为不属于判断范围的非法字符，后续直接返回 $\text{false}$ 。 + 2. **终止条件：** 若字符类型 `t` 不在哈希表 `states[p]` 中，说明无法转移至下一状态，因此直接返回 $\text{false}$ 。 + 3. **状态转移：** 状态 `p` 转移至 `states[p][t]` 。 + +3. **返回值：** 跳出循环后，若状态 `p` $\in {2, 3, 7, 8}$ ，说明结尾合法，返回 $\text{true}$ ，否则返回 $\text{false}$ 。 + + + +## 代码： + +Java 的状态转移表 `states` 使用 Map[] 数组存储。 + +```Python [] +class Solution: + def validNumber(self, s: str) -> bool: + states = [ + { ' ': 0, 's': 1, 'd': 2, '.': 4 }, # 0. start with 'blank' + { 'd': 2, '.': 4 } , # 1. 'sign' before 'e' + { 'd': 2, '.': 3, 'e': 5, ' ': 8 }, # 2. 'digit' before 'dot' + { 'd': 3, 'e': 5, ' ': 8 }, # 3. 'digit' after 'dot' + { 'd': 3 }, # 4. 'digit' after 'dot' (‘blank’ before 'dot') + { 's': 6, 'd': 7 }, # 5. 'e' + { 'd': 7 }, # 6. 'sign' after 'e' + { 'd': 7, ' ': 8 }, # 7. 'digit' after 'e' + { ' ': 8 } # 8. end with 'blank' + ] + p = 0 # start with state 0 + for c in s: + if '0' <= c <= '9': t = 'd' # digit + elif c in "+-": t = 's' # sign + elif c in "eE": t = 'e' # e or E + elif c in ". ": t = c # dot, blank + else: t = '?' # unknown + if t not in states[p]: return False + p = states[p][t] + return p in (2, 3, 7, 8) +``` + +```Java [] +class Solution { + public boolean validNumber(String s) { + Map[] states = { + new HashMap<>() {{ put(' ', 0); put('s', 1); put('d', 2); put('.', 4); }}, // 0. + new HashMap<>() {{ put('d', 2); put('.', 4); }}, // 1. + new HashMap<>() {{ put('d', 2); put('.', 3); put('e', 5); put(' ', 8); }}, // 2. + new HashMap<>() {{ put('d', 3); put('e', 5); put(' ', 8); }}, // 3. + new HashMap<>() {{ put('d', 3); }}, // 4. + new HashMap<>() {{ put('s', 6); put('d', 7); }}, // 5. + new HashMap<>() {{ put('d', 7); }}, // 6. + new HashMap<>() {{ put('d', 7); put(' ', 8); }}, // 7. + new HashMap<>() {{ put(' ', 8); }} // 8. + }; + int p = 0; + char t; + for(char c : s.toCharArray()) { + if(c >= '0' && c <= '9') t = 'd'; + else if(c == '+' || c == '-') t = 's'; + else if(c == 'e' || c == 'E') t = 'e'; + else if(c == '.' || c == ' ') t = c; + else t = '?'; + if(!states[p].containsKey(t)) return false; + p = (int)states[p].get(t); + } + return p == 2 || p == 3 || p == 7 || p == 8; + } +} +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为字符串 `s` 的长度，判断需遍历字符串，每轮状态转移的使用 $O(1)$ 时间。 +- **空间复杂度 $O(1)$ ：** `states` 和 `p` 使用常数大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 139. \350\256\255\347\273\203\350\256\241\345\210\222 I.md" "b/leetbook_ioa/docs/LCR 139. \350\256\255\347\273\203\350\256\241\345\210\222 I.md" new file mode 100755 index 0000000..792ef3f --- /dev/null +++ "b/leetbook_ioa/docs/LCR 139. \350\256\255\347\273\203\350\256\241\345\210\222 I.md" @@ -0,0 +1,77 @@ +## 解题思路： + +考虑定义双指针 $i$ , $j$ 分列数组左右两端，循环执行： + +1. 指针 $i$ 从左向右寻找偶数； +2. 指针 $j$ 从右向左寻找奇数； +3. 将 偶数 $actions[i]$ 和 奇数 $actions[j]$ 交换。 + +可始终保证： 指针 $i$ 左边都是奇数，指针 $j$ 右边都是偶数 。 + +> 下图中的 `nums` 对应本题的 `actions` 。 + +![Picture1.png](https://pic.leetcode-cn.com/43e965485da89efa688947bc108232f10b65b5ba5c0dbd6a68227a82c7e451e4-Picture1.png){:align=center width=450} + +### 算法流程： + +1. **初始化：** $i$ , $j$ 双指针，分别指向数组 $actions$ 左右两端； +2. **循环交换：** 当 $i = j$ 时跳出； + 1. 指针 $i$ 遇到奇数则执行 $i = i + 1$ 跳过，直到找到偶数； + 2. 指针 $j$ 遇到偶数则执行 $j = j - 1$ 跳过，直到找到奇数； + 3. 交换 $actions[i]$ 和 $actions[j]$ 值； +3. **返回值：** 返回已修改的 $actions$ 数组。 + + + +## 代码： + +$x \& 1$ 位运算 等价于 $x \mod 2$ 取余运算，即皆可用于判断数字奇偶性。 + +```Python [] +class Solution: + def trainingPlan(self, actions: List[int]) -> List[int]: + i, j = 0, len(actions) - 1 + while i < j: + while i < j and actions[i] & 1 == 1: i += 1 + while i < j and actions[j] & 1 == 0: j -= 1 + actions[i], actions[j] = actions[j], actions[i] + return actions +``` + +```Java [] +class Solution { + public int[] trainingPlan(int[] actions) { + int i = 0, j = actions.length - 1, tmp; + while(i < j) { + while(i < j && (actions[i] & 1) == 1) i++; + while(i < j && (actions[j] & 1) == 0) j--; + tmp = actions[i]; + actions[i] = actions[j]; + actions[j] = tmp; + } + return actions; + } +} +``` + +```C++ [] +class Solution { +public: + vector trainingPlan(vector& actions) + { + int i = 0, j = actions.size() - 1; + while (i < j) + { + while(i < j && (actions[i] & 1) == 1) i++; + while(i < j && (actions[j] & 1) == 0) j--; + swap(actions[i], actions[j]); + } + return actions; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为数组 $actions$ 长度，双指针 $i$, $j$ 共同遍历整个数组。 +- **空间复杂度 $O(1)$ ：** 双指针 $i$, $j$ 使用常数大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 140. \350\256\255\347\273\203\350\256\241\345\210\222 II.md" "b/leetbook_ioa/docs/LCR 140. \350\256\255\347\273\203\350\256\241\345\210\222 II.md" new file mode 100755 index 0000000..87288d5 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 140. \350\256\255\347\273\203\350\256\241\345\210\222 II.md" @@ -0,0 +1,119 @@ +## 解题思路： + +第一时间想到的解法： + +1. 先遍历统计链表长度，记为 $n$ ； +2. 设置一个指针走 $(n-cnt)$ 步，即可找到链表倒数第 $cnt$ 个节点； + +使用双指针则可以不用统计链表长度。 + +> 下图中的 `k` 对应本题的 `cnt` 。 + +![Picture1.png](https://pic.leetcode-cn.com/1600794523-AAMvoP-Picture1.png){:align=center width=450} + +### 算法流程： + +1. **初始化：** 前指针 `former` 、后指针 `latter` ，双指针都指向头节点 `head​` 。 +2. **构建双指针距离：** 前指针 `former` 先向前走 $cnt$ 步（结束后，双指针 `former` 和 `latter` 间相距 $cnt$ 步）。 +3. **双指针共同移动：** 循环中，双指针 `former` 和 `latter` 每轮都向前走一步，直至 `former` 走过链表 **尾节点** 时跳出（跳出后，`latter` 与尾节点距离为 $cnt-1$，即 `latter` 指向倒数第 $cnt$ 个节点）。 +4. **返回值：** 返回 `latter` 即可。 + + + +## 代码： + +```Python [] +class Solution: + def trainingPlan(self, head: ListNode, cnt: int) -> ListNode: + former, latter = head, head + for _ in range(cnt): + former = former.next + while former: + former, latter = former.next, latter.next + return latter +``` + +```Java [] +class Solution { + public ListNode trainingPlan(ListNode head, int cnt) { + ListNode former = head, latter = head; + for(int i = 0; i < cnt; i++) + former = former.next; + while(former != null) { + former = former.next; + latter = latter.next; + } + return latter; + } +} +``` + +```C++ [] +class Solution { +public: + ListNode* trainingPlan(ListNode* head, int cnt) { + ListNode *former = head, *latter = head; + for(int i = 0; i < cnt; i++) + former = former->next; + while(former != nullptr) { + former = former->next; + latter = latter->next; + } + return latter; + } +}; +``` + +本题没有 $cnt>$ 链表长度的测试样例 ，因此不用考虑越界。考虑越界问题的代码如下： + +```Python [] +class Solution: + def trainingPlan(self, head: ListNode, cnt: int) -> ListNode: + former, latter = head, head + for _ in range(cnt): + if not former: return + former = former.next + while former: + former, latter = former.next, latter.next + return latter +``` + +```Java [] +class Solution { + public ListNode trainingPlan(ListNode head, int cnt) { + ListNode former = head, latter = head; + for(int i = 0; i < cnt; i++) { + if(former == null) return null; + former = former.next; + } + while(former != null) { + former = former.next; + latter = latter.next; + } + return latter; + } +} +``` + +```C++ [] +class Solution { +public: + ListNode* trainingPlan(ListNode* head, int cnt) { + ListNode *former = head, *latter = head; + for(int i = 0; i < cnt; i++) { + if(former == nullptr) return nullptr; + former = former->next; + } + while(former != nullptr) { + former = former->next; + latter = latter->next; + } + return latter; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(n)$ ：** $n$ 为链表长度；总体看，`former` 走了 $n$ 步，`latter` 走了 $(-cnt)$ 步。 +- **空间复杂度 $O(1)$ ：** 双指针 `former` , `latter` 使用常数大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 141. \350\256\255\347\273\203\350\256\241\345\210\222 III.md" "b/leetbook_ioa/docs/LCR 141. \350\256\255\347\273\203\350\256\241\345\210\222 III.md" new file mode 100755 index 0000000..209fde8 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 141. \350\256\255\347\273\203\350\256\241\345\210\222 III.md" @@ -0,0 +1,127 @@ +## 解题思路： + +如下图所示，题目要求将链表反转。本文介绍迭代（双指针）、递归两种实现方法。 + +![Picture1.png](https://pic.leetcode-cn.com/1604779288-WXygqL-Picture1.png){:align=center width=450} + +## 方法一：迭代（双指针） + +考虑遍历链表，并在访问各节点时修改 `next` 引用指向，算法流程见注释。 + + + +### 代码： + +```Python [] +class Solution: + def trainningPlan(self, head: ListNode) -> ListNode: + cur, pre = head, None + while cur: + tmp = cur.next # 暂存后继节点 cur.next + cur.next = pre # 修改 next 引用指向 + pre = cur # pre 暂存 cur + cur = tmp # cur 访问下一节点 + return pre +``` + +```Java [] +class Solution { + public ListNode trainningPlan(ListNode head) { + ListNode cur = head, pre = null; + while(cur != null) { + ListNode tmp = cur.next; // 暂存后继节点 cur.next + cur.next = pre; // 修改 next 引用指向 + pre = cur; // pre 暂存 cur + cur = tmp; // cur 访问下一节点 + } + return pre; + } +} +``` + +```C++ [] +class Solution { +public: + ListNode* trainningPlan(ListNode* head) { + ListNode *cur = head, *pre = nullptr; + while(cur != nullptr) { + ListNode* tmp = cur->next; // 暂存后继节点 cur.next + cur->next = pre; // 修改 next 引用指向 + pre = cur; // pre 暂存 cur + cur = tmp; // cur 访问下一节点 + } + return pre; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 遍历链表使用线性大小时间。 +- **空间复杂度 $O(1)$ ：** 变量 `pre` 和 `cur` 使用常数大小额外空间。 + +## 方法二：递归 + +考虑使用递归遍历链表，当越过尾节点后终止递归，在回溯时修改各节点的 `next` 引用指向。 + +### `recur(cur, pre)` 递归函数： + +1. 终止条件：当 `cur` 为空，则返回尾节点 `pre` （即反转链表的头节点）； +2. 递归后继节点，记录返回值（即反转链表的头节点）为 `res` ； +3. 修改当前节点 `cur` 引用指向前驱节点 `pre` ； +4. 返回反转链表的头节点 `res` ； + +### `trainningPlan(head)` 函数： + +调用并返回 `recur(head, null)` 。传入 `null` 是因为反转链表后，`head` 节点指向 `null` ； + + + +### 代码： + +```Python [] +class Solution: + def trainningPlan(self, head: ListNode) -> ListNode: + def recur(cur, pre): + if not cur: return pre # 终止条件 + res = recur(cur.next, cur) # 递归后继节点 + cur.next = pre # 修改节点引用指向 + return res # 返回反转链表的头节点 + + return recur(head, None) # 调用递归并返回 +``` + +```Java [] +class Solution { + public ListNode trainningPlan(ListNode head) { + return recur(head, null); // 调用递归并返回 + } + private ListNode recur(ListNode cur, ListNode pre) { + if (cur == null) return pre; // 终止条件 + ListNode res = recur(cur.next, cur); // 递归后继节点 + cur.next = pre; // 修改节点引用指向 + return res; // 返回反转链表的头节点 + } +} +``` + +```C++ [] +class Solution { +public: + ListNode* trainningPlan(ListNode* head) { + return recur(head, nullptr); // 调用递归并返回 + } +private: + ListNode* recur(ListNode* cur, ListNode* pre) { + if (cur == nullptr) return pre; // 终止条件 + ListNode* res = recur(cur->next, cur); // 递归后继节点 + cur->next = pre; // 修改节点引用指向 + return res; // 返回反转链表的头节点 + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 遍历链表使用线性大小时间。 +- **空间复杂度 $O(N)$ ：** 遍历链表的递归深度达到 $N$ ，系统使用 $O(N)$ 大小额外空间。 diff --git "a/leetbook_ioa/docs/LCR 142. \350\256\255\347\273\203\350\256\241\345\210\222 IV.md" "b/leetbook_ioa/docs/LCR 142. \350\256\255\347\273\203\350\256\241\345\210\222 IV.md" new file mode 100755 index 0000000..9b11b8a --- /dev/null +++ "b/leetbook_ioa/docs/LCR 142. \350\256\255\347\273\203\350\256\241\345\210\222 IV.md" @@ -0,0 +1,88 @@ +## 解题思路： + +根据题目描述， 链表 `l1` , `l2` 是 **递增** 的，因此容易想到使用双指针 `l1` 和 `l2` 遍历两链表，根据 `l1.val` 和 `l2.val` 的大小关系确定节点添加顺序，两节点指针交替前进，直至遍历完毕。 + +**引入伪头节点：** 由于初始状态合并链表中无节点，因此循环第一轮时无法将节点添加到合并链表中。解决方案：初始化一个辅助节点 `dum` 作为合并链表的伪头节点，将各节点添加至 `dum` 之后。 + +![Picture1.png](https://pic.leetcode-cn.com/e4c8c97883da50d81498fd1f1e6cdd575429bd65f9f2babb00dc2b709f7ad8b2-Picture1.png){:align=center width=450} + +### 算法流程： + +1. **初始化：** 伪头节点 `dum` ，节点 `cur` 指向 `dum` 。 +2. **循环合并：** 当 `l1` 或 `l2` 为空时跳出； + 1. 当 `l1.val < l2.val` 时： `cur` 的后继节点指定为 `l1` ，并 `l1` 向前走一步； + 2. 当 `l1.val >= l2.val` 时： `cur` 的后继节点指定为 `l2` ，并 `l2` 向前走一步 ； + 3. 节点 `cur` 向前走一步，即 `cur = cur.next` 。 +3. **合并剩余尾部：** 跳出时有两种情况，即 `l1` 为空 **或** `l2` 为空。 + 1. 若 `l1 != null` ： 将 `l1` 添加至节点 `cur` 之后； + 2. 否则： 将 `l2` 添加至节点 `cur` 之后。 +4. **返回值：** 合并链表在伪头节点 `dum` 之后，因此返回 `dum.next` 即可。 + + + +## 代码： + +Python 三元表达式写法 `A if x else B` ，代表当 `x = True` 时执行 `A` ，否则执行 `B` 。 + +```Python [] +class Solution: + def trainningPlan(self, l1: ListNode, l2: ListNode) -> ListNode: + cur = dum = ListNode(0) + while l1 and l2: + if l1.val < l2.val: + cur.next, l1 = l1, l1.next + else: + cur.next, l2 = l2, l2.next + cur = cur.next + cur.next = l1 if l1 else l2 + return dum.next +``` + +```Java [] +class Solution { + public ListNode trainningPlan(ListNode l1, ListNode l2) { + ListNode dum = new ListNode(0), cur = dum; + while(l1 != null && l2 != null) { + if(l1.val < l2.val) { + cur.next = l1; + l1 = l1.next; + } + else { + cur.next = l2; + l2 = l2.next; + } + cur = cur.next; + } + cur.next = l1 != null ? l1 : l2; + return dum.next; + } +} +``` + +```C++ [] +class Solution { +public: + ListNode* trainningPlan(ListNode* l1, ListNode* l2) { + ListNode* dum = new ListNode(0); + ListNode* cur = dum; + while(l1 != nullptr && l2 != nullptr) { + if(l1->val < l2->val) { + cur->next = l1; + l1 = l1->next; + } + else { + cur->next = l2; + l2 = l2->next; + } + cur = cur->next; + } + cur->next = l1 != nullptr ? l1 : l2; + return dum->next; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(M+N)$ ：** $M$ , $N$ 分别为链表 `l1`, `l2` 的长度，合并操作需遍历两链表。 +- **空间复杂度 $O(1)$ ：** 节点引用 `dum` , `cur` 使用常数大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 143. \345\255\220\347\273\223\346\236\204\345\210\244\346\226\255.md" "b/leetbook_ioa/docs/LCR 143. \345\255\220\347\273\223\346\236\204\345\210\244\346\226\255.md" new file mode 100755 index 0000000..f9d9a06 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 143. \345\255\220\347\273\223\346\236\204\345\210\244\346\226\255.md" @@ -0,0 +1,78 @@ +## 解题思路： + +若树 `B` 是树 `A` 的子结构，则子结构的根节点可能为树 `A` 的任意一个节点。因此，判断树 `B` 是否是树 `A` 的子结构，需完成以下两步工作： + +1. 先序遍历树 `A` 中的每个节点 `node` ；（对应函数 `isSubStructure(A, B)`） +2. 判断树 `A` 中以 `node` 为根节点的子树是否包含树 `B` 。（对应函数 `recur(A, B)`） + +![Picture1.png](https://pic.leetcode-cn.com/1599290566-VhWsiQ-Picture1.png){:align=center width=500} + +### 算法流程： + +本文名词规定：**树 `A`** 的根节点记作 **节点 `A`** ，**树 `B`** 的根节点称为 **节点 `B`** 。 + +**`recur(A, B)` 函数：** + +1. **终止条件：** + 1. 当节点 `B` 为空：说明树 `B` 已匹配完成（越过叶子节点），因此返回 $\text{true}$ ； + 2. 当节点 `A` 为空：说明已经越过树 `A` 的叶节点，即匹配失败，返回 $\text{false}$ ； + 3. 当节点 `A` 和 `B` 的值不同：说明匹配失败，返回 $\text{false}$ ； +2. **返回值：** + 1. 判断 `A` 和 `B` 的 **左子节点** 是否相等，即 `recur(A.left, B.left)` ； + 2. 判断 `A` 和 `B` 的 **右子节点** 是否相等，即 `recur(A.right, B.right)` ； + +**`isSubStructure(A, B)` 函数：** + +1. **特例处理：** 当 树 `A` 为空 **或** 树 `B` 为空 时，直接返回 $\text{false}$ ； +2. **返回值：** 若树 `B` 是树 `A` 的子结构，则必满足以下三种情况之一，因此用或 `||` 连接； + 1. 以 **节点 `A` 为根节点的子树** 包含树 `B` ，对应 `recur(A, B)`； + 2. 树 `B` 是 **树 `A` 左子树** 的子结构，对应 `isSubStructure(A.left, B)`； + 3. 树 `B` 是 **树 `A` 右子树** 的子结构，对应 `isSubStructure(A.right, B)`； + + + +## 代码： + +```Python [] +class Solution: + def isSubStructure(self, A: TreeNode, B: TreeNode) -> bool: + def recur(A, B): + if not B: return True + if not A or A.val != B.val: return False + return recur(A.left, B.left) and recur(A.right, B.right) + + return bool(A and B) and (recur(A, B) or self.isSubStructure(A.left, B) or self.isSubStructure(A.right, B)) +``` + +```Java [] +class Solution { + public boolean isSubStructure(TreeNode A, TreeNode B) { + return (A != null && B != null) && (recur(A, B) || isSubStructure(A.left, B) || isSubStructure(A.right, B)); + } + boolean recur(TreeNode A, TreeNode B) { + if(B == null) return true; + if(A == null || A.val != B.val) return false; + return recur(A.left, B.left) && recur(A.right, B.right); + } +} +``` + +```C++ [] +class Solution { +public: + bool isSubStructure(TreeNode* A, TreeNode* B) { + return (A != nullptr && B != nullptr) && (recur(A, B) || isSubStructure(A->left, B) || isSubStructure(A->right, B)); + } +private: + bool recur(TreeNode* A, TreeNode* B) { + if(B == nullptr) return true; + if(A == nullptr || A->val != B->val) return false; + return recur(A->left, B->left) && recur(A->right, B->right); + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(MN)$ ：** 其中 $M, N$ 分别为树 `A` 和 树 `B` 的节点数量；先序遍历树 `A` 占用 $O(M)$ ，每次调用 `recur(A, B)` 判断占用 $O(N)$ 。 +- **空间复杂度 $O(M)$ ：** 当树 `A` 和树 `B` 都退化为链表时，递归调用深度最大。当 $M \leq N$ 时，遍历树 `A` 与递归判断的总递归深度为 $M$ ；当 $M>N$ 时，最差情况为遍历至树 `A` 的叶节点，此时总递归深度为 $M$。 diff --git "a/leetbook_ioa/docs/LCR 144. \347\277\273\350\275\254\344\272\214\345\217\211\346\240\221.md" "b/leetbook_ioa/docs/LCR 144. \347\277\273\350\275\254\344\272\214\345\217\211\346\240\221.md" new file mode 100755 index 0000000..0dacee3 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 144. \347\277\273\350\275\254\344\272\214\345\217\211\346\240\221.md" @@ -0,0 +1,153 @@ +## 解题思路： + +**二叉树镜像定义：** 对于二叉树中任意节点 `root` ，设其左 / 右子节点分别为 `left` , `right` ；则在二叉树的镜像中的对应 `root` 节点，其左 / 右子节点分别为 `right` , `left` 。 + +![Picture1.png](https://pic.leetcode-cn.com/20717714d97fa04d509e4f0525a3089efefc2ea02cc08ce92b77978e9b51f15f-Picture1.png){:align=center width=450} + +## 方法一：递归 + +根据二叉树镜像的定义，考虑递归遍历（dfs）二叉树，交换每个节点的左 / 右子节点，即可生成二叉树的镜像。 + +### 递归解析： + +1. **终止条件：** 当节点 `root` 为空时（即越过叶节点），则返回 $\text{null}$ ； +2. **递推工作：** + 1. 初始化节点 `tmp` ，用于暂存 `root` 的左子节点； + 2. 开启递归 **右子节点** `mirrorTree(root.right)` ，并将返回值作为 `root` 的 **左子节点** 。 + 3. 开启递归 **左子节点** `mirrorTree(tmp)` ，并将返回值作为 `root` 的 **右子节点** 。 +3. **返回值：** 返回当前节点 `root` ； + +> **Q：** 为何需要暂存 `root` 的左子节点？ +> **A：** 在递归右子节点 “`root.left = mirrorTree(root.right)`” 执行完毕后，`root.left` 的值已经发生改变，此时递归左子节点 `mirrorTree(root.left)` 则会出问题。 + + + +## 代码： + +```Python [] +class Solution: + def mirrorTree(self, root: TreeNode) -> TreeNode: + if not root: return + tmp = root.left + root.left = self.mirrorTree(root.right) + root.right = self.mirrorTree(tmp) + return root +``` + +```Java [] +class Solution { + public TreeNode mirrorTree(TreeNode root) { + if(root == null) return null; + TreeNode tmp = root.left; + root.left = mirrorTree(root.right); + root.right = mirrorTree(tmp); + return root; + } +} +``` + +```C++ [] +class Solution { +public: + TreeNode* mirrorTree(TreeNode* root) { + if (root == nullptr) return nullptr; + TreeNode* tmp = root->left; + root->left = mirrorTree(root->right); + root->right = mirrorTree(tmp); + return root; + } +}; +``` + +Python 利用平行赋值的写法（即 `a, b = b, a` ），可省略暂存操作。其原理是先将等号右侧打包成元组 `(b,a)` ，再序列地分给等号左侧的 `a, b` 序列。 + +```Python [] +class Solution: + def mirrorTree(self, root: TreeNode) -> TreeNode: + if not root: return + root.left, root.right = self.mirrorTree(root.right), self.mirrorTree(root.left) + return root +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为二叉树的节点数量，建立二叉树镜像需要遍历树的所有节点，占用 $O(N)$ 时间。 +- **空间复杂度 $O(N)$ ：** 最差情况下（当二叉树退化为链表），递归时系统需使用 $O(N)$ 大小的栈空间。 + +## 方法二：辅助栈（或队列） + +利用栈（或队列）遍历树的所有节点 `node` ，并交换每个 `node` 的左 / 右子节点。 + +### 算法流程： + +1. **特例处理：** 当 `root` 为空时，直接返回 $null$ ； +2. **初始化：** 栈（或队列），本文用栈，并加入根节点 `root` 。 +3. **循环交换：** 当栈 `stack` 为空时跳出； + 1. **出栈：** 记为 `node` ； + 2. **添加子节点：** 将 `node` 左和右子节点入栈； + 3. **交换：** 交换 `node` 的左 / 右子节点。 +4. **返回值：** 返回根节点 `root` 。 + + + +### 代码： + +```Python [] +class Solution: + def mirrorTree(self, root: TreeNode) -> TreeNode: + if not root: return + stack = [root] + while stack: + node = stack.pop() + if node.left: stack.append(node.left) + if node.right: stack.append(node.right) + node.left, node.right = node.right, node.left + return root +``` + +```Java [] +class Solution { + public TreeNode mirrorTree(TreeNode root) { + if(root == null) return null; + Stack stack = new Stack<>() {{ add(root); }}; + while(!stack.isEmpty()) { + TreeNode node = stack.pop(); + if(node.left != null) stack.add(node.left); + if(node.right != null) stack.add(node.right); + TreeNode tmp = node.left; + node.left = node.right; + node.right = tmp; + } + return root; + } +} +``` + +```C++ [] +class Solution { +public: + TreeNode* mirrorTree(TreeNode* root) { + if(root == nullptr) return nullptr; + stack stack; + stack.push(root); + while (!stack.empty()) + { + TreeNode* node = stack.top(); + stack.pop(); + if (node->left != nullptr) stack.push(node->left); + if (node->right != nullptr) stack.push(node->right); + TreeNode* tmp = node->left; + node->left = node->right; + node->right = tmp; + } + return root; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为二叉树的节点数量，建立二叉树镜像需要遍历树的所有节点，占用 $O(N)$ 时间。 +- **空间复杂度 $O(N)$ ：** 如下图所示，最差情况下，栈 `stack` 最多同时存储 $\frac{N + 1}{2}$ 个节点，占用 $O(N)$ 额外空间。 + +![Picture0.png](https://pic.leetcode-cn.com/1614450330-bTAcyj-Picture0.png){:align=center width=450} diff --git "a/leetbook_ioa/docs/LCR 145. \345\210\244\346\226\255\345\257\271\347\247\260\344\272\214\345\217\211\346\240\221.md" "b/leetbook_ioa/docs/LCR 145. \345\210\244\346\226\255\345\257\271\347\247\260\344\272\214\345\217\211\346\240\221.md" new file mode 100755 index 0000000..35b07a9 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 145. \345\210\244\346\226\255\345\257\271\347\247\260\344\272\214\345\217\211\346\240\221.md" @@ -0,0 +1,79 @@ +## 解题思路： + +**对称二叉树定义：** 对于树中 **任意两个对称节点** `L` 和 `R `，一定有： + +- `L.val = R.val` ：即此两对称节点值相等。 +- `L.left.val = R.right.val` ：即 $L$ 的 左子节点 和 $R$ 的 右子节点 对称； +- `L.right.val = R.left.val` ：即 $L$ 的 右子节点 和 $R$ 的 左子节点 对称。 + +根据以上规律，考虑从顶至底递归，判断每对左右节点是否对称，从而判断树是否为对称二叉树。 + +![Picture1.png](https://pic.leetcode-cn.com/1599398062-PbkpuX-Picture1.png){:align=center width=450} + +### 算法流程： + +**`checkSymmetricTree(root)` ：** + +- **特例处理：** 若根节点 `root` 为空，则直接返回 $\text{true}$ 。 +- **返回值：** 即 `recur(root.left, root.right)` ; + +**`recur(L, R)` ：** + +- **终止条件：** + - 当 `L` 和 `R` 同时越过叶节点： 此树从顶至底的节点都对称，因此返回 $\text{true}$ ； + - 当 `L` 或 `R` 中只有一个越过叶节点： 此树不对称，因此返回 $\text{false}$ ； + - 当节点 `L` 值 $\ne$ 节点 `R` 值： 此树不对称，因此返回 $\text{false}$ ； +- **递推工作：** + - 判断两节点 `L.left` 和 `R.right` 是否对称，即 `recur(L.left, R.right)` ； + - 判断两节点 `L.right` 和 `R.left` 是否对称，即 `recur(L.right, R.left)` ； +- **返回值：** 两对节点都对称时，才是对称树，因此用与逻辑符 `&&` 连接。 + + + +## 代码： + +```Python [] +class Solution: + def checkSymmetricTree(self, root: TreeNode) -> bool: + def recur(L, R): + if not L and not R: return True + if not L or not R or L.val != R.val: return False + return recur(L.left, R.right) and recur(L.right, R.left) + + return not root or recur(root.left, root.right) +``` + +```Java [] +class Solution { + public boolean checkSymmetricTree(TreeNode root) { + return root == null || recur(root.left, root.right); + } + boolean recur(TreeNode L, TreeNode R) { + if(L == null && R == null) return true; + if(L == null || R == null || L.val != R.val) return false; + return recur(L.left, R.right) && recur(L.right, R.left); + } +} +``` + +```C++ [] +class Solution { +public: + bool checkSymmetricTree(TreeNode* root) { + return root == nullptr || recur(root->left, root->right); + } +private: + bool recur(TreeNode* L, TreeNode* R) { + if(L == nullptr && R == nullptr) return true; + if(L == nullptr || R == nullptr || L->val != R->val) return false; + return recur(L->left, R->right) && recur(L->right, R->left); + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为二叉树的节点数量，每次执行 `recur()` 可以判断一对节点是否对称，因此最多调用 $N/2$ 次 `recur()` 方法。 +- **空间复杂度 $O(N)$ ：** 如下图所示，最差情况下（二叉树退化为链表），系统使用 $O(N)$ 大小的空间。 + +![Picture2.png](https://pic.leetcode-cn.com/1599398062-LmPbix-Picture2.png){:align=center width=450} diff --git "a/leetbook_ioa/docs/LCR 146. \350\236\272\346\227\213\351\201\215\345\216\206\344\272\214\347\273\264\346\225\260\347\273\204.md" "b/leetbook_ioa/docs/LCR 146. \350\236\272\346\227\213\351\201\215\345\216\206\344\272\214\347\273\264\346\225\260\347\273\204.md" new file mode 100755 index 0000000..b6614ee --- /dev/null +++ "b/leetbook_ioa/docs/LCR 146. \350\236\272\346\227\213\351\201\215\345\216\206\344\272\214\347\273\264\346\225\260\347\273\204.md" @@ -0,0 +1,106 @@ +## 解题思路： + +根据题目示例 `array = [[1,2,3],[4,5,6],[7,8,9]]` 的对应输出 `[1,2,3,6,9,8,7,4,5]` 可以发现，顺时针打印矩阵的顺序是 **“从左向右、从上向下、从右向左、从下向上”** 循环。 + +因此，考虑设定矩阵的 “左、上、右、下” 四个边界，模拟以上矩阵遍历顺序。 + +![Picture1.png](https://pic.leetcode-cn.com/7605d807782923e4ad3c7995dc2485f538f202ac326bb330fe997f449123a548-Picture1.png){:align=center width=450} + +### 算法流程： + +1. **空值处理：** 当 `array` 为空时，直接返回空列表 `[]` 即可。 +2. **初始化：** 矩阵 左、右、上、下 四个边界 `l` , `r` , `t` , `b` ，用于打印的结果列表 `res` 。 +3. **循环打印：** “从左向右、从上向下、从右向左、从下向上” 四个方向循环打印； + 1. 根据边界打印，即将元素按顺序添加至列表 `res` 尾部； + 2. 边界向内收缩 1 （代表已被打印）； + 3. 判断边界是否相遇（是否打印完毕），若打印完毕则跳出。 +4. **返回值：** 返回 `res` 即可。 + +| 打印方向 | 1. 根据边界打印 | 2. 边界向内收缩 | 3. 是否打印完毕 | +| -------- | ---------------------- | ----------------- | --------------- | +| 从左向右 | 左边界`l` ，右边界 `r` | 上边界 `t` 加 $1$ | 是否 `t > b` | +| 从上向下 | 上边界 `t` ，下边界`b` | 右边界 `r` 减 $1$ | 是否 `l > r` | +| 从右向左 | 右边界 `r` ，左边界`l` | 下边界 `b` 减 $1$ | 是否 `t > b` | +| 从下向上 | 下边界 `b` ，上边界`t` | 左边界 `l` 加 $1$ | 是否 `l > r` | + + + +## 代码： + +Java/C++ 代码利用了 `++` 操作的便利性，详情可见 [++i 和 i++ 的区别](https://www.jianshu.com/p/b62eac216499) ； + +- `res[x++]` 等价于先给 `res[x]` 赋值，再给 `x` 自增 $1$ ； +- `++t > b` 等价于先给 `t` 自增 $1$ ，再判断 `t > b` 逻辑表达式。 + +> TIPS： 请注意区分数字 `1` 和变量 `l` 。 + +```Python [] +class Solution: + def spiralArray(self, array: List[List[int]]) -> List[int]: + if not array: return [] + l, r, t, b, res = 0, len(array[0]) - 1, 0, len(array) - 1, [] + while True: + for i in range(l, r + 1): res.append(array[t][i]) # left to right + t += 1 + if t > b: break + for i in range(t, b + 1): res.append(array[i][r]) # top to bottom + r -= 1 + if l > r: break + for i in range(r, l - 1, -1): res.append(array[b][i]) # right to left + b -= 1 + if t > b: break + for i in range(b, t - 1, -1): res.append(array[i][l]) # bottom to top + l += 1 + if l > r: break + return res +``` + +```Java [] +class Solution { + public int[] spiralArray(int[][] array) { + if(array.length == 0) return new int[0]; + int l = 0, r = array[0].length - 1, t = 0, b = array.length - 1, x = 0; + int[] res = new int[(r + 1) * (b + 1)]; + while(true) { + for(int i = l; i <= r; i++) res[x++] = array[t][i]; // left to right + if(++t > b) break; + for(int i = t; i <= b; i++) res[x++] = array[i][r]; // top to bottom + if(l > --r) break; + for(int i = r; i >= l; i--) res[x++] = array[b][i]; // right to left + if(t > --b) break; + for(int i = b; i >= t; i--) res[x++] = array[i][l]; // bottom to top + if(++l > r) break; + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + vector spiralArray(vector>& array) + { + if (array.empty()) return {}; + int l = 0, r = array[0].size() - 1, t = 0, b = array.size() - 1; + vector res; + while(true) + { + for (int i = l; i <= r; i++) res.push_back(array[t][i]); // left to right + if (++t > b) break; + for (int i = t; i <= b; i++) res.push_back(array[i][r]); // top to bottom + if (l > --r) break; + for (int i = r; i >= l; i--) res.push_back(array[b][i]); // right to left + if (t > --b) break; + for (int i = b; i >= t; i--) res.push_back(array[i][l]); // bottom to top + if (++l > r) break; + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(MN)$ ：** $M, N$ 分别为矩阵行数和列数。 +- **空间复杂度 $O(1)$ ：** 四个边界 `l` , `r` , `t` , `b` 使用常数大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 147. \346\234\200\345\260\217\346\240\210.md" "b/leetbook_ioa/docs/LCR 147. \346\234\200\345\260\217\346\240\210.md" new file mode 100755 index 0000000..b7296ee --- /dev/null +++ "b/leetbook_ioa/docs/LCR 147. \346\234\200\345\260\217\346\240\210.md" @@ -0,0 +1,117 @@ +## 解题思路： + +普通栈的 `push()` 和 `pop()` 函数的复杂度为 $O(1)$ ；而获取栈最小值 `getMin()` 函数需要遍历整个栈，复杂度为 $O(N)$ 。 + +**本题难点：** 将 `getMin()` 函数复杂度降为 $O(1)$ 。可借助辅助栈实现： + +- **数据栈 `A` ：** 栈 `A` 用于存储所有元素，保证入栈 `push()` 函数、出栈 `pop()` 函数、获取栈顶 `top()` 函数的正常逻辑。 +- **辅助栈 `B` ：** 栈 `B` 中存储栈 `A` 中所有 **非严格降序** 元素的子序列，则栈 `A` 中的最小元素始终对应栈 `B` 的栈顶元素。此时，`getMin()` 函数只需返回栈 `B` 的栈顶元素即可。 + +因此，只需设法维护好 栈 `B` 的元素，使其保持是栈 `A` 的非严格降序元素的子序列，即可实现 `getMin()` 函数的 $O(1)$ 复杂度。 + +![Picture1.png](https://pic.leetcode-cn.com/1599880866-aLaPYz-Picture1.png){:align=center width=450} + +### 函数设计： + +**`push(x)` 函数：** 重点为保持栈 `B` 的元素是 **非严格降序** 的； + +1. 执行「元素 `x` 压入栈 `A`」 ； +2. 若「栈 `B` 为空」**或**「`x` $\leq$ 栈 `B` 的栈顶元素」，则执行「元素 `x` 压入栈 `B`」 ； + +**`pop()` 函数：** 重点为保持栈 `A` , `B` 的 **元素一致性** ； + +1. 执行「栈 `A` 元素出栈」，将出栈元素记为 `y` ； +2. 若 「`y` 等于栈 `B` 的栈顶元素」，则执行「栈 `B` 元素出栈」； + +**`top()` 函数：** 直接返回栈 `A` 的栈顶元素，即返回 `A.peek()` ； + +**`getMin()` 函数：** 直接返回栈 `B` 的栈顶元素，即返回 `B.peek()` ； + +> 下图中的 `min()` 对应本题的 `getMin()` 。 + + + +### 采用 “非严格” 降序原因： + +在栈 `A` 具有 **重复** 最小值元素时，非严格降序可防止栈 `B` 提前弹出最小值元素，示例如下： + +![Picture2.png](https://pic.leetcode-cn.com/1600086305-BSfBJu-Picture2.png){:align=center width=550} + +## 代码： + +Java 代码中，由于 Stack 中存储的是 int 的包装类 Integer ，因此需要使用 `equals()` 代替 `==` ，以比较对象的值。 + +```Python [] +class MinStack: + def __init__(self): + self.A, self.B = [], [] + + def push(self, x: int) -> None: + self.A.append(x) + if not self.B or self.B[-1] >= x: + self.B.append(x) + + def pop(self) -> None: + if self.A.pop() == self.B[-1]: + self.B.pop() + + def top(self) -> int: + return self.A[-1] + + def getMin(self) -> int: + return self.B[-1] +``` + +```Java [] +class MinStack { + Stack A, B; + public MinStack() { + A = new Stack<>(); + B = new Stack<>(); + } + public void push(int x) { + A.add(x); + if(B.empty() || B.peek() >= x) + B.add(x); + } + public void pop() { + if(A.pop().equals(B.peek())) + B.pop(); + } + public int top() { + return A.peek(); + } + public int getMin() { + return B.peek(); + } +} +``` + +```C++ [] +class MinStack { +public: + stack A, B; + MinStack() {} + void push(int x) { + A.push(x); + if(B.empty() || B.top() >= x) + B.push(x); + } + void pop() { + if(A.top() == B.top()) + B.pop(); + A.pop(); + } + int top() { + return A.top(); + } + int getMin() { + return B.top(); + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(1)$ ：** `push()`, `pop()`, `top()`, `getMin()` 四个函数的时间复杂度均为常数级别。 +- **空间复杂度 $O(N)$ ：** 当共有 $N$ 个待入栈元素时，辅助栈 `B` 最差情况下存储 $N$ 个元素，使用 $O(N)$ 额外空间。 diff --git "a/leetbook_ioa/docs/LCR 148. \351\252\214\350\257\201\345\233\276\344\271\246\345\217\226\345\207\272\351\241\272\345\272\217.md" "b/leetbook_ioa/docs/LCR 148. \351\252\214\350\257\201\345\233\276\344\271\246\345\217\226\345\207\272\351\241\272\345\272\217.md" new file mode 100755 index 0000000..ee75cc4 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 148. \351\252\214\350\257\201\345\233\276\344\271\246\345\217\226\345\207\272\351\241\272\345\272\217.md" @@ -0,0 +1,84 @@ +## 解题思路： + +如下图所示，给定一个放入序列 `putIn` 和拿取序列 `takeOut` ，则放入（压栈）和拿取（弹出）操作的顺序是 **唯一确定** 的。 + +> 下图中 `pushed` 和 `popped` 分别对应本题的 `putIn` 和 `takeOut` 。 + +![Picture1.png](https://pic.leetcode-cn.com/1188474dc6a377fc258004bca84e5a130b663eeb24bf22c4fa4eb998a0249f97-Picture1.png){:align=center width=500} + +如下图所示，栈的数据操作具有 **先入后出** 的特性，因此某些拿取序列是无法实现的。 + +![Picture2.png](https://pic.leetcode-cn.com/3f43b224553bf3a37f9758dbb41655e547795e293524a148380c6f335af315e4-Picture2.png){:align=center width=500} + +考虑借用一个辅助栈 `stack` ，**模拟** 放入 / 拿取操作的排列。根据是否模拟成功，即可得到结果。 + +- **入栈操作：** 按照压栈序列的顺序执行。 +- **出栈操作：** 每次入栈后，循环判断 “栈顶元素 $=$ 拿取序列的当前元素” 是否成立，将符合拿取序列顺序的栈顶元素全部拿取。 + +> 由于题目规定 “栈的所有数字均不相等” ，因此在循环入栈中，每个元素出栈的位置的可能性是唯一的（若有重复数字，则具有多个可出栈的位置）。因而，在遇到 “栈顶元素 $=$ 拿取序列的当前元素” 就应立即执行出栈。 + +### 算法流程： + +1. **初始化：** 辅助栈 `stack` ，拿取序列的索引 `i` ； +2. **遍历压栈序列：** 各元素记为 `num` ； + 1. 元素 `num` 入栈； + 2. 循环出栈：若 `stack` 的栈顶元素 $=$ 拿取序列元素 `takeOut[i]` ，则执行出栈与 `i++` ； +3. **返回值：** 若 `stack` 为空，则此拿取序列合法。 + + + +## 代码： + +题目指出 “putIn 是 takeOut 的排列” 。因此，无需考虑 `putIn` 和 `takeOut` **长度不同** 或 **包含元素不同** 的情况。 + +```Python [] +class Solution: + def validateBookSequences(self, putIn: List[int], takeOut: List[int]) -> bool: + stack, i = [], 0 + for num in putIn: + stack.append(num) # num 入栈 + while stack and stack[-1] == takeOut[i]: # 循环判断与出栈 + stack.pop() + i += 1 + return not stack +``` + +```Java [] +class Solution { + public boolean validateBookSequences(int[] putIn, int[] takeOut) { + Stack stack = new Stack<>(); + int i = 0; + for(int num : putIn) { + stack.push(num); // num 入栈 + while(!stack.isEmpty() && stack.peek() == takeOut[i]) { // 循环判断与出栈 + stack.pop(); + i++; + } + } + return stack.isEmpty(); + } +} +``` + +```C++ [] +class Solution { +public: + bool validateBookSequences(vector& putIn, vector& takeOut) { + stack stk; + int i = 0; + for(int num : putIn) { + stk.push(num); // num 入栈 + while(!stk.empty() && stk.top() == takeOut[i]) { // 循环判断与出栈 + stk.pop(); + i++; + } + } + return stk.empty(); + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为列表 `putIn` 的长度；每个元素最多入栈与出栈一次，即最多共 $2N$ 次出入栈操作。 +- **空间复杂度 $O(N)$ ：** 辅助栈 `stack` 最多同时存储 $N$ 个元素。 diff --git "a/leetbook_ioa/docs/LCR 149. \345\275\251\347\201\257\350\243\205\351\245\260\350\256\260\345\275\225 I.md" "b/leetbook_ioa/docs/LCR 149. \345\275\251\347\201\257\350\243\205\351\245\260\350\256\260\345\275\225 I.md" new file mode 100755 index 0000000..ae905ea --- /dev/null +++ "b/leetbook_ioa/docs/LCR 149. \345\275\251\347\201\257\350\243\205\351\245\260\350\256\260\345\275\225 I.md" @@ -0,0 +1,80 @@ +## 解题思路： + +题目要求按层打印二叉树，即二叉树的 **广度优先遍历** ，其通常借助 **队列** 的先入先出特性来实现。 + +![Picture1.png](https://pic.leetcode-cn.com/a872b50fa42011748437ec9123d8f77a104b3d528880efca8b212f91d115f835-Picture1.png){:align=center width=450} + +### 算法流程： + +1. **特例处理：** 当树的根节点为空，则直接返回空列表 `[]` ； +2. **初始化：** 打印结果列表 `res = []` ，包含根节点的队列 `queue = [root]` ； +3. **BFS 循环：** 当队列 `queue` 为空时跳出； + 1. **出队：** 队首元素出队，记为 `node`； + 2. **打印：** 将 `node.val` 添加至列表 `tmp` 尾部； + 3. **添加子节点：** 若 `node` 的左（右）子节点不为空，则将左（右）子节点加入队列 `queue` ； +4. **返回值：** 返回打印结果列表 `res` 即可。 + + + +## 代码： + +Python 中使用 collections 中的双端队列 `deque()` ，其 `popleft()` 方法可达到 $O(1)$ 时间复杂度；列表 list 的 `pop(0)` 方法时间复杂度为 $O(N)$ 。 + +```Python [] +class Solution: + def decorateRecord(self, root: TreeNode) -> List[int]: + if not root: return [] + res, queue = [], collections.deque() + queue.append(root) + while queue: + node = queue.popleft() + res.append(node.val) + if node.left: queue.append(node.left) + if node.right: queue.append(node.right) + return res +``` + +```Java [] +class Solution { + public int[] decorateRecord(TreeNode root) { + if(root == null) return new int[0]; + Queue queue = new LinkedList<>(){{ add(root); }}; + ArrayList ans = new ArrayList<>(); + while(!queue.isEmpty()) { + TreeNode node = queue.poll(); + ans.add(node.val); + if(node.left != null) queue.add(node.left); + if(node.right != null) queue.add(node.right); + } + int[] res = new int[ans.size()]; + for(int i = 0; i < ans.size(); i++) + res[i] = ans.get(i); + return res; + } +} +``` + +```C++ [] +class Solution { +public: + vector decorateRecord(TreeNode* root) { + vector res; + if(!root) return res; + queue que; + que.push(root); + while(!que.empty()){ + TreeNode* node = que.front(); + que.pop(); + res.push_back(node->val); + if(node->left) que.push(node->left); + if(node->right) que.push(node->right); + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数量，即 BFS 需循环 $N$ 次。 +- **空间复杂度 $O(N)$ ：** 最差情况下，即当树为平衡二叉树时，最多有 $N/2$ 个树节点**同时**在 `queue` 中，使用 $O(N)$ 大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 150. \345\275\251\347\201\257\350\243\205\351\245\260\350\256\260\345\275\225 II.md" "b/leetbook_ioa/docs/LCR 150. \345\275\251\347\201\257\350\243\205\351\245\260\350\256\260\345\275\225 II.md" new file mode 100755 index 0000000..4169b1b --- /dev/null +++ "b/leetbook_ioa/docs/LCR 150. \345\275\251\347\201\257\350\243\205\351\245\260\350\256\260\345\275\225 II.md" @@ -0,0 +1,90 @@ +## 解题思路： + +在上一题层序遍历的基础上，本题要求将 **每层打印到一行**。考虑将当前全部节点打印到一行，并将下一层全部节点加入队列，以此类推，即可分为多行打印。 + +![Picture1.png](https://pic.leetcode-cn.com/a58477c74c96779c265ce3028def7625d16042895d2c21f7fb0293df7b213276-Picture1.png){:align=center width=450} + +### 算法流程： + +1. **特例处理：** 当根节点为空，则返回空列表 `[]` ； +2. **初始化：** 打印结果列表 `res = []` ，包含根节点的队列 `queue = [root]` ； +3. **BFS 循环：** 当队列 `queue` 为空时跳出； + 1. 新建一个临时列表 `tmp` ，用于存储当前层打印结果； + 2. **当前层打印循环：** 循环次数为当前层节点数（即队列 `queue` 长度）； + 1. **出队：** 队首元素出队，记为 `node`； + 2. **打印：** 将 `node.val` 添加至 `tmp` 尾部； + 3. **添加子节点：** 若 `node` 的左（右）子节点不为空，则将左（右）子节点加入队列 `queue` ； + 3. 将当前层结果 `tmp` 添加入 `res` 。 +4. **返回值：** 返回打印结果列表 `res` 即可。 + + + +## 代码： + +Python 中使用 collections 中的双端队列 `deque()` ，其 `popleft()` 方法可达到 $O(1)$ 时间复杂度；列表 list 的 `pop(0)` 方法时间复杂度为 $O(N)$ 。 + +```Python [] +class Solution: + def decorateRecord(self, root: TreeNode) -> List[List[int]]: + if not root: return [] + res, queue = [], collections.deque() + queue.append(root) + while queue: + tmp = [] + for _ in range(len(queue)): + node = queue.popleft() + tmp.append(node.val) + if node.left: queue.append(node.left) + if node.right: queue.append(node.right) + res.append(tmp) + return res +``` + +```Java [] +class Solution { + public List> decorateRecord(TreeNode root) { + Queue queue = new LinkedList<>(); + List> res = new ArrayList<>(); + if(root != null) queue.add(root); + while(!queue.isEmpty()) { + List tmp = new ArrayList<>(); + for(int i = queue.size(); i > 0; i--) { + TreeNode node = queue.poll(); + tmp.add(node.val); + if(node.left != null) queue.add(node.left); + if(node.right != null) queue.add(node.right); + } + res.add(tmp); + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + vector> decorateRecord(TreeNode* root) { + queue que; + vector> res; + if(root != NULL) que.push(root); + while(!que.empty()) { + vector tmp; + for(int i = que.size(); i > 0; --i) { + root = que.front(); + que.pop(); + tmp.push_back(root->val); + if(root->left != NULL) que.push(root->left); + if(root->right != NULL) que.push(root->right); + } + res.push_back(tmp); + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数量，即 BFS 需循环 $N$ 次。 +- **空间复杂度 $O(N)$ ：** 最差情况下，即当树为平衡二叉树时，最多有 $N/2$ 个树节点**同时**在 `queue` 中，使用 $O(N)$ 大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 151. \345\275\251\347\201\257\350\243\205\351\245\260\350\256\260\345\275\225 III.md" "b/leetbook_ioa/docs/LCR 151. \345\275\251\347\201\257\350\243\205\351\245\260\350\256\260\345\275\225 III.md" new file mode 100755 index 0000000..726a708 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 151. \345\275\251\347\201\257\350\243\205\351\245\260\350\256\260\345\275\225 III.md" @@ -0,0 +1,277 @@ +## 方法一：层序遍历 + 双端队列 + +利用双端队列的两端皆可添加元素的特性，设打印列表（双端队列） `tmp` ，并规定： + +- 奇数层 则添加至 `tmp` **尾部** 。 +- 偶数层 则添加至 `tmp` **头部** 。 + +![Picture1.png](https://pic.leetcode-cn.com/9513dcb034f5dcdea947a2f667b3d380df4f8732da6397778e00718b77584010-Picture1.png){:align=center width=450} + +### 算法流程： + +1. **特例处理：** 当树的根节点为空，则直接返回空列表 `[]` ； +2. **初始化：** 打印结果空列表 `res` ，包含根节点的双端队列 `deque` ； +3. **BFS 循环：** 当 `deque` 为空时跳出； + 1. 新建列表 `tmp` ，用于临时存储当前层打印结果； + 2. **当前层打印循环：** 循环次数为当前层节点数（即 `deque` 长度）； + 1. **出队：** 队首元素出队，记为 `node`； + 2. **打印：** 若为奇数层，将 `node.val` 添加至 `tmp` 尾部；否则，添加至 `tmp` 头部； + 3. **添加子节点：** 若 `node` 的左（右）子节点不为空，则加入 `deque` ； + 3. 将当前层结果 `tmp` 转化为 list 并添加入 `res` ； +4. **返回值：** 返回打印结果列表 `res` 即可； + + + +### 代码： + +- Python 中使用 collections 中的双端队列 `deque()` ，其 `popleft()` 方法可达到 $O(1)$ 时间复杂度；列表 list 的 `pop(0)` 方法时间复杂度为 $O(N)$ 。 +- Java 中将链表 LinkedList 作为双端队列使用。 + +```Python [] +class Solution: + def decorateRecord(self, root: TreeNode) -> List[List[int]]: + if not root: return [] + res, deque = [], collections.deque([root]) + while deque: + tmp = collections.deque() + for _ in range(len(deque)): + node = deque.popleft() + if len(res) % 2 == 0: tmp.append(node.val) # 奇数层 -> 插入队列尾部 + else: tmp.appendleft(node.val) # 偶数层 -> 插入队列头部 + if node.left: deque.append(node.left) + if node.right: deque.append(node.right) + res.append(list(tmp)) + return res +``` + +```Java [] +class Solution { + public List> decorateRecord(TreeNode root) { + Queue queue = new LinkedList<>(); + List> res = new ArrayList<>(); + if(root != null) queue.add(root); + while(!queue.isEmpty()) { + LinkedList tmp = new LinkedList<>(); + for(int i = queue.size(); i > 0; i--) { + TreeNode node = queue.poll(); + if(res.size() % 2 == 0) tmp.addLast(node.val); + else tmp.addFirst(node.val); + if(node.left != null) queue.add(node.left); + if(node.right != null) queue.add(node.right); + } + res.add(tmp); + } + return res; + } +} +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数量，即 BFS 需循环 $N$ 次，占用 $O(N)$ ；双端队列的队首和队尾的添加和删除操作的时间复杂度均为 $O(1)$ 。 +- **空间复杂度 $O(N)$ ：** 最差情况下，即当树为满二叉树时，最多有 $N/2$ 个树节点 **同时** 在 `deque` 中，使用 $O(N)$ 大小的额外空间。 + +## 方法二：层序遍历 + 双端队列（奇偶层逻辑分离） + +- 方法一代码简短、容易实现；但需要判断每个节点的所在层奇偶性，即冗余了 $N$ 次判断。 +- 通过将奇偶层逻辑拆分，可以消除冗余的判断。 + +### 算法流程： + +> 与方法一对比，仅 BFS 循环不同。 + +**BFS 循环：** 循环打印奇 / 偶数层，当 `deque` 为空时跳出； + +1. **打印奇数层：** **从左向右** 打印，**先左后右** 加入下层节点； +2. 若 `deque` 为空，说明向下无偶数层，则跳出； +3. **打印偶数层：** **从右向左** 打印，**先右后左** 加入下层节点； + +### 代码： + +```Python [] +class Solution: + def decorateRecord(self, root: TreeNode) -> List[List[int]]: + if not root: return [] + res, deque = [], collections.deque() + deque.append(root) + while deque: + tmp = [] + # 打印奇数层 + for _ in range(len(deque)): + # 从左向右打印 + node = deque.popleft() + tmp.append(node.val) + # 先左后右加入下层节点 + if node.left: deque.append(node.left) + if node.right: deque.append(node.right) + res.append(tmp) + if not deque: break # 若为空则提前跳出 + # 打印偶数层 + tmp = [] + for _ in range(len(deque)): + # 从右向左打印 + node = deque.pop() + tmp.append(node.val) + # 先右后左加入下层节点 + if node.right: deque.appendleft(node.right) + if node.left: deque.appendleft(node.left) + res.append(tmp) + return res +``` + +```Java [] +class Solution { + public List> decorateRecord(TreeNode root) { + Deque deque = new LinkedList<>(); + List> res = new ArrayList<>(); + if(root != null) deque.add(root); + while(!deque.isEmpty()) { + // 打印奇数层 + List tmp = new ArrayList<>(); + for(int i = deque.size(); i > 0; i--) { + // 从左向右打印 + TreeNode node = deque.removeFirst(); + tmp.add(node.val); + // 先左后右加入下层节点 + if(node.left != null) deque.addLast(node.left); + if(node.right != null) deque.addLast(node.right); + } + res.add(tmp); + if(deque.isEmpty()) break; // 若为空则提前跳出 + // 打印偶数层 + tmp = new ArrayList<>(); + for(int i = deque.size(); i > 0; i--) { + // 从右向左打印 + TreeNode node = deque.removeLast(); + tmp.add(node.val); + // 先右后左加入下层节点 + if(node.right != null) deque.addFirst(node.right); + if(node.left != null) deque.addFirst(node.left); + } + res.add(tmp); + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + vector> decorateRecord(TreeNode* root) { + deque deque; + vector> res; + if(root != NULL) deque.push_back(root); + while(!deque.empty()) { + // 打印奇数层 + vector tmp; + for(int i = deque.size(); i > 0; i--) { + // 从左向右打印 + TreeNode* node = deque.front(); + deque.pop_front(); + tmp.push_back(node->val); + // 先左后右加入下层节点 + if(node->left != NULL) deque.push_back(node->left); + if(node->right != NULL) deque.push_back(node->right); + } + res.push_back(tmp); + if(deque.empty()) break; // 若为空则提前跳出 + // 打印偶数层 + tmp.clear(); + for(int i = deque.size(); i > 0; i--) { + // 从右向左打印 + TreeNode* node = deque.back(); + deque.pop_back(); + tmp.push_back(node->val); + // 先右后左加入下层节点 + if(node->right != NULL) deque.push_front(node->right); + if(node->left != NULL) deque.push_front(node->left); + } + res.push_back(tmp); + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 同方法一。 +- **空间复杂度 $O(N)$ ：** 同方法一。 + +## 方法三：层序遍历 + 倒序 + +- 此方法的优点是只用列表即可，无需其他数据结构。 +- **偶数层倒序：** 若 `res` 的长度为 **奇数** ，说明当前是偶数层，则对 `tmp` 执行 **倒序** 操作。 + + + +### 代码： + +```Python [] +class Solution: + def decorateRecord(self, root: TreeNode) -> List[List[int]]: + if not root: return [] + res, queue = [], collections.deque() + queue.append(root) + while queue: + tmp = [] + for _ in range(len(queue)): + node = queue.popleft() + tmp.append(node.val) + if node.left: queue.append(node.left) + if node.right: queue.append(node.right) + res.append(tmp[::-1] if len(res) % 2 else tmp) + return res +``` + +```Java [] +class Solution { + public List> decorateRecord(TreeNode root) { + Queue queue = new LinkedList<>(); + List> res = new ArrayList<>(); + if(root != null) queue.add(root); + while(!queue.isEmpty()) { + List tmp = new ArrayList<>(); + for(int i = queue.size(); i > 0; i--) { + TreeNode node = queue.poll(); + tmp.add(node.val); + if(node.left != null) queue.add(node.left); + if(node.right != null) queue.add(node.right); + } + if(res.size() % 2 == 1) Collections.reverse(tmp); + res.add(tmp); + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + vector> decorateRecord(TreeNode* root) { + queue que; + vector> res; + if(root != NULL) que.push(root); + while(!que.empty()) { + vector tmp; + for(int i = que.size(); i > 0; i--) { + TreeNode* node = que.front(); + que.pop(); + tmp.push_back(node->val); + if(node->left != NULL) que.push(node->left); + if(node->right != NULL) que.push(node->right); + } + if(res.size() % 2 == 1) reverse(tmp.begin(),tmp.end()); + res.push_back(tmp); + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数量，即 BFS 需循环 $N$ 次，占用 $O(N)$ 。**共完成** 少于 $N$ 个节点的倒序操作，占用 $O(N)$ 。 +- **空间复杂度 $O(N)$ ：** 最差情况下，即当树为满二叉树时，最多有 $N/2$ 个树节点**同时**在 `queue` 中，使用 $O(N)$ 大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 152. \351\252\214\350\257\201\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\347\232\204\345\220\216\345\272\217\351\201\215\345\216\206\345\272\217\345\210\227.md" "b/leetbook_ioa/docs/LCR 152. \351\252\214\350\257\201\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\347\232\204\345\220\216\345\272\217\351\201\215\345\216\206\345\272\217\345\210\227.md" new file mode 100755 index 0000000..3fffa92 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 152. \351\252\214\350\257\201\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\347\232\204\345\220\216\345\272\217\351\201\215\345\216\206\345\272\217\345\210\227.md" @@ -0,0 +1,175 @@ +## 解题思路： + +**后序遍历定义：** `[ 左子树 | 右子树 | 根节点 ]` ，即遍历顺序为 “左、右、根” 。 + +**二叉搜索树定义：** 左子树中所有节点的值 $<$ 根节点的值；右子树中所有节点的值 $>$ 根节点的值；其左、右子树也分别为二叉搜索树。 + +![Picture1.png](https://pic.leetcode-cn.com/1599753507-JrFBjm-Picture1.png){:align=center width=500} + +## 方法一：递归分治 + +根据二叉搜索树的定义，可以通过递归，判断所有子树的 **正确性** （即其后序遍历是否满足二叉搜索树的定义） ，若所有子树都正确，则此序列为二叉搜索树的后序遍历。 + +### 递归解析： + +**终止条件：** 当 $i \geq j$ ，说明此子树节点数量 $\leq 1$ ，无需判别正确性，因此直接返回 $\text{true}$ ； + +**递推工作：** + +1. **划分左右子树：** 遍历后序遍历的 $[i, j]$ 区间元素，寻找 **第一个大于根节点** 的节点，索引记为 $m$ 。此时，可划分出左子树区间 $[i,m-1]$ 、右子树区间 $[m, j - 1]$ 、根节点索引 $j$ 。 +2. **判断是否为二叉搜索树：** + - **左子树区间** $[i, m - 1]$ 内的所有节点都应 $<$ $postorder[j]$ 。而第 `1.划分左右子树` 步骤已经保证左子树区间的正确性，因此只需要判断右子树区间即可。 + - **右子树区间** $[m, j-1]$ 内的所有节点都应 $>$ $postorder[j]$ 。实现方式为遍历，当遇到 $\leq postorder[j]$ 的节点则跳出；则可通过 $p = j$ 判断是否为二叉搜索树。 + +**返回值：** 所有子树都需正确才可判定正确，因此使用 **与逻辑符** $\&\&$ 连接。 + +1. **$p = j$ ：** 判断 **此树** 是否正确。 +2. **$recur(i, m - 1)$ ：** 判断 **此树的左子树** 是否正确。 +3. **$recur(m, j - 1)$ ：** 判断 **此树的右子树** 是否正确。 + + + +### 代码： + +```Python [] +class Solution: + def verifyTreeOrder(self, postorder: List[int]) -> bool: + def recur(i, j): + if i >= j: return True + p = i + while postorder[p] < postorder[j]: p += 1 + m = p + while postorder[p] > postorder[j]: p += 1 + return p == j and recur(i, m - 1) and recur(m, j - 1) + + return recur(0, len(postorder) - 1) +``` + +```Java [] +class Solution { + public boolean verifyTreeOrder(int[] postorder) { + return recur(postorder, 0, postorder.length - 1); + } + boolean recur(int[] postorder, int i, int j) { + if(i >= j) return true; + int p = i; + while(postorder[p] < postorder[j]) p++; + int m = p; + while(postorder[p] > postorder[j]) p++; + return p == j && recur(postorder, i, m - 1) && recur(postorder, m, j - 1); + } +} +``` + +```C++ [] +class Solution { +public: + bool verifyTreeOrder(vector& postorder) { + return recur(postorder, 0, postorder.size() - 1); + } +private: + bool recur(vector& postorder, int i, int j) { + if(i >= j) return true; + int p = i; + while(postorder[p] < postorder[j]) p++; + int m = p; + while(postorder[p] > postorder[j]) p++; + return p == j && recur(postorder, i, m - 1) && recur(postorder, m, j - 1); + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N^2)$ ：** 每次调用 $recur(i,j)$ 减去一个根节点，因此递归占用 $O(N)$ ；最差情况下（即当树退化为链表），每轮递归都需遍历树所有节点，占用 $O(N)$ 。 +- **空间复杂度 $O(N)$ ：** 最差情况下（即当树退化为链表），递归深度将达到 $N$ 。 + +## 方法二：辅助单调栈 + +**后序遍历倒序：** `[ 根节点 | 右子树 | 左子树 ]` 。类似 **先序遍历的镜像** ，即先序遍历为 “根、左、右” 的顺序，而后序遍历的倒序为 “根、右、左” 顺序。 + +![Picture10.png](https://pic.leetcode-cn.com/1599753507-KaeaWl-Picture10.png){:align=center width=500} + +设后序遍历倒序列表为 $[r_{n}, r_{n-1},...,r_1]$，遍历此列表，设索引为 $i$ ，若为 **二叉搜索树** ，则有： + +- **当节点值 $r_i > r_{i+1}$ 时：** 节点 $r_i$ 一定是节点 $r_{i+1}$ 的右子节点。 +- **当节点值 $r_i < r_{i+1}$ 时：** 节点 $r_i$ 一定是某节点 $root$ 的左子节点，且 $root$ 为节点 $r_{i+1}, r_{i+2},..., r_{n}$ 中值大于且最接近 $r_i$ 的节点（∵ $root$ **直接连接** 左子节点 $r_i$ ）。 + +当遍历时遇到递减节点 $r_i < r_{i+1}$ ，若为二叉搜索树，则对于后序遍历中节点 $r_i$ 右边的任意节点 $r_x \in [r_{i-1}, r_{i-2}, ..., r_1]$ ，必有节点值 $r_x < root$ 。 + +> 节点 $r_x$ 只可能为以下两种情况：(1) $r_x$ 为 $r_i$ 的左、右子树的各节点；(2) $r_x$ 为 $root$ 的父节点或更高层父节点的左子树的各节点。在二叉搜索树中，以上节点都应小于 $root$ 。 + +![Picture11.png](https://pic.leetcode-cn.com/1599753507-mgqkoF-Picture11.png){:align=center width=500} + +遍历 “后序遍历的倒序” 会多次遇到递减节点 $r_i$ ，若所有的递减节点 $r_i$ 对应的父节点 $root$ 都满足以上条件，则可判定为二叉搜索树。根据以上特点，考虑借助 **单调栈** 实现： + +1. 借助一个单调栈 $stack$ 存储值递增的节点； +2. 每当遇到值递减的节点 $r_i$ ，则通过出栈来更新节点 $r_i$ 的父节点 $root$ ； +3. 每轮判断 $r_i$ 和 $root$ 的值关系： + 1. 若 $r_i > root$ 则说明不满足二叉搜索树定义，直接返回 $\text{false}$ 。 + 2. 若 $r_i < root$ 则说明满足二叉搜索树定义，则继续遍历。 + +### 算法流程： + +1. **初始化：** 单调栈 $stack$ ，父节点值 $root = +\infin$ （初始值为正无穷大，可把树的根节点看为此无穷大节点的左孩子）； +2. **倒序遍历 $postorder$** ：记每个节点为 $r_i$； + 1. **判断：** 若 $r_i>root$ ，说明此后序遍历序列不满足二叉搜索树定义，直接返回 $\text{false}$ ； + 2. **更新父节点 $root$ ：** 当栈不为空 **且** $r_i + +### 代码： + +```Python [] +class Solution: + def verifyTreeOrder(self, postorder: List[int]) -> bool: + stack, root = [], float("+inf") + for i in range(len(postorder) - 1, -1, -1): + if postorder[i] > root: return False + while(stack and postorder[i] < stack[-1]): + root = stack.pop() + stack.append(postorder[i]) + return True +``` + +```Java [] +class Solution { + public boolean verifyTreeOrder(int[] postorder) { + Stack stack = new Stack<>(); + int root = Integer.MAX_VALUE; + for(int i = postorder.length - 1; i >= 0; i--) { + if(postorder[i] > root) return false; + while(!stack.isEmpty() && stack.peek() > postorder[i]) + root = stack.pop(); + stack.add(postorder[i]); + } + return true; + } +} +``` + +```C++ [] +class Solution { +public: + bool verifyTreeOrder(vector& postorder) { + stack stk; + int root = INT_MAX; + for(int i = postorder.size() - 1; i >= 0; i--) { + if(postorder[i] > root) return false; + while(!stk.empty() && stk.top() > postorder[i]) { + root = stk.top(); + stk.pop(); + } + stk.push(postorder[i]); + } + return true; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 遍历 $postorder$ 所有节点，各节点均入栈 / 出栈一次，使用 $O(N)$ 时间。 +- **空间复杂度 $O(N)$ ：** 最差情况下，单调栈 $stack$ 存储所有节点，使用 $O(N)$ 额外空间。 diff --git "a/leetbook_ioa/docs/LCR 153. \344\272\214\345\217\211\346\240\221\344\270\255\345\222\214\344\270\272\347\233\256\346\240\207\345\200\274\347\232\204\350\267\257\345\276\204.md" "b/leetbook_ioa/docs/LCR 153. \344\272\214\345\217\211\346\240\221\344\270\255\345\222\214\344\270\272\347\233\256\346\240\207\345\200\274\347\232\204\350\267\257\345\276\204.md" new file mode 100755 index 0000000..884434c --- /dev/null +++ "b/leetbook_ioa/docs/LCR 153. \344\272\214\345\217\211\346\240\221\344\270\255\345\222\214\344\270\272\347\233\256\346\240\207\345\200\274\347\232\204\350\267\257\345\276\204.md" @@ -0,0 +1,105 @@ +## 解题思路： + +本题是典型的二叉树方案搜索问题，使用回溯法解决，其包含 **先序遍历 + 路径记录** 两部分。 + +- **先序遍历：** 按照 “根、左、右” 的顺序，遍历树的所有节点。 +- **路径记录：** 在先序遍历中，记录从根节点到当前节点的路径。当路径满足 (1) 根节点到叶节点形成的路径 **且** (2) 各节点值的和等于目标值 `target` 时，将此路径加入结果列表。 + +![Picture1.png](https://pic.leetcode-cn.com/1599400747-BuGhCT-Picture1.png){:align=center width=500} + +### 算法流程： + +**`pathTarget(root, target)` 函数：** + +- **初始化：** 结果列表 `res` ，路径列表 `path` 。 +- **返回值：** 返回 `res` 即可。 + +**`recur(root, tar) 函数：`** + +- **递推参数：** 当前节点 `root` ，当前目标值 `tar` 。 +- **终止条件：** 若节点 `root` 为空，则直接返回。 +- **递推工作：** + 1. 路径更新： 将当前节点值 `root.val` 加入路径 `path` 。 + 2. 目标值更新： `tar = tar - root.val`（即目标值 `tar` 从 `target` 减至 $0$ ）。 + 3. 路径记录： 当 “`root` 为叶节点” **且** “路径和等于目标值” ，则将此路径 `path` 加入 `res` 。 + 4. 先序遍历： 递归左 / 右子节点。 + 5. 路径恢复： 向上回溯前，需要将当前节点从路径 `path` 中删除，即执行 `path.pop()` 。 + + + +## 代码： + +以 Python 语言为例，记录路径时若直接执行 `res.append(path)` ，则是将此 `path` 对象加入了 `res` ；后续 `path` 改变时，`res` 中的 `path` 对象也会随之改变，因此无法实现结果记录。正确做法为： + +- Python: `res.append(list(path))` ； +- Java: `res.add(new LinkedList(path))` ； +- C++: `res.push_back(path)` ； + +> 三者的原理都是避免直接添加 `path` 对象，而是 **拷贝** 了一个 `path` 对象并加入到 `res` 。 + +```Python [] +class Solution: + def pathTarget(self, root: TreeNode, target: int) -> List[List[int]]: + res, path = [], [] + def recur(root, tar): + if not root: return + path.append(root.val) + tar -= root.val + if tar == 0 and not root.left and not root.right: + res.append(list(path)) + recur(root.left, tar) + recur(root.right, tar) + path.pop() + + recur(root, target) + return res +``` + +```Java [] +class Solution { + LinkedList> res = new LinkedList<>(); + LinkedList path = new LinkedList<>(); + public List> pathTarget(TreeNode root, int target) { + recur(root, target); + return res; + } + void recur(TreeNode root, int tar) { + if(root == null) return; + path.add(root.val); + tar -= root.val; + if(tar == 0 && root.left == null && root.right == null) + res.add(new LinkedList(path)); + recur(root.left, tar); + recur(root.right, tar); + path.removeLast(); + } +} +``` + +```C++ [] +class Solution { +public: + vector> pathTarget(TreeNode* root, int target) { + recur(root, target); + return res; + } +private: + vector> res; + vector path; + void recur(TreeNode* root, int tar) { + if(root == nullptr) return; + path.push_back(root->val); + tar -= root->val; + if(tar == 0 && root->left == nullptr && root->right == nullptr) + res.push_back(path); + recur(root->left, tar); + recur(root->right, tar); + path.pop_back(); + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数，先序遍历需要遍历所有节点。 +- **空间复杂度 $O(N)$ ：** 最差情况下，即树退化为链表时，`path` 存储所有树节点，使用 $O(N)$ 额外空间。 diff --git "a/leetbook_ioa/docs/LCR 154. \351\232\217\346\234\272\351\223\276\350\241\250\347\232\204\345\244\215\345\210\266.md" "b/leetbook_ioa/docs/LCR 154. \351\232\217\346\234\272\351\223\276\350\241\250\347\232\204\345\244\215\345\210\266.md" new file mode 100755 index 0000000..297d35a --- /dev/null +++ "b/leetbook_ioa/docs/LCR 154. \351\232\217\346\234\272\351\223\276\350\241\250\347\232\204\345\244\215\345\210\266.md" @@ -0,0 +1,359 @@ +## 解题思路： + +普通链表的节点定义如下： + +```Python [] +# Definition for a Node. +class Node: + def __init__(self, x: int, next: 'Node' = None): + self.val = int(x) + self.next = next +``` + +```Java [] +// Definition for a Node. +class Node { + int val; + Node next; + public Node(int val) { + this.val = val; + this.next = null; + } +} +``` + +```C++ [] +// Definition for a Node. +class Node { +public: + int val; + Node* next; + Node(int _val) { + val = _val; + next = NULL; + } +}; +``` + +本题链表的节点定义如下： + +```Python [] +# Definition for a Node. +class Node: + def __init__(self, x: int, next: 'Node' = None, random: 'Node' = None): + self.val = int(x) + self.next = next + self.random = random +``` + +```Java [] +// Definition for a Node. +class Node { + int val; + Node next, random; + public Node(int val) { + this.val = val; + this.next = null; + this.random = null; + } +} +``` + +```C++ [] +// Definition for a Node. +class Node { +public: + int val; + Node* next; + Node* random; + Node(int _val) { + val = _val; + next = NULL; + random = NULL; + } +}; +``` + +给定链表的头节点 `head` ，复制普通链表很简单，只需遍历链表，每轮建立新节点 + 构建前驱节点 `pre` 和当前节点 `node` 的引用指向即可。 + +本题链表的节点新增了 `random` 指针，指向链表中的 **任意节点** 或者 $\text{null}$ 。这个 `random` 指针意味着在复制过程中，除了构建前驱节点和当前节点的引用指向 `pre.next` ，还要构建前驱节点和其随机节点的引用指向 `pre.random` 。 + +**本题难点：** 在复制链表的过程中构建新链表各节点的 `random` 引用指向。 + +![Picture1.png](https://pic.leetcode-cn.com/1604747285-ELUgCd-Picture1.png){:align=center width=450} + +```Python [] +class Solution: + def copyRandomList(self, head: 'Node') -> 'Node': + cur = head + dum = pre = Node(0) + while cur: + node = Node(cur.val) # 复制节点 cur + pre.next = node # 新链表的 前驱节点 -> 当前节点 + # pre.random = '???' # 新链表的 「 前驱节点 -> 当前节点 」 无法确定 + cur = cur.next # 遍历下一节点 + pre = node # 保存当前新节点 + return dum.next +``` + +```Java [] +class Solution { + public Node copyRandomList(Node head) { + Node cur = head; + Node dum = new Node(0), pre = dum; + while(cur != null) { + Node node = new Node(cur.val); // 复制节点 cur + pre.next = node; // 新链表的 前驱节点 -> 当前节点 + // pre.random = "???"; // 新链表的 「 前驱节点 -> 当前节点 」 无法确定 + cur = cur.next; // 遍历下一节点 + pre = node; // 保存当前新节点 + } + return dum.next; + } +} +``` + +```C++ [] +class Solution { +public: + Node* copyRandomList(Node* head) { + Node* cur = head; + Node* dum = new Node(0), *pre = dum; + while(cur != nullptr) { + Node* node = new Node(cur->val); // 复制节点 cur + pre->next = node; // 新链表的 前驱节点 -> 当前节点 + // pre->random = "???"; // 新链表的 「 前驱节点 -> 当前节点 」 无法确定 + cur = cur->next; // 遍历下一节点 + pre = node; // 保存当前新节点 + } + return dum->next; + } +}; +``` + +> 本文介绍「哈希表」和「拼接 + 拆分」两种方法。哈希表方法比较直观；拼接 + 拆分方法的空间复杂度更低。 + +## 方法一：哈希表 + +利用哈希表的查询特点，考虑构建 **原链表节点** 和 **新链表对应节点** 的键值对映射关系，再遍历构建新链表各节点的 `next` 和 `random` 引用指向即可。 + +### 算法流程： + +1. 若头节点 `head` 为空节点，直接返回 $\text{null}$ ； +2. **初始化：** 哈希表 `hmap` ， 节点 `cur` 指向头节点； +3. **复制链表：** + 1. 建立新节点，并向 `hmap` 添加键值对 `(原 cur 节点, 新 cur 节点）` ； + 2. `cur` 遍历至原链表下一节点； +4. **构建新链表的引用指向：** + 1. 构建新节点的 `next` 和 `random` 引用指向； + 2. `cur` 遍历至原链表下一节点； +5. **返回值：** 新链表的头节点 `hmap[cur]` ； + + + +### 代码： + +```Python [] +class Solution: + def copyRandomList(self, head: 'Node') -> 'Node': + if not head: return + hmap = {} + # 3. 复制各节点，并建立 “原节点 -> 新节点” 的 Map 映射 + cur = head + while cur: + hmap[cur] = Node(cur.val) + cur = cur.next + cur = head + # 4. 构建新节点的 next 和 random 指向 + while cur: + hmap[cur].next = hmap.get(cur.next) + hmap[cur].random = hmap.get(cur.random) + cur = cur.next + # 5. 返回新链表的头节点 + return hmap[head] +``` + +```Java [] +class Solution { + public Node copyRandomList(Node head) { + if(head == null) return null; + Node cur = head; + Map map = new HashMap<>(); + // 3. 复制各节点，并建立 “原节点 -> 新节点” 的 Map 映射 + while(cur != null) { + map.put(cur, new Node(cur.val)); + cur = cur.next; + } + cur = head; + // 4. 构建新链表的 next 和 random 指向 + while(cur != null) { + map.get(cur).next = map.get(cur.next); + map.get(cur).random = map.get(cur.random); + cur = cur.next; + } + // 5. 返回新链表的头节点 + return map.get(head); + } +} +``` + +```C++ [] +class Solution { +public: + Node* copyRandomList(Node* head) { + if(head == nullptr) return nullptr; + Node* cur = head; + unordered_map map; + // 3. 复制各节点，并建立 “原节点 -> 新节点” 的 Map 映射 + while(cur != nullptr) { + map[cur] = new Node(cur->val); + cur = cur->next; + } + cur = head; + // 4. 构建新链表的 next 和 random 指向 + while(cur != nullptr) { + map[cur]->next = map[cur->next]; + map[cur]->random = map[cur->random]; + cur = cur->next; + } + // 5. 返回新链表的头节点 + return map[head]; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 两轮遍历链表，使用 $O(N)$ 时间。 +- **空间复杂度 $O(N)$ ：** 哈希表 `hmap` 使用线性大小的额外空间。 + +## 方法二：拼接 + 拆分 + +考虑构建 `原节点 1 -> 新节点 1 -> 原节点 2 -> 新节点 2 -> ……` 的拼接链表，如此便可在访问原节点的 `random` 指向节点的同时找到新对应新节点的 `random` 指向节点。 + +### 算法流程： + +1. **复制各节点，构建拼接链表:** + + - 设原链表为 $node1 \rightarrow node2 \rightarrow \cdots$ ，构建的拼接链表如下所示： + +$$ +node1 \rightarrow node1_{new} \rightarrow node2 \rightarrow node2_{new} \rightarrow \cdots +$$ + +2. **构建新链表各节点的 `random` 指向：** + + - 当访问原节点 `cur` 的随机指向节点 `cur.random` 时，对应新节点 `cur.next` 的随机指向节点为 `cur.random.next` 。 + +3. **拆分原 / 新链表：** + + - 设置 `pre` / `cur` 分别指向原 / 新链表头节点，遍历执行 `pre.next = pre.next.next` 和 `cur.next = cur.next.next` 将两链表拆分开。 + +4. 返回新链表的头节点 `res` 即可。 + + + +## 代码： + +```Python [] +class Solution: + def copyRandomList(self, head: 'Node') -> 'Node': + if not head: return + cur = head + # 1. 复制各节点，并构建拼接链表 + while cur: + tmp = Node(cur.val) + tmp.next = cur.next + cur.next = tmp + cur = tmp.next + # 2. 构建各新节点的 random 指向 + cur = head + while cur: + if cur.random: + cur.next.random = cur.random.next + cur = cur.next.next + # 3. 拆分两链表 + cur = res = head.next + pre = head + while cur.next: + pre.next = pre.next.next + cur.next = cur.next.next + pre = pre.next + cur = cur.next + pre.next = None # 单独处理原链表尾节点 + return res # 返回新链表头节点 +``` + +```Java [] +class Solution { + public Node copyRandomList(Node head) { + if(head == null) return null; + Node cur = head; + // 1. 复制各节点，并构建拼接链表 + while(cur != null) { + Node tmp = new Node(cur.val); + tmp.next = cur.next; + cur.next = tmp; + cur = tmp.next; + } + // 2. 构建各新节点的 random 指向 + cur = head; + while(cur != null) { + if(cur.random != null) + cur.next.random = cur.random.next; + cur = cur.next.next; + } + // 3. 拆分两链表 + cur = head.next; + Node pre = head, res = head.next; + while(cur.next != null) { + pre.next = pre.next.next; + cur.next = cur.next.next; + pre = pre.next; + cur = cur.next; + } + pre.next = null; // 单独处理原链表尾节点 + return res; // 返回新链表头节点 + } +} +``` + +```C++ [] +class Solution { +public: + Node* copyRandomList(Node* head) { + if(head == nullptr) return nullptr; + Node* cur = head; + // 1. 复制各节点，并构建拼接链表 + while(cur != nullptr) { + Node* tmp = new Node(cur->val); + tmp->next = cur->next; + cur->next = tmp; + cur = tmp->next; + } + // 2. 构建各新节点的 random 指向 + cur = head; + while(cur != nullptr) { + if(cur->random != nullptr) + cur->next->random = cur->random->next; + cur = cur->next->next; + } + // 3. 拆分两链表 + cur = head->next; + Node* pre = head, *res = head->next; + while(cur->next != nullptr) { + pre->next = pre->next->next; + cur->next = cur->next->next; + pre = pre->next; + cur = cur->next; + } + pre->next = nullptr; // 单独处理原链表尾节点 + return res; // 返回新链表头节点 + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 三轮遍历链表，使用 $O(N)$ 时间。 +- **空间复杂度 $O(1)$ ：** 节点引用变量使用常数大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 155. \345\260\206\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\350\275\254\345\214\226\344\270\272\346\216\222\345\272\217\347\232\204\345\217\214\345\220\221\351\223\276\350\241\250.md" "b/leetbook_ioa/docs/LCR 155. \345\260\206\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\350\275\254\345\214\226\344\270\272\346\216\222\345\272\217\347\232\204\345\217\214\345\220\221\351\223\276\350\241\250.md" new file mode 100755 index 0000000..a519e7f --- /dev/null +++ "b/leetbook_ioa/docs/LCR 155. \345\260\206\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\350\275\254\345\214\226\344\270\272\346\216\222\345\272\217\347\232\204\345\217\214\345\220\221\351\223\276\350\241\250.md" @@ -0,0 +1,138 @@ +## 解题思路： + +本文解法基于性质：二叉搜索树的中序遍历为 **递增序列** 。 +将 二叉搜索树 转换成一个 “排序的循环双向链表” ，其中包含三个要素： + +1. **排序链表：** 节点应从小到大排序，因此应使用 **中序遍历** “从小到大”访问树的节点。 +2. **双向链表：** 在构建相邻节点的引用关系时，设前驱节点 `pre` 和当前节点 `cur` ，不仅应构建 `pre.right = cur` ，也应构建 `cur.left = pre` 。 +3. **循环链表：** 设链表头节点 `head` 和尾节点 `tail` ，则应构建 `head.left = tail` 和 `tail.right = head` 。 + +![Picture1.png](https://pic.leetcode-cn.com/1599401091-PKIjds-Picture1.png){:align=center width=500} + +**中序遍历** 为对二叉树作 “左、根、右” 顺序遍历，递归实现如下： + +```Python [] +# 打印中序遍历 +def dfs(root): + if not root: return + dfs(root.left) # 左 + print(root.val) # 根 + dfs(root.right) # 右 +``` + +```Java [] +// 打印中序遍历 +void dfs(Node root) { + if(root == null) return; + dfs(root.left); // 左 + System.out.println(root.val); // 根 + dfs(root.right); // 右 +} +``` + +```C++ [] +// 打印中序遍历 +void dfs(Node* root) { + if(root == nullptr) return; + dfs(root->left); // 左 + cout << root->val << endl; // 根 + dfs(root->right); // 右 +} +``` + +根据以上分析，考虑使用中序遍历访问树的各节点 `cur` ；并在访问每个节点时构建 `cur` 和前驱节点 `pre` 的引用指向；中序遍历完成后，最后构建头节点和尾节点的引用指向即可。 + +### 算法流程： + +**`dfs(cur):`** 递归中序遍历； + +1. **终止条件：** 当节点 `cur` 为空，代表越过叶节点，直接返回； +2. 递归左子树，即 `dfs(cur.left)` ； +3. **构建链表：** + 1. **当 `pre` 为空时：** 代表正在访问链表头节点，记为 `head` ； + 2. **当 `pre` 不为空时：** 修改双向节点引用，即 `pre.right = cur` ，`cur.left = pre` ； + 3. **保存 `cur` ：** 更新 `pre = cur` ，即节点 `cur` 是后继节点的 `pre` ； +4. 递归右子树，即 `dfs(cur.right)` ； + +**`treeToDoublyList(root)：`** + +1. **特例处理：** 若节点 `root` 为空，则直接返回； +2. **初始化：** 空节点 `pre` ； +3. **转化为双向链表：** 调用 `dfs(root)` ； +4. **构建循环链表：** 中序遍历完成后，`head` 指向头节点，`pre` 指向尾节点，因此修改 `head` 和 `pre` 的双向节点引用即可； +5. **返回值：** 返回链表的头节点 `head` 即可； + + + +## 代码： + +```Python [] +class Solution: + def treeToDoublyList(self, root: 'Node') -> 'Node': + def dfs(cur): + if not cur: return + dfs(cur.left) # 递归左子树 + if self.pre: # 修改节点引用 + self.pre.right, cur.left = cur, self.pre + else: # 记录头节点 + self.head = cur + self.pre = cur # 保存 cur + dfs(cur.right) # 递归右子树 + + if not root: return + self.pre = None + dfs(root) + self.head.left, self.pre.right = self.pre, self.head + return self.head +``` + +```Java [] +class Solution { + Node pre, head; + public Node treeToDoublyList(Node root) { + if(root == null) return null; + dfs(root); + head.left = pre; + pre.right = head; + return head; + } + void dfs(Node cur) { + if(cur == null) return; + dfs(cur.left); + if(pre != null) pre.right = cur; + else head = cur; + cur.left = pre; + pre = cur; + dfs(cur.right); + } +} +``` + +```C++ [] +class Solution { +public: + Node* treeToDoublyList(Node* root) { + if(root == nullptr) return nullptr; + dfs(root); + head->left = pre; + pre->right = head; + return head; + } +private: + Node *pre, *head; + void dfs(Node* cur) { + if(cur == nullptr) return; + dfs(cur->left); + if(pre != nullptr) pre->right = cur; + else head = cur; + cur->left = pre; + pre = cur; + dfs(cur->right); + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数，中序遍历需要访问所有节点。 +- **空间复杂度 $O(N)$ ：** 最差情况下，即树退化为链表时，递归深度达到 $N$，系统使用 $O(N)$ 栈空间。 diff --git "a/leetbook_ioa/docs/LCR 156. \345\272\217\345\210\227\345\214\226\344\270\216\345\217\215\345\272\217\345\210\227\345\214\226\344\272\214\345\217\211\346\240\221.md" "b/leetbook_ioa/docs/LCR 156. \345\272\217\345\210\227\345\214\226\344\270\216\345\217\215\345\272\217\345\210\227\345\214\226\344\272\214\345\217\211\346\240\221.md" new file mode 100755 index 0000000..221d488 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 156. \345\272\217\345\210\227\345\214\226\344\270\216\345\217\215\345\272\217\345\210\227\345\214\226\344\272\214\345\217\211\346\240\221.md" @@ -0,0 +1,155 @@ +## 解题思路： + +通常使用的前序、中序、后序、层序遍历记录的二叉树的信息不完整，即唯一的输出序列可能对应着多种二叉树可能性。题目要求的 序列化 和 反序列化 是 **可逆操作** 。因此，序列化的字符串应携带 **完整的二叉树信息** 。 + +> 观察题目示例，序列化的字符串实际上是二叉树的 “层序遍历”（BFS）结果，本文也采用层序遍历。 + +为完整表示二叉树，考虑将叶节点下的 $\text{null}$ 也记录。在此基础上，对于列表中任意某节点 `node` ，其左子节点 `node.left` 和右子节点 `node.right` 在序列中的位置都是 **唯一确定** 的。如下图所示： + +![Picture1.png](https://pic.leetcode-cn.com/1603117385-ehAGsP-Picture1.png){:align=center width=550} + +上图规律可总结为下表： + +| `node.val` | `node` 的索引 | `node.left` 的索引 | `node.right` 的索引 | +| :--------: | :-----------: | :----------------: | :-----------------: | +| $1$ | $0$ | $1$ | $2$ | +| $2$ | $1$ | $3$ | $4$ | +| $3$ | $2$ | $5$ | $6$ | +| $4$ | $5$ | $7$ | $8$ | +| $5$ | $6$ | $9$ | $10$ | + +设 $m$ 为列表区间 $[0, n]$ 中的 $\text{null}$ 节点个数，则可总结出根节点、左子节点、右子节点的列表索引的递推公式： + +| `node.val` | `node` 的列表索引 | `node.left` 的列表索引 | `node.right` 的列表索引 | +| :-----------------: | :---------------: | :--------------------: | :---------------------: | +| $\ne$ $\text{null}$ | $n$ | $2(n-m) + 1$ | $2(n-m) + 2$ | +| $=$ $\text{null}$ | $n$ | 无 | 无 | + +**序列化** 使用层序遍历实现。**反序列化** 通过以上递推公式反推各节点在序列中的索引，进而实现。 + +## 序列化 Serialize ： + +借助队列，对二叉树做层序遍历，并将越过叶节点的 $\text{null}$ 也打印出来。 + +### 算法流程： + +1. **特例处理：** 若 `root` 为空，则直接返回空列表 `"[]"` ； +2. **初始化：** 队列 `queue` （包含根节点 `root` ）；序列化列表 `res` ； +3. **层序遍历：** 当 `queue` 为空时跳出； + 1. 节点出队，记为 `node` ； + 2. 若 `node` 不为空：(1) 打印字符串 `node.val` ，(2) 将左、右子节点加入 `queue` ； + 3. 否则（若 `node` 为空）：打印字符串 `"null"` ； +4. **返回值：** 拼接列表，用 `','` 隔开，首尾添加中括号； + + + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数，层序遍历需要访问所有节点，最差情况下需要访问 $N + 1$ 个 $\text{null}$ ，总体复杂度为 $O(2N + 1) = O(N)$ 。 +- **空间复杂度 $O(N)$ ：** 最差情况下，队列 `queue` 同时存储 $\frac{N + 1}{2}$ 个节点（或 $N+1$ 个 $\text{null}$ ），使用 $O(N)$ ；列表 `res` 使用 $O(N)$ 。 + +## 反序列化 Deserialize ： + +基于本文开始推出的 `node` , `node.left` , `node.right` 在序列化列表中的位置关系，可实现反序列化。 + +利用队列按层构建二叉树，借助一个指针 `i` 指向节点 `node` 的左、右子节点，每构建一个 `node` 的左、右子节点，指针 `i` 就向右移动 $1$ 位。 + +### 算法流程： + +1. **特例处理：** 若 `data` 为空，直接返回 $\text{null}$ ； +2. **初始化：** 序列化列表 `vals` （先去掉首尾中括号，再用逗号隔开），指针 `i = 1` ，根节点 `root` （值为 `vals[0]` ），队列 `queue`（包含 `root` ）； +3. **按层构建：** 当 `queue` 为空时跳出； + 1. 节点出队，记为 `node` ； + 2. 构建 `node` 的左子节点：`node.left` 的值为 `vals[i]` ，并将 `node.left` 入队； + 3. 执行 `i += 1` ； + 4. 构建 `node` 的右子节点：`node.right` 的值为 `vals[i]` ，并将 `node.right` 入队； + 5. 执行 `i += 1` ； +4. **返回值：** 返回根节点 `root` 即可； + + + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数，按层构建二叉树需要遍历整个 $vals$ ，其长度最大为 $2N+1$ 。 +- **空间复杂度 $O(N)$ ：** 最差情况下，队列 `queue` 同时存储 $\frac{N + 1}{2}$ 个节点，因此使用 $O(N)$ 额外空间。 + +## 代码： + +```Python [] +class Codec: + def serialize(self, root): + if not root: return "[]" + queue = collections.deque() + queue.append(root) + res = [] + while queue: + node = queue.popleft() + if node: + res.append(str(node.val)) + queue.append(node.left) + queue.append(node.right) + else: res.append("null") + return '[' + ','.join(res) + ']' + + def deserialize(self, data): + if data == "[]": return + vals, i = data[1:-1].split(','), 1 + root = TreeNode(int(vals[0])) + queue = collections.deque() + queue.append(root) + while queue: + node = queue.popleft() + if vals[i] != "null": + node.left = TreeNode(int(vals[i])) + queue.append(node.left) + i += 1 + if vals[i] != "null": + node.right = TreeNode(int(vals[i])) + queue.append(node.right) + i += 1 + return root +``` + +```Java [] +public class Codec { + public String serialize(TreeNode root) { + if(root == null) return "[]"; + StringBuilder res = new StringBuilder("["); + Queue queue = new LinkedList<>() {{ add(root); }}; + while(!queue.isEmpty()) { + TreeNode node = queue.poll(); + if(node != null) { + res.append(node.val + ","); + queue.add(node.left); + queue.add(node.right); + } + else res.append("null,"); + } + res.deleteCharAt(res.length() - 1); + res.append("]"); + return res.toString(); + } + + public TreeNode deserialize(String data) { + if(data.equals("[]")) return null; + String[] vals = data.substring(1, data.length() - 1).split(","); + TreeNode root = new TreeNode(Integer.parseInt(vals[0])); + Queue queue = new LinkedList<>() {{ add(root); }}; + int i = 1; + while(!queue.isEmpty()) { + TreeNode node = queue.poll(); + if(!vals[i].equals("null")) { + node.left = new TreeNode(Integer.parseInt(vals[i])); + queue.add(node.left); + } + i++; + if(!vals[i].equals("null")) { + node.right = new TreeNode(Integer.parseInt(vals[i])); + queue.add(node.right); + } + i++; + } + return root; + } +} +``` diff --git "a/leetbook_ioa/docs/LCR 157. \345\245\227\351\244\220\345\206\205\345\225\206\345\223\201\347\232\204\346\216\222\345\210\227\351\241\272\345\272\217.md" "b/leetbook_ioa/docs/LCR 157. \345\245\227\351\244\220\345\206\205\345\225\206\345\223\201\347\232\204\346\216\222\345\210\227\351\241\272\345\272\217.md" new file mode 100755 index 0000000..851c53c --- /dev/null +++ "b/leetbook_ioa/docs/LCR 157. \345\245\227\351\244\220\345\206\205\345\225\206\345\223\201\347\232\204\346\216\222\345\210\227\351\241\272\345\272\217.md" @@ -0,0 +1,117 @@ +## 解题思路： + +对于一个长度为 $n$ 的字符串（假设字符互不重复），其排列方案数共有： + +$$ +n \times (n-1) \times (n-2) … \times 2 \times 1 +$$ + +**排列方案的生成：** + +根据字符串排列的特点，考虑深度优先搜索所有排列方案。即通过字符交换，先固定第 $1$ 位字符（ $n$ 种情况）、再固定第 $2$ 位字符（ $n-1$ 种情况）、... 、最后固定第 $n$ 位字符（ $1$ 种情况）。 + +![Picture1.png](https://pic.leetcode-cn.com/1599403497-KXKQcp-Picture1.png){:align=center width=500} + +**重复排列方案与剪枝：** + +当字符串存在重复字符时，排列方案中也存在重复的排列方案。为排除重复方案，需在固定某位字符时，保证 “每种字符只在此位固定一次” ，即遇到重复字符时不交换，直接跳过。从 DFS 角度看，此操作称为 “剪枝” 。 + +![Picture2.png](https://pic.leetcode-cn.com/1599403497-GATdFr-Picture2.png){:align=center width=500} + +### 递归解析： + +1. **终止条件：** 当 `x = len(arr) - 1` 时，代表所有位已固定（最后一位只有 $1$ 种情况），则将当前组合 `arr` 转化为字符串并加入 `res` ，并返回； +2. **递推参数：** 当前固定位 `x` ； +3. **递推工作：** 初始化一个 Set ，用于排除重复的字符；将第 `x` 位字符与 `i` $\in$ `[x, len(arr)]` 字符分别交换，并进入下层递归； + 1. **剪枝：** 若 `arr[i]` 在 Set​ 中，代表其是重复字符，因此 “剪枝” ； + 2. 将 `arr[i]` 加入 Set​ ，以便之后遇到重复字符时剪枝； + 3. **固定字符：** 将字符 `arr[i]` 和 `arr[x]` 交换，即固定 `arr[i]` 为当前位字符； + 4. **开启下层递归：** 调用 `dfs(x + 1)` ，即开始固定第 `x + 1` 个字符； + 5. **还原交换：** 将字符 `arr[i]` 和 `arr[x]` 交换（还原之前的交换）； + +> 下图的测试样例为 `goods = "abc"` 。 + + + +## 代码： + +```Python [] +class Solution: + def goodsOrder(self, goods: str) -> List[str]: + arr, res = list(goods), [] + def dfs(x): + if x == len(arr) - 1: + res.append(''.join(arr)) # 添加排列方案 + return + hmap = set() + for i in range(x, len(arr)): + if arr[i] in hmap: continue # 重复，因此剪枝 + hmap.add(arr[i]) + arr[i], arr[x] = arr[x], arr[i] # 交换，将 arr[i] 固定在第 x 位 + dfs(x + 1) # 开启固定第 x + 1 位字符 + arr[i], arr[x] = arr[x], arr[i] # 恢复交换 + dfs(0) + return res +``` + +```Java [] +class Solution { + List res = new LinkedList<>(); + char[] arr; + public String[] goodsOrder(String goods) { + arr = goods.toCharArray(); + dfs(0); + return res.toArray(new String[res.size()]); + } + void dfs(int x) { + if(x == arr.length - 1) { + res.add(String.valueOf(arr)); // 添加排列方案 + return; + } + HashSet set = new HashSet<>(); + for(int i = x; i < arr.length; i++) { + if(set.contains(arr[i])) continue; // 重复，因此剪枝 + set.add(arr[i]); + swap(i, x); // 交换，将 arr[i] 固定在第 x 位 + dfs(x + 1); // 开启固定第 x + 1 位字符 + swap(i, x); // 恢复交换 + } + } + void swap(int a, int b) { + char tmp = arr[a]; + arr[a] = arr[b]; + arr[b] = tmp; + } +} +``` + +```C++ [] +class Solution { +public: + vector goodsOrder(string goods) { + dfs(goods, 0); + return res; + } +private: + vector res; + void dfs(string goods, int x) { + if(x == goods.size() - 1) { + res.push_back(goods); // 添加排列方案 + return; + } + set st; + for(int i = x; i < goods.size(); i++) { + if(st.find(goods[i]) != st.end()) continue; // 重复，因此剪枝 + st.insert(goods[i]); + swap(goods[i], goods[x]); // 交换，将 goods[i] 固定在第 x 位 + dfs(goods, x + 1); // 开启固定第 x + 1 位字符 + swap(goods[i], goods[x]); // 恢复交换 + } + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N!N)$ ：** $N$ 为字符串 `goods` 的长度；时间复杂度和字符串排列的方案数成线性关系，方案数为 $N \times (N-1) \times (N-2) … \times 2 \times 1$ ，即复杂度为 $O(N!)$ ；字符串拼接操作 `join()` 使用 $O(N)$ ；因此总体时间复杂度为 $O(N!N)$ 。 +- **空间复杂度 $O(N^2)$ ：** 全排列的递归深度为 $N$ ，系统累计使用栈空间大小为 $O(N)$ ；递归中辅助 Set 累计存储的字符数量最多为 $N + (N-1) + ... + 2 + 1 = (N+1)N/2$ ，即占用 $O(N^2)$ 的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 158. \345\272\223\345\255\230\347\256\241\347\220\206 II.md" "b/leetbook_ioa/docs/LCR 158. \345\272\223\345\255\230\347\256\241\347\220\206 II.md" new file mode 100755 index 0000000..72eead2 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 158. \345\272\223\345\255\230\347\256\241\347\220\206 II.md" @@ -0,0 +1,135 @@ +## 解题思路： + +> 请注意，数学中众数的定义为 “数组中出现次数最多的数字” ，与本文定义不同。本文将 “数组中出现次数超过一半的数字” 称为 **“众数”**。 + +本题常见的三种解法： + +1. **哈希表统计法：** 遍历数组 `stock` ，用 HashMap 统计各数字的数量，即可找出 众数 。此方法时间和空间复杂度均为 $O(N)$ 。 +2. **数组排序法：** 将数组 `stock` 排序，**数组中点的元素** 一定为众数。 +3. **摩尔投票法：** 核心理念为 **票数正负抵消** 。此方法时间和空间复杂度分别为 $O(N)$ 和 $O(1)$ ，为本题的最佳解法。 + +### 摩尔投票法： + +> 设输入数组 `stock` 的众数为 $x$ ，数组长度为 $n$ 。 + +**推论一：** 若记 **众数** 的票数为 $+1$ ，**非众数** 的票数为 $-1$ ，则一定有所有数字的 **票数和 $> 0$** 。 + +**推论二：** 若数组的前 $a$ 个数字的 **票数和 $= 0$** ，则 数组剩余 $(n-a)$ 个数字的 **票数和一定仍 $>0$** ，即后 $(n-a)$ 个数字的 **众数仍为 $x$** 。 + +> 下图中的 `nums` 对应本题的 `stock` 。 + +![Picture1.png](https://pic.leetcode-cn.com/1603612327-bOQxzq-Picture1.png){:align=center width=500} + +根据以上推论，记数组首个元素为 $n_1$ ，众数为 $x$ ，遍历并统计票数。当发生 **票数和 $= 0$** 时，**剩余数组的众数一定不变** ，这是由于： + +- **当 $n_1 = x$ ：** 抵消的所有数字中，有一半是众数 $x$ 。 +- **当 $n_1 \neq x$ ：** 抵消的所有数字中，众数 $x$ 的数量最少为 0 个，最多为一半。 + +利用此特性，每轮假设发生 **票数和 $= 0$** 都可以 **缩小剩余数组区间** 。当遍历完成时，最后一轮假设的数字即为众数。 + +### 算法流程: + +1. **初始化：** 票数统计 `votes = 0` ， 众数 `x`； +2. **循环：** 遍历数组 `stock` 中的每个数字 `num` ； + 1. 当 票数 `votes` 等于 0 ，则假设当前数字 `num` 是众数； + 2. 当 `num = x` 时，票数 `votes` 自增 1 ；当 `num != x` 时，票数 `votes` 自减 1 ； +3. **返回值：** 返回 `x` 即可； + + + +## 代码： + +```Python [] +class Solution: + def inventoryManagement(self, stock: List[int]) -> int: + votes = 0 + for num in stock: + if votes == 0: x = num + votes += 1 if num == x else -1 + return x +``` + +```Java [] +class Solution { + public int inventoryManagement(int[] stock) { + int x = 0, votes = 0; + for(int num : stock){ + if(votes == 0) x = num; + votes += num == x ? 1 : -1; + } + return x; + } +} +``` + +```C++ [] +class Solution { +public: + int inventoryManagement(vector& stock) { + int x = 0, votes = 0; + for(int num : stock){ + if(votes == 0) x = num; + votes += num == x ? 1 : -1; + } + return x; + } +}; +``` + +**拓展：** 由于题目说明 “给定的数组总是存在多数元素” ，因此本题不用考虑 **数组不存在众数** 的情况。若考虑，需要加入一个 “验证环节” ，遍历数组 `stock` 统计 `x` 的数量。 + +- 若 `x` 的数量超过数组长度一半，则返回 `x` ； +- 否则，返回未找到众数； + +时间和空间复杂度不变，仍为 $O(N)$ 和 $O(1)$ 。 + +```Python [] +class Solution: + def inventoryManagement(self, stock: List[int]) -> int: + votes, count = 0, 0 + for num in stock: + if votes == 0: x = num + votes += 1 if num == x else -1 + # 验证 x 是否为众数 + for num in stock: + if num == x: count += 1 + return x if count > len(stock) // 2 else 0 # 当无众数时返回 0 +``` + +```Java [] +class Solution { + public int inventoryManagement(int[] stock) { + int x = 0, votes = 0, count = 0; + for(int num : stock){ + if(votes == 0) x = num; + votes += num == x ? 1 : -1; + } + // 验证 x 是否为众数 + for(int num : stock) + if(num == x) count++; + return count > stock.length / 2 ? x : 0; // 当无众数时返回 0 + } +} +``` + +```C++ [] +class Solution { +public: + int inventoryManagement(vector& stock) { + int x = 0, votes = 0, count = 0; + for(int num : stock){ + if(votes == 0) x = num; + votes += num == x ? 1 : -1; + } + // 验证 x 是否为众数 + for(int num : stock) + if(num == x) count++; + return count > stock.size() / 2 ? x : 0; // 当无众数时返回 0 + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为数组 `stock` 长度。 +- **空间复杂度 $O(1)$ ：** `votes` 变量使用常数大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 159. \345\272\223\345\255\230\347\256\241\347\220\206 III.md" "b/leetbook_ioa/docs/LCR 159. \345\272\223\345\255\230\347\256\241\347\220\206 III.md" new file mode 100755 index 0000000..6e72b2d --- /dev/null +++ "b/leetbook_ioa/docs/LCR 159. \345\272\223\345\255\230\347\256\241\347\220\206 III.md" @@ -0,0 +1,213 @@ +## 方法一：快速排序 + +本题使用排序算法解决最直观，对数组 `stock` 执行排序，再返回前 $cnt$ 个元素即可。使用任意排序算法皆可，本文采用并介绍 **快速排序** ，为下文 **方法二** 做铺垫。 + +### 快速排序原理： + +快速排序算法有两个核心点，分别为 “哨兵划分” 和 “递归” 。 + +**哨兵划分操作：** 以数组某个元素（一般选取首元素）为 **基准数** ，将所有小于基准数的元素移动至其左边，大于基准数的元素移动至其右边。 + +> 如下图所示，为哨兵划分操作流程。通过一轮 **哨兵划分** ，可将数组排序问题拆分为 **两个较短数组的排序问题** （本文称之为左（右）子数组）。 + + + +**递归：** 对 **左子数组** 和 **右子数组** 递归执行 **哨兵划分**，直至子数组长度为 1 时终止递归，即可完成对整个数组的排序。 + +> 如下图所示，为示例数组 `[2,4,1,0,3,5]` 的快速排序流程。观察发现，快速排序和 **二分法** 的原理类似，都是以 $\log$ 时间复杂度实现搜索区间缩小。 + +![Picture1.png](https://pic.leetcode-cn.com/1612615552-rifQwI-Picture1.png){:width=550} + +### 代码： + +```Python [] +class Solution: + def inventoryManagement(self, stock: List[int], cnt: int) -> List[int]: + def quick_sort(stock, l, r): + # 子数组长度为 1 时终止递归 + if l >= r: return + # 哨兵划分操作（以 stock[l] 作为基准数） + i, j = l, r + while i < j: + while i < j and stock[j] >= stock[l]: j -= 1 + while i < j and stock[i] <= stock[l]: i += 1 + stock[i], stock[j] = stock[j], stock[i] + stock[l], stock[i] = stock[i], stock[l] + # 递归左（右）子数组执行哨兵划分 + quick_sort(stock, l, i - 1) + quick_sort(stock, i + 1, r) + + quick_sort(stock, 0, len(stock) - 1) + return stock[:cnt] +``` + +```Java [] +class Solution { + public int[] inventoryManagement(int[] stock, int cnt) { + quickSort(stock, 0, stock.length - 1); + return Arrays.copyOf(stock, cnt); + } + private void quickSort(int[] stock, int l, int r) { + // 子数组长度为 1 时终止递归 + if (l >= r) return; + // 哨兵划分操作（以 stock[l] 作为基准数） + int i = l, j = r; + while (i < j) { + while (i < j && stock[j] >= stock[l]) j--; + while (i < j && stock[i] <= stock[l]) i++; + swap(stock, i, j); + } + swap(stock, i, l); + // 递归左（右）子数组执行哨兵划分 + quickSort(stock, l, i - 1); + quickSort(stock, i + 1, r); + } + private void swap(int[] stock, int i, int j) { + int tmp = stock[i]; + stock[i] = stock[j]; + stock[j] = tmp; + } +} +``` + +```C++ [] +class Solution { +public: + vector inventoryManagement(vector& stock, int cnt) { + quickSort(stock, 0, stock.size() - 1); + vector res; + res.assign(stock.begin(), stock.begin() + cnt); + return res; + } +private: + void quickSort(vector& stock, int l, int r) { + // 子数组长度为 1 时终止递归 + if (l >= r) return; + // 哨兵划分操作（以 stock[l] 作为基准数） + int i = l, j = r; + while (i < j) { + while (i < j && stock[j] >= stock[l]) j--; + while (i < j && stock[i] <= stock[l]) i++; + swap(stock[i], stock[j]); + } + swap(stock[i], stock[l]); + // 递归左（右）子数组执行哨兵划分 + quickSort(stock, l, i - 1); + quickSort(stock, i + 1, r); + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N \log N)$ ：** 库函数、快排等排序算法的平均时间复杂度为 $O(N \log N)$ 。 +- **空间复杂度 $O(N)$ ：** 快速排序的递归深度最好（平均）为 $O(\log N)$ ，最差情况（即输入数组完全倒序）为 $O(N)$。 + +## 方法二：快速选择 + +题目只要求返回最小的 cnt 个数，对这 cnt 个数的顺序并没有要求。因此，只需要将数组划分为 **最小的 $cnt$ 个数** 和 **其他数字** 两部分即可，而快速排序的哨兵划分可完成此目标。 + +根据快速排序原理，如果某次哨兵划分后 **基准数正好是第 $cnt+1$ 小的数字** ，那么此时基准数左边的所有数字便是题目所求的 **最小的 cnt 个数** 。 + +根据此思路，考虑在每次哨兵划分后，判断基准数在数组中的索引是否等于 $cnt$ ，若 $\text{true}$ 则直接返回此时数组的前 $cnt$ 个数字即可。 + +### 算法流程： + +**`inventoryManagement() 函数：`** + +1. 若 $cnt$ 大于数组长度，则直接返回整个数组； +2. 执行并返回 `quick_sort()` 即可； + +**`quick_sort() 函数：`** + +> 注意，此时 `quick_sort()` 的功能不是排序整个数组，而是搜索并返回最小的 $cnt$ 个数。 + +1. **哨兵划分**： + +- 划分完毕后，基准数为 `stock[i]` ，左 / 右子数组区间分别为 $[l, i - 1]$ , $[i + 1, r]$ ； + +2. **递归或返回：** + +- 若 $cnt < i$ ，代表第 $cnt + 1$ 小的数字在 **左子数组** 中，则递归左子数组； +- 若 $cnt > i$ ，代表第 $cnt + 1$ 小的数字在 **右子数组** 中，则递归右子数组； +- 若 $cnt = i$ ，代表此时 `stock[cnt]` 即为第 $cnt + 1$ 小的数字，则直接返回数组前 $cnt$ 个数字即可； + + + +## 代码： + +```Python [] +class Solution: + def inventoryManagement(self, stock: List[int], cnt: int) -> List[int]: + if cnt >= len(stock): return stock + def quick_sort(l, r): + i, j = l, r + while i < j: + while i < j and stock[j] >= stock[l]: j -= 1 + while i < j and stock[i] <= stock[l]: i += 1 + stock[i], stock[j] = stock[j], stock[i] + stock[l], stock[i] = stock[i], stock[l] + if cnt < i: return quick_sort(l, i - 1) + if cnt > i: return quick_sort(i + 1, r) + return stock[:cnt] + + return quick_sort(0, len(stock) - 1) +``` + +```Java [] +class Solution { + public int[] inventoryManagement(int[] stock, int cnt) { + if (cnt >= stock.length) return stock; + return quickSort(stock, cnt, 0, stock.length - 1); + } + private int[] quickSort(int[] stock, int cnt, int l, int r) { + int i = l, j = r; + while (i < j) { + while (i < j && stock[j] >= stock[l]) j--; + while (i < j && stock[i] <= stock[l]) i++; + swap(stock, i, j); + } + swap(stock, i, l); + if (i > cnt) return quickSort(stock, cnt, l, i - 1); + if (i < cnt) return quickSort(stock, cnt, i + 1, r); + return Arrays.copyOf(stock, cnt); + } + private void swap(int[] stock, int i, int j) { + int tmp = stock[i]; + stock[i] = stock[j]; + stock[j] = tmp; + } +} +``` + +```C++ [] +class Solution { +public: + vector inventoryManagement(vector& stock, int cnt) { + if (cnt >= stock.size()) return stock; + return quickSort(stock, cnt, 0, stock.size() - 1); + } +private: + vector quickSort(vector& stock, int cnt, int l, int r) { + int i = l, j = r; + while (i < j) { + while (i < j && stock[j] >= stock[l]) j--; + while (i < j && stock[i] <= stock[l]) i++; + swap(stock[i], stock[j]); + } + swap(stock[i], stock[l]); + if (i > cnt) return quickSort(stock, cnt, l, i - 1); + if (i < cnt) return quickSort(stock, cnt, i + 1, r); + vector res; + res.assign(stock.begin(), stock.begin() + cnt); + return res; + } +}; +``` + +### 复杂度分析： + +本方法优化时间复杂度的本质是通过判断舍去了不必要的递归（哨兵划分）。 + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为数组元素数量；对于长度为 $N$ 的数组执行哨兵划分操作的时间复杂度为 $O(N)$ ；每轮哨兵划分后根据 $cnt$ 和 $i$ 的大小关系选择递归，由于 $i$ 分布的随机性，则向下递归子数组的平均长度为 $\frac{N}{2}$ ；因此平均情况下，哨兵划分操作一共有 $N + \frac{N}{2} + \frac{N}{4} + ... + \frac{N}{N} = \frac{N - \frac{1}{2}}{1 - \frac{1}{2}} = 2N - 1$ （等比数列求和），即总体时间复杂度为 $O(N)$ 。 +- **空间复杂度 $O(\log N)$ ：** 划分函数的平均递归深度为 $O(\log N)$ 。 diff --git "a/leetbook_ioa/docs/LCR 160. \346\225\260\346\215\256\346\265\201\344\270\255\347\232\204\344\270\255\344\275\215\346\225\260.md" "b/leetbook_ioa/docs/LCR 160. \346\225\260\346\215\256\346\265\201\344\270\255\347\232\204\344\270\255\344\275\215\346\225\260.md" new file mode 100755 index 0000000..c9bfc9f --- /dev/null +++ "b/leetbook_ioa/docs/LCR 160. \346\225\260\346\215\256\346\265\201\344\270\255\347\232\204\344\270\255\344\275\215\346\225\260.md" @@ -0,0 +1,134 @@ +## 解题思路： + +> 给定一长度为 $N$ 的无序数组，其中位数的计算方法：首先对数组执行排序（使用 $O(N \log N)$ 时间），然后返回中间元素即可（使用 $O(1)$ 时间）。 + +针对本题，根据以上思路，可以将数据流保存在一个列表中，并在添加元素时 **保持数组有序** 。此方法的时间复杂度为 $O(N)$ ，其中包括： 查找元素插入位置 $O(\log N)$ （二分查找）、向数组某位置插入元素 $O(N)$ （插入位置之后的元素都需要向后移动一位）。 + +> 借助 **堆** 可进一步优化时间复杂度。 + +建立一个 **小顶堆** $A$ 和 **大顶堆** $B$ ，各保存列表的一半元素，且规定： + +- $A$ 保存 **较大** 的一半，长度为 $\frac{N}{2}$（ $N$ 为偶数）或 $\frac{N+1}{2}$（ $N$ 为奇数）； +- $B$ 保存 **较小** 的一半，长度为 $\frac{N}{2}$（ $N$ 为偶数）或 $\frac{N-1}{2}$（ $N$ 为奇数）； + +随后，中位数可仅根据 $A, B$ 的堆顶元素计算得到。 + +![Picture1.png](https://pic.leetcode-cn.com/bcfaca2b1920d2dd6bbb01aeff990698eb36d53830c38ed499ea3239a15296b3-Picture1.png){:align=center width=500} + +### 算法流程： + +> 设元素总数为 $N = m + n$ ，其中 $m$ 和 $n$ 分别为 $A$ 和 $B$ 中的元素个数。 + +**`addNum(num)` 函数：** + +1. 当 $m = n$（即 $N$ 为 **偶数**）：需向 $A$ 添加一个元素。实现方法：将新元素 $num$ 插入至 $B$ ，再将 $B$ 堆顶元素插入至 $A$ ； +2. 当 $m \ne n$（即 $N$ 为 **奇数**）：需向 $B$ 添加一个元素。实现方法：将新元素 $num$ 插入至 $A$ ，再将 $A$ 堆顶元素插入至 $B$ ； + +> 假设插入数字 $num$ 遇到情况 `1.` 。由于 $num$ 可能属于 “较小的一半” （即属于 $B$ ），因此不能将 $nums$ 直接插入至 $A$ 。而应先将 $num$ 插入至 $B$ ，再将 $B$ 堆顶元素插入至 $A$ 。这样就可以始终保持 $A$ 保存较大一半、 $B$ 保存较小一半。 + +**`findMedian()` 函数：** + +1. 当 $m = n$（ $N$ 为 **偶数**）：则中位数为 $($ $A$ 的堆顶元素 + $B$ 的堆顶元素 $)/2$。 +2. 当 $m \ne n$（ $N$ 为 **奇数**）：则中位数为 $A$ 的堆顶元素。 + + + +## 代码： + +Python 中 heapq 模块是小顶堆。实现 **大顶堆** 方法： 小顶堆的插入和弹出操作均将元素 **取反** 即可。 +Java 使用 `PriorityQueue<>((x, y) -> (y - x))` 可方便实现大顶堆。 +C++ 中 `greater` 为小顶堆，`less` 为大顶堆。 + +```Python [] +from heapq import * + +class MedianFinder: + def __init__(self): + self.A = [] # 小顶堆，保存较大的一半 + self.B = [] # 大顶堆，保存较小的一半 + + def addNum(self, num: int) -> None: + if len(self.A) != len(self.B): + heappush(self.A, num) + heappush(self.B, -heappop(self.A)) + else: + heappush(self.B, -num) + heappush(self.A, -heappop(self.B)) + + def findMedian(self) -> float: + return self.A[0] if len(self.A) != len(self.B) else (self.A[0] - self.B[0]) / 2.0 +``` + +```Java [] +class MedianFinder { + Queue A, B; + public MedianFinder() { + A = new PriorityQueue<>(); // 小顶堆，保存较大的一半 + B = new PriorityQueue<>((x, y) -> (y - x)); // 大顶堆，保存较小的一半 + } + public void addNum(int num) { + if(A.size() != B.size()) { + A.add(num); + B.add(A.poll()); + } else { + B.add(num); + A.add(B.poll()); + } + } + public double findMedian() { + return A.size() != B.size() ? A.peek() : (A.peek() + B.peek()) / 2.0; + } +} +``` + +```C++ [] +class MedianFinder { +public: + priority_queue, greater> A; // 小顶堆，保存较大的一半 + priority_queue, less> B; // 大顶堆，保存较小的一半 + MedianFinder() { } + void addNum(int num) { + if(A.size() != B.size()) { + A.push(num); + B.push(A.top()); + A.pop(); + } else { + B.push(num); + A.push(B.top()); + B.pop(); + } + } + double findMedian() { + return A.size() != B.size() ? A.top() : (A.top() + B.top()) / 2.0; + } +}; +``` + +> Push item on the heap, then pop and return the smallest item from the heap. The combined action runs more efficiently than heappush() followed by a separate call to heappop(). + +根据以上文档说明，可将 Python 代码优化为： + +```Python [] +from heapq import * + +class MedianFinder: + def __init__(self): + self.A = [] # 小顶堆，保存较大的一半 + self.B = [] # 大顶堆，保存较小的一半 + + def addNum(self, num: int) -> None: + if len(self.A) != len(self.B): + heappush(self.B, -heappushpop(self.A, num)) + else: + heappush(self.A, -heappushpop(self.B, -num)) + + def findMedian(self) -> float: + return self.A[0] if len(self.A) != len(self.B) else (self.A[0] - self.B[0]) / 2.0 +``` + +### 复杂度分析： + +- **时间复杂度：** + - **查找中位数 $O(1)$ ：** 获取堆顶元素使用 $O(1)$ 时间； + - **添加数字 $O(\log N)$ ：** 堆的插入和弹出操作使用 $O(\log N)$ 时间。 +- **空间复杂度 $O(N)$ ：** 其中 $N$ 为数据流中的元素数量，小顶堆 $A$ 和大顶堆 $B$ 最多同时保存 $N$ 个元素。 diff --git "a/leetbook_ioa/docs/LCR 161. \350\277\236\347\273\255\345\244\251\346\225\260\347\232\204\346\234\200\351\253\230\351\224\200\345\224\256\351\242\235.md" "b/leetbook_ioa/docs/LCR 161. \350\277\236\347\273\255\345\244\251\346\225\260\347\232\204\346\234\200\351\253\230\351\224\200\345\224\256\351\242\235.md" new file mode 100755 index 0000000..48c707c --- /dev/null +++ "b/leetbook_ioa/docs/LCR 161. \350\277\236\347\273\255\345\244\251\346\225\260\347\232\204\346\234\200\351\253\230\351\224\200\345\224\256\351\242\235.md" @@ -0,0 +1,81 @@ +## 解题思路： + +观察不同解法的复杂度，可知动态规划是本题的最优解法。 + +| 常见解法 | 时间复杂度 | 空间复杂度 | +| -------- | ------------- | ----------- | +| 暴力搜索 | $O(N^2)$ | $O(1)$ | +| 分治思想 | $O(N \log N)$ | $O(\log N)$ | +| 动态规划 | $O(N)$ | $O(1)$ | + +### 动态规划解析： + +**状态定义：** 设动态规划列表 $dp$ ，$dp[i]$ 代表以元素 $sales[i]$ 为结尾的连续子数组最大和。 + +**转移方程：** 若 $dp[i-1] \leq 0$ ，说明 $dp[i - 1]$ 对 $dp[i]$ 产生负贡献，即 $dp[i-1] + sales[i]$ 还不如 $sales[i]$ 本身大。 + +$$ +dp[i] = +\begin{cases} +dp[i-1] + sales[i] & , dp[i - 1] > 0 \\ +sales[i] & , dp[i - 1] \leq 0 \\ +\end{cases} +$$ + +**初始状态：** $dp[0] = sales[0]$，即以 $sales[0]$ 结尾的连续子数组最大和为 $sales[0]$ 。 + +**返回值：** 返回 $dp$ 列表中的最大值，代表全局最大值。 + +> 下图中的 `nums` 对应本题的 `sales` 。 + +![Picture1.png](https://pic.leetcode-cn.com/77d1aa6a444743d3c8606ac951cd7fc38faf68a62064fd2639df517cd666a4d0-Picture1.png){:align=center width=500} + +### 空间优化： + +由于 $dp[i]$ 只与 $dp[i-1]$ 和 $sales[i]$ 有关系，因此可以将原数组 $sales$ 用作 $dp$ 列表，即直接在 $sales$ 上修改即可。 + +由于省去 $dp$ 列表使用的额外空间，因此空间复杂度从 $O(N)$ 降至 $O(1)$ 。 + + + +## 代码： + +```Python [] +class Solution: + def maxSales(self, sales: List[int]) -> int: + for i in range(1, len(sales)): + sales[i] += max(sales[i - 1], 0) + return max(sales) +``` + +```Java [] +class Solution { + public int maxSales(int[] sales) { + int res = sales[0]; + for(int i = 1; i < sales.length; i++) { + sales[i] += Math.max(sales[i - 1], 0); + res = Math.max(res, sales[i]); + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int maxSales(vector& sales) { + int res = sales[0]; + for(int i = 1; i < sales.size(); i++) { + if(sales[i - 1] > 0) sales[i] += sales[i - 1]; + if(sales[i] > res) res = sales[i]; + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 线性遍历数组 $sales$ 即可获得结果，使用 $O(N)$ 时间。 +- **空间复杂度 $O(1)$ ：** 使用常数大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 162. \346\225\260\345\255\227 1 \347\232\204\344\270\252\346\225\260.md" "b/leetbook_ioa/docs/LCR 162. \346\225\260\345\255\227 1 \347\232\204\344\270\252\346\225\260.md" new file mode 100755 index 0000000..6fc1cbe --- /dev/null +++ "b/leetbook_ioa/docs/LCR 162. \346\225\260\345\255\227 1 \347\232\204\344\270\252\346\225\260.md" @@ -0,0 +1,163 @@ +## 解题思路： + +> 为简化篇幅，本文将 $num$ 记为 $n$ 。 + +将 $1$ ~ $n$ 的个位、十位、百位、...的 $1$ 出现次数相加，即为 $1$ 出现的总次数。 + +设数字 $n$ 是个 $x$ 位数，记 $n$ 的第 $i$ 位为 $n_i$ ，则可将 $n$ 写为 $n_{x} n_{x-1} \cdots n_{2} n_{1}$ ；本文名词规定如下： + +- 称 「 $n_i$ 」称为 **当前位** ，记为 $cur$ ； +- 将 「 $n_{i-1} n_{i-2} \cdots n_{2} n_{1}$ 」称为 **低位** ，记为 $low$ ； +- 将 「 $n_{x} n_{x-1} \cdots n_{i+2} n_{i+1}$ 」称为 **高位** ，记为 $high$ ； +- 将 「 $10^i$ 」称为 **位因子** ，记为 $digit$ ； + +### 某位中 $1$ 出现次数的计算方法： + +根据当前位 $cur$ 值的不同，分为以下三种情况： + +1. 当 **$cur = 0$ 时：** 此位 $1$ 的出现次数只由高位 $high$ 决定，计算公式为： + +$$ +high \times digit +$$ + +> 如下图所示，以 $n = 2304$ 为例，求 $digit = 10$ （即十位）的 $1$ 出现次数。 + +![Picture1.png](https://pic.leetcode-cn.com/1599887431-cVmcVA-Picture1.png){:align=center width=450} + +2. 当 **$cur = 1$ 时：** 此位 $1$ 的出现次数由高位 $high$ 和低位 $low$ 决定，计算公式为： + +$$ +high \times digit + low + 1 +$$ + +> 如下图所示，以 $n = 2314$ 为例，求 $digit = 10$ （即十位）的 $1$ 出现次数。 + +![Picture2.png](https://pic.leetcode-cn.com/1599887431-HAAvVp-Picture2.png){:align=center width=450} + +3. 当 **$cur = 2, 3, \cdots, 9$ 时：** 此位 $1$ 的出现次数只由高位 $high$ 决定，计算公式为： + +$$ +(high + 1) \times digit +$$ + +> 如下图所示，以 $n = 2324$ 为例，求 $digit = 10$ （即十位）的 $1$ 出现次数。 + +![Picture3.png](https://pic.leetcode-cn.com/1599887431-djUZTe-Picture3.png){:align=center width=450} + +### 变量递推公式： + +设计按照 “个位、十位、...” 的顺序计算，则 $high / cur / low / digit$ 应初始化为： + +```Python [] +high = n // 10 +cur = n % 10 +low = 0 +digit = 1 # 个位 +``` + +```Java [] +int high = n / 10; +int cur = n % 10; +int low = 0; +int digit = 1; // 个位 +``` + +```C++ [] +int high = n / 10; +int cur = n % 10; +int low = 0; +int digit = 1; // 个位 +``` + +因此，从个位到最高位的变量递推公式为： + +```Python [] +while high != 0 or cur != 0: # 当 high 和 cur 同时为 0 时，说明已经越过最高位，因此跳出 + low += cur * digit # 将 cur 加入 low ，组成下轮 low + cur = high % 10 # 下轮 cur 是本轮 high 的最低位 + high //= 10 # 将本轮 high 最低位删除，得到下轮 high + digit *= 10 # 位因子每轮 × 10 +``` + +```Java [] +while(high != 0 || cur != 0) { // 当 high 和 cur 同时为 0 时，说明已经越过最高位，因此跳出 + low += cur * digit; // 将 cur 加入 low ，组成下轮 low + cur = high % 10; // 下轮 cur 是本轮 high 的最低位 + high /= 10; // 将本轮 high 最低位删除，得到下轮 high + digit *= 10; // 位因子每轮 × 10 +} +``` + +```C++ [] +while(high != 0 || cur != 0) { // 当 high 和 cur 同时为 0 时，说明已经越过最高位，因此跳出 + low += cur * digit; // 将 cur 加入 low ，组成下轮 low + cur = high % 10; // 下轮 cur 是本轮 high 的最低位 + high /= 10; // 将本轮 high 最低位删除，得到下轮 high + digit *= 10; // 位因子每轮 × 10 +} +``` + + + +## 代码： + +```Python [] +class Solution: + def digitOneInNumber(self, n: int) -> int: + digit, res = 1, 0 + high, cur, low = n // 10, n % 10, 0 + while high != 0 or cur != 0: + if cur == 0: res += high * digit + elif cur == 1: res += high * digit + low + 1 + else: res += (high + 1) * digit + low += cur * digit + cur = high % 10 + high //= 10 + digit *= 10 + return res +``` + +```Java [] +class Solution { + public int digitOneInNumber(int n) { + int digit = 1, res = 0; + int high = n / 10, cur = n % 10, low = 0; + while(high != 0 || cur != 0) { + if(cur == 0) res += high * digit; + else if(cur == 1) res += high * digit + low + 1; + else res += (high + 1) * digit; + low += cur * digit; + cur = high % 10; + high /= 10; + digit *= 10; + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int digitOneInNumber(int n) { + long digit = 1; + int high = n / 10, cur = n % 10, low = 0, res = 0; + while(high != 0 || cur != 0) { + if(cur == 0) res += high * digit; + else if(cur == 1) res += high * digit + low + 1; + else res += (high + 1) * digit; + low += cur * digit; + cur = high % 10; + high /= 10; + digit *= 10; + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(\log n)$ ：** 循环内的计算操作使用 $O(1)$ 时间；循环次数为数字 $n$ 的位数，即 $\log_{10}{n}$ ，因此循环使用 $O(\log n)$ 时间。 +- **空间复杂度 $O(1)$ ：** 几个变量使用常数大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 163. \346\211\276\345\210\260\347\254\254 k \344\275\215\346\225\260\345\255\227.md" "b/leetbook_ioa/docs/LCR 163. \346\211\276\345\210\260\347\254\254 k \344\275\215\346\225\260\345\255\227.md" new file mode 100755 index 0000000..427f413 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 163. \346\211\276\345\210\260\347\254\254 k \344\275\215\346\225\260\345\255\227.md" @@ -0,0 +1,169 @@ +## 解题思路： + +文名词规定如下： + +1. 将 $101112 \cdots$ 中的每一位称为 **数位** ，记为 $k$ ； +2. 将 $10, 11, 12, \cdots$ 称为 **数字** ，记为 $num$ ； +3. 数字 $10$ 是一个两位数，称此数字的 **位数** 为 $2$ ，记为 $digit$ ； +4. 每 $digit$ 位数的起始数字（即：$1, 10, 100, \cdots$），记为 $start$ ； + +![Picture1.png](https://pic.leetcode-cn.com/1599888213-CYhLfm-Picture1.png){:align=center width=450} + +观察上表，可推出各 $digit$ 下的数位数量 $count$ 的计算公式： + +$$ +count = 9 \times start \times digit +$$ + +根据以上分析，可将求解分为三步： + +1. 确定 $k$ 所在 **数字** 的 **位数** ，记为 $digit$ ； +2. 确定 $k$ 所在的 **数字** ，记为 $num$ ； +3. 确定 $k$ 是 $num$ 中的哪一数位，并返回结果； + +### 1. 确定所求数位的所在数字的位数 + +如下图所示，循环执行 $k$ 减去 一位数、两位数、... 的数位数量 $count$ ，直至 $k \leq count$ 时跳出。 + +由于 $k$ 已经减去了一位数、两位数、...、$(digit-1)$ 位数的 **数位数量** $count$ ，因而此时的 $k$ 是从起始数字 $start$ 开始计数的。 + +```Python [] +digit, start, count = 1, 1, 9 +while k > count: + k -= count + start *= 10 # 1, 10, 100, ... + digit += 1 # 1, 2, 3, ... + count = 9 * start * digit # 9, 180, 2700, ... +``` + +```Java [] +int digit = 1; +long start = 1; +long count = 9; +while (k > count) { + k -= count; + start *= 10; // 1, 10, 100, ... + digit += 1; // 1, 2, 3, ... + count = digit * start * 9; // 9, 180, 2700, ... +} +``` + +```C++ [] +int digit = 1; +long start = 1; +long count = 9; +while (k > count) { // 1. + k -= count; + start *= 10; // 1, 10, 100, ... + digit += 1; // 1, 2, 3, ... + count = digit * start * 9; // 9, 180, 2700, ... +} +``` + +**结论：** 所求数位 (1) 在某个 $digit$ 位数中； (2) 为从数字 $start$ 开始的第 $k$ 个数位。 + +![Picture2.png](https://pic.leetcode-cn.com/1599888496-HivJvS-Picture2.png){:align=center width=500} + +### 2. 确定所求数位所在的数字 + +如下图所示，所求数位 在从数字 $start$ 开始的第 $[(k - 1) / digit]$ 个 **数字** 中（ $start$ 为第 0 个数字）。 + +```Python [] +num = start + (k - 1) // digit +``` + +```Java [] +long num = start + (k - 1) / digit; +``` + +```C++ [] +long num = start + (k - 1) / digit; +``` + +**结论：** 所求数位在数字 $num$ 中。 + +![Picture3.png](https://pic.leetcode-cn.com/1599888213-gCcnEA-Picture3.png){:align=center width=500} + +### 3. 确定所求数位在 $num$ 的哪一数位 + +如下图所示，所求数位为数字 $num$ 的第 $(k - 1) \mod digit$ 位（ 数字的首个数位为第 0 位）。 + +```Python [] +s = str(num) # 转化为 string +res = int(s[(k - 1) % digit]) # 获得 num 的 第 (k - 1) % digit 个数位，并转化为 int +``` + +```Java [] +String s = Long.toString(num); // 转化为 string +int res = s.charAt((k - 1) % digit) - '0'; // 获得 num 的 第 (k - 1) % digit 个数位，并转化为 int +``` + +```C++ [] +string s = to_string(num); // 转化为 string +int res = s[(k - 1) % digit] - '0'; // 获得 num 的 第 (k - 1) % digit 个数位，并转化为 int +``` + +**结论：** 所求数位是 $res$ 。 + +![Picture4.png](https://pic.leetcode-cn.com/1599888395-oeWGAH-Picture4.png){:align=center width=500} + +整体流程如下图所示。 + + + +## 代码： + +```Python [] +class Solution: + def findKthNumber(self, k: int) -> int: + digit, start, count = 1, 1, 9 + while k > count: # 1. + k -= count + start *= 10 + digit += 1 + count = 9 * start * digit + num = start + (k - 1) // digit # 2. + return int(str(num)[(k - 1) % digit]) # 3. +``` + +```Java [] +class Solution { + public int findKthNumber(int k) { + int digit = 1; + long start = 1; + long count = 9; + while (k > count) { // 1. + k -= count; + start *= 10; + digit += 1; + count = digit * start * 9; + } + long num = start + (k - 1) / digit; // 2. + return Long.toString(num).charAt((k - 1) % digit) - '0'; // 3. + } +} +``` + +```C++ [] +class Solution { +public: + int findKthNumber(int k) { + int digit = 1; + long start = 1; + long count = 9; + while (k > count) { // 1. + k -= count; + start *= 10; + digit += 1; + count = digit * start * 9; + } + long num = start + (k - 1) / digit; // 2. + return to_string(num)[(k - 1) % digit] - '0'; // 3. + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(\log k)$ ：** 所求数位 $k$ 对应数字 $num$ 的位数 $digit$ 最大为 $O(\log k)$ ；第一步最多循环 $O(\log k)$ 次；第三步中将 $num$ 转化为字符串使用 $O(\log k)$ 时间；因此总体为 $O(\log k)$ 。 +- **空间复杂度 $O(\log k)$ ：** 将数字 $num$ 转化为字符串 `str(num)` ，占用 $O(\log k)$ 的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 164. \347\240\264\350\247\243\351\227\257\345\205\263\345\257\206\347\240\201.md" "b/leetbook_ioa/docs/LCR 164. \347\240\264\350\247\243\351\227\257\345\205\263\345\257\206\347\240\201.md" new file mode 100755 index 0000000..3b41e3a --- /dev/null +++ "b/leetbook_ioa/docs/LCR 164. \347\240\264\350\247\243\351\227\257\345\205\263\345\257\206\347\240\201.md" @@ -0,0 +1,170 @@ +## 解题思路： + +此题求拼接起来的最小数字，本质上是一个排序问题。设数组 $password$ 中任意两数字的字符串为 $x$ 和 $y$ ，则规定 **排序判断规则** 为： + +- 若拼接字符串 $x + y > y + x$ ，则 $x$ “大于” $y$ ； +- 反之，若 $x + y < y + x$ ，则 $x$ “小于” $y$ ； + +> $x$ “小于” $y$ 代表：排序完成后，数组中 $x$ 应在 $y$ 左边；“大于” 则反之。 + +根据以上规则，套用任何排序方法对 $password$ 执行排序即可。 + +![Picture1.png](https://pic.leetcode-cn.com/95e81dbccc44f26292d88c509afd68204a86b37d342f83d109fa7aa0cd4a6049-Picture1.png){:align=center width=450} + +### 算法流程： + +1. **初始化：** 字符串列表 $strs$ ，保存各数字的字符串格式； +2. **列表排序：** 应用以上 “排序判断规则” ，对 $strs$ 执行排序； +3. **返回值：** 拼接 $strs$ 中的所有字符串，并返回。 + +> 下图中 `nums` 对应本题的 `password` 。 + + + +## 代码： + +本文列举 **快速排序** 和 **内置函数** 两种排序方法，其他排序方法也可实现。 + +### 快速排序： + +需修改快速排序函数中的排序判断规则。字符串大小（字典序）对比的实现方法： + +- Python/C++ 中可直接用 `<` , `>`； +- Java 中使用函数 `A.compareTo(B)`； + +```Python [] +class Solution: + def crackPassword(self, password: List[int]) -> str: + def quick_sort(l , r): + if l >= r: return + i, j = l, r + while i < j: + while strs[j] + strs[l] >= strs[l] + strs[j] and i < j: j -= 1 + while strs[i] + strs[l] <= strs[l] + strs[i] and i < j: i += 1 + strs[i], strs[j] = strs[j], strs[i] + strs[i], strs[l] = strs[l], strs[i] + quick_sort(l, i - 1) + quick_sort(i + 1, r) + + strs = [str(num) for num in password] + quick_sort(0, len(strs) - 1) + return ''.join(strs) +``` + +```Java [] +class Solution { + public String crackPassword(int[] password) { + String[] strs = new String[password.length]; + for(int i = 0; i < password.length; i++) + strs[i] = String.valueOf(password[i]); + quickSort(strs, 0, strs.length - 1); + StringBuilder res = new StringBuilder(); + for(String s : strs) + res.append(s); + return res.toString(); + } + void quickSort(String[] strs, int l, int r) { + if(l >= r) return; + int i = l, j = r; + String tmp = strs[i]; + while(i < j) { + while((strs[j] + strs[l]).compareTo(strs[l] + strs[j]) >= 0 && i < j) j--; + while((strs[i] + strs[l]).compareTo(strs[l] + strs[i]) <= 0 && i < j) i++; + tmp = strs[i]; + strs[i] = strs[j]; + strs[j] = tmp; + } + strs[i] = strs[l]; + strs[l] = tmp; + quickSort(strs, l, i - 1); + quickSort(strs, i + 1, r); + } +} +``` + +```C++ [] +class Solution { +public: + string crackPassword(vector& password) { + vector strs; + for(int i = 0; i < password.size(); i++) + strs.push_back(to_string(password[i])); + quickSort(strs, 0, strs.size() - 1); + string res; + for(string s : strs) + res.append(s); + return res; + } +private: + void quickSort(vector& strs, int l, int r) { + if(l >= r) return; + int i = l, j = r; + while(i < j) { + while(strs[j] + strs[l] >= strs[l] + strs[j] && i < j) j--; + while(strs[i] + strs[l] <= strs[l] + strs[i] && i < j) i++; + swap(strs[i], strs[j]); + } + swap(strs[i], strs[l]); + quickSort(strs, l, i - 1); + quickSort(strs, i + 1, r); + } +}; +``` + +### 内置函数： + +需定义排序规则： + +- Python 定义在函数 `sort_rule(x, y)` 中； +- Java 定义为 `(x, y) -> (x + y).compareTo(y + x)` ； +- C++ 定义为 `(string& x, string& y){ return x + y < y + x; }` ； + +```Python [] +class Solution: + def crackPassword(self, password: List[int]) -> str: + def sort_rule(x, y): + a, b = x + y, y + x + if a > b: return 1 + elif a < b: return -1 + else: return 0 + + strs = [str(num) for num in password] + strs.sort(key = functools.cmp_to_key(sort_rule)) + return ''.join(strs) +``` + +```Java [] +class Solution { + public String crackPassword(int[] password) { + String[] strs = new String[password.length]; + for(int i = 0; i < password.length; i++) + strs[i] = String.valueOf(password[i]); + Arrays.sort(strs, (x, y) -> (x + y).compareTo(y + x)); + StringBuilder res = new StringBuilder(); + for(String s : strs) + res.append(s); + return res.toString(); + } +} +``` + +```C++ [] +class Solution { +public: + string crackPassword(vector& password) { + vector strs; + string res; + for(int i = 0; i < password.size(); i++) + strs.push_back(to_string(password[i])); + sort(strs.begin(), strs.end(), [](string& x, string& y){ return x + y < y + x; }); + for(int i = 0; i < strs.size(); i++) + res.append(strs[i]); + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N \log N)$ ：** $N$ 为最终返回值的字符数量（ $strs$ 列表的长度 $\leq N$ ）；使用快排或内置函数的平均时间复杂度为 $O(N \log N)$ ，最差为 $O(N^2)$ 。 +- **空间复杂度 $O(N)$ ：** 字符串列表 $strs$ 占用线性大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 165. \350\247\243\345\257\206\346\225\260\345\255\227.md" "b/leetbook_ioa/docs/LCR 165. \350\247\243\345\257\206\346\225\260\345\255\227.md" new file mode 100755 index 0000000..dfbbb34 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 165. \350\247\243\345\257\206\346\225\260\345\255\227.md" @@ -0,0 +1,212 @@ +## 解题思路： + +根据题意，可按照下图的思路，总结出 “递推公式” （即转移方程）。 + +> 下图中的 `num` 对应本题的 `ciphertext` 。 + +![Picture1.png](https://pic.leetcode-cn.com/1603462412-iUcKzA-Picture1.png){:align=center width=600} + +因此，此题可用动态规划解决，以下按照流程解题。 + +### 动态规划解析： + +> 记数字 $ciphertext$ 第 $i$ 位数字为 $x_i$ ，数字 $ciphertext$ 的位数为 $n$ ； +> 例如： $ciphertext = 12258$ 的 $n = 5$ , $x_1 = 1$ 。 + +- **状态定义：** 设动态规划列表 $dp$ ，$dp[i]$ 代表以 $x_i$ 为结尾的数字的翻译方案数量。 + +- **转移方程：** 若 $x_i$ 和 $x_{i-1}$ 组成的两位数字可被整体翻译，则 $dp[i] = dp[i - 1] + dp[i - 2]$ ，否则 $dp[i] = dp[i - 1]$ 。 + +$$ +dp[i] = +\begin{cases} +dp[i - 1] + dp[i - 2] & {, (10 x_{i-1} + x_i) \in [10,25]} \\ +dp[i - 1] & {, (10 x_{i-1} + x_i) \in [0, 10) \cup (25, 99]} +\end{cases} +$$ + +> **可被整体翻译的两位数区间分析：** 当 $x_{i-1} = 0$ 时，组成的两位数无法被整体翻译（例如 $00, 01, 02, \cdots$ ），大于 $25$ 的两位数也无法被整体翻译（例如 $26, 27, \cdots$ ），因此区间为 $[10, 25]$ 。 + +- **初始状态：** $dp[0] = dp[1] = 1$ ，即 “无数字” 和 “第 $1$ 位数字” 的翻译方法数量均为 $1$ ； + +- **返回值：** $dp[n]$ ，即此数字的翻译方案数量； + +> **Q：** 无数字情况 $dp[0] = 1$ 从何而来？ +> **A：** 当 $ciphertext$ 第 $1, 2$ 位的组成的数字 $\in [10,25]$ 时，显然应有 $2$ 种翻译方法，即 $dp[2] = dp[1] + dp[0] = 2$ ，而显然 $dp[1] = 1$ ，因此推出 $dp[0] = 1$ 。 + +## 方法一：字符串遍历 + +- 为方便获取数字的各位 $x_i$ ，考虑先将数字 $ciphertext$ 转化为字符串 $s$ ，通过遍历 $s$ 实现动态规划。 +- 通过字符串切片 $s[i - 2:i]$ 获取数字组合 $10 x_{i-1} + x_i$ ，通过对比字符串 ASCII 码判断字符串对应的数字区间。 +- **空间使用优化：** 由于 $dp[i]$ 只与 $dp[i - 1]$ 有关，因此可使用两个变量 $a, b$ 分别记录 $dp[i]$ , $dp[i - 1]$ ，两变量交替前进即可。此方法可省去 $dp$ 列表使用的 $O(N)$ 的额外空间。 + + + +### 代码： + +```Python [] +class Solution: + def crackNumber(self, ciphertext: int) -> int: + s = str(ciphertext) + a = b = 1 + for i in range(2, len(s) + 1): + tmp = s[i - 2:i] + c = a + b if "10" <= tmp <= "25" else a + b = a + a = c + return a +``` + +```Java [] +class Solution { + public int crackNumber(int ciphertext) { + String s = String.valueOf(ciphertext); + int a = 1, b = 1; + for(int i = 2; i <= s.length(); i++) { + String tmp = s.substring(i - 2, i); + int c = tmp.compareTo("10") >= 0 && tmp.compareTo("25") <= 0 ? a + b : a; + b = a; + a = c; + } + return a; + } +} +``` + +```C++ [] +class Solution { +public: + int crackNumber(int ciphertext) { + string s = to_string(ciphertext); + int a = 1, b = 1, len = s.size(); + for(int i = 2; i <= len; i++) { + string tmp = s.substr(i - 2, 2); + int c = tmp.compare("10") >= 0 && tmp.compare("25") <= 0 ? a + b : a; + b = a; + a = c; + } + return a; + } +}; +``` + +此题的动态规划计算是 **对称的** ，即 **从左向右** 遍历（从第 $dp[2]$ 计算至 $dp[n]$ ）和 **从右向左** 遍历（从第 $dp[n - 2]$ 计算至 $dp[0]$ ）所得方案数一致。从右向左遍历的代码如下所示。 + +```Python [] +class Solution: + def crackNumber(self, ciphertext: int) -> int: + s = str(ciphertext) + a = b = 1 + for i in range(len(s) - 2, -1, -1): + a, b = (a + b if "10" <= s[i:i + 2] <= "25" else a), a + return a +``` + +```Java [] +class Solution { + public int crackNumber(int ciphertext) { + String s = String.valueOf(ciphertext); + int a = 1, b = 1; + for(int i = s.length() - 2; i > -1; i--) { + String tmp = s.substring(i, i + 2); + int c = tmp.compareTo("10") >= 0 && tmp.compareTo("25") <= 0 ? a + b : a; + b = a; + a = c; + } + return a; + } +} +``` + +```C++ [] +class Solution { +public: + int crackNumber(int ciphertext) { + string s = to_string(ciphertext); + int a = 1, b = 1, len = s.size(); + for(int i = len - 2; i > -1; i--) { + string tmp = s.substr(i, 2); + int c = tmp.compare("10") >= 0 && tmp.compare("25") <= 0 ? a + b : a; + b = a; + a = c; + } + return a; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为字符串 $s$ 的长度（即数字 $ciphertext$ 的位数 $\log(ciphertext)$ ），其决定了循环次数。 +- **空间复杂度 $O(N)$ ：** 字符串 $s$ 使用 $O(N)$ 大小的额外空间。 + +## 方法二：数字求余 + +上述方法虽然已经节省了 $dp$ 列表的空间占用，但字符串 $s$ 仍使用了 $O(N)$ 大小的额外空间。 + +### 空间优化： + +- 利用求余运算 $ciphertext \mod 10$ 和求整运算 $ciphertext // 10$ ，可获取数字 $ciphertext$ 的各位数字（获取顺序为个位、十位、百位…）。 +- 运用 **求余** 和 **求整** 运算实现，可实现 **从右向左** 的动态规划计算。而根据上述动态规划 “对称性” ，可知从右向左计算是正确的。 +- 自此，字符串 $s$ 的空间占用也被省去，空间复杂度从 $O(N)$ 降至 $O(1)$ 。 + + + +### 代码： + +```Python [] +class Solution: + def crackNumber(self, ciphertext: int) -> int: + a = b = 1 + y = ciphertext % 10 + while ciphertext > 9: + ciphertext //= 10 + x = ciphertext % 10 + tmp = 10 * x + y + c = a + b if 10 <= tmp <= 25 else a + a, b = c, a + y = x + return a +``` + +```Java [] +class Solution { + public int crackNumber(int ciphertext) { + int a = 1, b = 1, x, y = ciphertext % 10; + while(ciphertext > 9) { + ciphertext /= 10; + x = ciphertext % 10; + int tmp = 10 * x + y; + int c = (tmp >= 10 && tmp <= 25) ? a + b : a; + b = a; + a = c; + y = x; + } + return a; + } +} +``` + +```C++ [] +class Solution { +public: + int crackNumber(int ciphertext) { + int a = 1, b = 1, x, y = ciphertext % 10; + while(ciphertext > 9) { + ciphertext /= 10; + x = ciphertext % 10; + int tmp = 10 * x + y; + int c = (tmp >= 10 && tmp <= 25) ? a + b : a; + b = a; + a = c; + y = x; + } + return a; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为字符串 $s$ 的长度，即数字 $ciphertext$ 的位数 $\log(ciphertext)$ ，其决定了循环次数。 +- **空间复杂度 $O(1)$ ：** 几个变量使用常数大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 166. \347\217\240\345\256\235\347\232\204\346\234\200\351\253\230\344\273\267\345\200\274.md" "b/leetbook_ioa/docs/LCR 166. \347\217\240\345\256\235\347\232\204\346\234\200\351\253\230\344\273\267\345\200\274.md" new file mode 100755 index 0000000..44818dd --- /dev/null +++ "b/leetbook_ioa/docs/LCR 166. \347\217\240\345\256\235\347\232\204\346\234\200\351\253\230\344\273\267\345\200\274.md" @@ -0,0 +1,152 @@ +## 解题思路： + +题目说明：从棋盘的左上角开始拿格子里的珠宝，并每次 **向右** 或者 **向下** 移动一格、直到到达棋盘的右下角。 +根据题目说明，易得某单元格只可能从上边单元格或左边单元格到达。 + +设 $f(i, j)$ 为从棋盘左上角走至单元格 $(i ,j)$ 的珠宝最大累计价值，易得到以下递推关系：$f(i,j)$ 等于 $f(i,j-1)$ 和 $f(i-1,j)$ 中的较大值加上当前单元格珠宝价值 $frame(i,j)$ 。 + +$$ +f(i,j) = \max[f(i,j-1), f(i-1,j)] + frame(i,j) +$$ + +因此，可用动态规划解决此问题，以上公式便为转移方程。 + +> 下图中的 `grid` 对应本题的 `frame` 。 + +![Picture1.png](https://pic.leetcode-cn.com/1662cdf7aafd8c9ed6e1eadc41bfc9adf58ea808e11f1a3dd2e8ba4632b9d1ac-Picture1.png){:align=center width=450} + +### 动态规划解析： + +**状态定义：** 设动态规划矩阵 $dp$ ，$dp(i,j)$ 代表从棋盘的左上角开始，到达单元格 $(i,j)$ 时能拿到珠宝的最大累计价值。 + +**转移方程：** + +1. 当 $i = 0$ 且 $j = 0$ 时，为起始元素； +2. 当 $i = 0$ 且 $j \ne 0$ 时，为矩阵第一行元素，只可从左边到达； +3. 当 $i \ne 0$ 且 $j = 0$ 时，为矩阵第一列元素，只可从上边到达； +4. 当 $i \ne 0$ 且 $j \ne 0$ 时，可从左边或上边到达； + +$$ +dp(i,j)= +\begin{cases} +frame(i,j) & {,i=0, j=0}\\ +frame(i,j) + dp(i,j-1) & {,i=0, j \ne 0}\\ +frame(i,j) + dp(i-1,j) & {,i \ne 0, j=0}\\ +frame(i,j) + \max[dp(i-1,j),dp(i,j-1)]& ,{i \ne 0, j \ne 0} +\end{cases} +$$ + +**初始状态：** $dp[0][0] = frame[0][0]$ ，即到达单元格 $(0,0)$ 时能拿到珠宝的最大累计价值为 $frame[0][0]$ ； + +**返回值：** $dp[m-1][n-1]$ ，$m, n$ 分别为矩阵的行高和列宽，即返回 $dp$ 矩阵右下角元素。 + +### 空间优化： + +由于 $dp[i][j]$ 只与 $dp[i-1][j]$ , $dp[i][j-1]$ , $frame[i][j]$ 有关系，因此可以将原矩阵 $frame$ 用作 $dp$ 矩阵，即直接在 $frame$ 上修改即可。 + +应用此方法可省去 $dp$ 矩阵使用的额外空间，因此空间复杂度从 $O(MN)$ 降至 $O(1)$ 。 + + + +## 代码： + +```Python [] +class Solution: + def jewelleryValue(self, frame: List[List[int]]) -> int: + for i in range(len(frame)): + for j in range(len(frame[0])): + if i == 0 and j == 0: continue + if i == 0: frame[i][j] += frame[i][j - 1] + elif j == 0: frame[i][j] += frame[i - 1][j] + else: frame[i][j] += max(frame[i][j - 1], frame[i - 1][j]) + return frame[-1][-1] +``` + +```Java [] +class Solution { + public int jewelleryValue(int[][] frame) { + int m = frame.length, n = frame[0].length; + for(int i = 0; i < m; i++) { + for(int j = 0; j < n; j++) { + if(i == 0 && j == 0) continue; + if(i == 0) frame[i][j] += frame[i][j - 1] ; + else if(j == 0) frame[i][j] += frame[i - 1][j]; + else frame[i][j] += Math.max(frame[i][j - 1], frame[i - 1][j]); + } + } + return frame[m - 1][n - 1]; + } +} +``` + +```C++ [] +class Solution { +public: + int jewelleryValue(vector>& frame) { + int m = frame.size(), n = frame[0].size(); + for(int i = 0; i < m; i++) { + for(int j = 0; j < n; j++) { + if(i == 0 && j == 0) continue; + if(i == 0) frame[i][j] += frame[i][j - 1] ; + else if(j == 0) frame[i][j] += frame[i - 1][j]; + else frame[i][j] += max(frame[i][j - 1], frame[i - 1][j]); + } + } + return frame[m - 1][n - 1]; + } +}; +``` + +以上代码逻辑清晰，和转移方程直接对应，但仍可提升效率，这是因为：当 $frame$ 矩阵很大时，$i = 0$ 或 $j = 0$ 的情况仅占极少数，相当循环每轮都冗余了一次判断。因此，可先初始化矩阵第一行和第一列，再开始遍历递推。 + +```Python [] +class Solution: + def jewelleryValue(self, frame: List[List[int]]) -> int: + m, n = len(frame), len(frame[0]) + for j in range(1, n): # 初始化第一行 + frame[0][j] += frame[0][j - 1] + for i in range(1, m): # 初始化第一列 + frame[i][0] += frame[i - 1][0] + for i in range(1, m): + for j in range(1, n): + frame[i][j] += max(frame[i][j - 1], frame[i - 1][j]) + return frame[-1][-1] +``` + +```Java [] +class Solution { + public int jewelleryValue(int[][] frame) { + int m = frame.length, n = frame[0].length; + for(int j = 1; j < n; j++) // 初始化第一行 + frame[0][j] += frame[0][j - 1]; + for(int i = 1; i < m; i++) // 初始化第一列 + frame[i][0] += frame[i - 1][0]; + for(int i = 1; i < m; i++) + for(int j = 1; j < n; j++) + frame[i][j] += Math.max(frame[i][j - 1], frame[i - 1][j]); + return frame[m - 1][n - 1]; + } +} +``` + +```C++ [] +class Solution { +public: + int jewelleryValue(vector>& frame) { + int m = frame.size(), n = frame[0].size(); + for(int j = 1; j < n; j++) // 初始化第一行 + frame[0][j] += frame[0][j - 1]; + for(int i = 1; i < m; i++) // 初始化第一列 + frame[i][0] += frame[i - 1][0]; + for(int i = 1; i < m; i++) + for(int j = 1; j < n; j++) + frame[i][j] += max(frame[i][j - 1], frame[i - 1][j]); + return frame[m - 1][n - 1]; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(MN)$ ：** $M, N$ 分别为矩阵行高、列宽；动态规划需遍历整个 $frame$ 矩阵，使用 $O(MN)$ 时间。 +- **空间复杂度 $O(1)$ ：** 原地修改使用常数大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 167. \346\213\233\345\274\217\346\213\206\350\247\243 I.md" "b/leetbook_ioa/docs/LCR 167. \346\213\233\345\274\217\346\213\206\350\247\243 I.md" new file mode 100755 index 0000000..1daf14c --- /dev/null +++ "b/leetbook_ioa/docs/LCR 167. \346\213\233\345\274\217\346\213\206\350\247\243 I.md" @@ -0,0 +1,169 @@ +## 解题思路： + +长度为 $N$ 的字符串共有 $\frac{(1 + N)N}{2}$ 个子字符串（复杂度为 $O(N^2)$ ），判断长度为 $N$ 的字符串是否有重复字符的复杂度为 $O(N)$ ，因此本题使用暴力法解决的复杂度为 $O(N^3)$ 。 + +本题有滑动窗口和动态规划两种解法。 + +## 方法一：滑动窗口 + 哈希表 + +**哈希表 $dic$ 统计：** 指针 $j$ 遍历字符 $arr$ ，哈希表统计字符 $arr[j]$ **最后一次出现的索引** 。 + +**更新左指针 $i$ ：** 根据上轮左指针 $i$ 和 $dic[arr[j]]$ ，每轮更新左边界 $i$ ，保证区间 $[i + 1, j]$ 内无重复字符且最大。 + +$$ +i = \max(dic[arr[j]], i) +$$ + +**更新结果 $res$ ：** 取上轮 $res$ 和本轮双指针区间 $[i + 1,j]$ 的宽度（即 $j - i$ ）中的最大值。 + +$$ +res = \max(res, j - i) +$$ + +> 下图中的 `s` 对应本题中的 `arr` 。 + + + +### 代码： + +```Python [] +class Solution: + def dismantlingAction(self, arr: str) -> int: + dic, res, i = {}, 0, -1 + for j in range(len(arr)): + if arr[j] in dic: + i = max(dic[arr[j]], i) # 更新左指针 i + dic[arr[j]] = j # 哈希表记录 + res = max(res, j - i) # 更新结果 + return res +``` + +```Java [] +class Solution { + public int dismantlingAction(String arr) { + Map dic = new HashMap<>(); + int i = -1, res = 0, len = arr.length(); + for(int j = 0; j < len; j++) { + if (dic.containsKey(arr.charAt(j))) + i = Math.max(i, dic.get(arr.charAt(j))); // 更新左指针 i + dic.put(arr.charAt(j), j); // 哈希表记录 + res = Math.max(res, j - i); // 更新结果 + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int dismantlingAction(string arr) { + unordered_map dic; + int i = -1, res = 0, len = arr.size(); + for(int j = 0; j < len; j++) { + if (dic.find(arr[j]) != dic.end()) + i = max(i, dic.find(arr[j])->second); // 更新左指针 + dic[arr[j]] = j; // 哈希表记录 + res = max(res, j - i); // 更新结果 + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为字符串长度，动态规划需遍历计算 $dp$ 列表。 +- **空间复杂度 $O(1)$ ：** 字符的 ASCII 码范围为 $0$ ~ $127$ ，哈希表 $dic$ 最多使用 $O(128) = O(1)$ 大小的额外空间。 + +## 方法二：动态规划 + 哈希表 + +- **状态定义：** 设动态规划列表 $dp$ ，$dp[j]$ 代表以字符 $arr[j]$ 为结尾的 “最长不重复子字符串” 的长度。 +- **转移方程：** 固定右边界 $j$ ，设字符 $arr[j]$ 左边距离最近的相同字符为 $arr[i]$ ，即 $arr[i] = arr[j]$ 。 + 1. 当 $i < 0$ ，即 $arr[j]$ 左边无相同字符，则 $dp[j] = dp[j-1] + 1$ 。 + 2. 当 $dp[j - 1] < j - i$ ，说明字符 $arr[i]$ 在子字符串 $dp[j-1]$ **区间之外** ，则 $dp[j] = dp[j - 1] + 1$ 。 + 3. 当 $dp[j - 1] \geq j - i$ ，说明字符 $arr[i]$ 在子字符串 $dp[j-1]$ **区间之中** ，则 $dp[j]$ 的左边界由 $arr[i]$ 决定，即 $dp[j] = j - i$ 。 + + > 当 $i < 0$ 时，由于 $dp[j - 1] \leq j$ 恒成立，因而 $dp[j - 1] < j - i$ 恒成立，因此分支 `1.` 和 `2.` 可被合并。 + +$$ +dp[j] = +\begin{cases} +dp[j - 1] + 1 & , dp[j-1] < j - i \\ +j - i & , dp[j-1] \geq j - i +\end{cases} +$$ + +- **返回值：** $\max(dp)$ ，即全局的 “最长不重复子字符串” 的长度。 + +![Picture1.png](https://pic.leetcode-cn.com/1599287290-mTdFye-Picture1.png){:align=center width=500} + +### 状态压缩： + +- 由于返回值是取 $dp$ 列表最大值，因此可借助变量 $tmp$ 存储 $dp[j]$ ，变量 $res$ 每轮更新最大值即可。 +- 此优化可节省 $dp$ 列表使用的 $O(N)$ 大小的额外空间。 + +### 哈希表记录： + +观察转移方程，可知关键问题：每轮遍历字符 $arr[j]$ 时，如何计算索引 $i$ ？ + +- **哈希表统计：** 遍历字符串 $arr$ 时，使用哈希表（记为 $dic$ ）统计 **各字符最后一次出现的索引位置** 。 +- **左边界 $i$ 获取方式：** 遍历到 $arr[j]$ 时，可通过访问哈希表 $dic[arr[j]]$ 获取最近的相同字符的索引 $i$ 。 + + + +### 代码： + +Python 的 `get(key, default)` 方法和 Java 的 `getOrDefault(key, default)` ，代表当哈希表包含键 `key` 时返回对应 `value` ，不包含时返回默认值 `default` 。 + +```Python [] +class Solution: + def dismantlingAction(self, arr: str) -> int: + dic = {} + res = tmp = 0 + for j in range(len(arr)): + i = dic.get(arr[j], -1) # 获取索引 i + dic[arr[j]] = j # 更新哈希表 + tmp = tmp + 1 if tmp < j - i else j - i # dp[j - 1] -> dp[j] + res = max(res, tmp) # max(dp[j - 1], dp[j]) + return res +``` + +```Java [] +class Solution { + public int dismantlingAction(String arr) { + Map dic = new HashMap<>(); + int res = 0, tmp = 0, len = arr.length(); + for(int j = 0; j < len; j++) { + int i = dic.getOrDefault(arr.charAt(j), -1); // 获取索引 i + dic.put(arr.charAt(j), j); // 更新哈希表 + tmp = tmp < j - i ? tmp + 1 : j - i; // dp[j - 1] -> dp[j] + res = Math.max(res, tmp); // max(dp[j - 1], dp[j]) + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int dismantlingAction(string arr) { + unordered_map dic; + int res = 0, tmp = 0, len = arr.size(), i; + for(int j = 0; j < len; j++) { + if (dic.find(arr[j]) == dic.end()) i = - 1; + else i = dic.find(arr[j])->second; // 获取索引 i + dic[arr[j]] = j; // 更新哈希表 + tmp = tmp < j - i ? tmp + 1 : j - i; // dp[j - 1] -> dp[j] + res = max(res, tmp); // max(dp[j - 1], dp[j]) + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为字符串长度，动态规划需遍历计算 $dp$ 列表。 +- **空间复杂度 $O(1)$ ：** 字符的 ASCII 码范围为 $0$ ~ $127$ ，哈希表 $dic$ 最多使用 $O(128) = O(1)$ 大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 168. \344\270\221\346\225\260.md" "b/leetbook_ioa/docs/LCR 168. \344\270\221\346\225\260.md" new file mode 100755 index 0000000..cd5205b --- /dev/null +++ "b/leetbook_ioa/docs/LCR 168. \344\270\221\346\225\260.md" @@ -0,0 +1,82 @@ +## 解题思路： + +根据题意，每个丑数都可以由其他较小的丑数通过乘以 $2$ 或 $3$ 或 $5$ 得到。 + +所以，可以考虑使用一个优先队列保存所有的丑数，每次取出最小的那个，然后乘以 $2$ , $3$ , $5$ 后放回队列。然而，**这样做会出现重复的丑数**。例如： + +```shell +初始化丑数列表 [1] +第一轮： 1 -> 2, 3, 5 ，丑数列表变为 [1, 2, 3, 5] +第二轮： 2 -> 4, 6, 10 ，丑数列表变为 [1, 2, 3, 4, 6, 10] +第三轮： 3 -> 6, 9, 15 ，出现重复的丑数 6 +``` + +为了避免重复，我们可以用三个指针 $a$ , $b$, $c$ ，分别表示下一个丑数是当前指针指向的丑数乘以 $2$ , $3$ , $5$ 。 + +利用三个指针生成丑数的算法流程： + +1. 初始化丑数列表 $res$ ，首个丑数为 $1$ ，三个指针 $a$ , $b$, $c$ 都指向首个丑数。 +2. 开启循环生成丑数： + 1. 计算下一个丑数的候选集 $res[a] \cdot 2$ , $res[b] \cdot 3$ , $res[c] \cdot 5$ 。 + 2. 选择丑数候选集中最小的那个作为下一个丑数，填入 $res$ 。 + 3. 将被选中的丑数对应的指针向右移动一格。 +3. 返回 $res$ 的最后一个元素即可。 + + + +## 代码： + +```Python [] +class Solution: + def nthUglyNumber(self, n: int) -> int: + res, a, b, c = [1] * n, 0, 0, 0 + for i in range(1, n): + n2, n3, n5 = res[a] * 2, res[b] * 3, res[c] * 5 + res[i] = min(n2, n3, n5) + if res[i] == n2: a += 1 + if res[i] == n3: b += 1 + if res[i] == n5: c += 1 + return res[-1] +``` + +```Java [] +class Solution { + public int nthUglyNumber(int n) { + int a = 0, b = 0, c = 0; + int[] res = new int[n]; + res[0] = 1; + for(int i = 1; i < n; i++) { + int n2 = res[a] * 2, n3 = res[b] * 3, n5 = res[c] * 5; + res[i] = Math.min(Math.min(n2, n3), n5); + if (res[i] == n2) a++; + if (res[i] == n3) b++; + if (res[i] == n5) c++; + } + return res[n - 1]; + } +} +``` + +```C++ [] +class Solution { +public: + int nthUglyNumber(int n) { + int a = 0, b = 0, c = 0; + int res[n]; + res[0] = 1; + for(int i = 1; i < n; i++) { + int n2 = res[a] * 2, n3 = res[b] * 3, n5 = res[c] * 5; + res[i] = min(min(n2, n3), n5); + if (res[i] == n2) a++; + if (res[i] == n3) b++; + if (res[i] == n5) c++; + } + return res[n - 1]; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(n)$ ：** 计算 $res$ 列表需遍历 $n-1$ 轮。 +- **空间复杂度 $O(n)$ ：** 长度为 $n$ 的 $res$ 列表使用 $O(n)$ 的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 169. \346\213\233\345\274\217\346\213\206\350\247\243 II.md" "b/leetbook_ioa/docs/LCR 169. \346\213\233\345\274\217\346\213\206\350\247\243 II.md" new file mode 100755 index 0000000..30b047a --- /dev/null +++ "b/leetbook_ioa/docs/LCR 169. \346\213\233\345\274\217\346\213\206\350\247\243 II.md" @@ -0,0 +1,146 @@ +## 解题思路： + +本题考察 **哈希表** 的使用，本文介绍 **哈希表** 和 **有序哈希表** 两种解法。其中，在字符串长度较大、重复字符很多时，“有序哈希表” 解法理论上效率更高。 + +## 方法一：哈希表 + +1. 遍历字符串 `arr` ，使用哈希表统计 “各字符数量是否 $> 1$ ”。 +2. 再遍历字符串 `arr` ，在哈希表中找到首个 “数量为 $1$ 的字符”，并返回。 + +![Picture1.png](https://pic.leetcode-cn.com/ed093aabc9195caff6d088454eaebe3cad875e8ca4a643c004ef25e4e5e9e174-Picture1.png){:align=center width=450} + +### 算法流程： + +1. **初始化：** 字典 (Python)、HashMap(Java)、map(C++)，记为 `hmap` ； +2. **字符统计：** 遍历字符串 `arr` 中的每个字符 `c` ； + 1. 若 `hmap` 中 **不包含** 键(key) `c` ：则向 `hmap` 中添加键值对 `(c, True)` ，代表字符 `c` 的数量为 $1$ ； + 2. 若 `hmap` 中 **包含** 键(key) `c` ：则修改键 `c` 的键值对为 `(c, False)` ，代表字符 `c` 的数量 $> 1$ 。 +3. **查找数量为 $1$ 的字符：** 遍历字符串 `arr` 中的每个字符 `c` ； + 1. 若 `hmap`中键 `c` 对应的值为 `True` ：，则返回 `c` 。 +4. 返回 `' '` ，代表字符串无数量为 $1$ 的字符。 + +> 下图中的 `s` 对应本题的 `arr` 。 + + + +### 代码： + +Python 代码中的 `not c in hmap` 整体为一个布尔值；`c in hmap` 为判断字典中是否含有键 `c` 。 + +```Python [] +class Solution: + def dismantlingAction(self, arr: str) -> str: + hmap = {} + for c in arr: + hmap[c] = not c in hmap + for c in arr: + if hmap[c]: return c + return ' ' +``` + +```Java [] +class Solution { + public char dismantlingAction(String arr) { + HashMap hmap = new HashMap<>(); + char[] sc = arr.toCharArray(); + for(char c : sc) + hmap.put(c, !hmap.containsKey(c)); + for(char c : sc) + if(hmap.get(c)) return c; + return ' '; + } +} +``` + +```C++ [] +class Solution { +public: + char dismantlingAction(string arr) { + unordered_map hmap; + for(char c : arr) + hmap[c] = hmap.find(c) == hmap.end(); + for(char c : arr) + if(hmap[c]) return c; + return ' '; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为字符串 `arr` 的长度；需遍历 `arr` 两轮，使用 $O(N)$ ；HashMap 查找操作的复杂度为 $O(1)$ ； +- **空间复杂度 $O(1)$ ：** 由于题目指出 `arr` 只包含小写字母，因此最多有 26 个不同字符，HashMap 存储需占用 $O(26) = O(1)$ 的额外空间。 + +## 方法二：有序哈希表 + +在哈希表的基础上，有序哈希表中的键值对是 **按照插入顺序排序** 的。基于此，可通过遍历有序哈希表，实现搜索首个 “数量为 $1$ 的字符”。 + +哈希表是 **去重** 的，即哈希表中键值对数量 $\leq$ 字符串 `arr` 的长度。因此，相比于方法一，方法二减少了第二轮遍历的循环次数。当字符串很长（重复字符很多）时，方法二则效率更高。 + +### 代码： + +Python 3.6 后，默认字典就是有序的，因此无需使用 `OrderedDict()` ，详情可见：[为什么Python 3.6以后字典有序并且效率更高？](https://www.cnblogs.com/xieqiankun/p/python_dict.html) + +Java 使用 `LinkedHashMap` 实现有序哈希表。 + +由于 C++ 未提供自带的链式哈希表，因此借助一个 vector 按序存储哈希表 hmap 中的 key ，第二轮遍历此 vector 即可。 + +```Python [] +class Solution: + def dismantlingAction(self, arr: str) -> str: + hmap = collections.OrderedDict() + for c in arr: + hmap[c] = not c in hmap + for k, v in hmap.items(): + if v: return k + return ' ' +``` + +```Python [] +class Solution: + def dismantlingAction(self, arr: str) -> str: + hmap = {} + for c in arr: + hmap[c] = not c in hmap + for k, v in hmap.items(): + if v: return k + return ' ' +``` + +```Java [] +class Solution { + public char dismantlingAction(String arr) { + Map hmap = new LinkedHashMap<>(); + char[] sc = arr.toCharArray(); + for(char c : sc) + hmap.put(c, !hmap.containsKey(c)); + for(Map.Entry d : hmap.entrySet()){ + if(d.getValue()) return d.getKey(); + } + return ' '; + } +} +``` + +```C++ [] +class Solution { +public: + char dismantlingAction(string arr) { + vector keys; + unordered_map hmap; + for(char c : arr) { + if(hmap.find(c) == hmap.end()) + keys.push_back(c); + hmap[c] = hmap.find(c) == hmap.end(); + } + for(char c : keys) { + if(hmap[c]) return c; + } + return ' '; + } +}; +``` + +### 复杂度分析： + +时间和空间复杂度均与 “方法一” 相同，而具体分析：方法一 需遍历 `arr` 两轮；方法二 遍历 `arr` 一轮，遍历 `hmap` 一轮（ `hmap` 的长度不大于 26 ）。 diff --git "a/leetbook_ioa/docs/LCR 170. \344\272\244\346\230\223\351\200\206\345\272\217\345\257\271\347\232\204\346\200\273\346\225\260.md" "b/leetbook_ioa/docs/LCR 170. \344\272\244\346\230\223\351\200\206\345\272\217\345\257\271\347\232\204\346\200\273\346\225\260.md" new file mode 100755 index 0000000..d19b1e9 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 170. \344\272\244\346\230\223\351\200\206\345\272\217\345\257\271\347\232\204\346\200\273\346\225\260.md" @@ -0,0 +1,148 @@ +## 解题思路： + +直观来看，使用暴力统计法即可，即遍历数组的所有数字对并统计逆序对数量。此方法时间复杂度为 $O(N^2)$ ，观察题目给定的数组长度范围 $0 \leq N \leq 50000$ ，可知此复杂度是不能接受的。 + +「归并排序」与「逆序对」是息息相关的。归并排序体现了 “分而治之” 的算法思想，具体为： + +- **分：** 不断将数组从中点位置划分开（即二分法），将整个数组的排序问题转化为子数组的排序问题； +- **治：** 划分到子数组长度为 1 时，开始向上合并，不断将 **较短排序数组** 合并为 **较长排序数组**，直至合并至原数组时完成排序； + +> 如下图所示，为数组 $[7, 3, 2, 6, 0, 1, 5, 4]$ 的归并排序过程。 + +![Picture1.png](https://pic.leetcode-cn.com/1614274007-nBQbZZ-Picture1.png){:align=center width=500} + +**合并阶段** 本质上是 **合并两个排序数组** 的过程，而每当遇到 左子数组当前元素 > 右子数组当前元素 时，意味着 「左子数组当前元素 至 末尾元素」 与 「右子数组当前元素」 构成了若干 「逆序对」 。 + +> 如下图所示，为左子数组 $[2, 3, 6, 7]$ 与 右子数组 $[0, 1, 4, 5]$ 的合并与逆序对统计过程。 + + + +因此，考虑在归并排序的合并阶段统计「逆序对」数量，完成归并排序时，也随之完成所有逆序对的统计。 + +### 算法流程： + +**`merge_sort()` 归并排序与逆序对统计：** + +1. **终止条件：** 当 $l \geq r$ 时，代表子数组长度为 1 ，此时终止划分； +2. **递归划分：** 计算数组中点 $m$ ，递归划分左子数组 `merge_sort(l, m)` 和右子数组 `merge_sort(m + 1, r)` ； +3. **合并与逆序对统计：** + 1. 暂存数组 $record$ 闭区间 $[l, r]$ 内的元素至辅助数组 $tmp$ ； + 2. **循环合并：** 设置双指针 $i$ , $j$ 分别指向左 / 右子数组的首元素； + - **当 $i = m + 1$ 时：** 代表左子数组已合并完，因此添加右子数组当前元素 $tmp[j]$ ，并执行 $j = j + 1$ ； + - **否则，当 $j = r + 1$ 时：** 代表右子数组已合并完，因此添加左子数组当前元素 $tmp[i]$ ，并执行 $i = i + 1$ ； + - **否则，当 $tmp[i] \leq tmp[j]$ 时：** 添加左子数组当前元素 $tmp[i]$ ，并执行 $i = i + 1$； + - **否则（即 $tmp[i] > tmp[j]$）时：** 添加右子数组当前元素 $tmp[j]$ ，并执行 $j = j + 1$ ；此时构成 $m - i + 1$ 个「逆序对」，统计添加至 $res$ ； +4. **返回值：** 返回直至目前的逆序对总数 $res$ ； + +**`reversePairs()` 主函数：** + +1. **初始化：** 辅助数组 $tmp$ ，用于合并阶段暂存元素； +2. **返回值：** 执行归并排序 `merge_sort()` ，并返回逆序对总数即可； + +> 如下图所示，为数组 $[7, 3, 2, 6, 0, 1, 5, 4]$ 的归并排序与逆序对统计过程。 + +![Picture2.png](https://pic.leetcode-cn.com/1614274007-rtFHbG-Picture2.png){:align=center width=500} + +## 代码： + +为简化代码，可将“当 $j = r + 1$ 时”与“当 $tmp[i] \leq tmp[j]$ 时”两判断项合并。 + +```Python [] +class Solution: + def reversePairs(self, record: List[int]) -> int: + def merge_sort(l, r): + # 终止条件 + if l >= r: return 0 + # 递归划分 + m = (l + r) // 2 + res = merge_sort(l, m) + merge_sort(m + 1, r) + # 合并阶段 + i, j = l, m + 1 + tmp[l:r + 1] = record[l:r + 1] + for k in range(l, r + 1): + if i == m + 1: + record[k] = tmp[j] + j += 1 + elif j == r + 1 or tmp[i] <= tmp[j]: + record[k] = tmp[i] + i += 1 + else: + record[k] = tmp[j] + j += 1 + res += m - i + 1 # 统计逆序对 + return res + + tmp = [0] * len(record) + return merge_sort(0, len(record) - 1) +``` + +```Java [] +class Solution { + int[] record, tmp; + public int reversePairs(int[] record) { + this.record = record; + tmp = new int[record.length]; + return mergeSort(0, record.length - 1); + } + private int mergeSort(int l, int r) { + // 终止条件 + if (l >= r) return 0; + // 递归划分 + int m = (l + r) / 2; + int res = mergeSort(l, m) + mergeSort(m + 1, r); + // 合并阶段 + int i = l, j = m + 1; + for (int k = l; k <= r; k++) + tmp[k] = record[k]; + for (int k = l; k <= r; k++) { + if (i == m + 1) + record[k] = tmp[j++]; + else if (j == r + 1 || tmp[i] <= tmp[j]) + record[k] = tmp[i++]; + else { + record[k] = tmp[j++]; + res += m - i + 1; // 统计逆序对 + } + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int reversePairs(vector& record) { + vector tmp(record.size()); + return mergeSort(0, record.size() - 1, record, tmp); + } +private: + int mergeSort(int l, int r, vector& record, vector& tmp) { + // 终止条件 + if (l >= r) return 0; + // 递归划分 + int m = (l + r) / 2; + int res = mergeSort(l, m, record, tmp) + mergeSort(m + 1, r, record, tmp); + // 合并阶段 + int i = l, j = m + 1; + for (int k = l; k <= r; k++) + tmp[k] = record[k]; + for (int k = l; k <= r; k++) { + if (i == m + 1) + record[k] = tmp[j++]; + else if (j == r + 1 || tmp[i] <= tmp[j]) + record[k] = tmp[i++]; + else { + record[k] = tmp[j++]; + res += m - i + 1; // 统计逆序对 + } + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N \log N)$ ：** 其中 $N$ 为数组长度；归并排序使用 $O(N \log N)$ 时间； +- **空间复杂度 $O(N)$ ：** 辅助数组 $tmp$ 占用 $O(N)$ 大小的额外空间； diff --git "a/leetbook_ioa/docs/LCR 171. \350\256\255\347\273\203\350\256\241\345\210\222 V.md" "b/leetbook_ioa/docs/LCR 171. \350\256\255\347\273\203\350\256\241\345\210\222 V.md" new file mode 100755 index 0000000..3cd80c6 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 171. \350\256\255\347\273\203\350\256\241\345\210\222 V.md" @@ -0,0 +1,81 @@ +## 解题思路： + +设第一个公共节点为 `node` ，链表 `headA` 的节点数量为 $a$ ，链表 `headB` 的节点数量为 $b$ ，两链表的公共尾部的节点数量为 $c$ ，则有： + +- 头节点 `headA` 到 `node` 前，共有 $a - c$ 个节点； +- 头节点 `headB` 到 `node` 前，共有 $b - c$ 个节点； + +![Picture1.png](https://pic.leetcode-cn.com/1615224578-EBRtwv-Picture1.png){:align=center width=500} + +考虑构建两个节点指针 `A​` , `B` 分别指向两链表头节点 `headA` , `headB` ，做如下操作： + +- 指针 `A` 先遍历完链表 `headA` ，再开始遍历链表 `headB` ，当走到 `node` 时，共走步数为： + +$$ +a + (b - c) +$$ + +- 指针 `B` 先遍历完链表 `headB` ，再开始遍历链表 `headA` ，当走到 `node` 时，共走步数为： + +$$ +b + (a - c) +$$ + +如下式所示，此时指针 `A` , `B` 重合，并有两种情况： + +$$ +a + (b - c) = b + (a - c) +$$ + +1. 若两链表 **有** 公共尾部 (即 $c > 0$ ) ：指针 `A` , `B` 同时指向「第一个公共节点」`node` 。 +2. 若两链表 **无** 公共尾部 (即 $c = 0$ ) ：指针 `A` , `B` 同时指向 $\text{null}$ 。 + +因此返回 `A` 即可。 + +> 下图展示了 $a = 5$ , $b = 3$ , $c = 2$ 示例的算法执行过程。 + + + +## 代码： + +```Python [] +class Solution: + def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode: + A, B = headA, headB + while A != B: + A = A.next if A else headB + B = B.next if B else headA + return A +``` + +```Java [] +public class Solution { + public ListNode getIntersectionNode(ListNode headA, ListNode headB) { + ListNode A = headA, B = headB; + while (A != B) { + A = A != null ? A.next : headB; + B = B != null ? B.next : headA; + } + return A; + } +} +``` + +```C++ [] +class Solution { +public: + ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { + ListNode *A = headA, *B = headB; + while (A != B) { + A = A != nullptr ? A->next : headB; + B = B != nullptr ? B->next : headA; + } + return A; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(a + b)$ ：** 最差情况下（即 $|a - b| = 1$ , $c = 0$ ），此时需遍历 $a + b$ 个节点。 +- **空间复杂度 $O(1)$ ：** 节点指针 `A` , `B` 使用常数大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 172. \347\273\237\350\256\241\347\233\256\346\240\207\346\210\220\347\273\251\347\232\204\345\207\272\347\216\260\346\254\241\346\225\260.md" "b/leetbook_ioa/docs/LCR 172. \347\273\237\350\256\241\347\233\256\346\240\207\346\210\220\347\273\251\347\232\204\345\207\272\347\216\260\346\254\241\346\225\260.md" new file mode 100755 index 0000000..0f8fddc --- /dev/null +++ "b/leetbook_ioa/docs/LCR 172. \347\273\237\350\256\241\347\233\256\346\240\207\346\210\220\347\273\251\347\232\204\345\207\272\347\216\260\346\254\241\346\225\260.md" @@ -0,0 +1,172 @@ +## 解题思路： + +> 排序数组中的搜索问题，首先想到 **二分法** 解决。 + +排序数组 $scores$ 中的所有数字 $target$ 形成一个窗口，记窗口的 **左 / 右边界** 索引分别为 $left$ 和 $right$ ，分别对应窗口左边 / 右边的首个元素。 + +本题要求统计数字 $target$ 的出现次数，可转化为：使用二分法分别找到 **左边界 $left$** 和 **右边界 $right$** ，易得数字 $target$ 的数量为 $right - left - 1$ 。 + +> 下图中的 `nums` 对应本题的 `scores` 。 + +![Picture1.png](https://pic.leetcode-cn.com/1600793982-pWqZGf-Picture1.png){:align=center width=500} + +### 算法解析： + +1. **初始化：** 左边界 $i = 0$ ，右边界 $j = len(scores) - 1$ 。 +2. **循环二分：** 当闭区间 $[i, j]$ 无元素时跳出； + 1. 计算中点 $m = (i + j) / 2$ （向下取整）； + 2. 若 $scores[m] < target$ ，则 $target$ 在闭区间 $[m + 1, j]$ 中，因此执行 $i = m + 1$； + 3. 若 $scores[m] > target$ ，则 $target$ 在闭区间 $[i, m - 1]$ 中，因此执行 $j = m - 1$； + 4. 若 $scores[m] = target$ ，则右边界 $right$ 在闭区间 $[m+1, j]$ 中；左边界 $left$ 在闭区间 $[i, m-1]$ 中。因此分为以下两种情况： + 1. 若查找 **右边界 $right$** ，则执行 $i = m + 1$ ；（跳出时 $i$ 指向右边界） + 2. 若查找 **左边界 $left$** ，则执行 $j = m - 1$ ；（跳出时 $j$ 指向左边界） +3. **返回值：** 应用两次二分，分别查找 $right$ 和 $left$ ，最终返回 $right - left - 1$ 即可。 + +### 效率优化： + +> 以下优化基于：查找完右边界 $right = i$ 后，则 $scores[j]$ 指向最右边的 $target$ （若存在）。 + +1. 查找完右边界后，可用 $scores[j] = target$ 判断数组中是否包含 $target$ ，若不包含则直接提前返回 $0$ ，无需后续查找左边界。 +2. 查找完右边界后，左边界 $left$ 一定在闭区间 $[0, j]$ 中，因此直接从此区间开始二分查找即可。 + + + +## 代码： + +可将 $scores[m] = target$ 情况合并至其他两种情况中。 + +```Python [] +class Solution: + def countTarget(self, scores: List[int], target: int) -> int: + # 搜索右边界 right + i, j = 0, len(scores) - 1 + while i <= j: + m = (i + j) // 2 + if scores[m] <= target: i = m + 1 + else: j = m - 1 + right = i + # 若数组中无 target ，则提前返回 + if j >= 0 and scores[j] != target: return 0 + # 搜索左边界 left + i = 0 + while i <= j: + m = (i + j) // 2 + if scores[m] < target: i = m + 1 + else: j = m - 1 + left = j + return right - left - 1 +``` + +```Java [] +class Solution { + public int countTarget(int[] scores, int target) { + // 搜索右边界 right + int i = 0, j = scores.length - 1; + while(i <= j) { + int m = (i + j) / 2; + if(scores[m] <= target) i = m + 1; + else j = m - 1; + } + int right = i; + // 若数组中无 target ，则提前返回 + if(j >= 0 && scores[j] != target) return 0; + // 搜索左边界 right + i = 0; j = scores.length - 1; + while(i <= j) { + int m = (i + j) / 2; + if(scores[m] < target) i = m + 1; + else j = m - 1; + } + int left = j; + return right - left - 1; + } +} +``` + +```C++ [] +class Solution { +public: + int countTarget(vector& scores, int target) { + // 搜索右边界 right + int i = 0, j = scores.size() - 1; + while(i <= j) { + int m = (i + j) / 2; + if(scores[m] <= target) i = m + 1; + else j = m - 1; + } + int right = i; + // 若数组中无 target ，则提前返回 + if(j >= 0 && scores[j] != target) return 0; + // 搜索左边界 right + i = 0; j = scores.size() - 1; + while(i <= j) { + int m = (i + j) / 2; + if(scores[m] < target) i = m + 1; + else j = m - 1; + } + int left = j; + return right - left - 1; + } +}; +``` + +以上代码显得比较臃肿（两轮二分查找代码冗余）。为简化代码，可将**二分查找右边界 $right$ 的代码** 封装至函数 `helper()` 。 +如下图所示，由于数组 $scores$ 中元素都为整数，因此可以分别二分查找 $target$ 和 $target - 1$ 的右边界，将两结果相减并返回即可。 + +![Picture2.png](https://pic.leetcode-cn.com/1600793982-ikEYZs-Picture2.png){:align=center width=450} + +本质上看，`helper()` 函数旨在查找数字 $tar$ 在数组 $scores$ 中的 **插入点** ，且若数组中存在值相同的元素，则插入到这些元素的右边。 + +```Python [] +class Solution: + def countTarget(self, scores: List[int], target: int) -> int: + def helper(tar): + i, j = 0, len(scores) - 1 + while i <= j: + m = (i + j) // 2 + if scores[m] <= tar: i = m + 1 + else: j = m - 1 + return i + return helper(target) - helper(target - 1) +``` + +```Java [] +class Solution { + public int countTarget(int[] scores, int target) { + return helper(scores, target) - helper(scores, target - 1); + } + int helper(int[] scores, int tar) { + int i = 0, j = scores.length - 1; + while(i <= j) { + int m = (i + j) / 2; + if(scores[m] <= tar) i = m + 1; + else j = m - 1; + } + return i; + } +} +``` + +```C++ [] +class Solution { +public: + int countTarget(vector& scores, int target) { + return helper(scores, target) - helper(scores, target - 1); + } +private: + int helper(vector& scores, int tar) { + int i = 0, j = scores.size() - 1; + while(i <= j) { + int m = (i + j) / 2; + if(scores[m] <= tar) i = m + 1; + else j = m - 1; + } + return i; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(\log N)$ ：** 二分法为对数级别复杂度。 +- **空间复杂度 $O(1)$ ：** 几个变量使用常数大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 173. \347\202\271\345\220\215.md" "b/leetbook_ioa/docs/LCR 173. \347\202\271\345\220\215.md" new file mode 100755 index 0000000..0bc578b --- /dev/null +++ "b/leetbook_ioa/docs/LCR 173. \347\202\271\345\220\215.md" @@ -0,0 +1,70 @@ +## 解题思路： + +排序数组中的搜索问题，首先想到 **二分法** 解决。根据题意，数组可以按照以下规则划分为两部分。 + +- **左子数组：** $records[i] = i$ ； +- **右子数组：** $records[i] \ne i$ ； + +缺失的数字等于 **“右子数组的首位元素”** 对应的索引；因此考虑使用二分法查找 “右子数组的首位元素” 。 + +> 下图中的 `nums` 对应本题的 `records` 。 + +![Picture1.png](https://pic.leetcode-cn.com/1600794300-lLZwAz-Picture1.png){:align=center width=500} + +### 算法解析： + +1. **初始化：** 左边界 $i = 0$ ，右边界 $j = len(records) - 1$ ；代表闭区间 $[i, j]$ 。 +2. **循环二分：** 当 $i \leq j$ 时循环 *（即当闭区间 $[i, j]$ 为空时跳出）* ； + 1. 计算中点 $m = (i + j) // 2$ ，其中 "$//$" 为向下取整除法； + 2. 若 $records[m] = m$ ，则 “右子数组的首位元素” 一定在闭区间 $[m + 1, j]$ 中，因此执行 $i = m + 1$； + 3. 若 $records[m] \ne m$ ，则 “左子数组的末位元素” 一定在闭区间 $[i, m - 1]$ 中，因此执行 $j = m - 1$； +3. **返回值：** 跳出时，变量 $i$ 和 $j$ 分别指向 “右子数组的首位元素” 和 “左子数组的末位元素” 。因此返回 $i$ 即可。 + + + +## 代码： + +```Python [] +class Solution: + def takeAttendance(self, records: List[int]) -> int: + i, j = 0, len(records) - 1 + while i <= j: + m = (i + j) // 2 + if records[m] == m: i = m + 1 + else: j = m - 1 + return i +``` + +```Java [] +class Solution { + public int takeAttendance(int[] records) { + int i = 0, j = records.length - 1; + while(i <= j) { + int m = (i + j) / 2; + if(records[m] == m) i = m + 1; + else j = m - 1; + } + return i; + } +} +``` + +```C++ [] +class Solution { +public: + int takeAttendance(vector& records) { + int i = 0, j = records.size() - 1; + while(i <= j) { + int m = (i + j) / 2; + if(records[m] == m) i = m + 1; + else j = m - 1; + } + return i; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(\log N)$：** 二分法为对数级别复杂度。 +- **空间复杂度 $O(1)$：** 几个变量使用常数大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 174. \345\257\273\346\211\276\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\344\270\255\347\232\204\347\233\256\346\240\207\350\212\202\347\202\271.md" "b/leetbook_ioa/docs/LCR 174. \345\257\273\346\211\276\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\344\270\255\347\232\204\347\233\256\346\240\207\350\212\202\347\202\271.md" new file mode 100755 index 0000000..7cfe64f --- /dev/null +++ "b/leetbook_ioa/docs/LCR 174. \345\257\273\346\211\276\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\344\270\255\347\232\204\347\233\256\346\240\207\350\212\202\347\202\271.md" @@ -0,0 +1,151 @@ +## 解题思路： + +本文解法基于性质：二叉搜索树的中序遍历为递增序列。根据此性质，易得二叉搜索树的 **中序遍历倒序** 为 **递减序列** 。 + +因此，我们可将求 “二叉搜索树第 $cnt$ 大的节点” 可转化为求 “此树的中序遍历倒序的第 $cnt$ 个节点”。 + +> 下图中的 `k` 对应本题的 `cnt` 。 + +![Picture1.png](https://pic.leetcode-cn.com/1600793852-IaPwtP-Picture1.png){:align=center width=450} + +**中序遍历** 为 “左、根、右” 顺序，递归代码如下： + +```Python [] +# 打印中序遍历 +def dfs(root): + if not root: return + dfs(root.left) # 左 + print(root.val) # 根 + dfs(root.right) # 右 +``` + +```Java [] +// 打印中序遍历 +void dfs(TreeNode root) { + if(root == null) return; + dfs(root.left); // 左 + System.out.println(root.val); // 根 + dfs(root.right); // 右 +} +``` + +```C++ [] +void dfs(TreeNode* root) { + if(root == nullptr) return; + dfs(root->left); + cout << root->val; + dfs(root->right); +} +``` + +**中序遍历的倒序** 为 “右、根、左” 顺序，递归法代码如下： + +```Python [] +# 打印中序遍历倒序 +def dfs(root): + if not root: return + dfs(root.right) # 右 + print(root.val) # 根 + dfs(root.left) # 左 +``` + +```Java [] +// 打印中序遍历倒序 +void dfs(TreeNode root) { + if(root == null) return; + dfs(root.right); // 右 + System.out.println(root.val); // 根 + dfs(root.left); // 左 +} +``` + +```C++ [] +void dfs(TreeNode* root) { + if(root == nullptr) return; + dfs(root->right); + cout << root->val; + dfs(root->left); +} +``` + +为求第 $cnt$ 个节点，需要实现以下三项工作： + +1. 递归遍历时计数，统计当前节点的序号； +2. 递归到第 $cnt$ 个节点时，应记录结果 $res$ ； +3. 记录结果后，后续的遍历即失去意义，应提前终止（即返回）； + +### 递归解析： + +1. **终止条件：** 当节点 $root$ 为空（越过叶节点），则直接返回； +2. **递归右子树：** 即 $dfs(root.right)$ ； +3. **递推工作：** + 1. 提前返回： 若 $cnt = 0$ ，代表已找到目标节点，无需继续遍历，因此直接返回； + 2. 统计序号： 执行 $cnt = cnt - 1$ （即从 $cnt$ 减至 $0$ ）； + 3. 记录结果： 若 $cnt = 0$ ，代表当前节点为第 $cnt$ 大的节点，因此记录 $res = root.val$ ； +4. **递归左子树：** 即 $dfs(root.left)$ ； + + + +## 代码： + +题目指出：$1 \leq cnt \leq N$ （二叉搜索树节点个数）；因此无需考虑 $cnt > N$ 的情况。 +若考虑，可以在中序遍历完成后判断 $cnt > 0$ 是否成立，若成立则说明 $cnt > N$ 。 + +```Python [] +class Solution: + def findTargetNode(self, root: TreeNode, cnt: int) -> int: + def dfs(root): + if not root: return + dfs(root.right) + if self.cnt == 0: return + self.cnt -= 1 + if self.cnt == 0: self.res = root.val + dfs(root.left) + + self.cnt = cnt + dfs(root) + return self.res +``` + +```Java [] +class Solution { + int res, cnt; + public int findTargetNode(TreeNode root, int cnt) { + this.cnt = cnt; + dfs(root); + return res; + } + void dfs(TreeNode root) { + if(root == null) return; + dfs(root.right); + if(cnt == 0) return; + if(--cnt == 0) res = root.val; + dfs(root.left); + } +} +``` + +```C++ [] +class Solution { +public: + int findTargetNode(TreeNode* root, int cnt) { + this->cnt = cnt; + dfs(root); + return res; + } +private: + int res, cnt; + void dfs(TreeNode* root) { + if(root == nullptr) return; + dfs(root->right); + if(cnt == 0) return; + if(--cnt == 0) res = root->val; + dfs(root->left); + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 当树退化为链表时（全部为右子节点），无论 $cnt$ 的值大小，递归深度都为 $N$ ，占用 $O(N)$ 时间。 +- **空间复杂度 $O(N)$ ：** 当树退化为链表时（全部为右子节点），系统使用 $O(N)$ 大小的栈空间。 diff --git "a/leetbook_ioa/docs/LCR 175. \350\256\241\347\256\227\344\272\214\345\217\211\346\240\221\347\232\204\346\267\261\345\272\246.md" "b/leetbook_ioa/docs/LCR 175. \350\256\241\347\256\227\344\272\214\345\217\211\346\240\221\347\232\204\346\267\261\345\272\246.md" new file mode 100755 index 0000000..fce8ba0 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 175. \350\256\241\347\256\227\344\272\214\345\217\211\346\240\221\347\232\204\346\267\261\345\272\246.md" @@ -0,0 +1,142 @@ +## 解题思路： + +树的遍历方式总体分为两类： + +- **深度优先搜索（DFS）：** 先序遍历、中序遍历、后序遍历； +- **广度优先搜索（BFS）：** 层序遍历； + +求树的深度需要遍历树的所有节点，本文将介绍基于 **后序遍历（DFS）** 和 **层序遍历（BFS）** 的两种解法。 + +## 方法一：后序遍历（DFS） + +树的后序遍历 / 深度优先搜索往往利用 **递归** 或 **栈** 实现，本文使用递归实现。 + +**关键点：** 此树的深度和其左（右）子树的深度之间的关系。显然，**此树的深度** 等于 **左子树的深度** 与 **右子树的深度** 中的 **最大值** $+1$ 。 + +![Picture1.png](https://pic.leetcode-cn.com/1603024336-lXVRDd-Picture1.png){:align=center width=450} + +### 算法解析： + +1. **终止条件：** 当 `root​` 为空，说明已越过叶节点，因此返回 深度 $0$ 。 +2. **递推工作：** 本质上是对树做后序遍历。 + 1. 计算节点 `root​` 的 **左子树的深度** ，即调用 `calculateDepth(root.left)`； + 2. 计算节点 `root​` 的 **右子树的深度** ，即调用 `calculateDepth(root.right)`； +3. **返回值：** 返回 **此树的深度** ，即 `max(calculateDepth(root.left), calculateDepth(root.right)) + 1`。 + + + +### 代码： + +```Python [] +class Solution: + def calculateDepth(self, root: TreeNode) -> int: + if not root: return 0 + return max(self.calculateDepth(root.left), self.calculateDepth(root.right)) + 1 +``` + +```Java [] +class Solution { + public int calculateDepth(TreeNode root) { + if(root == null) return 0; + return Math.max(calculateDepth(root.left), calculateDepth(root.right)) + 1; + } +} +``` + +```C++ [] +class Solution { +public: + int calculateDepth(TreeNode* root) { + if(root == nullptr) return 0; + return max(calculateDepth(root->left), calculateDepth(root->right)) + 1; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为树的节点数量，计算树的深度需要遍历所有节点。 +- **空间复杂度 $O(N)$ ：** 最差情况下（当树退化为链表时），递归深度可达到 $N$ 。 + +## 方法二：层序遍历（BFS） + +树的层序遍历 / 广度优先搜索往往利用 **队列** 实现。 + +**关键点：** 每遍历一层，则计数器 $+1$ ，直到遍历完成，则可得到树的深度。 + +### 算法解析： + +1. **特例处理：** 当 `root​` 为空，直接返回 深度 $0$ 。 +2. **初始化：** 队列 `queue` （加入根节点 `root` ），计数器 `res = 0`。 +3. **循环遍历：** 当 `queue` 为空时跳出。 + 1. 初始化一个空列表 `tmp` ，用于临时存储下一层节点； + 2. 遍历队列： 遍历 `queue` 中的各节点 `node` ，并将其左子节点和右子节点加入 `tmp`； + 3. 更新队列： 执行 `queue = tmp` ，将下一层节点赋值给 `queue`； + 4. 统计层数： 执行 `res += 1` ，代表层数加 $1$； +4. **返回值：** 返回 `res` 即可。 + + + +### 代码： + +```Python [] +class Solution: + def calculateDepth(self, root: TreeNode) -> int: + if not root: return 0 + queue, res = [root], 0 + while queue: + tmp = [] + for node in queue: + if node.left: tmp.append(node.left) + if node.right: tmp.append(node.right) + queue = tmp + res += 1 + return res +``` + +```Java [] +class Solution { + public int calculateDepth(TreeNode root) { + if(root == null) return 0; + List queue = new LinkedList<>() {{ add(root); }}, tmp; + int res = 0; + while(!queue.isEmpty()) { + tmp = new LinkedList<>(); + for(TreeNode node : queue) { + if(node.left != null) tmp.add(node.left); + if(node.right != null) tmp.add(node.right); + } + queue = tmp; + res++; + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int calculateDepth(TreeNode* root) { + if(root == nullptr) return 0; + vector que; + que.push_back(root); + int res = 0; + while(!que.empty()) { + vector tmp; + for(TreeNode* node : que) { + if(node->left != nullptr) tmp.push_back(node->left); + if(node->right != nullptr) tmp.push_back(node->right); + } + que = tmp; + res++; + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为树的节点数量，计算树的深度需要遍历所有节点。 +- **空间复杂度 $O(N)$ ：** 最差情况下（当树平衡时），队列 `queue` 同时存储 $N/2$ 个节点。 diff --git "a/leetbook_ioa/docs/LCR 176. \345\210\244\346\226\255\346\230\257\345\220\246\344\270\272\345\271\263\350\241\241\344\272\214\345\217\211\346\240\221.md" "b/leetbook_ioa/docs/LCR 176. \345\210\244\346\226\255\346\230\257\345\220\246\344\270\272\345\271\263\350\241\241\344\272\214\345\217\211\346\240\221.md" new file mode 100755 index 0000000..7444bf4 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 176. \345\210\244\346\226\255\346\230\257\345\220\246\344\270\272\345\271\263\350\241\241\344\272\214\345\217\211\346\240\221.md" @@ -0,0 +1,162 @@ +## 解题思路： + +以下两种方法均基于以下性质推出： **此树的深度** 等于 **左子树的深度** 与 **右子树的深度** 中的 **最大值** $+1$ 。 + +![Picture1.png](https://pic.leetcode-cn.com/1603024695-GYNvjf-Picture1.png){:align=center width=450} + +## 方法一：后序遍历 + 剪枝 （从底至顶） + +> 此方法为本题的最优解法，但剪枝的方法不易第一时间想到。 + +思路是对二叉树做后序遍历，从底至顶返回子树深度，若判定某子树不是平衡树则 “剪枝” ，直接向上返回。 + +### 算法流程： + +**`recur(root)` 函数：** + +- **返回值：** + 1. 当节点`root` 左 / 右子树的深度差 $\leq 1$ ：则返回当前子树的深度，即节点 `root` 的左 / 右子树的深度最大值 $+1$ （ `max(left, right) + 1` ）； + 2. 当节点`root` 左 / 右子树的深度差 $> 1$ ：则返回 $-1$ ，代表 **此子树不是平衡树** 。 +- **终止条件：** + 1. 当 `root` 为空：说明越过叶节点，因此返回高度 $0$ ； + 2. 当左（右）子树深度为 $-1$ ：代表此树的 **左（右）子树** 不是平衡树，因此剪枝，直接返回 $-1$ ； + +**`isBalanced(root)` 函数：** + +- **返回值：** 若 `recur(root) != -1` ，则说明此树平衡，返回 $\text{true}$ ； 否则返回 $\text{false}$ 。 + + + +### 代码： + +```Python [] +class Solution: + def isBalanced(self, root: Optional[TreeNode]) -> bool: + def recur(root): + if not root: return 0 + left = recur(root.left) + if left == -1: return -1 + right = recur(root.right) + if right == -1: return -1 + return max(left, right) + 1 if abs(left - right) <= 1 else -1 + + return recur(root) != -1 +``` + +```Java [] +class Solution { + public boolean isBalanced(TreeNode root) { + return recur(root) != -1; + } + + private int recur(TreeNode root) { + if (root == null) return 0; + int left = recur(root.left); + if(left == -1) return -1; + int right = recur(root.right); + if(right == -1) return -1; + return Math.abs(left - right) < 2 ? Math.max(left, right) + 1 : -1; + } +} +``` + +```C++ [] +class Solution { +public: + bool isBalanced(TreeNode* root) { + return recur(root) != -1; + } +private: + int recur(TreeNode* root) { + if (root == nullptr) return 0; + int left = recur(root->left); + if(left == -1) return -1; + int right = recur(root->right); + if(right == -1) return -1; + return abs(left - right) < 2 ? max(left, right) + 1 : -1; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$：** $N$ 为树的节点数；最差情况下，需要递归遍历树的所有节点。 +- **空间复杂度 $O(N)$：** 最差情况下（树退化为链表时），系统递归需要使用 $O(N)$ 的栈空间。 + +## 方法二：先序遍历 + 判断深度 （从顶至底） + +> 此方法容易想到，但会产生大量重复计算，时间复杂度较高。 + +思路是构造一个获取当前子树的深度的函数 `depth(root)` （即 [面试题55 - I. 二叉树的深度](https://leetcode-cn.com/problems/er-cha-shu-de-shen-du-lcof/solution/mian-shi-ti-55-i-er-cha-shu-de-shen-du-xian-xu-bia/) ），通过比较某子树的左右子树的深度差 `abs(depth(root.left) - depth(root.right)) <= 1` 是否成立，来判断某子树是否是二叉平衡树。若所有子树都平衡，则此树平衡。 + +### 算法流程： + +**`isBalanced(root)` 函数：** 判断树 `root` 是否平衡 + +- **特例处理：** 若树根节点 `root` 为空，则直接返回 $\text{true}$ ； +- **返回值：** 所有子树都需要满足平衡树性质，因此以下三者使用与逻辑 $\&\&$ 连接； + 1. `abs(self.depth(root.left) - self.depth(root.right)) <= 1` ：判断 **当前子树** 是否是平衡树； + 2. `self.isBalanced(root.left)` ： 先序遍历递归，判断 **当前子树的左子树** 是否是平衡树； + 3. `self.isBalanced(root.right)` ： 先序遍历递归，判断 **当前子树的右子树** 是否是平衡树； + +**`depth(root)` 函数：** 计算树 `root` 的深度 + +- **终止条件：** 当 `root` 为空，即越过叶子节点，则返回高度 $0$ ； +- **返回值：** 返回左 / 右子树的深度的最大值 $+1$ 。 + + + +### 代码： + +```Python [] +class Solution: + def isBalanced(self, root: Optional[TreeNode]) -> bool: + if not root: return True + return abs(self.depth(root.left) - self.depth(root.right)) <= 1 and \ + self.isBalanced(root.left) and self.isBalanced(root.right) + + def depth(self, root): + if not root: return 0 + return max(self.depth(root.left), self.depth(root.right)) + 1 +``` + +```Java [] +class Solution { + public boolean isBalanced(TreeNode root) { + if (root == null) return true; + return Math.abs(depth(root.left) - depth(root.right)) <= 1 && isBalanced(root.left) && isBalanced(root.right); + } + + private int depth(TreeNode root) { + if (root == null) return 0; + return Math.max(depth(root.left), depth(root.right)) + 1; + } +} +``` + +```C++ [] +class Solution { +public: + bool isBalanced(TreeNode* root) { + if (root == nullptr) return true; + return abs(depth(root->left) - depth(root->right)) <= 1 && isBalanced(root->left) && isBalanced(root->right); + } +private: + int depth(TreeNode* root) { + if (root == nullptr) return 0; + return max(depth(root->left), depth(root->right)) + 1; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N \log N)$：** 最差情况下（为 “满二叉树” 时），`isBalanced(root)` 遍历树所有节点，判断每个节点的深度 `depth(root)` 需要遍历 **各子树的所有节点** 。 + - 满二叉树高度的复杂度 $O(log N)$ ，将满二叉树按层分为 $log (N+1)$ 层； + - 通过调用 `depth(root)` ，判断二叉树各层的节点的对应子树的深度，需遍历节点数量为 $N \times 1, \frac{N-1}{2} \times 2, \frac{N-3}{4} \times 4, \frac{N-7}{8} \times 8, ..., 1 \times \frac{N+1}{2}$ 。因此各层执行 `depth(root)` 的时间复杂度为 $O(N)$ （每层开始，最多遍历 $N$ 个节点，最少遍历 $\frac{N+1}{2}$ 个节点）。 + > 其中，$\frac{N-3}{4} \times 4$ 代表从此层开始总共需遍历 $N-3$ 个节点，此层共有 $4$ 个节点，因此每个子树需遍历 $\frac{N-3}{4}$ 个节点。 + - 因此，总体时间复杂度 $=$ 每层执行复杂度 $\times$ 层数复杂度 = $O(N \times \log N)$ 。 + +![Picture2.png](https://pic.leetcode-cn.com/1603024695-yyFsRH-Picture2.png){:align=center width=550} + +- **空间复杂度 $O(N)$：** 最差情况下（树退化为链表时），系统递归需要使用 $O(N)$ 的栈空间。 diff --git "a/leetbook_ioa/docs/LCR 177. \346\222\236\350\211\262\346\220\255\351\205\215.md" "b/leetbook_ioa/docs/LCR 177. \346\222\236\350\211\262\346\220\255\351\205\215.md" new file mode 100755 index 0000000..4dd9132 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 177. \346\222\236\350\211\262\346\220\255\351\205\215.md" @@ -0,0 +1,190 @@ +## 解题思路： + +题目要求时间复杂度 $O(N)$ ，空间复杂度 $O(1)$ ，因此首先排除 **暴力法** 和 **哈希表统计法** 。 + +> **简化问题：** 一个整型数组 `sockets` 里除 **一个** 数字之外，其他数字都出现了两次。 + +设整型数组 $sockets$ 中出现一次的数字为 $x$ ，出现两次的数字为 $a, a, b, b, ...$ ，即： + +$$ +sockets = [a, a, b, b, ..., x] +$$ + +异或运算有个重要的性质，两个相同数字异或为 $0$ ，即对于任意整数 $a$ 有 $a \oplus a = 0$ 。因此，若将 $sockets$ 中所有数字执行异或运算，留下的结果则为 **出现一次的数字 $x$** ，即： + +$$ +\begin{aligned} +& \ \ a \oplus a \oplus b \oplus b \oplus ... \oplus x \\ += & \ \ 0 \oplus 0 \oplus ... \oplus x \\ += & \ \ x +\end{aligned} +$$ + +异或运算满足交换律 $a \oplus b = b \oplus a$ ，即以上运算结果与 $sockets$ 的元素顺序无关。代码如下： + +```Python [] +def singleNumber(self, sockets: List[int]) -> List[int]: + x = 0 + for num in sockets: # 1. 遍历 sockets 执行异或运算 + x ^= num + return x; # 2. 返回出现一次的数字 x +``` + +```Java [] +public int[] singleNumber(int[] sockets) { + int x = 0; + for(int num : sockets) // 1. 遍历 sockets 执行异或运算 + x ^= num; + return x; // 2. 返回出现一次的数字 x +} +``` + +```C++ [] +vector singleNumber(vector& sockets) { + int x = 0; + for(int num : sockets) // 1. 遍历 sockets 执行异或运算 + x ^= num; + return x; // 2. 返回出现一次的数字 x +} +``` + +> 下图中的 `nums` 对应本题的 `sockets` 。 + +![Picture1.png](https://pic.leetcode-cn.com/1611393960-EnUIaQ-Picture1.png){:align=center width=500} + +> **本题难点：** 数组 $sockets$ 有 **两个** 只出现一次的数字，因此无法通过异或直接得到这两个数字。 + +设两个只出现一次的数字为 $x$ , $y$ ，由于 $x \ne y$ ，则 $x$ 和 $y$ 二进制至少有一位不同（即分别为 $0$ 和 $1$ ），根据此位可以将 $sockets$ 拆分为分别包含 $x$ 和 $y$ 的两个子数组。 + +易知两子数组都满足 「除一个数字之外，其他数字都出现了两次」。因此，仿照以上简化问题的思路，分别对两子数组遍历执行异或操作，即可得到两个只出现一次的数字 $x$, $y$ 。 + +### 算法流程： + +1. **遍历 $sockets$ 执行异或：** + +- 设整型数组 $sockets = [a, a, b, b, ..., x, y]$ ，对 $sockets$ 中所有数字执行异或，得到的结果为 $x \oplus y$ ，即： + +$$ +\begin{aligned} +& \ \ a \oplus a \oplus b \oplus b \oplus ... \oplus x \oplus y \\ += & \ \ 0 \oplus 0 \oplus ... \oplus x \oplus y \\ += & \ \ x \oplus y +\end{aligned} +$$ + +2. **循环左移计算 $m$ ：** + +- 根据异或运算定义，若整数 $x \oplus y$ 某二进制位为 $1$ ，则 $x$ 和 $y$ 的此二进制位一定不同。换言之，找到 $x \oplus y$ 某为 $1$ 的二进制位，即可将数组 $sockets$ 拆分为上述的两个子数组。根据与运算特点，可知对于任意整数 $a$ 有： + + - 若 $a \& 0001 \ne 0$ ，则 $a$ 的第一位为 $1$ ； + - 若 $a \& 0010 \ne 0$ ，则 $a$ 的第二位为 $1$ ； + - 以此类推…… + +- 因此，初始化一个辅助变量 $m = 1$ ，通过与运算从右向左循环判断，可 **获取整数 $x \oplus y$ 首位 $1$** ，记录于 $m$ 中，代码如下： + +```Python [] +while z & m == 0: # m 循环左移一位，直到 z & m ！= 0 + m <<= 1 +``` + +```Java [] +while(z & m == 0) // m 循环左移一位，直到 z & m ！= 0 + m <<= 1 +``` + +```C++ [] +while(z & m == 0) // m 循环左移一位，直到 z & m ！= 0 + m <<= 1 +``` + +3. **拆分 $sockets$ 为两个子数组：** +4. **分别遍历两个子数组执行异或：** + +- 通过遍历判断 $sockets$ 中各数字和 $m$ 做与运算的结果，可将数组拆分为两个子数组，并分别对两个子数组遍历求异或，则可得到两个只出现一次的数字，代码如下： + +```Python [] +for num in sockets: + if num & m: x ^= num # 若 num & m != 0 , 划分至子数组 1 ，执行遍历异或 + else: y ^= num # 若 num & m == 0 , 划分至子数组 2 ，执行遍历异或 +return x, y # 遍历异或完毕，返回只出现一次的数字 x 和 y +``` + +```Java [] +for(int num: sockets) { + if((num & m) != 0) x ^= num; // 若 num & m != 0 , 划分至子数组 1 ，执行遍历异或 + else y ^= num; // 若 num & m == 0 , 划分至子数组 2 ，执行遍历异或 +} +return new int[] {x, y}; // 遍历异或完毕，返回只出现一次的数字 x 和 y +``` + +```C++ [] +for(int num : sockets) { + if(num & m) x ^= num; // 若 num & m != 0 , 划分至子数组 1 ，执行遍历异或 + else y ^= num; // 若 num & m == 0 , 划分至子数组 2 ，执行遍历异或 +} +return vector {x, y}; // 遍历异或完毕，返回只出现一次的数字 x 和 y +``` + +5. **返回值**： + +- 返回只出现一次的数字 x, y 即可。 + +> 下图中的 `nums` 对应本题的 `sockets` 。 + +![Picture2.png](https://pic.leetcode-cn.com/1614836837-oygHyk-Picture2.png) + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 线性遍历 $sockets$ 使用 $O(N)$ 时间，遍历 $x \oplus y$ 二进制位使用 $O(32) = O(1)$ 时间。 +- **空间复杂度 $O(1)$ ：** 辅助变量 $a$ , $b$ , $x$ , $y$ 使用常数大小额外空间。 + +## 代码： + +```Python [] +class Solution: + def sockCollocation(self, sockets: List[int]) -> List[int]: + x, y, n, m = 0, 0, 0, 1 + for num in sockets: # 1. 遍历异或 + n ^= num + while n & m == 0: # 2. 循环左移，计算 m + m <<= 1 + for num in sockets: # 3. 遍历 sockets 分组 + if num & m: x ^= num # 4. 当 num & m != 0 + else: y ^= num # 4. 当 num & m == 0 + return x, y # 5. 返回出现一次的数字 +``` + +```Java [] +class Solution { + public int[] sockCollocation(int[] sockets) { + int x = 0, y = 0, n = 0, m = 1; + for(int num : sockets) // 1. 遍历异或 + n ^= num; + while((n & m) == 0) // 2. 循环左移，计算 m + m <<= 1; + for(int num: sockets) { // 3. 遍历 sockets 分组 + if((num & m) != 0) x ^= num; // 4. 当 num & m != 0 + else y ^= num; // 4. 当 num & m == 0 + } + return new int[] {x, y}; // 5. 返回出现一次的数字 + } +} +``` + +```C++ [] +class Solution { +public: + vector sockCollocation(vector& sockets) { + int x = 0, y = 0, n = 0, m = 1; + for(int num : sockets) // 1. 遍历异或 + n ^= num; + while((n & m) == 0) // 2. 循环左移，计算 m + m <<= 1; + for(int num : sockets) { // 3. 遍历 sockets 分组 + if(num & m) x ^= num; // 4. 当 num & m != 0 + else y ^= num; // 4. 当 num & m == 0 + } + return vector {x, y}; // 5. 返回出现一次的数字 + } +}; +``` diff --git "a/leetbook_ioa/docs/LCR 178. \350\256\255\347\273\203\350\256\241\345\210\222 VI.md" "b/leetbook_ioa/docs/LCR 178. \350\256\255\347\273\203\350\256\241\345\210\222 VI.md" new file mode 100755 index 0000000..e08bb6b --- /dev/null +++ "b/leetbook_ioa/docs/LCR 178. \350\256\255\347\273\203\350\256\241\345\210\222 VI.md" @@ -0,0 +1,287 @@ +## 解题思路： + +如下图所示，考虑数字的二进制形式，对于出现三次的数字，各 **二进制位** 出现的次数都是 $3$ 的倍数。 +因此，统计所有数字的各二进制位中 $1$ 的出现次数，并对 $3$ 求余，结果则为只出现一次的数字。 + +> 下图中的 `nums` 对应本题的 `actions` 。 + +![Picture1.png](https://pic.leetcode-cn.com/1603022900-quEtJr-Picture1.png){:align=center width=450} + +## 方法一：有限状态自动机 + +各二进制位的 **位运算规则相同** ，因此只需考虑一位即可。如下图所示，对于所有数字中的某二进制位 $1$ 的个数，存在 3 种状态，即对 3 余数为 $0, 1, 2$ 。 + +- 若输入二进制位 $1$ ，则状态按照以下顺序转换； +- 若输入二进制位 $0$ ，则状态不变。 + +$$ +0 \rightarrow 1 \rightarrow 2 \rightarrow 0 \rightarrow \cdots +$$ + +![Picture2.png](https://pic.leetcode-cn.com/1603022900-GNKGMP-Picture2.png){:align=center width=450} + +如下图所示，由于二进制只能表示 $0, 1$ ，因此需要使用两个二进制位来表示 $3$ 个状态。设此两位分别为 $two$ , $one$ ，则状态转换变为： + +$$ +00 \rightarrow 01 \rightarrow 10 \rightarrow 00 \rightarrow \cdots +$$ + +![Picture3.png](https://pic.leetcode-cn.com/1603022900-nPXQLk-Picture3.png){:align=center width=450} + +接下来，需要通过 **状态转换表** 导出 **状态转换的计算公式** 。首先回忆一下位运算特点，对于任意二进制位 $x$ ，有： + +- 异或运算：`x ^ 0 = x`​ ，`x ^ 1 = ~x` +- 与运算：`x & 0 = 0` ，`x & 1 = x` + +**计算 $one$ 方法：** + +设当前状态为 $two$ $one$ ，此时输入二进制位 $n$ 。如下图所示，通过对状态表的情况拆分，可推出 $one$ 的计算方法为： + +```Python +if two == 0: + if n == 0: + one = one + if n == 1: + one = ~one +if two == 1: + one = 0 +``` + +引入 **异或运算** ，可将以上拆分简化为： + +```Python +if two == 0: + one = one ^ n +if two == 1: + one = 0 +``` + +引入 **与运算** ，可继续简化为： + +```Python +one = one ^ n & ~two +``` + +![Picture4.png](https://pic.leetcode-cn.com/1603022900-qIFpAR-Picture4.png){:align=center width=550} + +**计算 $two$ 方法：** + +由于是先计算 $one$ ，因此应在新 $one$ 的基础上计算 $two$ 。 +如下图所示，修改为新 $one$ 后，得到了新的状态图。观察发现，可以使用同样的方法计算 $two$ ，即： + +```Python +two = two ^ n & ~one +``` + +![Picture5.png](https://pic.leetcode-cn.com/1603022900-hnUxBz-Picture5.png){:align=center width=450} + +**返回值：** + +以上是对数字的二进制中 “一位” 的分析，而 `int` 类型的其他 31 位具有相同的运算规则，因此可将以上公式直接套用在 32 位数上。 + +遍历完所有数字后，各二进制位都处于状态 $00$ 和状态 $01$ （取决于 “只出现一次的数字” 的各二进制位是 $1$ 还是 $0$ ），而此两状态是由 $one$ 来记录的（此两状态下 $twos$ 恒为 $0$ ），因此返回 $ones$ 即可。 + + + +### 代码： + +```Python [] +class Solution: + def trainingPlan(self, actions: List[int]) -> int: + ones, twos = 0, 0 + for action in actions: + ones = ones ^ action & ~twos + twos = twos ^ action & ~ones + return ones +``` + +```Java [] +class Solution { + public int trainingPlan(int[] actions) { + int ones = 0, twos = 0; + for(int action : actions){ + ones = ones ^ action & ~twos; + twos = twos ^ action & ~ones; + } + return ones; + } +} +``` + +```C++ [] +class Solution { +public: + int trainingPlan(vector& actions) { + int ones = 0, twos = 0; + for(int action : actions){ + ones = ones ^ action & ~twos; + twos = twos ^ action & ~ones; + } + return ones; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 位数组 $actions$ 的长度；遍历数组占用 $O(N)$ ，每轮中的常数个位运算操作占用 $O(32 \times3 \times 2) = O(1)$ 。 +- **空间复杂度 $O(1)$ ：** 变量 $ones$ , $twos$ 使用常数大小的额外空间。 + +## 方法二：遍历统计 + +> 此方法相对容易理解，但效率较低，总体推荐方法一。 + +使用 **与运算** ，可获取二进制数字 $action$ 的最右一位 $n_1$ ： + +$$ +n_1 = action \& i +$$ + +配合 **右移操作** ，可从低位至高位，获取 $action$ 所有位的值（设 int 类型从低至高的位数为第 0 位 至第 31 位，即 $n_0$ ~ $n_{31}$ ）： + +$$ +action = action >> 1 +$$ + +建立一个长度为 32 的数组 $counts$ ，通过以上方法可记录所有数字的各二进制位的 $1$ 的出现次数之和。 + +```Python [] +counts = [0] * 32 +for action in actions: + for i in range(32): + counts[i] += action & 1 # 更新第 i 位 1 的个数之和 + action >>= 1 # 第 i 位 --> 第 i + 1 位 +``` + +```Java [] +int[] counts = new int[32]; +for(int action : actions) { + for(int i = 0; i < 32; i++) { + counts[i] += action & 1; // 更新第 i 位 1 的个数之和 + action >>= 1; // 第 i 位 --> 第 i + 1 位 + } +} +``` + +```C++ [] +int counts[32] = {0}; // C++ 初始化数组需要写明初始值 0 +for(int action : actions) { + for(int i = 0; i < 32; i++) { + counts[i] += action & 1; // 更新第 i 位 1 的个数之和 + action >>= 1; // 第 i 位 --> 第 i + 1 位 + } +} +``` + +将 $counts$ 各元素对 $3$ 求余，则结果为 “只出现一次的数字” 的各二进制位。 + +```Python [] +for i in range(31, -1, -1): + x = counts[i] %= 3 # 得到 “只出现一次的数字” 的第 i 位 +``` + +```Java [] +for(int i = 31; i >= 0; i--) { + int x = counts[i] %= 3; // 得到 “只出现一次的数字” 的第 i 位 +} +``` + +```C++ [] +for(int i = 31; i >= 0; i--) { + int x = counts[i] % 3; // 得到 “只出现一次的数字” 的第 i 位 +} +``` + +利用 **左移操作** 和 **或运算** ，可将 $counts$ 数组中各二进位的值恢复到数字 $res$ 上。 + +```Python [] +for i in range(31, -1, -1): + res <<= 1 + res |= counts[i] % 3 # 恢复第 i 位 +``` + +```Java [] +for(int i = 31; i >= 0; i--) { + res <<= 1; + res |= counts[i] % 3; // 恢复第 i 位 +} +``` + +```C++ [] +for(int i = 31; i >= 0; i--) { + res <<= 1; + res |= counts[i] % 3; // 恢复第 i 位 +} +``` + +最终返回 $res$ 即可。 + +> 由于 Python 的存储负数的特殊性，需要先将 $0$ - $31$ 位取反（即 `res ^ 0xffffffff` ），再将所有位取反（即 `~` ）。 +> **此组合操作含义：** 将数字 $31$ 以上位取反，$0$ - $31$ 位不变。 + +### 代码： + +实际上，只需要修改求余数值 $m$ ，即可实现解决 **除了一个数字以外，其余数字都出现 $m$ 次** 的通用问题。 + +> 设 int 类型从低至高的位数为第 0 位 至第 31 位。 + +```Python [] +class Solution: + def trainingPlan(self, actions: List[int]) -> int: + counts = [0] * 32 + for action in actions: + for i in range(32): + counts[i] += action & 1 # 更新第 i 位 1 的个数之和 + action >>= 1 # 第 i 位 --> 第 i 位 + res, m = 0, 3 + for i in range(31, -1, -1): + res <<= 1 + res |= counts[i] % m # 恢复第 i 位 + return res if counts[31] % m == 0 else ~(res ^ 0xffffffff) +``` + +```Java [] +class Solution { + public int trainingPlan(int[] actions) { + int[] counts = new int[32]; + for(int action : actions) { + for(int i = 0; i < 32; i++) { + counts[i] += action & 1; // 更新第 i 位 1 的个数之和 + action >>= 1; // 第 i 位 --> 第 i 位 + } + } + int res = 0, m = 3; + for(int i = 31; i >= 0; i--) { + res <<= 1; + res |= counts[i] % m; // 恢复第 i 位 + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int trainingPlan(vector& actions) { + int counts[32] = {0}; // C++ 初始化数组需要写明初始值 0 + for(int action : actions) { + for(int i = 0; i < 32; i++) { + counts[i] += action & 1; // 更新第 i 位 1 的个数之和 + action >>= 1; // 第 i 位 --> 第 i 位 + } + } + int res = 0, m = 3; + for(int i = 31; i >= 0; i--) { + res <<= 1; + res |= counts[i] % m; // 恢复第 i 位 + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 位数组 $actions$ 的长度；遍历数组占用 $O(N)$ ，每轮中的常数个位运算操作占用 $O(1)$ 。 +- **空间复杂度 $O(1)$ ：** 数组 $counts$ 长度恒为 $32$ ，占用常数大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 179. \346\237\245\346\211\276\346\200\273\344\273\267\346\240\274\344\270\272\347\233\256\346\240\207\345\200\274\347\232\204\344\270\244\344\270\252\345\225\206\345\223\201.md" "b/leetbook_ioa/docs/LCR 179. \346\237\245\346\211\276\346\200\273\344\273\267\346\240\274\344\270\272\347\233\256\346\240\207\345\200\274\347\232\204\344\270\244\344\270\252\345\225\206\345\223\201.md" new file mode 100755 index 0000000..78448a2 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 179. \346\237\245\346\211\276\346\200\273\344\273\267\346\240\274\344\270\272\347\233\256\346\240\207\345\200\274\347\232\204\344\270\244\344\270\252\345\225\206\345\223\201.md" @@ -0,0 +1,80 @@ +## 解题思路： + +利用 HashMap 可以通过遍历数组找到数字组合，时间和空间复杂度均为 $O(N)$ 。 +注意本题的 $price$ 是 **排序数组** ，因此可使用 **双指针法** 将空间优化至 $O(1)$ 。 + +### 算法流程： + +1. **初始化：** 双指针 $i$ , $j$ 分别指向数组 $price$ 的左右两端。 +2. **循环搜索：** 当双指针相遇时跳出； + 1. 计算和 $s = price[i] + price[j]$ ； + 2. 若 $s > target$ ，则指针 $j$ 向左移动，即执行 $j = j - 1$ ； + 3. 若 $s < target$ ，则指针 $i$ 向右移动，即执行 $i = i + 1$ ； + 4. 若 $s = target$ ，立即返回数组 $[price[i], price[j]]$ ； +3. 若循环结束，则返回空数组，代表无和为 $target$ 的数字组合。 + +> 下图中的 `nums` 对应本题的 `price` 。 + + + +### 正确性证明： + +> 记每个状态为 $S(i, j)$ ，即 $S(i, j) = price[i] + price[j]$ 。假设 $S(i, j) < target$ ，则执行 $i = i + 1$ ，即状态切换至 $S(i + 1, j)$ 。 + +状态 $S(i, j)$ 切换至 $S(i + 1, j)$ ，则会消去一行元素，相当于 **消去了状态集合** {$S(i, i + 1), S(i, i + 2), ..., S(i, j - 2), S(i, j - 1), S(i, j)$ } 。（由于双指针都是向中间收缩，因此这些状态之后不可能再遇到）。 + +由于 $price$ 是排序数组，因此这些 **消去的状态** 都一定满足 $S(i, j) < target$ ，即这些状态都 **不是解** 。 + +**结论：** 以上分析已证明 “每次指针 $i$ 的移动操作，都不会导致解的丢失” ，即指针 $i$ 的移动操作是安全的；同理，对于指针 $j$ 可得出同样推论；因此，此双指针法是正确的。 + +![Picture1.png](https://pic.leetcode-cn.com/1600794717-VSmNyQ-Picture1.png){:align=center width=550} + +## 代码： + +```Python [] +class Solution: + def twoSum(self, price: List[int], target: int) -> List[int]: + i, j = 0, len(price) - 1 + while i < j: + s = price[i] + price[j] + if s > target: j -= 1 + elif s < target: i += 1 + else: return price[i], price[j] + return [] +``` + +```Java [] +class Solution { + public int[] twoSum(int[] price, int target) { + int i = 0, j = price.length - 1; + while(i < j) { + int s = price[i] + price[j]; + if(s < target) i++; + else if(s > target) j--; + else return new int[] { price[i], price[j] }; + } + return new int[0]; + } +} +``` + +```C++ [] +class Solution { +public: + vector twoSum(vector& price, int target) { + int i = 0, j = price.size() - 1; + while(i < j) { + int s = price[i] + price[j]; + if(s < target) i++; + else if(s > target) j--; + else return { price[i], price[j] }; + } + return {}; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为数组 $price$ 的长度；双指针共同线性遍历整个数组。 +- **空间复杂度 $O(1)$ ：** 变量 $i$, $j$ 使用常数大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 180. \346\226\207\344\273\266\347\273\204\345\220\210.md" "b/leetbook_ioa/docs/LCR 180. \346\226\207\344\273\266\347\273\204\345\220\210.md" new file mode 100755 index 0000000..117186a --- /dev/null +++ "b/leetbook_ioa/docs/LCR 180. \346\226\207\344\273\266\347\273\204\345\220\210.md" @@ -0,0 +1,205 @@ +## 方法一：求和公式 + +设连续正整数序列的左边界 $i$ 和右边界 $j$ ，则此序列的 **元素和** $target$ 等于 **元素平均值 $\frac{i + j}{2}$** 乘以 **元素数量 $(j - i + 1)$** ，即： + +$$ +target = \frac{(i + j) \times (j - i + 1)}{2} +$$ + +观察发现，当确定 元素和 $target$ 与 左边界 $i$ 时，可通过 **解一元二次方程** ，直接计算出右边界 $j$ ，公式推导如下： + +$$ +\begin{aligned} +target & = \frac{(i + j) \times (j - i + 1)}{2} \\ +& = \frac{j^2 + j - i^2 + i}{2} \\ +\end{aligned} +$$ + +整理上式得： + +$$ +0 = j^2 + j - (2 \times target + i^2 - i) +$$ + +根据一元二次方程求根公式得： + +$$ +j = \frac{-1 \pm \sqrt{1 + 4(2 \times target + i^2 - i)}}{2} +$$ + +由于 $j > i$ 恒成立，因此直接 **舍去必为负数的解** ，即 $j$ 的唯一解求取公式为： + +$$ +\begin{aligned} +j & = \frac{-1 + \sqrt{1 + 4(2 \times target + i^2 - i)}}{2} +\end{aligned} +$$ + +因此，通过从小到大遍历左边界 $i$ 来计算 **以 $i$ 为起始数字的连续正整数序列** 。每轮中，由以上公式计算得到右边界 $j$ ，当 $j$ 满足以下两个条件时记录结果： + +1. $j$ 为 **整数** ：符合题目所求「连续正整数序列」； +2. $i < j$ ：满足题目要求「至少含有两个数」； + +> 当 $target = 9$ 时，以上求解流程如下图所示。 + +![Picture1.png](https://pic.leetcode-cn.com/1611494538-VUzxtS-Picture1.png){:align=center width=550xl} + +### 代码： + +计算公式中 $i^2$ 项可能超过 int 类型取值范围，因此在 Java, C++ 中需要转化成 long 类型。 + +```Python [] +class Solution: + def fileCombination(self, target: int): + i, j, res = 1, 2, [] + while i < j: + j = (-1 + (1 + 4 * (2 * target + i * i - i)) ** 0.5) / 2 + if i < j and j == int(j): + res.append(list(range(i, int(j) + 1))) + i += 1 + return res +``` + +```Java [] +class Solution { + public int[][] fileCombination(int target) { + int i = 1; + double j = 2.0; + List res = new ArrayList<>(); + while(i < j) { + j = (-1 + Math.sqrt(1 + 4 * (2 * target + (long) i * i - i))) / 2; + if(i < j && j == (int)j) { + int[] ans = new int[(int)j - i + 1]; + for(int k = i; k <= (int)j; k++) + ans[k - i] = k; + res.add(ans); + } + i++; + } + return res.toArray(new int[0][]); + } +} +``` + +```C++ [] +class Solution { +public: + vector> fileCombination(int target) { + int i = 1; + double j = 2.0; + vector> res; + while(i < j) { + j = (-1 + sqrt(1 + 4 * (2 * target + (long) i * i - i))) / 2; + if(i < j && j == (int)j) { + vector ans; + for(int k = i; k <= (int)j; k++) + ans.push_back(k); + res.push_back(ans); + } + i++; + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N = target$ ；连续整数序列至少有两个数字，而 $i < j$ 恒成立，因此至多循环 $\frac{target}{2}$ 次，使用 $O(N)$ 时间；循环内，计算 $j$ 使用 $O(1)$ 时间；当 $i = 1$ 时，达到最大序列长度 $\frac{-1 + \sqrt{1 + 8s}}{2}$ ，考虑到解的稀疏性，将列表构建时间简化考虑为 $O(1)$ ； +- **空间复杂度 $O(1)$ ：** 变量 $i$ , $j$ 使用常数大小的额外空间。 + +## 方法二：滑动窗口 + +设连续正整数序列的左边界 $i$ 和右边界 $j$ ，则可构建滑动窗口从左向右滑动。循环中，每轮判断滑动窗口内元素和与目标值 $target$ 的大小关系，若相等则记录结果，若大于 $target$ 则移动左边界 $i$ （以减小窗口内的元素和），若小于 $target$ 则移动右边界 $j$ （以增大窗口内的元素和）。 + +### 算法流程： + +1. **初始化：** 左边界 $i = 1$ ，右边界 $j = 2$ ，元素和 $s = 3$ ，结果列表 $res$ ； + +2. **循环：** 当 $i \geq j$ 时跳出； + + - 当 $s > target$ 时： 向右移动左边界 $i = i + 1$ ，并更新元素和 $s$ ； + - 当 $s < target$ 时： 向右移动右边界 $j = j + 1$ ，并更新元素和 $s$ ； + - 当 $s = target$ 时： 记录连续整数序列，并向右移动左边界 $i = i + 1$ ； + +3. **返回值：** 返回结果列表 $res$ ； + +> 当 $target = 9$ 时，以上求解流程如下图所示： + +![Picture2.png](https://pic.leetcode-cn.com/1611495306-LsrxgS-Picture2.png){:align=center width=600} + +### 代码： + +观察本文的算法流程发现，当 $s = target$ 和 $s > target$ 的移动边界操作相同，因此可以合并，代码如下所示。 + +```Python [] +class Solution: + def fileCombination(self, target: int) -> List[List[int]]: + i, j, s, res = 1, 2, 3, [] + while i < j: + if s == target: + res.append(list(range(i, j + 1))) + if s >= target: + s -= i + i += 1 + else: + j += 1 + s += j + return res +``` + +```Java [] +class Solution { + public int[][] fileCombination(int target) { + int i = 1, j = 2, s = 3; + List res = new ArrayList<>(); + while(i < j) { + if(s == target) { + int[] ans = new int[j - i + 1]; + for(int k = i; k <= j; k++) + ans[k - i] = k; + res.add(ans); + } + if(s >= target) { + s -= i; + i++; + } else { + j++; + s += j; + } + } + return res.toArray(new int[0][]); + } +} +``` + +```C++ [] +class Solution { +public: + vector> fileCombination(int target) { + int i = 1, j = 2, s = 3; + vector> res; + while(i < j) { + if(s == target) { + vector ans; + for(int k = i; k <= j; k++) + ans.push_back(k); + res.push_back(ans); + } + if(s >= target) { + s -= i; + i++; + } else { + j++; + s += j; + } + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N = target$ ；连续整数序列至少有两个数字，而 $i < j$ 恒成立，因此至多循环 $target$ 次（ $i$ , $j$ 都移动到 $\frac{target}{2}$ ），使用 $O(N)$ 时间；当 $i = 1$ 时，达到最大序列长度 $\frac{-1 + \sqrt{1 + 8s}}{2}$ ，考虑到解的稀疏性，将列表构建时间简化考虑为 $O(1)$ ； +- **空间复杂度 $O(1)$ ：** 变量 $i$ , $j$ , $s$ 使用常数大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 181. \345\255\227\347\254\246\344\270\262\344\270\255\347\232\204\345\215\225\350\257\215\345\217\215\350\275\254.md" "b/leetbook_ioa/docs/LCR 181. \345\255\227\347\254\246\344\270\262\344\270\255\347\232\204\345\215\225\350\257\215\345\217\215\350\275\254.md" new file mode 100755 index 0000000..af6e758 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 181. \345\255\227\347\254\246\344\270\262\344\270\255\347\232\204\345\215\225\350\257\215\345\217\215\350\275\254.md" @@ -0,0 +1,105 @@ +## 方法一：双指针 + +### 算法解析： + +- 倒序遍历字符串 $message$ ，记录单词左右索引边界 $i$ , $j$ ； +- 每确定一个单词的边界，则将其添加至单词列表 $res$ ； +- 最终，将单词列表拼接为字符串，并返回即可。 + +> 下图中的 `s` 对应本题的 `message` 。 + + + +### 代码： + +```Python [] +class Solution: + def reverseMessage(self, message: str) -> str: + message = message.strip() # 删除首尾空格 + i = j = len(message) - 1 + res = [] + while i >= 0: + while i >= 0 and message[i] != ' ': i -= 1 # 搜索首个空格 + res.append(message[i + 1: j + 1]) # 添加单词 + while i >= 0 and message[i] == ' ': i -= 1 # 跳过单词间空格 + j = i # j 指向下个单词的尾字符 + return ' '.join(res) # 拼接并返回 +``` + +```Java [] +class Solution { + public String reverseMessage(String message) { + message = message.trim(); // 删除首尾空格 + int j = message.length() - 1, i = j; + StringBuilder res = new StringBuilder(); + while (i >= 0) { + while (i >= 0 && message.charAt(i) != ' ') i--; // 搜索首个空格 + res.append(message.substring(i + 1, j + 1) + " "); // 添加单词 + while (i >= 0 && message.charAt(i) == ' ') i--; // 跳过单词间空格 + j = i; // j 指向下个单词的尾字符 + } + return res.toString().trim(); // 转化为字符串并返回 + } +} +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为字符串 $message$ 的长度，线性遍历字符串。 +- **空间复杂度 $O(N)$ ：** 新建的 list(Python) 或 StringBuilder(Java) 中的字符串总长度 $\leq N$ ，占用 $O(N)$ 大小的额外空间。 + +## 方法二：分割 + 倒序 + +利用 “字符串分割”、“列表倒序” 的内置函数 *（面试时不建议使用）* ，可简便地实现本题的字符串翻转要求。 + +### 算法解析： + +- **Python ：** 由于 $split()$ 方法将单词间的 “多个空格看作一个空格” （参考自 [split()和split(' ')的区别](https://www.cnblogs.com/python-coder/p/10073329.html) ），因此不会出现多余的 “空单词” 。因此，直接利用 $reverse()$ 方法翻转单词列表 $strs$ ，拼接为字符串并返回即可。 + +![Picture1.png](https://pic.leetcode-cn.com/1600795186-pzCvCm-Picture1.png){:align=center width=500} + +- **Java ：** 以空格为分割符完成字符串分割后，若两单词间有 $x > 1$ 个空格，则在单词列表 $strs$ 中，此两单词间会多出 $x - 1$ 个 “空单词” （即 `""` ）。解决方法：倒序遍历单词列表，并将单词逐个添加至 StringBuilder ，遇到空单词时跳过。 + +![Picture2.png](https://pic.leetcode-cn.com/1600795186-RmKJXL-Picture2.png){:align=center width=500} + +### 代码： + +```Python [] +class Solution: + def reverseMessage(self, message: str) -> str: + message = message.strip() # 删除首尾空格 + strs = message.split() # 分割字符串 + strs.reverse() # 翻转单词列表 + return ' '.join(strs) # 拼接为字符串并返回 +``` + +```Java [] +class Solution { + public String reverseMessage(String message) { + String[] strs = message.trim().split(" "); // 删除首尾空格，分割字符串 + StringBuilder res = new StringBuilder(); + for (int i = strs.length - 1; i >= 0; i--) { // 倒序遍历单词列表 + if(strs[i].equals("")) continue; // 遇到空单词则跳过 + res.append(strs[i] + " "); // 将单词拼接至 StringBuilder + } + return res.toString().trim(); // 转化为字符串，删除尾部空格，并返回 + } +} +``` + +Python 可一行实现： + +```Python [] +class Solution: + def reverseMessage(self, message: str) -> str: + return ' '.join(message.strip().split()[::-1]) +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 总体为线性时间复杂度，各函数时间复杂度和参考资料链接如下。 + - [`split()` 方法：](https://softwareengineering.stackexchange.com/questions/331909/whats-the-complexity-of-javas-string-split-function) 为 $O(N)$ ； + - [`trim()` 和 `strip()` 方法：](https://stackoverflow.com/questions/51110114/is-string-trim-faster-than-string-replace) 最差情况下（当字符串全为空格时），为 $O(N)$ ； + - [`join()` 方法：](https://stackoverflow.com/questions/37133547/time-complexity-of-string-concatenation-in-python) 为 $O(N)$ ； + - [`reverse()` 方法：](https://stackoverflow.com/questions/37606159/what-is-the-time-complexity-of-python-list-reverse) 为 $O(N)$ ； +- **空间复杂度 $O(N)$ ：** 单词列表 $strs$ 占用线性大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 182. \345\212\250\346\200\201\345\217\243\344\273\244.md" "b/leetbook_ioa/docs/LCR 182. \345\212\250\346\200\201\345\217\243\344\273\244.md" new file mode 100755 index 0000000..19e4991 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 182. \345\212\250\346\200\201\345\217\243\344\273\244.md" @@ -0,0 +1,278 @@ +## 解题思路： + +本题解法较多，本文主要介绍 **字符串切片** ， **列表遍历拼接** ， **字符串遍历拼接** 三种方法，适用于 Python 和 Java 语言。同时，介绍了 **三次翻转法** ，适用于 C++ 语言。 + +## 方法一：字符串切片 + +获取字符串 `password[target:]` 切片和 `password[:target]` 切片，使用 "$+$" 运算符拼接并返回即可。 + +> 下图中的 `s` 对应本题的 `password` 。 + +![Picture1.png](https://pic.leetcode-cn.com/1600793170-eyvDTJ-Picture1.png){:align=center width=500} + +### 代码： + +```Python [] +class Solution: + def dynamicPassword(self, password: str, target: int) -> str: + return password[target:] + password[:target] +``` + +```Java [] +class Solution { + public String dynamicPassword(String password, int target) { + return password.substring(target, password.length()) + password.substring(0, target); + } +} +``` + +```C++ [] +class Solution { +public: + string dynamicPassword(string password, int target) { + return password.substr(target, password.size()) + password.substr(0, target); + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为字符串 `password` 的长度，字符串切片函数为线性时间复杂度（[参考资料](https://stackoverflow.com/questions/4679746/time-complexity-of-javas-substring)）。 +- **空间复杂度 $O(N)$ ：** 两个字符串切片的总长度为 $N$ 。 + +## 方法二：列表遍历拼接 + +> 若面试规定不允许使用 **切片函数** ，则使用此方法。 + +### 算法流程： + +1. 新建一个 list (Python) 、StringBuilder (Java) ，记为 `res` ； +2. 先向 `res` 添加 “第 $target + 1$ 位至末位的字符” ； +3. 再向 `res` 添加 “首位至第 $target$ 位的字符” ； +4. 将 `res` 转化为字符串并返回； + +![Picture2.png](https://pic.leetcode-cn.com/1600793170-ViWBNV-Picture2.png){:align=center width=550} + +### 代码： + +```Python [] +class Solution: + def dynamicPassword(self, password: str, target: int) -> str: + res = [] + for i in range(target, len(password)): + res.append(password[i]) + for i in range(target): + res.append(password[i]) + return ''.join(res) +``` + +```Java [] +class Solution { + public String dynamicPassword(String password, int target) { + StringBuilder res = new StringBuilder(); + for(int i = target; i < password.length(); i++) + res.append(password.charAt(i)); + for(int i = 0; i < target; i++) + res.append(password.charAt(i)); + return res.toString(); + } +} +``` + +利用求余运算，可以简化代码。 + +```Python [] +class Solution: + def dynamicPassword(self, password: str, target: int) -> str: + res = [] + for i in range(target, target + len(password)): + res.append(password[i % len(password)]) + return ''.join(res) +``` + +```Java [] +class Solution { + public String dynamicPassword(String password, int target) { + StringBuilder res = new StringBuilder(); + for(int i = target; i < target + password.length(); i++) + res.append(password.charAt(i % password.length())); + return res.toString(); + } +} +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 线性遍历 `password` 并添加，使用线性时间。 +- **空间复杂度 $O(N)$ ：** 新建的辅助 `res` 使用 $O(N)$ 大小的额外空间。 + +## 方法三：字符串遍历拼接 + +> 若规定 Python 不能使用 `join()` 函数，或规定 Java 只能用 String ，则使用此方法。 + +此方法与 **方法二** 思路一致，区别是使用字符串代替列表。 + +![Picture3.png](https://pic.leetcode-cn.com/1600793170-uasqXO-Picture3.png){:align=center width=550} + +```Python [] +class Solution: + def dynamicPassword(self, password: str, target: int) -> str: + res = "" + for i in range(target, len(password)): + res += password[i] + for i in range(target): + res += password[i] + return res +``` + +```Java [] +class Solution { + public String dynamicPassword(String password, int target) { + String res = ""; + for(int i = target; i < password.length(); i++) + res += password.charAt(i); + for(int i = 0; i < target; i++) + res += password.charAt(i); + return res; + } +} +``` + +同理，利用求余运算，可以简化代码。 + +```Python [] +class Solution: + def dynamicPassword(self, password: str, target: int) -> str: + res = "" + for i in range(target, target + len(password)): + res += password[i % len(password)] + return res +``` + +```Java [] +class Solution { + public String dynamicPassword(String password, int target) { + String res = ""; + for(int i = target; i < target + password.length(); i++) + res += password.charAt(i % password.length()); + return res; + } +} +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 线性遍历 `password` 并添加，使用线性时间。 +- **空间复杂度 $O(N)$ ：** 假设循环过程中内存会被及时回收，内存中至少同时存在长度为 $N$ 和 $N-1$ 的两个字符串（新建长度为 $N$ 的 `res` 需要使用前一个长度 $N-1$ 的 `res` ），因此至少使用 $O(N)$ 的额外空间。 + +## 效率对比： + +由于本题的多解法涉及到了 **字符串为不可变对象** 的相关概念，导致效率区别较大。以上三种方法的空间使用如下图所示。 + +> 详细分析请参考 [Efficient String Concatenation in Python](https://waymoot.org/home/python_string/) 。 + +以 Python 为例开展三种方法的效率测试，结论同样适用于 Java 语言。 + +![Picture4.png](https://pic.leetcode-cn.com/1600793170-xExsDE-Picture4.png){:align=center width=650} + +### 测试数据： + +长度为 $10000000$ 的全为 `'1'` 的字符串。 + +```Python +password = "1" * 10000000 +``` + +**方法一测试：** + +新建两切片字符串，并将两切片拼接为结果字符串，无冗余操作，效率最高。 + +```Python [] +# 运行时间: 0.01 秒 +def func1(password): + cut = len(password) // 3 + return password[:cut] + password[cut:] +``` + +**方法二测试：** + +列表(Python) 和 StringBuilder(Java) 都是可变对象，每轮遍历拼接字符时，只是向列表尾部添加一个新的字符元素。最终拼接转化为字符串时，系统 **仅申请一次内存** 。 + +```Python [] +# 运行时间: 1.86 秒 +def func2(password): + res = [] + for i in range(len(password)): + res.append(password[i]) # 仅需在列表尾部添加元素 + return ''.join(res) +``` + +**方法三测试：** + +在 Python 和 Java 中，字符串是 “不可变对象” 。因此，每轮遍历拼接字符时，都需要新建一个字符串；因此，系统 **需申请 $N$ 次内存** ，数据量较大时效率低下。 + +```Python [] +# 运行时间: 6.31 秒 +def func3(password): + res = "" + for i in range(len(password)): + res += password[i] # 每次拼接都需要新建一个字符串 + return res +``` + +## 方法四：三次翻转（C++） + +由于 C++ 中的字符串是 **可变类型** ，因此可在原字符串上直接操作实现字符串旋转，实现 $O(1)$ 的空间复杂度。 + +设字符串 $password = s_1 s_2$ ，字符串 $password$ 的反转字符串为 $\hat password$ ，则左旋转字符串 $s_2 s_1$ 计算方法为： + +$$ +s_2 s_1 = \hat{\hat{s_1} \hat{s_2}} +$$ + +> 例如，$password = "abcdefg"$ , $s_1 = "ab"$ , $s_2 = "cdefg"$ ，则有： +> $$ +> \hat{s_1} = "ba" \\ +> \hat{s_2} = "gfedc" \\ +> \hat{\hat{s_1} \hat{s_2}} = \hat{"bagfedc"} = "cdefgba" +> $$ +> 即 $"cdefgba"$ 为所求字符串 $password$ 的左旋转结果。 + +### 代码： + +自行实现字符串翻转函数 `reverseString()` ，代码如下： + +```C++ [] +class Solution { +public: + string dynamicPassword(string password, int target) { + reverseString(password, 0, target - 1); + reverseString(password, target, password.size() - 1); + reverseString(password, 0, password.size() - 1); + return password; + } +private: + void reverseString(string& password, int i, int j) { + while(i < j) swap(password[i++], password[j--]); + } +}; +``` + +也可使用库函数实现，代码如下： + +```C++ [] +class Solution { +public: + string dynamicPassword(string password, int target) { + reverse(password.begin(), password.begin() + target); + reverse(password.begin() + target, password.end()); + reverse(password.begin(), password.end()); + return password; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 共线性遍历两轮 `password` 。 +- **空间复杂度 $O(1)$ ：** C++ 原地字符串操作，使用常数大小额外空间。 diff --git "a/leetbook_ioa/docs/LCR 183. \346\234\233\350\277\234\351\225\234\344\270\255\346\234\200\351\253\230\347\232\204\346\265\267\346\213\224.md" "b/leetbook_ioa/docs/LCR 183. \346\234\233\350\277\234\351\225\234\344\270\255\346\234\200\351\253\230\347\232\204\346\265\267\346\213\224.md" new file mode 100755 index 0000000..fbec2e4 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 183. \346\234\233\350\277\234\351\225\234\344\270\255\346\234\200\351\253\230\347\232\204\346\265\267\346\213\224.md" @@ -0,0 +1,147 @@ +## 解题思路： + +设窗口区间为 $[i, j]$ ，最大值为 $x_j$ 。当窗口向前移动一格，则区间变为 $[i+1,j+1]$ ，即添加了 $heights[j + 1]$ ，删除了 $heights[i]$ 。 + +若只向窗口 $[i, j]$ 右边添加数字 $heights[j + 1]$ ，则新窗口最大值可以 **通过一次对比** 使用 $O(1)$ 时间得到，即： + +$$ +x_{j+1} = \max(x_{j}, heights[j + 1]) +$$ + +而由于删除的 $heights[i]$ 可能恰好是窗口内唯一的最大值 $x_j$ ，因此不能通过以上方法计算 $x_{j+1}$ ，而必须使用 $O(j-i)$ 时间， **遍历整个窗口区间** 获取最大值，即： + +$$ +x_{j+1} = \max(heights(i+1), \cdots , heights(j+1)) +$$ + +根据以上分析，可得 **暴力法** 的时间复杂度为 $O((n-limit+1)limit) \approx O(nk)$ 。 + +- 设数组 $heights$ 的长度为 $n$ ，则共有 $(n-limit+1)$ 个窗口； +- 获取每个窗口最大值需线性遍历，时间复杂度为 $O(limit)$ 。 + +> 下图中的 `nums` 对应本题的 `heights` 。 + +![Picture1.png](https://pic.leetcode-cn.com/1600878237-pBiBdf-Picture1.png){:align=center width=650} + +> **本题难点：** 如何在每次窗口滑动后，将 “获取窗口内最大值” 的时间复杂度从 $O(limit)$ 降低至 $O(1)$ 。 + +回忆“最小栈”问题，其使用 **单调栈** 实现了随意入栈、出栈情况下的 $O(1)$ 时间获取 “栈内最小值” 。本题同理，不同点在于 “出栈操作” 删除的是 “列表尾部元素” ，而 “窗口滑动” 删除的是 “列表首部元素” 。 + +窗口对应的数据结构为 **双端队列** ，本题使用 **单调队列** 即可解决以上问题。遍历数组时，每轮保证单调队列 $deque$ ： + +1. $deque$ 内 **仅包含窗口内的元素** $\Rightarrow$ 每轮窗口滑动移除了元素 $heights[i - 1]$ ，需将 $deque$ 内的对应元素一起删除。 +2. $deque$ 内的元素 **非严格递减** $\Rightarrow$ 每轮窗口滑动添加了元素 $heights[j + 1]$ ，需将 $deque$ 内所有 $< heights[j + 1]$ 的元素删除。 + +### 算法流程： + +1. **初始化：** 双端队列 $deque$ ，结果列表 $res$ ，数组长度 $n$ ； +2. **滑动窗口：** 左边界范围 $i \in [1 - limit, n - limit]$ ，右边界范围 $j \in [0, n - 1]$ ； + 1. 若 $i > 0$ 且 队首元素 $deque[0]$ $=$ 被删除元素 $heights[i - 1]$ ：则队首元素出队； + 2. 删除 $deque$ 内所有 $< heights[j]$ 的元素，以保持 $deque$ 递减； + 3. 将 $heights[j]$ 添加至 $deque$ 尾部； + 4. 若已形成窗口（即 $i \geq 0$ ）：将窗口最大值（即队首元素 $deque[0]$ ）添加至列表 $res$ ； +3. **返回值：** 返回结果列表 $res$ ； + + + +## 代码： + +Python 通过 `zip(range(), range())` 可实现滑动窗口的左右边界 `i, j` 同时遍历。 + +```Python [] +class Solution: + def maxAltitude(self, heights: List[int], limit: int) -> List[int]: + deque = collections.deque() + res, n = [], len(heights) + for i, j in zip(range(1 - limit, n + 1 - limit), range(n)): + # 删除 deque 中对应的 heights[i-1] + if i > 0 and deque[0] == heights[i - 1]: + deque.popleft() + # 保持 deque 递减 + while deque and deque[-1] < heights[j]: + deque.pop() + deque.append(heights[j]) + # 记录窗口最大值 + if i >= 0: + res.append(deque[0]) + return res +``` + +```Java [] +class Solution { + public int[] maxAltitude(int[] heights, int limit) { + if(heights.length == 0 || limit == 0) return new int[0]; + Deque deque = new LinkedList<>(); + int[] res = new int[heights.length - limit + 1]; + for(int j = 0, i = 1 - limit; j < heights.length; i++, j++) { + // 删除 deque 中对应的 heights[i-1] + if(i > 0 && deque.peekFirst() == heights[i - 1]) + deque.removeFirst(); + // 保持 deque 递减 + while(!deque.isEmpty() && deque.peekLast() < heights[j]) + deque.removeLast(); + deque.addLast(heights[j]); + // 记录窗口最大值 + if(i >= 0) + res[i] = deque.peekFirst(); + } + return res; + } +} +``` + +可以将 “未形成窗口” 和 “形成窗口后” 两个阶段拆分到两个循环里实现。代码虽变长，但减少了冗余的判断操作。 + +```Python [] +class Solution: + def maxAltitude(self, heights: List[int], limit: int) -> List[int]: + if not heights or limit == 0: return [] + deque = collections.deque() + # 未形成窗口 + for i in range(limit): + while deque and deque[-1] < heights[i]: + deque.pop() + deque.append(heights[i]) + res = [deque[0]] + # 形成窗口后 + for i in range(limit, len(heights)): + if deque[0] == heights[i - limit]: + deque.popleft() + while deque and deque[-1] < heights[i]: + deque.pop() + deque.append(heights[i]) + res.append(deque[0]) + return res +``` + +```Java [] +class Solution { + public int[] maxAltitude(int[] heights, int limit) { + if(heights.length == 0 || limit == 0) return new int[0]; + Deque deque = new LinkedList<>(); + int[] res = new int[heights.length - limit + 1]; + // 未形成窗口 + for(int i = 0; i < limit; i++) { + while(!deque.isEmpty() && deque.peekLast() < heights[i]) + deque.removeLast(); + deque.addLast(heights[i]); + } + res[0] = deque.peekFirst(); + // 形成窗口后 + for(int i = limit; i < heights.length; i++) { + if(deque.peekFirst() == heights[i - limit]) + deque.removeFirst(); + while(!deque.isEmpty() && deque.peekLast() < heights[i]) + deque.removeLast(); + deque.addLast(heights[i]); + res[i - limit + 1] = deque.peekFirst(); + } + return res; + } +} +``` + +### 复杂度分析： + +- **时间复杂度 $O(n)$ ：** 其中 $n$ 为数组 $heights$ 长度；线性遍历 $heights$ 占用 $O(n)$ ；每个元素最多仅入队和出队一次，因此单调队列 $deque$ 占用 $O(2n)$ 。 +- **空间复杂度 $O(limit)$ ：** 双端队列 $deque$ 中最多同时存储 $limit$ 个元素（即窗口大小）。 diff --git "a/leetbook_ioa/docs/LCR 184. \350\256\276\350\256\241\350\207\252\345\212\251\347\273\223\347\256\227\347\263\273\347\273\237.md" "b/leetbook_ioa/docs/LCR 184. \350\256\276\350\256\241\350\207\252\345\212\251\347\273\223\347\256\227\347\263\273\347\273\237.md" new file mode 100755 index 0000000..c8312a3 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 184. \350\256\276\350\256\241\350\207\252\345\212\251\347\273\223\347\256\227\347\263\273\347\273\237.md" @@ -0,0 +1,129 @@ +## 解题思路： + +> 对于普通队列，入队 `add()` 和出队 `remove()` 的时间复杂度均为 $O(1)$ ；本题难点为实现查找最大值 `get_max()` 的 $O(1)$ 时间复杂度。 +> 假设队列中存储 $N$ 个元素，从中获取最大值需要遍历队列，时间复杂度为 $O(N)$ ，单从算法上无优化空间。 + +如下图所示，最直观的想法是 **维护一个最大值变量** ，在元素入队时更新此变量即可；但当最大值出队后，并无法确定下一个 **次最大值** ，因此不可行。 + +![Picture1.png](https://pic.leetcode-cn.com/1609261470-WanZuG-Picture1.png){:align=center width=500} + +考虑利用 **数据结构** 来实现，即经常使用的 “空间换时间” 。如下图所示，考虑构建一个递减列表来保存队列 **所有递减的元素** ，递减链表随着入队和出队操作实时更新，这样队列最大元素就始终对应递减列表的首元素，实现了获取最大值 $O(1)$ 时间复杂度。 + +![Picture2.png](https://pic.leetcode-cn.com/1609261470-gMTEAf-Picture2.png){:align=center width=500} + +为了实现此递减列表，需要使用 **双向队列** ，假设队列已经有若干元素： + +1. 当执行入队 `add()` 时： 若入队一个比队列某些元素更大的数字 $x$ ，则为了保持此列表递减，需要将双向队列 **尾部所有小于 $x$ 的元素** 弹出。 +2. 当执行出队 `remove()` 时： 若出队的元素是最大元素，则 双向队列 需要同时 **将首元素出队** ，以保持队列和双向队列的元素一致性。 + +> 使用双向队列原因：维护递减列表需要元素队首弹出、队尾插入、队尾弹出操作皆为 $O(1)$ 时间复杂度。 + +### 函数设计： + +初始化队列 `queue` ，双向队列 `deque` ； + +**最大值 `get_max()` ：** + +- 当双向队列 `deque` 为空，则返回 $-1$ ； +- 否则，返回 `deque` 首元素； + +**入队 `add()` ：** + +1. 将元素 `value` 入队 `queue` ； +2. 将双向队列中队尾 **所有** 小于 `value` 的元素弹出（以保持 `deque` 非单调递减），并将元素 `value` 入队 `deque` ； + +**出队 `remove()` ：** + +1. 若队列 `queue` 为空，则直接返回 $-1$ ； +2. 否则，将 `queue` 首元素出队； +3. 若 `deque` 首元素和 `queue` 首元素 **相等** ，则将 `deque` 首元素出队（以保持两队列 **元素一致** ） ； + +> 设计双向队列为 **单调不增** 的原因：若队列 `queue` 中存在两个 **值相同的最大元素** ，此时 `queue` 和 `deque` 同时弹出一个最大元素，而 `queue` 中还有一个此最大元素；即采用单调递减将导致两队列中的元素不一致。 + +> 下图中的 `push_back()` , `pop_front()` , `max_value()` 分别对应本题的 `add()` , `remove()` , `get_max()` 。 + + + +## 代码： + +```Python [] +import queue + +class Checkout: + def __init__(self): + self.queue = queue.Queue() + self.deque = queue.deque() + + def get_max(self) -> int: + return self.deque[0] if self.deque else -1 + + def add(self, value: int) -> None: + self.queue.put(value) + while self.deque and self.deque[-1] < value: + self.deque.pop() + self.deque.append(value) + + def remove(self) -> int: + if self.queue.empty(): return -1 + val = self.queue.get() + if val == self.deque[0]: + self.deque.popleft() + return val +``` + +```Java [] +class Checkout { + Queue queue; + Deque deque; + public Checkout() { + queue = new LinkedList<>(); + deque = new LinkedList<>(); + } + public int get_max() { + return deque.isEmpty() ? -1 : deque.peekFirst(); + } + public void add(int value) { + queue.offer(value); + while(!deque.isEmpty() && deque.peekLast() < value) + deque.pollLast(); + deque.offerLast(value); + } + public int remove() { + if(queue.isEmpty()) return -1; + if(queue.peek().equals(deque.peekFirst())) + deque.pollFirst(); + return queue.poll(); + } +} +``` + +```C++ [] +class Checkout { + queue que; + deque deq; +public: + Checkout() { } + int get_max() { + return deq.empty() ? -1 : deq.front(); + } + void add(int value) { + que.push(value); + while(!deq.empty() && deq.back() < value) + deq.pop_back(); + deq.push_back(value); + } + int remove() { + if(que.empty()) return -1; + int val = que.front(); + if(val == deq.front()) + deq.pop_front(); + que.pop(); + return val; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(1)$ ：** `get_max()`, `add()`, `remove()` 方法的均摊时间复杂度均为 $O(1)$ ； +- **空间复杂度 $O(N)$ ：** 当元素个数为 $N$ 时，最差情况下`deque` 中保存 $N$ 个元素，使用 $O(N)$ 的额外空间； diff --git "a/leetbook_ioa/docs/LCR 185. \347\273\237\350\256\241\347\273\223\346\236\234\346\246\202\347\216\207.md" "b/leetbook_ioa/docs/LCR 185. \347\273\237\350\256\241\347\273\223\346\236\234\346\246\202\347\216\207.md" new file mode 100755 index 0000000..fee9c5c --- /dev/null +++ "b/leetbook_ioa/docs/LCR 185. \347\273\237\350\256\241\347\273\223\346\236\234\346\246\202\347\216\207.md" @@ -0,0 +1,113 @@ +## 方法一：暴力法 + +> 此方法超时，但为便于理解「方法二」，建议先理解此方法。 +> +> 为简化篇幅，本文使用 $n$ 代替题目中的 $num$ 。 + +给定 $n$ 个骰子，可得： + +- 每个骰子摇到 $1$ 至 $6$ 的概率相等，都为 $\frac{1}{6}$ 。 +- 将每个骰子的点数看作独立情况，共有 $6^n$ 种「**点数组合**」。例如 $n = 2$ 时的点数组合为： + +$$ +(1,1), (1,2), \cdots, (2, 1), (2, 2), \cdots, (6,1), \cdots, (6, 6) +$$ + +- $n$ 个骰子「**点数和**」的范围为 $[n, 6n]$ ，数量为 $6n - n + 1 = 5n + 1$ 种。 + +**暴力统计：** 每个「点数组合」都对应一个「点数和」，考虑遍历所有点数组合，统计每个点数和的出现次数，最后除以点数组合的总数（即除以 $6^n$ ），即可得到每个点数和的出现概率。 + +> 如下图所示，为输入 $n = 2$ 时，点数组合、点数和、各点数概率的计算过程。 + +![Picture1.png](https://pic.leetcode-cn.com/1615223242-EMOnIR-Picture1.png){:align=center width=550} + +暴力法需要遍历所有点数组合，因此时间复杂度为 $O(6^n)$ ，观察本题输入取值范围 $1 \leq n \leq 11$ ，可知此复杂度是无法接受的。 + +## 方法二：动态规划 + +> 设输入 $n$ 个骰子的解（即概率列表）为 $f(n)$ ，其中「点数和」 $x$ 的概率为 $f(n, x)$ 。 + +假设已知 $n - 1$ 个骰子的解 $f(n - 1)$ ，此时**添加**一枚骰子，求 $n$ 个骰子的点数和为 $x$ 的概率 $f(n, x)$ 。 + +当添加骰子的点数为 $1$ 时，前 $n - 1$ 个骰子的点数和应为 $x - 1$ ，方可组成点数和 $x$ ；同理，当此骰子为 $2$ 时，前 $n - 1$ 个骰子应为 $x - 2$ ；以此类推，直至此骰子点数为 $6$ 。将这 $6$ 种情况的概率相加，即可得到概率 $f(n, x)$ 。递推公式如下所示： +$$ +f(n, x) = \sum_{i=1}^6 f(n - 1, x - i) \times \frac{1}{6} +$$ + +根据以上分析，得知通过子问题的解 $f(n - 1)$ 可递推计算出 $f(n)$ ，而输入一个骰子的解 $f(1)$ 已知，因此可通过解 $f(1)$ 依次递推出任意解 $f(n)$ 。 + +> 如下图所示，为 $n = 2$ , $x = 7$ 的递推计算示例。 + +![Picture2.png](https://pic.leetcode-cn.com/1614960989-tpJNRQ-Picture2.png){:align=center width=550} + +观察发现，以上递推公式虽然可行，但 $f(n - 1, x - i)$ 中的 $x - i$ 会有越界问题。例如，若希望递推计算 $f(2, 2)$ ，由于一个骰子的点数和范围为 $[1, 6]$ ，因此只应求和 $f(1, 1)$ ，即 $f(1, 0)$ , $f(1, -1)$ , ... , $f(1, -4)$ 皆无意义。此越界问题导致代码编写的难度提升。 + +> 如下图所示，以上递推公式是 “逆向” 的，即为了计算 $f(n, x)$ ，将所有与之有关的情况求和；而倘若改换为 “正向” 的递推公式，便可解决越界问题。 + +![Picture3.png](https://pic.leetcode-cn.com/1614960989-mMonMs-Picture3.png){:align=center width=550} + +具体来看，由于新增骰子的点数只可能为 $1$ 至 $6$ ，因此概率 $f(n - 1, x)$ 仅与 $f(n, x + 1)$ , $f(n, x + 2)$, ... , $f(n, x + 6)$ 相关。因而，遍历 $f(n - 1)$ 中各点数和的概率，并将其相加至 $f(n)$ 中所有相关项，即可完成 $f(n - 1)$ 至 $f(n)$ 的递推。 + +> 将 $f(i)$ 记为动态规划列表形式 $dp[i]$ ，则 $i = 1, 2, ..., n$ 的状态转移过程如下图所示。 + + + +## 代码： + +通常做法是声明一个二维数组 $dp$ ，$dp[i][j]$ 代表前 $i$ 个骰子的点数和 $j$ 的概率，并执行状态转移。而由于 $dp[i]$ 仅由 $dp[i-1]$ 递推得出，为降低空间复杂度，只建立两个一维数组 $dp$ , $tmp$ 交替前进即可。 + +```Python [] +class Solution: + def statisticsProbability(self, n: int) -> List[float]: + dp = [1 / 6] * 6 + for i in range(2, n + 1): + tmp = [0] * (5 * i + 1) + for j in range(len(dp)): + for k in range(6): + tmp[j + k] += dp[j] / 6 + dp = tmp + return dp +``` + +```Java [] +class Solution { + public double[] statisticsProbability(int n) { + double[] dp = new double[6]; + Arrays.fill(dp, 1.0 / 6.0); + for (int i = 2; i <= n; i++) { + double[] tmp = new double[5 * i + 1]; + for (int j = 0; j < dp.length; j++) { + for (int k = 0; k < 6; k++) { + tmp[j + k] += dp[j] / 6.0; + } + } + dp = tmp; + } + return dp; + } +} +``` + +```C++ [] +class Solution { +public: + vector statisticsProbability(int n) { + vector dp(6, 1.0 / 6.0); + for (int i = 2; i <= n; i++) { + vector tmp(5 * i + 1, 0); + for (int j = 0; j < dp.size(); j++) { + for (int k = 0; k < 6; k++) { + tmp[j + k] += dp[j] / 6.0; + } + } + dp = tmp; + } + return dp; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(n ^ 2)$ ：** 状态转移循环 $n - 1$ 轮；每轮中，当 $i = 2, 3, ..., n$ 时，对应循环数量分别为 $6 \times 6, 11 \times 6, ..., [5(n - 1) + 1] \times 6$ ；因此总体复杂度为 $O((n - 1) \times \frac{6 + [5(n - 1) + 1]}{2} \times 6)$ ，即等价于 $O(n^2)$ 。 +- **空间复杂度 $O(n)$ ：** 状态转移过程中，辅助数组 `tmp` 最大长度为 $6(n-1) - [(n-1) - 1] = 5n - 4$ ，因此使用 $O(5n - 4) = O(n)$ 大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 186. \346\226\207\347\211\251\346\234\235\344\273\243\345\210\244\346\226\255.md" "b/leetbook_ioa/docs/LCR 186. \346\226\207\347\211\251\346\234\235\344\273\243\345\210\244\346\226\255.md" new file mode 100755 index 0000000..1a85470 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 186. \346\226\207\347\211\251\346\234\235\344\273\243\345\210\244\346\226\255.md" @@ -0,0 +1,135 @@ +## 解题思路： + +根据题意，此 $5$ 个朝代连续的 **充分条件** 如下： + +1. 除未知朝代外，所有朝代 **无重复** ； +2. 设此 $5$ 个朝代中最大的朝代为 $ma$ ，最小的朝代为 $mi$ （未知朝代除外），则需满足： + +$$ +ma - mi < 5 +$$ + +因此可将问题转化为：此 $5$ 个朝代是否满足以上两个条件？ + +> 下图中的“牌”对应本题的“朝代”。 + +![Picture1.png](https://pic.leetcode-cn.com/1599885716-MGMODX-Picture1.png){:align=center width=650} + +## 方法一： 辅助哈希表 + +- 遍历五个朝代，遇到未知朝代（即 $0$ ）直接跳过。 +- **判别重复：** 利用 Set 实现遍历判重， Set 的查找方法的时间复杂度为 $O(1)$ ； +- **获取最大 / 最小的朝代：** 借助辅助变量 $ma$ 和 $mi$ ，遍历统计即可。 + + + +### 代码： + +```Python [] +class Solution: + def checkDynasty(self, places: List[int]) -> bool: + repeat = set() + ma, mi = 0, 14 + for place in places: + if place == 0: continue # 跳过未知朝代 + ma = max(ma, place) # 最大编号朝代 + mi = min(mi, place) # 最小编号朝代 + if place in repeat: return False # 若有重复，提前返回 false + repeat.add(place) # 添加朝代至 Set + return ma - mi < 5 # 最大编号朝代 - 最小编号朝代 < 5 则连续 +``` + +```Java [] +class Solution { + public boolean checkDynasty(int[] places) { + Set repeat = new HashSet<>(); + int max = 0, min = 14; + for(int place : places) { + if(place == 0) continue; // 跳过未知朝代 + max = Math.max(max, place); // 最大编号朝代 + min = Math.min(min, place); // 最小编号朝代 + if(repeat.contains(place)) return false; // 若有重复，提前返回 false + repeat.add(place); // 添加此朝代至 Set + } + return max - min < 5; // 最大编号朝代 - 最小编号朝代 < 5 则连续 + } +} +``` + +```C++ [] +class Solution { +public: + bool checkDynasty(vector& places) { + unordered_set repeat; + int ma = 0, mi = 14; + for(int place : places) { + if(place == 0) continue; // 跳过未知朝代 + ma = max(ma, place); // 最大编号朝代 + mi = min(mi, place); // 最小编号朝代 + if(repeat.find(place) != repeat.end()) return false; // 若有重复，提前返回 false + repeat.insert(place); // 添加此朝代至 Set + } + return ma - mi < 5; // 最大编号朝代 - 最小编号朝代 < 5 则连续 + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(1)$ ：** 本题中给定朝代数量 $N \equiv 5$ ；遍历数组使用 $O(N) = O(5) = O(1)$ 时间。 +- **空间复杂度 $O(1)$ ：** 用于判重的辅助 Set 使用 $O(N) = O(1)$ 额外空间。 + +## 方法二：排序 + 遍历 + +- 先对数组执行排序。 +- **判别重复：** 排序数组中的相同元素位置相邻，因此可通过遍历数组，判断 $places[i] = places[i + 1]$ 是否成立来判重。 +- **获取最大 / 最小的朝代：** 排序后，数组末位元素 $places[4]$ 为最大编号朝代；元素 $places[unknown]$ 为最小编号朝代，其中 $unknown$ 为未知朝代的数量。 + + + +### 代码： + +```Python [] +class Solution: + def checkDynasty(self, places: List[int]) -> bool: + unknown = 0 + places.sort() # 数组排序 + for i in range(4): + if places[i] == 0: unknown += 1 # 统计未知朝代数量 + elif places[i] == places[i + 1]: return False # 若有重复，提前返回 false + return places[4] - places[unknown] < 5 # 最大编号朝代 - 最小编号朝代 < 5 则连续 +``` + +```Java [] +class Solution { + public boolean checkDynasty(int[] places) { + int unknown = 0; + Arrays.sort(places); // 数组排序 + for(int i = 0; i < 4; i++) { + if(places[i] == 0) unknown++; // 统计未知朝代数量 + else if(places[i] == places[i + 1]) return false; // 若有重复，提前返回 false + } + return places[4] - places[unknown] < 5; // 最大编号朝代 - 最小编号朝代 < 5 则连续 + } +} +``` + +```C++ [] +class Solution { +public: + bool checkDynasty(vector& places) { + int unknown = 0; + sort(places.begin(), places.end()); // 数组排序 + for(int i = 0; i < 4; i++) { + if(places[i] == 0) unknown++; // 统计未知朝代数量 + else if(places[i] == places[i + 1]) return false; // 若有重复，提前返回 false + } + return places[4] - places[unknown] < 5; // 最大编号朝代 - 最小编号朝代 < 5 则连续 + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(1)$ ：** 本题中给定朝代数量 $N \equiv 5$ ；数组排序使用 $O(N \log N) = O(5 \log 5) = O(1)$ 时间。 +- **空间复杂度 $O(1)$ ：** 变量 $unknown$ 使用 $O(1)$ 大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 187. \347\240\264\345\206\260\346\270\270\346\210\217.md" "b/leetbook_ioa/docs/LCR 187. \347\240\264\345\206\260\346\270\270\346\210\217.md" new file mode 100755 index 0000000..aa309b9 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 187. \347\240\264\345\206\260\346\270\270\346\210\217.md" @@ -0,0 +1,123 @@ +## 解题思路： + +> 为简化篇幅，本文将 $num$ 和 $target$ 分别记为 $n$ 和 $m$ 。 + +模拟整个删除过程最直观，即构建一个长度为 $n$ 的链表，各节点值为对应的顺序索引；每轮删除第 $m$ 个节点，直至链表长度为 1 时结束，返回最后剩余节点的值即可。 + +模拟法需要循环删除 $n - 1$ 轮，每轮在链表中寻找删除节点需要 $m$ 次访问操作（链表线性遍历），因此总体时间复杂度为 $O(nm)$ 。题目给定的 $m, n$ 取值范围如下所示，观察可知此时间复杂度是不可接受的。 + +$$ +1 \leq n \leq 10^5 \\ +1 \leq m \leq 10^6 +$$ + +> 实际上，本题是著名的 “约瑟夫环” 问题，可使用 **动态规划** 解决。 + +输入 $n, m$ ，记此约瑟夫环问题为 「$n, m$ 问题」 ，设解（即最后留下的数字）为 $f(n)$ ，则有： + +- 「$n, m$ 问题」：数字环为 $0, 1, 2, ..., n - 1$ ，解为 $f(n)$ ； +- 「$n-1, m$ 问题」：数字环为 $0, 1, 2, ..., n - 2$ ，解为 $f(n-1)$ ； +- 以此类推…… + +> 请注意，数字环是 **首尾相接** 的，为方便行文，本文使用列表形式表示。 + +对于「$n, m$ 问题」，首轮删除环中第 $m$ 个数字后，得到一个长度为 $n - 1$ 的数字环。由于有可能 $m > n$ ，因此删除的数字为 $(m - 1) \mod n$ ，删除后的数字环从下个数字（即 $m \mod n$ ）开始，设 $t = m \mod n$ ，可得数字环： + +$$ +t, t + 1, t + 2, ..., 0, 1, ..., t - 3, t - 2 +$$ + +删除一轮后的数字环也变为一个「$n-1, m$ 问题」，观察以下数字编号对应关系： + +$$ +\begin{aligned} +「n-1, m 问题」 && \rightarrow && 「n, m 问题」删除后 \\ +0 && \rightarrow && t + 0 \\ +1 && \rightarrow && t + 1 \\ +... && \rightarrow && ... \\ +n - 2 && \rightarrow && t - 2 \\ +\end{aligned} +$$ + +设「$n-1, m$ 问题」某数字为 $x$ ，则可得递推关系： + +$$ +x \rightarrow (x + t) \mod n \\ +$$ + +换而言之，若已知「$n-1, m$ 问题」的解 $f(n - 1)$ ，则可通过以上公式计算得到「$n, m$ 问题」的解 $f(n)$ ，即： + +$$ +\begin{aligned} +f(n) & = (f(n - 1) + t) \mod n \\ +& = (f(n - 1) + m \mod n) \mod n \\ +& = (f(n - 1) + m) \mod n +\end{aligned} +$$ + +> 下图中 `n` , `m` 分别对应本题的 `n` , `m` 。 + +![Picture1.png](https://pic.leetcode-cn.com/1615096532-kUoKUe-Picture1.png){:align=center width=550} + +$f(n)$ 可由 $f(n - 1)$ 得到，$f(n - 1)$ 可由 $f(n - 2)$ 得到，……，$f(2)$ 可由 $f(1)$ 得到；因此，若给定 $f(1)$ 的值，就可以递推至任意 $f(n)$ 。而「$1, m$ 问题」的解 $f(1) = 0$ 恒成立，即无论 $m$ 为何值，长度为 1 的数字环留下的是一定是数字 $0$ 。 + +> 以上数学推导本质是得出动态规划的 转移方程 和 初始状态 。 + +### 动态规划解析： + +1. **状态定义：** 设「$i, m$ 问题」的解为 $dp[i]$ ； +2. **转移方程：** 通过以下公式可从 $dp[i - 1]$ 递推得到 $dp[i]$ ； + +$$ +dp[i] = (dp[i - 1] + m) \mod i +$$ + +3. **初始状态：**「$1, m$ 问题」的解恒为 $0$ ，即 $dp[1] = 0$ ； +4. **返回值：** 返回「$n, m$ 问题」的解 $dp[n]$ ； + +> 如下图所示，为 $n = 5$ , $m = 3$ 时的状态转移和对应的模拟删除过程。 + +![Picture2.png](https://pic.leetcode-cn.com/1613584667-AQpTlK-Picture2.png) + +## 代码： + +根据状态转移方程的递推特性，无需建立状态列表 $dp$ ，而使用一个变量 $x$ 执行状态转移即可。 + +```Python [] +class Solution: + def iceBreakingGame(self, num: int, target: int) -> int: + x = 0 + for i in range(2, num + 1): + x = (x + target) % i + return x +``` + +```Java [] +class Solution { + public int iceBreakingGame(int num, int target) { + int x = 0; + for (int i = 2; i <= num; i++) { + x = (x + target) % i; + } + return x; + } +} +``` + +```C++ [] +class Solution { +public: + int iceBreakingGame(int num, int target) { + int x = 0; + for (int i = 2; i <= num; i++) { + x = (x + target) % i; + } + return x; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(n)$ ：** 状态转移循环 $n - 1$ 次使用 $O(n)$ 时间，状态转移方程计算使用 $O(1)$ 时间； +- **空间复杂度 $O(1)$ ：** 使用常数大小的额外空间； diff --git "a/leetbook_ioa/docs/LCR 188. \344\271\260\345\215\226\350\212\257\347\211\207\347\232\204\346\234\200\344\275\263\346\227\266\346\234\272.md" "b/leetbook_ioa/docs/LCR 188. \344\271\260\345\215\226\350\212\257\347\211\207\347\232\204\346\234\200\344\275\263\346\227\266\346\234\272.md" new file mode 100755 index 0000000..84d970e --- /dev/null +++ "b/leetbook_ioa/docs/LCR 188. \344\271\260\345\215\226\350\212\257\347\211\207\347\232\204\346\234\200\344\275\263\346\227\266\346\234\272.md" @@ -0,0 +1,87 @@ +## 解题思路： + +设共有 $n$ 天，第 $a$ 天买，第 $b$ 天卖，则需保证 $a < b$ ；可推出交易方案数共有： + +$$ +(n - 1) + (n - 2) + \cdots + 2 + 1 = n(n - 1) / 2 +$$ + +因此，暴力法的时间复杂度为 $O(n^2)$ 。考虑使用动态规划降低时间复杂度。 + +### 动态规划解析： + +- **状态定义：** 设动态规划列表 $dp$ ，$dp[i]$ 代表以 $prices[i]$ 为结尾的子数组的最大利润（以下简称为 **前 $i$ 日的最大利润** ）。 +- **转移方程：** 由于题目限定 “买卖该芯片一次” ，因此前 $i$ 日最大利润 $dp[i]$ 等于前 $i - 1$ 日最大利润 $dp[i-1]$ 和第 $i$ 日卖出的最大利润中的最大值。 + +$$ +dp[i] = \max(dp[i - 1], prices[i] - \min(prices[0:i])) \\ +\uparrow \\ +前 i 日最大利润 = \max(前 (i-1) 日最大利润, 第 i 日价格 - 前 i 日最低价格) +$$ + +- **初始状态：** $dp[0] = 0$ ，即首日利润为 $0$ ； +- **返回值：** $dp[n - 1]$ ，其中 $n$ 为 $dp$ 列表长度。 + +![Picture1.png](https://pic.leetcode-cn.com/1600880605-QGgqZW-Picture1.png){:align=center width=550} + +### 时间优化： + +前 $i$ 日的最低价格 $\min(prices[0:i])$ 时间复杂度为 $O(i)$ 。而在遍历 $prices$ 时，可以借助一个变量（记为成本 $cost$ ）每日更新最低价格。优化后的转移方程为： + +$$ +dp[i] = \max(dp[i - 1], prices[i] - \min(cost, prices[i]) +$$ + +### 空间优化： + +由于 $dp[i]$ 只与 $dp[i - 1]$ , $prices[i]$ , $cost$ 相关，因此可使用一个变量（记为利润 $profit$ ）代替 $dp$ 列表。优化后的转移方程为： + +$$ +profit = \max(profit, prices[i] - \min(cost, prices[i]) +$$ + + + +## 代码： + +```Python [] +class Solution: + def bestTiming(self, prices: List[int]) -> int: + cost, profit = float("+inf"), 0 + for price in prices: + cost = min(cost, price) + profit = max(profit, price - cost) + return profit +``` + +```Java [] +class Solution { + public int bestTiming(int[] prices) { + int cost = Integer.MAX_VALUE, profit = 0; + for(int price : prices) { + cost = Math.min(cost, price); + profit = Math.max(profit, price - cost); + } + return profit; + } +} +``` + +```C++ [] +class Solution { +public: + int bestTiming(vector& prices) { + int cost = INT_MAX, profit = 0; + for(int price : prices) { + cost = min(cost, price); + profit = max(profit, price - cost); + } + return profit; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为 $prices$ 列表长度，动态规划需遍历 $prices$ 。 +- **空间复杂度 $O(1)$ ：** 变量 $cost$ 和 $profit$ 使用常数大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 189. \350\256\276\350\256\241\346\234\272\346\242\260\347\264\257\345\212\240\345\231\250.md" "b/leetbook_ioa/docs/LCR 189. \350\256\276\350\256\241\346\234\272\346\242\260\347\264\257\345\212\240\345\231\250.md" new file mode 100755 index 0000000..c24ef54 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 189. \350\256\276\350\256\241\346\234\272\346\242\260\347\264\257\345\212\240\345\231\250.md" @@ -0,0 +1,153 @@ +## 解题思路： + +本题在简单问题上做了许多限制，需要使用排除法一步步导向答案。 +$1+2+...+(target-1)+target$ 的计算方法主要有三种：平均计算、迭代、递归。 + +**方法一：** 平均计算 +**问题：** 此计算必须使用 **乘除法** ，因此本方法不可取，直接排除。 + +```Java [] +public int mechanicalAccumulator(int target) { + return (1 + target) * target / 2; +} +``` + +```Python [] +def mechanicalAccumulator(target): + return (1 + target) * target // 2 +``` + +```C++ [] +int mechanicalAccumulator(int target) { + return (1 + target) * target / 2; +} +``` + +**方法二：** 迭代 +**问题：** 循环必须使用 $while$ 或 $for$ ，因此本方法不可取，直接排除。 + +```Java [] +public int mechanicalAccumulator(int target) { + int res = 0; + for(int i = 1; i <= target; i++) + res += i; + return res; +} +``` + +```Python [] +def mechanicalAccumulator(target): + res = 0 + for i in range(1, target + 1): + res += i + return res +``` + +```C++ [] +int mechanicalAccumulator(int target) { + int res = 0; + for(int i = 1; i <= target; i++) + res += i; + return res; +} +``` + +**方法三：** 递归 +**问题：** 终止条件需要使用 $if$ ，因此本方法不可取。 +**思考：** 除了 $if$ 和 $switch$ 等判断语句外，是否有其他方法可用来终止递归？ + +```Java [] +public int mechanicalAccumulator(int target) { + if(target == 1) return 1; + target += mechanicalAccumulator(target - 1); + return target; +} +``` + +```Python [] +def mechanicalAccumulator(target): + if target == 1: return 1 + target += mechanicalAccumulator(target - 1) + return target +``` + +```C++ [] +int mechanicalAccumulator(int target) { + if(target == 1) return 1; + target += mechanicalAccumulator(target - 1); + return target; +} +``` + +> 下图中的 `sumNums()` 对应本题的 `mechanicalAccumulator` 。 + +![Picture1.png](https://pic.leetcode-cn.com/1603023621-WFZwzS-Picture1.png){:align=center width=500} + +### 逻辑运算符的短路效应： + +常见的逻辑运算符有三种，即 “与 $\&\&$ ”，“或 $||$ ”，“非 $!$ ” ；而其有重要的短路效应，如下所示： + +```Java +if(A && B) // 若 A 为 false ，则 B 的判断不会执行（即短路），直接判定 A && B 为 false + +if(A || B) // 若 A 为 true ，则 B 的判断不会执行（即短路），直接判定 A || B 为 true +``` + +本题需要实现 “当 $target = 1$ 时终止递归” 的需求，可通过短路效应实现。 + +```Java +target > 1 && mechanicalAccumulator(target - 1) // 当 target = 1 时 target > 1 不成立 ，此时 “短路” ，终止后续递归 +``` + + + +## 代码： + +1. Java 中，为构成语句，需加一个辅助布尔量 $x$ ，否则会报错； +2. Java 中，开启递归函数需改写为 `mechanicalAccumulator(target - 1) > 0` ，此整体作为一个布尔量输出，否则会报错； +3. 初始化变量 $res$ 记录结果。（ Java 可使用第二栏的简洁写法，不用借助变量 $res$ ）。 + +```Java [] +class Solution { + int res = 0; + public int mechanicalAccumulator(int target) { + boolean x = target > 1 && mechanicalAccumulator(target - 1) > 0; + res += target; + return res; + } +} +``` + +```Java [] +class Solution { + public int mechanicalAccumulator(int target) { + boolean x = target > 1 && (target += mechanicalAccumulator(target - 1)) > 0; + return target; + } +} +``` + +```Python [] +class Solution: + def __init__(self): + self.res = 0 + def mechanicalAccumulator(self, target: int) -> int: + target > 1 and self.mechanicalAccumulator(target - 1) + self.res += target + return self.res +``` + +```C++ [] +class Solution { +public: + int mechanicalAccumulator(int target) { + target > 1 && (target += mechanicalAccumulator(target - 1)); + return target; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(target)$ ：** 计算 $target + (target-1) + ... + 2 + 1$ 需要开启 $target$ 个递归函数。 +- **空间复杂度 $O(target)$ ：** 递归深度达到 $target$ ，系统使用 $O(target)$ 大小的额外空间。 diff --git "a/leetbook_ioa/docs/LCR 190. \345\212\240\345\257\206\350\277\220\347\256\227.md" "b/leetbook_ioa/docs/LCR 190. \345\212\240\345\257\206\350\277\220\347\256\227.md" new file mode 100755 index 0000000..6c3866b --- /dev/null +++ "b/leetbook_ioa/docs/LCR 190. \345\212\240\345\257\206\350\277\220\347\256\227.md" @@ -0,0 +1,101 @@ +## 解题思路： + +本题考察对位运算的灵活使用，即使用位运算实现加法。 + +设两数字的二进制形式 $dataA, dataB$ ，其求和 $s = dataA + dataB$ ，$dataA(i)$ 代表 $dataA$ 的二进制第 $i$ 位，则分为以下四种情况： + +| $dataA(i)$ | $dataB(i)$ | 无进位和 $n(i)$ | 进位 $c(i+1)$ | +| :--------: | :--------: | :-------------: | :-----------: | +| $0$ | $0$ | $0$ | $0$ | +| $0$ | $1$ | $1$ | $0$ | +| $1$ | $0$ | $1$ | $0$ | +| $1$ | $1$ | $0$ | $1$ | + +观察发现，**无进位和** 与 **异或运算** 规律相同，**进位** 和 **与运算** 规律相同（并需左移一位）。因此，无进位和 $n$ 与进位 $c$ 的计算公式如下； + +$$ +\begin{cases} +n = dataA \oplus dataB & 非进位和：异或运算 \\ +c = dataA \space \& \space dataB << 1 & 进位：与运算 + 左移一位 +\end{cases} +$$ + +（和 $s$ ）$=$（非进位和 $n$ ）$+$（进位 $c$ ）。即可将 $s = dataA + dataB$ 转化为： + +$$ +s = dataA + dataB \Rightarrow s = n + c +$$ + +循环求 $n$ 和 $c$ ，直至进位 $c = 0$ ；此时 $s = n$ ，返回 $n$ 即可。 + +> 下图中的 `a` 和 `b` 对应本题的 `dataA` 和 `dataB` 。 + +![Picture1.png](https://pic.leetcode-cn.com/9716b1a1ead21824b8216c7d54910bee4d838c011581f4e3d82a14f71cb392a1-Picture1.png){:align=center width=500} + +> **Q ：** 若数字 $dataA$ 和 $dataB$ 中有负数，则变成了减法，如何处理？ +> **A ：** 在计算机系统中，数值一律用 **补码** 来表示和存储。**补码的优势：** 加法、减法可以统一处理（CPU只有加法器）。因此，以上方法 **同时适用于正数和负数的加法** 。 + + + +## 代码： + +```Java [] +class Solution { + public int encryptionCalculate(int dataA, int dataB) { + while(dataB != 0) { // 当进位为 0 时跳出 + int c = (dataA & dataB) << 1; // c = 进位 + dataA ^= dataB; // dataA = 非进位和 + dataB = c; // dataB = 进位 + } + return dataA; + } +} +``` + +```C++ [] +class Solution { +public: + int encryptionCalculate(int dataA, int dataB) { + while(dataB != 0) + { + int c = (unsigned int)(dataA & dataB) << 1; + dataA ^= dataB; + dataB = c; + } + return dataA; + } +}; +``` + +```Python [] +class Solution: + def encryptionCalculate(self, dataA: int, dataB: int) -> int: + x = 0xffffffff + dataA, dataB = dataA & x, dataB & x + while dataB != 0: + dataA, dataB = (dataA ^ dataB), (dataA & dataB) << 1 & x + return dataA if dataA <= 0x7fffffff else ~(dataA ^ x) +``` + +### 复杂度分析： + +- **时间复杂度 $O(1)$ ：** 最差情况下（例如 $dataA =$ $\text{0x7fffffff}$ , $dataB = 1$ 时），需循环 32 次，使用 $O(1)$ 时间；每轮中的常数次位操作使用 $O(1)$ 时间。 +- **空间复杂度 $O(1)$ ：** 使用常数大小的额外空间。 + +### Python 负数的存储： + +Python，Java, C++ 等语言中的数字都是以 **补码** 形式存储的。但 Python 没有 `int` , `long` 等不同长度变量，即在编程时无变量位数的概念。 + +**获取负数的补码：** 需要将数字与十六进制数 `0xffffffff` 相与。可理解为舍去此数字 32 位以上的数字（将 32 位以上都变为 $0$ ），从无限长度变为一个 32 位整数。 + +**返回前数字还原：** 若补码 $dataA$ 为负数（ `0x7fffffff` 是最大的正数的补码 ），需执行 `~(dataA ^ x)` 操作，将补码还原至 Python 的存储格式。`dataA ^ x` 运算将 1 至 32 位按位取反；`~` 运算是将整个数字取反；因此，`~(dataA ^ x)` 是将 32 位以上的位取反，1 至 32 位不变。 + +```Python +print(hex(1)) # = 0x1 补码 +print(hex(-1)) # = -0x1 负号 + 原码 （ Python 特色，Java 会直接输出补码） + +print(hex(1 & 0xffffffff)) # = 0x1 正数补码 +print(hex(-1 & 0xffffffff)) # = 0xffffffff 负数补码 + +print(-1 & 0xffffffff) # = 4294967295 （ Python 将其认为正数） +``` diff --git "a/leetbook_ioa/docs/LCR 191. \346\214\211\350\247\204\345\210\231\350\256\241\347\256\227\347\273\237\350\256\241\347\273\223\346\236\234.md" "b/leetbook_ioa/docs/LCR 191. \346\214\211\350\247\204\345\210\231\350\256\241\347\256\227\347\273\237\350\256\241\347\273\223\346\236\234.md" new file mode 100755 index 0000000..97187ca --- /dev/null +++ "b/leetbook_ioa/docs/LCR 191. \346\214\211\350\247\204\345\210\231\350\256\241\347\256\227\347\273\237\350\256\241\347\273\223\346\236\234.md" @@ -0,0 +1,78 @@ +## 解题思路： + +> 本文将 `arrayA` , `arrayB` 简写为 `A` , `B` 。 + +本题的难点在于 **不能使用除法** ，即需要 **只用乘法** 生成数组 $B$ 。根据题目对 $B[i]$ 的定义，可列如下图所示的表格。 + +根据表格的主对角线（全为 $1$ ），可将表格分为 **上三角** 和 **下三角** 两部分。分别迭代计算下三角和上三角两部分的乘积，即可 **不使用除法** 就获得结果。 + +![Picture1.png](https://pic.leetcode-cn.com/1624619180-vpyyqh-Picture1.png){:align=center width=500} + +### 算法流程： + +1. 初始化：数组 $B$ ，其中 $B[0] = 1$ ；辅助变量 $tmp = 1$ ； +2. 计算 $B[i]$ 的 **下三角** 各元素的乘积，直接乘入 $B[i]$ ； +3. 计算 $B[i]$ 的 **上三角** 各元素的乘积，记为 $tmp$ ，并乘入 $B[i]$ ； +4. 返回 $B$ 。 + + + +## 代码： + +```Python [] +class Solution: + def statisticalResult(self, arrayA: List[int]) -> List[int]: + arrayB, tmp = [1] * len(arrayA), 1 + for i in range(1, len(arrayA)): + arrayB[i] = arrayB[i - 1] * arrayA[i - 1] + for i in range(len(arrayA) - 2, -1, -1): + tmp *= arrayA[i + 1] + arrayB[i] *= tmp + return arrayB +``` + +```Java [] +class Solution { + public int[] statisticalResult(int[] arrayA) { + int len = arrayA.length; + if(len == 0) return new int[0]; + int[] arrayB = new int[len]; + arrayB[0] = 1; + int tmp = 1; + for(int i = 1; i < len; i++) { + arrayB[i] = arrayB[i - 1] * arrayA[i - 1]; + } + for(int i = len - 2; i >= 0; i--) { + tmp *= arrayA[i + 1]; + arrayB[i] *= tmp; + } + return arrayB; + } +} +``` + +```C++ [] +class Solution { +public: + vector statisticalResult(vector& arrayA) { + int len = arrayA.size(); + if(len == 0) return {}; + vector arrayB(len, 1); + arrayB[0] = 1; + int tmp = 1; + for(int i = 1; i < len; i++) { + arrayB[i] = arrayB[i - 1] * arrayA[i - 1]; + } + for(int i = len - 2; i >= 0; i--) { + tmp *= arrayA[i + 1]; + arrayB[i] *= tmp; + } + return arrayB; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为数组长度，两轮遍历数组 $A$ ，使用 $O(N)$ 时间。 +- **空间复杂度 $O(1)$ ：** 变量 $tmp$ 使用常数大小额外空间（数组 $B$ 作为返回值，不计入复杂度考虑）。 diff --git "a/leetbook_ioa/docs/LCR 192. \346\212\212\345\255\227\347\254\246\344\270\262\350\275\254\346\215\242\346\210\220\346\225\264\346\225\260 (atoi).md" "b/leetbook_ioa/docs/LCR 192. \346\212\212\345\255\227\347\254\246\344\270\262\350\275\254\346\215\242\346\210\220\346\225\264\346\225\260 (atoi).md" new file mode 100755 index 0000000..44d5cbc --- /dev/null +++ "b/leetbook_ioa/docs/LCR 192. \346\212\212\345\255\227\347\254\246\344\270\262\350\275\254\346\215\242\346\210\220\346\225\264\346\225\260 (atoi).md" @@ -0,0 +1,143 @@ +## 解题思路： + +根据题意，有以下四种字符需要考虑： + +1. **首部空格：** 删除之即可； +2. **符号位：** 三种情况，即 ''$+$'' , ''$-$'' , ''无符号" ；新建一个变量保存符号位，返回前判断正负即可； +3. **非数字字符：** 遇到首个非数字的字符时，应立即返回； +4. **数字字符：** + 1. **字符转数字：** “此数字的 ASCII 码” 与 “ $0$ 的 ASCII 码” 相减即可； + 2. **数字拼接：** 若从左向右遍历数字，设当前位字符为 $c$ ，当前位数字为 $x$ ，数字结果为 $res$ ，则数字拼接公式为： + +$$ +res = 10 \times res + x \\ +x = ascii(c) - ascii('0') +$$ + +![Picture1.png](https://pic.leetcode-cn.com/1600793383-jCgsGU-Picture1.png){:align=center width=450} + +**数字越界处理：** + +> 题目要求返回的数值范围应在 $[-2^{31}, 2^{31} - 1]$ ，因此需要考虑数字越界问题。而由于题目指出 `环境只能存储 32 位大小的有符号整数` ，因此判断数字越界时，要始终保持 $res$ 在 int 类型的取值范围内。 + +在每轮数字拼接前，判断 $res$ **在此轮拼接后是否超过 $2147483647$** ，若超过则加上符号位直接返回。 +设数字拼接边界 $bndry = 2147483647 // 10 = 214748364$ ，则以下两种情况越界： + +$$ +\begin{cases} + res > bndry & 情况一：执行拼接 10 \times res \geq 2147483650 越界 \\ + res = bndry, x > 7 & 情况二：拼接后是 2147483648 或 2147483649 越界 \\ +\end{cases} +$$ + +![Picture2.png](https://pic.leetcode-cn.com/1600793383-JZRYip-Picture2.png){:align=center width=450} + +解题的整体流程为： + + + +## 代码： + +```Python [] +class Solution: + def myAtoi(self, str: str) -> int: + str = str.strip() # 删除首尾空格 + if not str: return 0 # 字符串为空则直接返回 + res, i, sign = 0, 1, 1 + int_max, int_min, bndry = 2 ** 31 - 1, -2 ** 31, 2 ** 31 // 10 + if str[0] == '-': sign = -1 # 保存负号 + elif str[0] != '+': i = 0 # 若无符号位，则需从 i = 0 开始数字拼接 + for c in str[i:]: + if not '0' <= c <= '9' : break # 遇到非数字的字符则跳出 + if res > bndry or res == bndry and c > '7': return int_max if sign == 1 else int_min # 数字越界处理 + res = 10 * res + ord(c) - ord('0') # 数字拼接 + return sign * res +``` + +```Java [] +class Solution { + public int myAtoi(String str) { + char[] c = str.trim().toCharArray(); + if(c.length == 0) return 0; + int res = 0, bndry = Integer.MAX_VALUE / 10; + int i = 1, sign = 1; + if(c[0] == '-') sign = -1; + else if(c[0] != '+') i = 0; + for(int j = i; j < c.length; j++) { + if(c[j] < '0' || c[j] > '9') break; + if(res > bndry || res == bndry && c[j] > '7') return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE; + res = res * 10 + (c[j] - '0'); + } + return sign * res; + } +} +``` + +若不使用 `trim() / strip()` 删除首部空格，而采取遍历跳过空格的方式，则可以将空间优化至 $O(1)$ ，代码如下： + +```Python [] +class Solution: + def myAtoi(self, str: str) -> int: + res, i, sign, length = 0, 0, 1, len(str) + int_max, int_min, bndry = 2 ** 31 - 1, -2 ** 31, 2 ** 31 // 10 + if not str: return 0 # 空字符串，提前返回 + while str[i] == ' ': + i += 1 + if i == length: return 0 # 字符串全为空格，提前返回 + if str[i] == '-': sign = -1 + if str[i] in '+-': i += 1 + for j in range(i, length): + if not '0' <= str[j] <= '9' : break + if res > bndry or res == bndry and str[j] > '7': + return int_max if sign == 1 else int_min + res = 10 * res + ord(str[j]) - ord('0') + return sign * res +``` + +```Java [] +class Solution { + public int myAtoi(String str) { + int res = 0, bndry = Integer.MAX_VALUE / 10; + int i = 0, sign = 1, length = str.length(); + if(length == 0) return 0; + while(str.charAt(i) == ' ') + if(++i == length) return 0; + if(str.charAt(i) == '-') sign = -1; + if(str.charAt(i) == '-' || str.charAt(i) == '+') i++; + for(int j = i; j < length; j++) { + if(str.charAt(j) < '0' || str.charAt(j) > '9') break; + if(res > bndry || res == bndry && str.charAt(j) > '7') + return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE; + res = res * 10 + (str.charAt(j) - '0'); + } + return sign * res; + } +} +``` + +```C++ [] +class Solution { +public: + int myAtoi(string str) { + int res = 0, bndry = INT_MAX / 10; + int i = 0, sign = 1, length = str.size(); + if(length == 0) return 0; + while(str[i] == ' ') + if(++i == length) return 0; + if(str[i] == '-') sign = -1; + if(str[i] == '-' || str[i] == '+') i++; + for(int j = i; j < length; j++) { + if(str[j] < '0' || str[j] > '9') break; + if(res > bndry || res == bndry && str[j] > '7') + return sign == 1 ? INT_MAX : INT_MIN; + res = res * 10 + (str[j] - '0'); + } + return sign * res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为字符串长度，线性遍历字符串占用 $O(N)$ 时间。 +- **空间复杂度 $O(N)$ ：** 删除首尾空格后需建立新字符串，最差情况下占用 $O(N)$ 额外空间。 diff --git "a/leetbook_ioa/docs/LCR 193. \346\261\202\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\347\232\204\346\234\200\350\277\221\345\205\254\345\205\261\347\245\226\345\205\210.md" "b/leetbook_ioa/docs/LCR 193. \346\261\202\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\347\232\204\346\234\200\350\277\221\345\205\254\345\205\261\347\245\226\345\205\210.md" new file mode 100755 index 0000000..4706365 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 193. \346\261\202\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\347\232\204\346\234\200\350\277\221\345\205\254\345\205\261\347\245\226\345\205\210.md" @@ -0,0 +1,183 @@ +## 解题思路： + +**祖先的定义：** 若节点 `p` 在节点 `root` 的左（右）子树中，或 `p = root`，则称 `root` 是 `p` 的祖先。 + +**最近公共祖先的定义：** 设节点 `root` 为节点 `p` , `q` 的某公共祖先，若其左子节点 `root.left` 和右子节点 `root.right` 都不是 `p` , `q` 的公共祖先，则称 `root` 是 “最近的公共祖先” 。 + +![Picture1.png](https://pic.leetcode-cn.com/1599885085-LbAmPd-Picture1.png){:align=center width=450} + +根据以上定义，若 `root` 是 `p` , `q` 的 **最近公共祖先** ，则只可能为以下三种情况之一： + +1. `p` 和 `q` 在 `root` 的子树中，且分列 `root` 的 **异侧**（即分别在左、右子树中）； +2. `p = root` 且 `q` 在 `root` 的左或右子树中； +3. `q = root` 且 `p` 在 `root` 的左或右子树中； + +![Picture2.png](https://pic.leetcode-cn.com/1599885085-mTpblH-Picture2.png){:align=center width=450} + +本题给定了两个重要条件：(1) 树为 **二叉搜索树** ，(2) 树的所有节点的值都是 **唯一** 的。根据以上条件，可方便地判断 `p` , `q` 与 `root` 的子树关系，即： + +- 若 `root.val < p.val` ，则 `p` 在 `root` **右子树** 中； +- 若 `root.val > p.val` ，则 `p` 在 `root` **左子树** 中； +- 若 `root.val = p.val` ，则 `p` 和 `root` 指向 **同一节点** ； + +## 方法一：迭代 + +1. **循环搜索：** 当节点 `root` 为空时跳出； + 1. 当 `p, q` 都在 `root` 的 **右子树** 中，则遍历至 `root.right` ； + 2. 否则，当 `p` , `q` 都在 `root` 的 **左子树** 中，则遍历至 `root.left` ； + 3. 否则，说明找到了 **最近公共祖先** ，跳出； +2. **返回值：** 最近公共祖先 `root` ； + + + +### 代码： + +```Python [] +class Solution: + def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode': + while root: + if root.val < p.val and root.val < q.val: # p,q 都在 root 的右子树中 + root = root.right # 遍历至右子节点 + elif root.val > p.val and root.val > q.val: # p,q 都在 root 的左子树中 + root = root.left # 遍历至左子节点 + else: break + return root +``` + +```Java [] +class Solution { + public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { + while(root != null) { + if(root.val < p.val && root.val < q.val) // p,q 都在 root 的右子树中 + root = root.right; // 遍历至右子节点 + else if(root.val > p.val && root.val > q.val) // p,q 都在 root 的左子树中 + root = root.left; // 遍历至左子节点 + else break; + } + return root; + } +} +``` + +```C++ [] +class Solution { +public: + TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { + while(root != nullptr) { + if(root->val < p->val && root->val < q->val) // p,q 都在 root 的右子树中 + root = root->right; // 遍历至右子节点 + else if(root->val > p->val && root->val > q->val) // p,q 都在 root 的左子树中 + root = root->left; // 遍历至左子节点 + else break; + } + return root; + } +}; +``` + +代码优化：若可保证 `p.val < q.val` ，则在循环中可减少判断条件，提升计算效率。 + +```Python [] +class Solution: + def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode': + if p.val > q.val: p, q = q, p # 保证 p.val < q.val + while root: + if root.val < p.val: # p,q 都在 root 的右子树中 + root = root.right # 遍历至右子节点 + elif root.val > q.val: # p,q 都在 root 的左子树中 + root = root.left # 遍历至左子节点 + else: break + return root +``` + +```Java [] +class Solution { + public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { + if(p.val > q.val) { // 保证 p.val < q.val + TreeNode tmp = p; + p = q; + q = tmp; + } + while(root != null) { + if(root.val < p.val) // p,q 都在 root 的右子树中 + root = root.right; // 遍历至右子节点 + else if(root.val > q.val) // p,q 都在 root 的左子树中 + root = root.left; // 遍历至左子节点 + else break; + } + return root; + } +} +``` + +```C++ [] +class Solution { +public: + TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { + if(p->val > q->val) + swap(p, q); + while(root != nullptr) { + if(root->val < p->val) // p,q 都在 root 的右子树中 + root = root->right; // 遍历至右子节点 + else if(root->val > q->val) // p,q 都在 root 的左子树中 + root = root->left; // 遍历至左子节点 + else break; + } + return root; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为二叉树节点数；每循环一轮排除一层，二叉搜索树的层数最小为 $\log N$ （满二叉树），最大为 $N$ （退化为链表）。 +- **空间复杂度 $O(1)$ ：** 使用常数大小的额外空间。 + +## 方法二：递归 + +1. **递推工作：** + 1. 当 `p` , `q` 都在 `root` 的 **右子树** 中，则开启递归 `root.right` 并返回； + 2. 否则，当 `p` , `q` 都在 `root` 的 **左子树** 中，则开启递归 `root.left` 并返回； +2. **返回值：** 最近公共祖先 `root` ； + +### 代码： + +```Python [] +class Solution: + def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode': + if root.val < p.val and root.val < q.val: + return self.lowestCommonAncestor(root.right, p, q) + if root.val > p.val and root.val > q.val: + return self.lowestCommonAncestor(root.left, p, q) + return root +``` + +```Java [] +class Solution { + public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { + if(root.val < p.val && root.val < q.val) + return lowestCommonAncestor(root.right, p, q); + if(root.val > p.val && root.val > q.val) + return lowestCommonAncestor(root.left, p, q); + return root; + } +} +``` + +```C++ [] +class Solution { +public: + TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { + if(root->val < p->val && root->val < q->val) + return lowestCommonAncestor(root->right, p, q); + if(root->val > p->val && root->val > q->val) + return lowestCommonAncestor(root->left, p, q); + return root; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为二叉树节点数；每循环一轮排除一层，二叉搜索树的层数最小为 $\log N$ （满二叉树），最大为 $N$ （退化为链表）。 +- **空间复杂度 $O(N)$ ：** 最差情况下，即树退化为链表时，递归深度达到树的层数 $N$ 。 diff --git "a/leetbook_ioa/docs/LCR 194. \345\257\273\346\211\276\344\272\214\345\217\211\346\240\221\347\232\204\346\234\200\350\277\221\345\205\254\345\205\261\347\245\226\345\205\210.md" "b/leetbook_ioa/docs/LCR 194. \345\257\273\346\211\276\344\272\214\345\217\211\346\240\221\347\232\204\346\234\200\350\277\221\345\205\254\345\205\261\347\245\226\345\205\210.md" new file mode 100755 index 0000000..f82a0f8 --- /dev/null +++ "b/leetbook_ioa/docs/LCR 194. \345\257\273\346\211\276\344\272\214\345\217\211\346\240\221\347\232\204\346\234\200\350\277\221\345\205\254\345\205\261\347\245\226\345\205\210.md" @@ -0,0 +1,125 @@ +## 解题思路： + +**祖先的定义：** 若节点 `p` 在节点 `root` 的左（右）子树中，或 `p = root` ，则称 `root` 是 `p` 的祖先。 + +**最近公共祖先的定义：** 设节点 `root` 为节点 `p` , `q` 的某公共祖先，若其左子节点 `root.left` 和右子节点 `root.right` 都不是 `p` , `q` 的公共祖先，则称 `root` 是 “最近的公共祖先” 。 + +![Picture1.png](https://pic.leetcode-cn.com/1599885247-rxcHcZ-Picture1.png){:align=center width=450} + +根据以上定义，若 `root` 是 `p` , `q` 的 **最近公共祖先** ，则只可能为以下情况之一： + +1. `p` 和 `q` 在 `root` 的子树中，且分列 `root` 的 **异侧**（即分别在左、右子树中）； +2. `p = root` ，且 `q` 在 `root` 的左或右子树中； +3. `q = root` ，且 `p` 在 `root` 的左或右子树中； + +![Picture2.png](https://pic.leetcode-cn.com/1599885247-mgYjRv-Picture2.png){:align=center width=450} + +考虑通过递归对二叉树进行先序遍历，当遇到节点 `p` 或 `q` 时返回。从底至顶回溯，当节点 `p` , `q` 在节点 `root` 的异侧时，节点 `root` 即为最近公共祖先，则向上返回 `root` 。 + +### 递归解析： + +1. **终止条件：** + 1. 当越过叶节点，则直接返回 $\text{null}$ ； + 2. 当 `root` 等于 `p` , `q` ，则直接返回 `root` ； +2. **递推工作：** + 1. 开启递归左子节点，返回值记为 `left` ； + 2. 开启递归右子节点，返回值记为 `right` ； +3. **返回值：** 根据 `left` 和 `right` ，可展开为四种情况； + 1. 当 `left` 和 `right` **同时为空** ：说明 `root` 的左 / 右子树中都不包含 `p` , `q` ，返回 $\text{null}$ ； + 2. 当 `left` 和 `right` **同时不为空** ：说明 `p` , `q` 分列在 `root` 的 **异侧** （分别在 左 / 右子树），因此 `root` 为最近公共祖先，返回 `root` ； + 3. 当 `left` **为空** ，`right` **不为空** ：`p` , `q` 都不在 `root` 的左子树中，直接返回 `right` 。具体可分为两种情况： + 1. `p` , `q` 其中一个在 `root` 的 **右子树** 中，此时 `right` 指向 `p`（假设为 `p` ）； + 2. `p` , `q` 两节点都在 `root` 的 **右子树** 中，此时的 `right` 指向 **最近公共祖先节点** ； + 4. 当 `left` **不为空** ，`right` **为空** ：与情况 `3.` 同理； + +> 观察发现，情况 `1.` 可合并至 `3.` 和 `4.` 内，详见文章末尾代码。 + + + +### 代码： + +```Python [] +class Solution: + def lowestCommonAncestor(self, root: TreeNode, p: TreeNode, q: TreeNode) -> TreeNode: + if not root or root == p or root == q: return root + left = self.lowestCommonAncestor(root.left, p, q) + right = self.lowestCommonAncestor(root.right, p, q) + if not left: return right + if not right: return left + return root +``` + +```Java [] +class Solution { + public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { + if(root == null || root == p || root == q) return root; + TreeNode left = lowestCommonAncestor(root.left, p, q); + TreeNode right = lowestCommonAncestor(root.right, p, q); + if(left == null) return right; + if(right == null) return left; + return root; + } +} +``` + +```C++ [] +class Solution { +public: + TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { + if(root == nullptr || root == p || root == q) return root; + TreeNode *left = lowestCommonAncestor(root->left, p, q); + TreeNode *right = lowestCommonAncestor(root->right, p, q); + if(left == nullptr) return right; + if(right == nullptr) return left; + return root; + } +}; +``` + +情况 `1.` , `2.` , `3.` , `4.` 的展开写法如下。 + +```Python [] +class Solution: + def lowestCommonAncestor(self, root: TreeNode, p: TreeNode, q: TreeNode) -> TreeNode: + if not root or root == p or root == q: return root + left = self.lowestCommonAncestor(root.left, p, q) + right = self.lowestCommonAncestor(root.right, p, q) + if not left and not right: return # 1. + if not left: return right # 3. + if not right: return left # 4. + return root # 2. if left and right: +``` + +```Java [] +class Solution { + public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { + if(root == null || root == p || root == q) return root; + TreeNode left = lowestCommonAncestor(root.left, p, q); + TreeNode right = lowestCommonAncestor(root.right, p, q); + if(left == null && right == null) return null; // 1. + if(left == null) return right; // 3. + if(right == null) return left; // 4. + return root; // 2. if(left != null and right != null) + } +} +``` + +```C++ [] +class Solution { +public: + TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { + if(root == nullptr || root == p || root == q) return root; + TreeNode *left = lowestCommonAncestor(root->left, p, q); + TreeNode *right = lowestCommonAncestor(root->right, p, q); + if(left == nullptr && right == nullptr) return nullptr; // 1. + if(left == nullptr) return right; // 3. + if(right == nullptr) return left; // 4. + return root; // 2. if(left != null and right != null) + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为二叉树节点数；最差情况下，需要递归遍历树的所有节点。 +- **空间复杂度 $O(N)$ ：** 最差情况下，递归深度达到 $N$ ，系统使用 $O(N)$ 大小的额外空间。 diff --git "a/selected_coding_interview/docs/10. \346\255\243\345\210\231\350\241\250\350\276\276\345\274\217\345\214\271\351\205\215.md" "b/selected_coding_interview/docs/10. \346\255\243\345\210\231\350\241\250\350\276\276\345\274\217\345\214\271\351\205\215.md" new file mode 100644 index 0000000..0811314 --- /dev/null +++ "b/selected_coding_interview/docs/10. \346\255\243\345\210\231\350\241\250\350\276\276\345\274\217\345\214\271\351\205\215.md" @@ -0,0 +1,188 @@ +## 解题思路： + +设 $s$ 的长度为 $n$ ， $p$ 的长度为 $m$ ；将 $s$ 的第 $i$ 个字符记为 $s_i$ ，$p$ 的第 $j$ 个字符记为 $p_j$ ，将 $s$ 的前 $i$ 个字符组成的子字符串记为 $s[:i]$ , 同理将 $p$ 的前 $j$ 个字符组成的子字符串记为 $p[:j]$ 。本题可转化为求 $s[:n]$ 是否能和 $p[:m]$ 匹配。 + +总体思路是从 $s[:1]$ 和 $p[:1]$ 开始判断是否能匹配，每轮添加一个字符并判断是否能匹配，直至添加完整个字符串 $s$ 和 $p$ 。展开来看，假设 $s[:i]$ 与 $p[:j]$ 可以匹配，那么下一状态有两种： + +1. 添加一个字符 $s_{i+1}$ 后是否能匹配？ +2. 添加字符 $p_{j+1}$ 后是否能匹配？ + +![Picture1.png](https://pic.leetcode-cn.com/1614516402-HyzAil-Picture1.png){:width=500} + +因此，本题的状态共有 $m \times n$ 种，应定义状态矩阵 $dp$ ，$dp[i][j]$ 代表 $s[:i]$ 与 $p[:j]$ 是否可以匹配。 + +做好状态定义，接下来就是根据 「`普通字符`」 , 「`.`」 , 「`*`」三种字符的功能定义，分析出动态规划的转移方程。 + +- **状态定义：** 设动态规划矩阵 `dp` ， `dp[i][j]` 代表字符串 `s` 的前 `i` 个字符和 `p` 的前 `j` 个字符能否匹配。 + +- **转移方程：** 需要注意，由于 `dp[0][0]` 代表的是空字符的状态， 因此 `dp[i][j]` 对应的添加字符是 `s[i - 1]` 和 `p[j - 1]` 。 + + - 当 `p[j - 1] = '*'` 时， `dp[i][j]` 在当以下任一情况为 $true$ 时等于 $true$ ： + + 1. **`dp[i][j - 2]`：** 即将字符组合 `p[j - 2] *` 看作出现 0 次时，能否匹配。 + 2. **`dp[i - 1][j]` 且 `s[i - 1] = p[j - 2]`:** 即让字符 `p[j - 2]` 多出现 1 次时，能否匹配。 + 3. **`dp[i - 1][j]` 且 `p[j - 2] = '.'`:** 即让字符 `'.'` 多出现 1 次时，能否匹配。 + + - 当 `p[j - 1] != '*'` 时， `dp[i][j]` 在当以下任一情况为 $true$ 时等于 $true$ ： + + 1. **`dp[i - 1][j - 1]` 且 `s[i - 1] = p[j - 1]`：** 即让字符 `p[j - 1]` 多出现一次时，能否匹配。 + 2. **`dp[i - 1][j - 1]` 且 `p[j - 1] = '.'`：** 即将字符 `.` 看作字符 `s[i - 1]` 时，能否匹配。 + +- **初始化：** 需要先初始化 `dp` 矩阵首行，以避免状态转移时索引越界。 + + - **`dp[0][0] = true`：** 代表两个空字符串能够匹配。 + - **`dp[0][j] = dp[0][j - 2]` 且 `p[j - 1] = '*'`：** 首行 `s` 为空字符串，因此当 `p` 的偶数位为 `*` 时才能够匹配（即让 `p` 的奇数位出现 0 次，保持 `p` 是空字符串）。因此，循环遍历字符串 `p` ，步长为 2（即只看偶数位）。 + +- **返回值：** `dp` 矩阵右下角字符，代表字符串 `s` 和 `p` 能否匹配。 + + + +## 代码： + +```Python [] +class Solution: + def isMatch(self, s: str, p: str) -> bool: + m, n = len(s) + 1, len(p) + 1 + dp = [[False] * n for _ in range(m)] + dp[0][0] = True + for j in range(2, n, 2): + dp[0][j] = dp[0][j - 2] and p[j - 1] == '*' + for i in range(1, m): + for j in range(1, n): + dp[i][j] = dp[i][j - 2] or dp[i - 1][j] and (s[i - 1] == p[j - 2] or p[j - 2] == '.') \ + if p[j - 1] == '*' else \ + dp[i - 1][j - 1] and (p[j - 1] == '.' or s[i - 1] == p[j - 1]) + return dp[-1][-1] +``` + +```Java [] +class Solution { + public boolean isMatch(String s, String p) { + int m = s.length() + 1, n = p.length() + 1; + boolean[][] dp = new boolean[m][n]; + dp[0][0] = true; + for(int j = 2; j < n; j += 2) + dp[0][j] = dp[0][j - 2] && p.charAt(j - 1) == '*'; + for(int i = 1; i < m; i++) { + for(int j = 1; j < n; j++) { + dp[i][j] = p.charAt(j - 1) == '*' ? + dp[i][j - 2] || dp[i - 1][j] && (s.charAt(i - 1) == p.charAt(j - 2) || p.charAt(j - 2) == '.') : + dp[i - 1][j - 1] && (p.charAt(j - 1) == '.' || s.charAt(i - 1) == p.charAt(j - 1)); + } + } + return dp[m - 1][n - 1]; + } +} +``` + +```C++ [] +class Solution { +public: + bool isMatch(string s, string p) { + int m = s.size() + 1, n = p.size() + 1; + vector> dp(m, vector(n, false)); + dp[0][0] = true; + for(int j = 2; j < n; j += 2) + dp[0][j] = dp[0][j - 2] && p[j - 1] == '*'; + for(int i = 1; i < m; i++) { + for(int j = 1; j < n; j++) { + dp[i][j] = p[j - 1] == '*' ? + dp[i][j - 2] || dp[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == '.'): + dp[i - 1][j - 1] && (p[j - 1] == '.' || s[i - 1] == p[j - 1]); + } + } + return dp[m - 1][n - 1]; + } +}; +``` + +以上代码利用布尔运算实现简短长度，若阅读不畅，可先理解以下代码，与文中内容一一对应： + +```Python [] +class Solution: + def isMatch(self, s: str, p: str) -> bool: + m, n = len(s) + 1, len(p) + 1 + dp = [[False] * n for _ in range(m)] + dp[0][0] = True + # 初始化首行 + for j in range(2, n, 2): + dp[0][j] = dp[0][j - 2] and p[j - 1] == '*' + # 状态转移 + for i in range(1, m): + for j in range(1, n): + if p[j - 1] == '*': + if dp[i][j - 2]: dp[i][j] = True # 1. + elif dp[i - 1][j] and s[i - 1] == p[j - 2]: dp[i][j] = True # 2. + elif dp[i - 1][j] and p[j - 2] == '.': dp[i][j] = True # 3. + else: + if dp[i - 1][j - 1] and s[i - 1] == p[j - 1]: dp[i][j] = True # 1. + elif dp[i - 1][j - 1] and p[j - 1] == '.': dp[i][j] = True # 2. + return dp[-1][-1] +``` + +```Java [] +class Solution { + public boolean isMatch(String s, String p) { + int m = s.length() + 1, n = p.length() + 1; + boolean[][] dp = new boolean[m][n]; + dp[0][0] = true; + // 初始化首行 + for(int j = 2; j < n; j += 2) + dp[0][j] = dp[0][j - 2] && p.charAt(j - 1) == '*'; + // 状态转移 + for(int i = 1; i < m; i++) { + for(int j = 1; j < n; j++) { + if (p.charAt(j - 1) == '*') { + if (dp[i][j - 2]) dp[i][j] = true; // 1. + else if (dp[i - 1][j] && s.charAt(i - 1) == p.charAt(j - 2)) dp[i][j] = true; // 2. + else if (dp[i - 1][j] && p.charAt(j - 2) == '.') dp[i][j] = true; // 3. + } else { + if (dp[i - 1][j - 1] && s.charAt(i - 1) == p.charAt(j - 1)) dp[i][j] = true; // 1. + else if (dp[i - 1][j - 1] && p.charAt(j - 1) == '.') dp[i][j] = true; // 2. + } + } + } + return dp[m - 1][n - 1]; + } +} +``` + +```C++ [] +class Solution { +public: + bool isMatch(string s, string p) { + int m = s.size() + 1, n = p.size() + 1; + vector> dp(m, vector(n, false)); + dp[0][0] = true; + // 初始化首行 + for(int j = 2; j < n; j += 2) + dp[0][j] = dp[0][j - 2] && p[j - 1] == '*'; + // 状态转移 + for(int i = 1; i < m; i++) { + for(int j = 1; j < n; j++) { + if (p[j - 1] == '*') { + if (dp[i][j - 2]) dp[i][j] = true; // 1. + else if (dp[i - 1][j] && s[i - 1] == p[j - 2]) dp[i][j] = true; // 2. + else if (dp[i - 1][j] && p[j - 2] == '.') dp[i][j] = true; // 3. + } else { + if (dp[i - 1][j - 1] && s[i - 1] == p[j - 1]) dp[i][j] = true; // 1. + else if (dp[i - 1][j - 1] && p[j - 1] == '.') dp[i][j] = true; // 2. + } + } + } + return dp[m - 1][n - 1]; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(MN)$ ：** 其中 $M, N$ 分别为 `s` 和 `p` 的长度，状态转移需遍历整个 `dp` 矩阵。 +- **空间复杂度 $O(MN)$ ：** 状态矩阵 `dp` 使用 $O(MN)$ 的额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/101. \345\257\271\347\247\260\344\272\214\345\217\211\346\240\221.md" "b/selected_coding_interview/docs/101. \345\257\271\347\247\260\344\272\214\345\217\211\346\240\221.md" new file mode 100644 index 0000000..047cf26 --- /dev/null +++ "b/selected_coding_interview/docs/101. \345\257\271\347\247\260\344\272\214\345\217\211\346\240\221.md" @@ -0,0 +1,86 @@ +## 解题思路： + +**对称二叉树定义：** 对于树中 **任意两个对称节点** `L` 和 `R `，一定有： + +- `L.val = R.val` ：即此两对称节点值相等。 +- `L.left.val = R.right.val` ：即 $L$ 的 左子节点 和 $R$ 的 右子节点 对称。 +- `L.right.val = R.left.val` ：即 $L$ 的 右子节点 和 $R$ 的 左子节点 对称。 + +根据以上规律，考虑从顶至底递归，判断每对左右节点是否对称，从而判断树是否为对称二叉树。 + +![Picture1.png](https://pic.leetcode-cn.com/1599398062-PbkpuX-Picture1.png){:width=450} + +### 算法流程： + +**函数 `isSymmetric(root)` ：** + +- **特例处理：** 若根节点 `root` 为空，则直接返回 $true$ 。 +- **返回值：** 即 `recur(root.left, root.right)` ; + +**函数 `recur(L, R)` ：** + +- **终止条件：** + - 当 `L` 和 `R` 同时越过叶节点： 此树从顶至底的节点都对称，因此返回 $true$ 。 + - 当 `L` 或 `R` 中只有一个越过叶节点： 此树不对称，因此返回 $false$ 。 + - 当节点 `L` 值 $\ne$ 节点 `R` 值： 此树不对称，因此返回 $false$ 。 +- **递推工作：** + - 判断两节点 `L.left` 和 `R.right` 是否对称，即 `recur(L.left, R.right)` 。 + - 判断两节点 `L.right` 和 `R.left` 是否对称，即 `recur(L.right, R.left)` 。 +- **返回值：** 两对节点都对称时，才是对称树，因此用与逻辑符 `&&` 连接。 + + + +## 代码： + +```Python [] +class Solution: + def isSymmetric(self, root: Optional[TreeNode]) -> bool: + def recur(L, R): + if not L and not R: return True + if not L or not R or L.val != R.val: return False + return recur(L.left, R.right) and recur(L.right, R.left) + + return not root or recur(root.left, root.right) +``` + +```Java [] +class Solution { + public boolean isSymmetric(TreeNode root) { + return root == null || recur(root.left, root.right); + } + boolean recur(TreeNode L, TreeNode R) { + if (L == null && R == null) return true; + if (L == null || R == null || L.val != R.val) return false; + return recur(L.left, R.right) && recur(L.right, R.left); + } +} +``` + +```C++ [] +class Solution { +public: + bool isSymmetric(TreeNode* root) { + return root == nullptr || recur(root->left, root->right); + } +private: + bool recur(TreeNode* L, TreeNode* R) { + if (L == nullptr && R == nullptr) return true; + if (L == nullptr || R == nullptr || L->val != R->val) return false; + return recur(L->left, R->right) && recur(L->right, R->left); + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为二叉树的节点数量，每次执行 `recur()` 可以判断一对节点是否对称，因此最多调用 $N/2$ 次 `recur()` 方法。 +- **空间复杂度 $O(N)$ ：** 如下图所示，最差情况下（二叉树退化为链表），系统使用 $O(N)$ 大小的空间。 + +![Picture2.png](https://pic.leetcode-cn.com/1599398062-LmPbix-Picture2.png){:width=450} + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/102. \344\272\214\345\217\211\346\240\221\347\232\204\345\261\202\345\272\217\351\201\215\345\216\206.md" "b/selected_coding_interview/docs/102. \344\272\214\345\217\211\346\240\221\347\232\204\345\261\202\345\272\217\351\201\215\345\216\206.md" new file mode 100644 index 0000000..afab2d0 --- /dev/null +++ "b/selected_coding_interview/docs/102. \344\272\214\345\217\211\346\240\221\347\232\204\345\261\202\345\272\217\351\201\215\345\216\206.md" @@ -0,0 +1,99 @@ +## 解题思路： + +**I. 按层打印：** 题目要求的二叉树的 **从上至下** 打印（即按层打印），又称为二叉树的 **广度优先搜索**（BFS）。BFS 通常借助 **队列** 的先入先出特性来实现。 + +**II. 每层打印到一行：** 将本层全部节点打印到一行，并将下一层全部节点加入队列，以此类推，即可分为多行打印。 + +![Picture1.png](https://pic.leetcode-cn.com/a58477c74c96779c265ce3028def7625d16042895d2c21f7fb0293df7b213276-Picture1.png){:width=400} + +### 算法流程： + +1. **特例处理：** 当根节点为空，则返回空列表 `[]` 。 +2. **初始化：** 打印结果列表 `res = []` ，包含根节点的队列 `queue = [root]` 。 +3. **BFS 循环：** 当队列 `queue` 为空时跳出。 + 1. 新建一个临时列表 `tmp` ，用于存储当前层打印结果。 + 2. **当前层打印循环：** 循环次数为当前层节点数（即队列 `queue` 长度）。 + 1. **出队：** 队首元素出队，记为 `node`。 + 2. **打印：** 将 `node.val` 添加至 `tmp` 尾部。 + 3. **添加子节点：** 若 `node` 的左（右）子节点不为空，则将左（右）子节点加入队列 `queue` 。 + 3. 将当前层结果 `tmp` 添加入 `res` 。 +4. **返回值：** 返回打印结果列表 `res` 即可。 + + + +## 代码： + +Python 中使用 collections 中的双端队列 `deque()` ，其 `popleft()` 方法可达到 $O(1)$ 时间复杂度；列表 list 的 `pop(0)` 方法时间复杂度为 $O(N)$ 。 + +```Python [] +class Solution: + def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]: + if not root: return [] + res, queue = [], collections.deque() + queue.append(root) + while queue: + tmp = [] + for _ in range(len(queue)): + node = queue.popleft() + tmp.append(node.val) + if node.left: queue.append(node.left) + if node.right: queue.append(node.right) + res.append(tmp) + return res +``` + +```Java [] +class Solution { + public List> levelOrder(TreeNode root) { + Queue queue = new LinkedList<>(); + List> res = new ArrayList<>(); + if (root != null) queue.add(root); + while (!queue.isEmpty()) { + List tmp = new ArrayList<>(); + for(int i = queue.size(); i > 0; i--) { + TreeNode node = queue.poll(); + tmp.add(node.val); + if (node.left != null) queue.add(node.left); + if (node.right != null) queue.add(node.right); + } + res.add(tmp); + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + vector> levelOrder(TreeNode* root) { + queue que; + vector> res; + if (root != nullptr) que.push(root); + while (!que.empty()) { + vector tmp; + for(int i = que.size(); i > 0; --i) { + root = que.front(); + que.pop(); + tmp.push_back(root->val); + if (root->left != nullptr) que.push(root->left); + if (root->right != nullptr) que.push(root->right); + } + res.push_back(tmp); + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数量，即 BFS 需循环 $N$ 次。 +- **空间复杂度 $O(N)$ ：** 最差情况下，即当树为平衡二叉树时，最多有 $N/2$ 个树节点**同时**在 `queue` 中，使用 $O(N)$ 大小的额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/103. \344\272\214\345\217\211\346\240\221\347\232\204\351\224\257\351\275\277\345\275\242\345\261\202\345\272\217\351\201\215\345\216\206.md" "b/selected_coding_interview/docs/103. \344\272\214\345\217\211\346\240\221\347\232\204\351\224\257\351\275\277\345\275\242\345\261\202\345\272\217\351\201\215\345\216\206.md" new file mode 100644 index 0000000..7a9070e --- /dev/null +++ "b/selected_coding_interview/docs/103. \344\272\214\345\217\211\346\240\221\347\232\204\351\224\257\351\275\277\345\275\242\345\261\202\345\272\217\351\201\215\345\216\206.md" @@ -0,0 +1,287 @@ +## 解题思路： + +相比于上一题 [102. 二叉树的层序遍历](https://leetcode.cn/problems/binary-tree-level-order-traversal/)，这道题额外要求 **打印顺序交替变化**。 + +![Picture1.png](https://pic.leetcode-cn.com/9513dcb034f5dcdea947a2f667b3d380df4f8732da6397778e00718b77584010-Picture1.png){:width=400} + +## 方法一：层序遍历 + 双端队列 + +- 利用双端队列的两端皆可添加元素的特性，设打印列表（双端队列） `tmp` ，并规定： + - 奇数层 则添加至 `tmp` **尾部** ， + - 偶数层 则添加至 `tmp` **头部** 。 + +### 算法流程： + +1. **特例处理：** 当树的根节点为空，则直接返回空列表 `[]` 。 +2. **初始化：** 打印结果空列表 `res` ，包含根节点的双端队列 `deque` 。 +3. **BFS 循环：** 当 `deque` 为空时跳出。 + 1. 新建列表 `tmp` ，用于临时存储当前层打印结果。 + 2. **当前层打印循环：** 循环次数为当前层节点数（即 `deque` 长度）。 + 1. **出队：** 队首元素出队，记为 `node`。 + 2. **打印：** 若为奇数层，将 `node.val` 添加至 `tmp` 尾部；否则，添加至 `tmp` 头部。 + 3. **添加子节点：** 若 `node` 的左（右）子节点不为空，则加入 `deque` 。 + 3. 将当前层结果 `tmp` 转化为 list 并添加入 `res` 。 +4. **返回值：** 返回打印结果列表 `res` 即可。 + + + +### 代码： + +Python 中使用 collections 中的双端队列 `deque()` ，其 `popleft()` 方法可达到 $O(1)$ 时间复杂度；列表 list 的 `pop(0)` 方法时间复杂度为 $O(N)$ 。 + +Java 中将链表 LinkedList 作为双端队列使用。 + +```Python [] +class Solution: + def zigzagLevelOrder(self, root: Optional[TreeNode]) -> List[List[int]]: + if not root: return [] + res, deque = [], collections.deque([root]) + while deque: + tmp = collections.deque() + for _ in range(len(deque)): + node = deque.popleft() + if len(res) % 2 == 0: tmp.append(node.val) # 奇数层 -> 插入队列尾部 + else: tmp.appendleft(node.val) # 偶数层 -> 插入队列头部 + if node.left: deque.append(node.left) + if node.right: deque.append(node.right) + res.append(list(tmp)) + return res +``` + +```Java [] +class Solution { + public List> zigzagLevelOrder(TreeNode root) { + Queue queue = new LinkedList<>(); + List> res = new ArrayList<>(); + if (root != null) queue.add(root); + while (!queue.isEmpty()) { + LinkedList tmp = new LinkedList<>(); + for(int i = queue.size(); i > 0; i--) { + TreeNode node = queue.poll(); + if (res.size() % 2 == 0) tmp.addLast(node.val); + else tmp.addFirst(node.val); + if (node.left != null) queue.add(node.left); + if (node.right != null) queue.add(node.right); + } + res.add(tmp); + } + return res; + } +} +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数量，即 BFS 需循环 $N$ 次，占用 $O(N)$ ；双端队列的队首和队尾的添加和删除操作的时间复杂度均为 $O(1)$ 。 +- **空间复杂度 $O(N)$ ：** 最差情况下，即当树为满二叉树时，最多有 $N/2$ 个树节点 **同时** 在 `deque` 中，使用 $O(N)$ 大小的额外空间。 + +## 方法二：层序遍历 + 双端队列（奇偶层逻辑分离） + +- 方法一代码简短、容易实现；但需要判断每个节点的所在层奇偶性，即冗余了 $N$ 次判断。 +- 通过将奇偶层逻辑拆分，可以消除冗余的判断。 + +### 算法流程： + +> 与方法一对比，仅 BFS 循环不同。 + +- **BFS 循环：** 循环打印奇 / 偶数层，当 `deque` 为空时跳出。 + 1. **打印奇数层：** **从左向右** 打印，**先左后右** 加入下层节点。 + 2. 若 `deque` 为空，说明向下无偶数层，则跳出。 + 3. **打印偶数层：** **从右向左** 打印，**先右后左** 加入下层节点。 + +### 代码： + +```Python [] +class Solution: + def zigzagLevelOrder(self, root: Optional[TreeNode]) -> List[List[int]]: + if not root: return [] + res, deque = [], collections.deque() + deque.append(root) + while deque: + tmp = [] + # 打印奇数层 + for _ in range(len(deque)): + # 从左向右打印 + node = deque.popleft() + tmp.append(node.val) + # 先左后右加入下层节点 + if node.left: deque.append(node.left) + if node.right: deque.append(node.right) + res.append(tmp) + if not deque: break # 若为空则提前跳出 + # 打印偶数层 + tmp = [] + for _ in range(len(deque)): + # 从右向左打印 + node = deque.pop() + tmp.append(node.val) + # 先右后左加入下层节点 + if node.right: deque.appendleft(node.right) + if node.left: deque.appendleft(node.left) + res.append(tmp) + return res +``` + +```Java [] +class Solution { + public List> zigzagLevelOrder(TreeNode root) { + Deque deque = new LinkedList<>(); + List> res = new ArrayList<>(); + if (root != null) deque.add(root); + while (!deque.isEmpty()) { + // 打印奇数层 + List tmp = new ArrayList<>(); + for(int i = deque.size(); i > 0; i--) { + // 从左向右打印 + TreeNode node = deque.removeFirst(); + tmp.add(node.val); + // 先左后右加入下层节点 + if (node.left != null) deque.addLast(node.left); + if (node.right != null) deque.addLast(node.right); + } + res.add(tmp); + if (deque.isEmpty()) break; // 若为空则提前跳出 + // 打印偶数层 + tmp = new ArrayList<>(); + for(int i = deque.size(); i > 0; i--) { + // 从右向左打印 + TreeNode node = deque.removeLast(); + tmp.add(node.val); + // 先右后左加入下层节点 + if (node.right != null) deque.addFirst(node.right); + if (node.left != null) deque.addFirst(node.left); + } + res.add(tmp); + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + vector> zigzagLevelOrder(TreeNode* root) { + deque deque; + vector> res; + if (root != NULL) deque.push_back(root); + while (!deque.empty()) { + // 打印奇数层 + vector tmp; + for(int i = deque.size(); i > 0; i--) { + // 从左向右打印 + TreeNode* node = deque.front(); + deque.pop_front(); + tmp.push_back(node->val); + // 先左后右加入下层节点 + if (node->left != NULL) deque.push_back(node->left); + if (node->right != NULL) deque.push_back(node->right); + } + res.push_back(tmp); + if (deque.empty()) break; // 若为空则提前跳出 + // 打印偶数层 + tmp.clear(); + for(int i = deque.size(); i > 0; i--) { + // 从右向左打印 + TreeNode* node = deque.back(); + deque.pop_back(); + tmp.push_back(node->val); + // 先右后左加入下层节点 + if (node->right != NULL) deque.push_front(node->right); + if (node->left != NULL) deque.push_front(node->left); + } + res.push_back(tmp); + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 同方法一。 +- **空间复杂度 $O(N)$ ：** 同方法一。 + +## 方法三：层序遍历 + 倒序 + +- 此方法的优点是只用列表即可，无需其他数据结构。 +- **偶数层倒序：** 若 `res` 的长度为 **奇数** ，说明当前是偶数层，则对 `tmp` 执行 **倒序** 操作。 + + + +### 代码： + +```Python [] +class Solution: + def zigzagLevelOrder(self, root: Optional[TreeNode]) -> List[List[int]]: + if not root: return [] + res, queue = [], collections.deque() + queue.append(root) + while queue: + tmp = [] + for _ in range(len(queue)): + node = queue.popleft() + tmp.append(node.val) + if node.left: queue.append(node.left) + if node.right: queue.append(node.right) + res.append(tmp[::-1] if len(res) % 2 else tmp) + return res +``` + +```Java [] +class Solution { + public List> zigzagLevelOrder(TreeNode root) { + Queue queue = new LinkedList<>(); + List> res = new ArrayList<>(); + if (root != null) queue.add(root); + while (!queue.isEmpty()) { + List tmp = new ArrayList<>(); + for(int i = queue.size(); i > 0; i--) { + TreeNode node = queue.poll(); + tmp.add(node.val); + if (node.left != null) queue.add(node.left); + if (node.right != null) queue.add(node.right); + } + if (res.size() % 2 == 1) Collections.reverse(tmp); + res.add(tmp); + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + vector> zigzagLevelOrder(TreeNode* root) { + queue que; + vector> res; + if (root != NULL) que.push(root); + while (!que.empty()) { + vector tmp; + for(int i = que.size(); i > 0; i--) { + TreeNode* node = que.front(); + que.pop(); + tmp.push_back(node->val); + if (node->left != NULL) que.push(node->left); + if (node->right != NULL) que.push(node->right); + } + if (res.size() % 2 == 1) reverse(tmp.begin(),tmp.end()); + res.push_back(tmp); + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数量，即 BFS 需循环 $N$ 次，占用 $O(N)$ 。**共完成** 少于 $N$ 个节点的倒序操作，占用 $O(N)$ 。 +- **空间复杂度 $O(N)$ ：** 最差情况下，即当树为满二叉树时，最多有 $N/2$ 个树节点**同时**在 `queue` 中，使用 $O(N)$ 大小的额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/104. \344\272\214\345\217\211\346\240\221\347\232\204\346\234\200\345\244\247\346\267\261\345\272\246.md" "b/selected_coding_interview/docs/104. \344\272\214\345\217\211\346\240\221\347\232\204\346\234\200\345\244\247\346\267\261\345\272\246.md" new file mode 100644 index 0000000..7664055 --- /dev/null +++ "b/selected_coding_interview/docs/104. \344\272\214\345\217\211\346\240\221\347\232\204\346\234\200\345\244\247\346\267\261\345\272\246.md" @@ -0,0 +1,149 @@ +## 解题思路： + +树的遍历方式总体分为两类：深度优先搜索（DFS）、广度优先搜索（BFS）。 + +- **常见 DFS ：** 先序遍历、中序遍历、后序遍历。 +- **常见 BFS ：** 层序遍历（即按层遍历）。 + +求树的深度需要遍历树的所有节点，本文将介绍基于 **后序遍历（DFS）** 和 **层序遍历（BFS）** 的两种解法。 + +## 方法一：后序遍历（DFS） + +树的后序遍历 / 深度优先搜索往往利用 **递归** 或 **栈** 实现，本文使用递归实现。 + +**关键点：** 此树的深度和其左（右）子树的深度之间的关系。显然，**此树的深度** 等于 **左子树的深度** 与 **右子树的深度**中的 **最大值** $+1$ 。 + +![Picture1.png](https://pic.leetcode-cn.com/1603024336-lXVRDd-Picture1.png){:width=450} + +### 算法解析： + +1. **终止条件：** 当 `root​` 为空，说明已越过叶节点，因此返回 深度 $0$ 。 +2. **递推工作：** 本质上是对树做后序遍历。 + 1. 计算节点 `root​` 的 **左子树的深度** ，即调用 `maxDepth(root.left)`。 + 2. 计算节点 `root​` 的 **右子树的深度** ，即调用 `maxDepth(root.right)`。 +3. **返回值：** 返回 **此树的深度** ，即 `max(maxDepth(root.left), maxDepth(root.right)) + 1`。 + + + +### 代码： + +```Python [] +class Solution: + def maxDepth(self, root: Optional[TreeNode]) -> int: + if not root: return 0 + return max(self.maxDepth(root.left), self.maxDepth(root.right)) + 1 +``` + +```Java [] +class Solution { + public int maxDepth(TreeNode root) { + if (root == null) return 0; + return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1; + } +} +``` + +```C++ [] +class Solution { +public: + int maxDepth(TreeNode* root) { + if (root == nullptr) return 0; + return max(maxDepth(root->left), maxDepth(root->right)) + 1; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为树的节点数量，计算树的深度需要遍历所有节点。 +- **空间复杂度 $O(N)$ ：** 最差情况下（当树退化为链表时），递归深度可达到 $N$ 。 + +## 方法二：层序遍历（BFS） + +树的层序遍历 / 广度优先搜索往往利用 **队列** 实现。 + +**关键点：** 每遍历一层，则计数器 $+1$ ，直到遍历完成，则可得到树的深度。 + +### 算法解析： + +1. **特例处理：** 当 `root​` 为空，直接返回 深度 $0$ 。 +2. **初始化：** 队列 `queue` （加入根节点 `root` ），计数器 `res = 0`。 +3. **循环遍历：** 当 `queue` 为空时跳出。 + 1. 初始化一个空列表 `tmp` ，用于临时存储下一层节点。 + 2. 遍历队列： 遍历 `queue` 中的各节点 `node` ，并将其左子节点和右子节点加入 `tmp`。 + 3. 更新队列： 执行 `queue = tmp` ，将下一层节点赋值给 `queue`。 + 4. 统计层数： 执行 `res += 1` ，代表层数加 $1$。 +4. **返回值：** 返回 `res` 即可。 + + + +### 代码： + +```Python [] +class Solution: + def maxDepth(self, root: TreeNode) -> int: + if not root: return 0 + queue, res = [root], 0 + while queue: + tmp = [] + for node in queue: + if node.left: tmp.append(node.left) + if node.right: tmp.append(node.right) + queue = tmp + res += 1 + return res +``` + +```Java [] +class Solution { + public int maxDepth(TreeNode root) { + if (root == null) return 0; + List queue = new LinkedList<>() {{ add(root); }}, tmp; + int res = 0; + while (!queue.isEmpty()) { + tmp = new LinkedList<>(); + for(TreeNode node : queue) { + if (node.left != null) tmp.add(node.left); + if (node.right != null) tmp.add(node.right); + } + queue = tmp; + res++; + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int maxDepth(TreeNode* root) { + if (root == nullptr) return 0; + vector que; + que.push_back(root); + int res = 0; + while (!que.empty()) { + vector tmp; + for(TreeNode* node : que) { + if (node->left != nullptr) tmp.push_back(node->left); + if (node->right != nullptr) tmp.push_back(node->right); + } + que = tmp; + res++; + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为树的节点数量，计算树的深度需要遍历所有节点。 +- **空间复杂度 $O(N)$ ：** 最差情况下（当树平衡时），队列 `queue` 同时存储 $N/2$ 个节点。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/105. \344\273\216\345\211\215\345\272\217\344\270\216\344\270\255\345\272\217\351\201\215\345\216\206\345\272\217\345\210\227\346\236\204\351\200\240\344\272\214\345\217\211\346\240\221.md" "b/selected_coding_interview/docs/105. \344\273\216\345\211\215\345\272\217\344\270\216\344\270\255\345\272\217\351\201\215\345\216\206\345\272\217\345\210\227\346\236\204\351\200\240\344\272\214\345\217\211\346\240\221.md" new file mode 100644 index 0000000..16fce3f --- /dev/null +++ "b/selected_coding_interview/docs/105. \344\273\216\345\211\215\345\272\217\344\270\216\344\270\255\345\272\217\351\201\215\345\216\206\345\272\217\345\210\227\346\236\204\351\200\240\344\272\214\345\217\211\346\240\221.md" @@ -0,0 +1,123 @@ +## 解题思路： + +前序遍历性质： 节点按照 `[ 根节点 | 左子树 | 右子树 ]` 排序。 +中序遍历性质： 节点按照 `[ 左子树 | 根节点 | 右子树 ]` 排序。 + +> 以题目示例为例： +> +> - 前序遍历划分 `[ 3 | 9 | 20 15 7 ]` +> - 中序遍历划分 `[ 9 | 3 | 15 20 7 ]` + +根据以上性质，可得出以下推论： + +1. 前序遍历的首元素 为 树的根节点 `node` 的值。 +2. 在中序遍历中搜索根节点 `node` 的索引 ，可将 中序遍历 划分为 `[ 左子树 | 根节点 | 右子树 ]` 。 +3. 根据中序遍历中的左（右）子树的节点数量，可将 前序遍历 划分为 `[ 根节点 | 左子树 | 右子树 ]` 。 + +![Picture1.png](https://pic.leetcode-cn.com/1629825510-roByLr-Picture1.png){:width=550} + +通过以上三步，可确定 **三个节点** ：1.树的根节点、2.左子树根节点、3.右子树根节点。 + +根据分治思想，对于树的左、右子树，仍可复用以上方法划分子树的左右子树。 + +### 算法流程： + +**递推参数：** 根节点在前序遍历的索引 `root` 、子树在中序遍历的左边界 `left` 、子树在中序遍历的右边界 `right` 。 + +**终止条件：** 当 `left > right` ，代表已经越过叶节点，此时返回 $null$ 。 + +**递推工作：** + +1. **建立根节点 `node` ：** 节点值为 `preorder[root]` 。 +2. **划分左右子树：** 查找根节点在中序遍历 `inorder` 中的索引 `i` 。、 +3. **构建左右子树：** 开启左右子树递归。 + + | | 根节点索引 | 中序遍历左边界 | 中序遍历右边界 | + | ---------- | --------------------- | -------------- | -------------- | + | **左子树** | `root + 1` | `left` | `i - 1` | + | **右子树** | `i - left + root + 1` | `i + 1` | `right` | + + > **TIPS：** `i - left + root + 1`含义为 `根节点索引 + 左子树长度 + 1` + +- **返回值：** 回溯返回 `node` ，作为上一层递归中根节点的左 / 右子节点。 + + + +## 代码： + +注意，本文方法只适用于 “无重复节点值” 的二叉树。 + +为了提升效率，本文使用哈希表 `dic` 存储中序遍历的值与索引的映射，查找操作的时间复杂度为 $O(1)$ 。 + +```Python [] +class Solution: + def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode: + def recur(root, left, right): + if left > right: return # 递归终止 + node = TreeNode(preorder[root]) # 建立根节点 + i = dic[preorder[root]] # 划分根节点、左子树、右子树 + node.left = recur(root + 1, left, i - 1) # 开启左子树递归 + node.right = recur(i - left + root + 1, i + 1, right) # 开启右子树递归 + return node # 回溯返回根节点 + + dic, preorder = {}, preorder + for i in range(len(inorder)): + dic[inorder[i]] = i + return recur(0, 0, len(inorder) - 1) +``` + +```Java [] +class Solution { + int[] preorder; + HashMap dic = new HashMap<>(); + public TreeNode buildTree(int[] preorder, int[] inorder) { + this.preorder = preorder; + for(int i = 0; i < inorder.length; i++) + dic.put(inorder[i], i); + return recur(0, 0, inorder.length - 1); + } + TreeNode recur(int root, int left, int right) { + if (left > right) return null; // 递归终止 + TreeNode node = new TreeNode(preorder[root]); // 建立根节点 + int i = dic.get(preorder[root]); // 划分根节点、左子树、右子树 + node.left = recur(root + 1, left, i - 1); // 开启左子树递归 + node.right = recur(root + i - left + 1, i + 1, right); // 开启右子树递归 + return node; // 回溯返回根节点 + } +} +``` + +```C++ [] +class Solution { +public: + TreeNode* buildTree(vector& preorder, vector& inorder) { + this->preorder = preorder; + for(int i = 0; i < inorder.size(); i++) + dic[inorder[i]] = i; + return recur(0, 0, inorder.size() - 1); + } +private: + vector preorder; + unordered_map dic; + TreeNode* recur(int root, int left, int right) { + if (left > right) return nullptr; // 递归终止 + TreeNode* node = new TreeNode(preorder[root]); // 建立根节点 + int i = dic[preorder[root]]; // 划分根节点、左子树、右子树 + node->left = recur(root + 1, left, i - 1); // 开启左子树递归 + node->right = recur(root + i - left + 1, i + 1, right); // 开启右子树递归 + return node; // 回溯返回根节点 + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为树的节点数量。初始化 HashMap 需遍历 `inorder` ，占用 $O(N)$ 。递归共建立 $N$ 个节点，每层递归中的节点建立、搜索操作占用 $O(1)$ ，因此使用 $O(N)$ 时间。 +- **空间复杂度 $O(N)$ ：** HashMap 使用 $O(N)$ 额外空间；最差情况下（输入二叉树为链表时），递归深度达到 $N$ ，占用 $O(N)$ 的栈帧空间；因此总共使用 $O(N)$ 空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/11. \347\233\233\346\234\200\345\244\232\346\260\264\347\232\204\345\256\271\345\231\250.md" "b/selected_coding_interview/docs/11. \347\233\233\346\234\200\345\244\232\346\260\264\347\232\204\345\256\271\345\231\250.md" new file mode 100644 index 0000000..ce8d663 --- /dev/null +++ "b/selected_coding_interview/docs/11. \347\233\233\346\234\200\345\244\232\346\260\264\347\232\204\345\256\271\345\231\250.md" @@ -0,0 +1,85 @@ +设两指针 $i$ , $j$ ，指向的水槽板高度分别为 $h[i]$ , $h[j]$ ，此状态下水槽面积为 $S(i, j)$ 。由于可容纳水的高度由两板中的 **短板** 决定，因此可得如下 **面积公式** ： + +$$ +S(i, j) = min(h[i], h[j]) × (j - i) +$$ + +![Picture0.png](https://pic.leetcode-cn.com/1628780627-VtSmcP-Picture0.png){:width=500} + +在每个状态下，无论长板或短板向中间收窄一格，都会导致水槽 **底边宽度** $-1$​ 变短： + +- 若向内 **移动短板** ，水槽的短板 $min(h[i], h[j])$ 可能变大，因此下个水槽的面积 **可能增大** 。 +- 若向内 **移动长板** ，水槽的短板 $min(h[i], h[j])$​ 不变或变小，因此下个水槽的面积 **一定变小** 。 + +因此，初始化双指针分列水槽左右两端，循环每轮将短板向内移动一格，并更新面积最大值，直到两指针相遇时跳出；即可获得最大面积。 + +#### 算法流程： + +1. **初始化：** 双指针 $i$ , $j$ 分列水槽左右两端； +2. **循环收窄：** 直至双指针相遇时跳出； + 1. 更新面积最大值 $res$ ； + 2. 选定两板高度中的短板，向中间收窄一格； +3. **返回值：** 返回面积最大值 $res$ 即可； + +#### 正确性证明： + +若暴力枚举，水槽两板围成面积 $S(i, j)$ 的状态总数为 $C(n, 2)$ 。 + +假设状态 $S(i, j)$ 下 $h[i] < h[j]$ ，在向内移动短板至 $S(i + 1, j)$ ，则相当于消去了 ${S(i, j - 1), S(i, j - 2), ... , S(i, i + 1)}$ 状态集合。而所有消去状态的面积一定都小于当前面积（即 $< S(i, j)$），因为这些状态： + +- 短板高度：相比 $S(i, j)$ 相同或更短（即 $\leq h[i]$ ）； +- 底边宽度：相比 $S(i, j)$ 更短； + +因此，每轮向内移动短板，所有消去的状态都 **不会导致面积最大值丢失** ，证毕。 + + + +#### 复杂度分析： + +- **时间复杂度 $O(N)$​ ：** 双指针遍历一次底边宽度 $N$​​ 。 +- **空间复杂度 $O(1)$​ ：** 变量 $i$ , $j$ , $res$ 使用常数额外空间。 + +#### 代码： + +```Python [] +class Solution: + def maxArea(self, height: List[int]) -> int: + i, j, res = 0, len(height) - 1, 0 + while i < j: + if height[i] < height[j]: + res = max(res, height[i] * (j - i)) + i += 1 + else: + res = max(res, height[j] * (j - i)) + j -= 1 + return res +``` + +```Java [] +class Solution { + public int maxArea(int[] height) { + int i = 0, j = height.length - 1, res = 0; + while(i < j) { + res = height[i] < height[j] ? + Math.max(res, (j - i) * height[i++]): + Math.max(res, (j - i) * height[j--]); + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int maxArea(vector& height) { + int i = 0, j = height.size() - 1, res = 0; + while(i < j) { + res = height[i] < height[j] ? + max(res, (j - i) * height[i++]): + max(res, (j - i) * height[j--]); + } + return res; + } +}; +``` \ No newline at end of file diff --git "a/selected_coding_interview/docs/110. \345\271\263\350\241\241\344\272\214\345\217\211\346\240\221.md" "b/selected_coding_interview/docs/110. \345\271\263\350\241\241\344\272\214\345\217\211\346\240\221.md" new file mode 100644 index 0000000..e95bf06 --- /dev/null +++ "b/selected_coding_interview/docs/110. \345\271\263\350\241\241\344\272\214\345\217\211\346\240\221.md" @@ -0,0 +1,162 @@ +## 解题思路： + +以下两种方法均基于以下性质推出：**当前树的深度** 等于 **左子树的深度** 与 **右子树的深度** 中的 **最大值** $+1$ 。 + +![Picture1.png](https://pic.leetcode-cn.com/1603024695-GYNvjf-Picture1.png){:width=450} + +## 方法一：后序遍历 + 剪枝 （从底至顶） + +> 此方法为本题的最优解法，但剪枝的方法不易第一时间想到。 + +思路是对二叉树做后序遍历，从底至顶返回子树深度，若判定某子树不是平衡树则 “剪枝” ，直接向上返回。 + +### 算法流程： + +**函数 `recur(root)` ：** + +- **返回值：** + 1. 当节点`root` 左 / 右子树的深度差 $\leq 1$ ：则返回当前子树的深度，即节点 `root` 的左 / 右子树的深度最大值 $+1$ （ `max(left, right) + 1` ）。 + 2. 当节点`root` 左 / 右子树的深度差 $> 1$ ：则返回 $-1$ ，代表 **此子树不是平衡树** 。 +- **终止条件：** + 1. 当 `root` 为空：说明越过叶节点，因此返回高度 $0$ 。 + 2. 当左（右）子树深度为 $-1$ ：代表此树的 **左（右）子树** 不是平衡树，因此剪枝，直接返回 $-1$ 。 + +**函数 `isBalanced(root)` ：** + +- **返回值：** 若 `recur(root) != -1` ，则说明此树平衡，返回 $true$ ； 否则返回 $false$ 。 + + + +### 代码： + +```Python [] +class Solution: + def isBalanced(self, root: Optional[TreeNode]) -> bool: + def recur(root): + if not root: return 0 + left = recur(root.left) + if left == -1: return -1 + right = recur(root.right) + if right == -1: return -1 + return max(left, right) + 1 if abs(left - right) <= 1 else -1 + + return recur(root) != -1 +``` + +```Java [] +class Solution { + public boolean isBalanced(TreeNode root) { + return recur(root) != -1; + } + + private int recur(TreeNode root) { + if (root == null) return 0; + int left = recur(root.left); + if (left == -1) return -1; + int right = recur(root.right); + if (right == -1) return -1; + return Math.abs(left - right) < 2 ? Math.max(left, right) + 1 : -1; + } +} +``` + +```C++ [] +class Solution { +public: + bool isBalanced(TreeNode* root) { + return recur(root) != -1; + } +private: + int recur(TreeNode* root) { + if (root == nullptr) return 0; + int left = recur(root->left); + if (left == -1) return -1; + int right = recur(root->right); + if (right == -1) return -1; + return abs(left - right) < 2 ? max(left, right) + 1 : -1; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$：** $N$ 为树的节点数；最差情况下，需要递归遍历树的所有节点。 +- **空间复杂度 $O(N)$：** 最差情况下（树退化为链表时），系统递归需要使用 $O(N)$ 的栈空间。 + +## 方法二：先序遍历 + 判断深度 （从顶至底） + +> 此方法容易想到，但会产生大量重复计算，时间复杂度较高。 + +思路是构造一个获取当前子树的深度的函数 `depth(root)` ，通过比较某子树的左右子树的深度差 `abs(depth(root.left) - depth(root.right)) <= 1` 是否成立，来判断某子树是否是二叉平衡树。若所有子树都平衡，则此树平衡。 + +### 算法流程： + +**函数 `isBalanced(root)` ：** 判断树 `root` 是否平衡 + +- **特例处理：** 若树根节点 `root` 为空，则直接返回 $true$ 。 +- **返回值：** 所有子树都需要满足平衡树性质，因此以下三者使用与逻辑 $\&\&$ 连接。 + 1. `abs(self.depth(root.left) - self.depth(root.right)) <= 1` ：判断 **当前子树** 是否是平衡树。 + 2. `self.isBalanced(root.left)` ： 先序遍历递归，判断 **当前子树的左子树** 是否是平衡树。 + 3. `self.isBalanced(root.right)` ： 先序遍历递归，判断 **当前子树的右子树** 是否是平衡树。 + +**函数 `depth(root)` ：** 计算树 `root` 的深度 + +- **终止条件：** 当 `root` 为空，即越过叶子节点，则返回高度 $0$ 。 +- **返回值：** 返回左 / 右子树的深度的最大值 $+1$ 。 + + + +### 代码： + +```Python [] +class Solution: + def isBalanced(self, root: Optional[TreeNode]) -> bool: + if not root: return True + return abs(self.depth(root.left) - self.depth(root.right)) <= 1 and \ + self.isBalanced(root.left) and self.isBalanced(root.right) + + def depth(self, root): + if not root: return 0 + return max(self.depth(root.left), self.depth(root.right)) + 1 +``` + +```Java [] +class Solution { + public boolean isBalanced(TreeNode root) { + if (root == null) return true; + return Math.abs(depth(root.left) - depth(root.right)) <= 1 && isBalanced(root.left) && isBalanced(root.right); + } + + private int depth(TreeNode root) { + if (root == null) return 0; + return Math.max(depth(root.left), depth(root.right)) + 1; + } +} +``` + +```C++ [] +class Solution { +public: + bool isBalanced(TreeNode* root) { + if (root == nullptr) return true; + return abs(depth(root->left) - depth(root->right)) <= 1 && isBalanced(root->left) && isBalanced(root->right); + } +private: + int depth(TreeNode* root) { + if (root == nullptr) return 0; + return max(depth(root->left), depth(root->right)) + 1; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N \log N)$：** 最差情况下（为 “满二叉树” 时）， `isBalanced(root)` 遍历树所有节点，判断每个节点的深度 `depth(root)` 需要遍历 **各子树的所有节点** 。 + - 满二叉树高度的复杂度 $O(log N)$ ，将满二叉树按层分为 $log (N+1)$ 层。 + - 通过调用 `depth(root)` ，判断二叉树各层的节点的对应子树的深度，需遍历节点数量为 $N \times 1, \frac{N-1}{2} \times 2, \frac{N-3}{4} \times 4, \frac{N-7}{8} \times 8, ..., 1 \times \frac{N+1}{2}$ 。因此各层执行 `depth(root)` 的时间复杂度为 $O(N)$ （每层开始，最多遍历 $N$ 个节点，最少遍历 $\frac{N+1}{2}$ 个节点）。 + > 其中，$\frac{N-3}{4} \times 4$ 代表从此层开始总共需遍历 $N-3$ 个节点，此层共有 $4$ 个节点，因此每个子树需遍历 $\frac{N-3}{4}$ 个节点。 + - 因此，总体时间复杂度 $=$ 每层执行复杂度 $\times$ 层数复杂度 = $O(N \times \log N)$ 。 + +![Picture2.png](https://pic.leetcode-cn.com/1603024695-yyFsRH-Picture2.png){:width=550} + +- **空间复杂度 $O(N)$：** 最差情况下（树退化为链表时），系统递归需要使用 $O(N)$ 的栈空间。 diff --git "a/selected_coding_interview/docs/113. \350\267\257\345\276\204\346\200\273\345\222\214 II.md" "b/selected_coding_interview/docs/113. \350\267\257\345\276\204\346\200\273\345\222\214 II.md" new file mode 100644 index 0000000..6af756a --- /dev/null +++ "b/selected_coding_interview/docs/113. \350\267\257\345\276\204\346\200\273\345\222\214 II.md" @@ -0,0 +1,112 @@ +## 解题思路： + +本题是典型的回溯问题，解法包含**先序遍历 + 路径记录**两部分： + +- **先序遍历：** 按照 “根、左、右” 的顺序，遍历树的所有节点。 +- **路径记录：** 在先序遍历中，记录从根节点到当前节点的路径。当路径满足 (1) 根节点到叶节点形成的路径 **且** (2) 各节点值的和等于目标值 `targetSum` 时，将此路径加入结果列表。 + +![Picture1.png](https://pic.leetcode-cn.com/1599400747-BuGhCT-Picture1.png){:width=500} + +### 算法流程： + +**函数 `pathSum(root, targetSum)` ：** + +- **初始化：** 结果列表 `res` ，路径列表 `path` 。 +- **返回值：** 返回 `res` 即可。 + +**函数 `recur(root, tar)` ：** + +- **递推参数：** 当前节点 `root` ，当前目标值 `tar` 。 +- **终止条件：** 若节点 `root` 为空，则直接返回。 +- **递推工作：** + 1. 路径更新： 将当前节点值 `root.val` 加入路径 `path` 。 + 2. 目标值更新： `tar = tar - root.val`（即目标值 `tar` 从 `targetSum` 减至 $0$ ）。 + 3. 路径记录： 当 (1) `root` 为叶节点 **且** (2) 路径和等于目标值 ，则将此路径 `path` 加入 `res` 。 + 4. 先序遍历： 递归左 / 右子节点。 + 5. 路径恢复： 向上回溯前，需要将当前节点从路径 `path` 中删除，即执行 `path.pop()` 。 + + + +## 代码： + +以 Python 语言为例，记录路径时若直接执行 `res.append(path)` ，则是将此 `path` 对象加入了 `res` ；后续 `path` 改变时， `res` 中的 `path` 对象也会随之改变，因此无法实现结果记录。正确做法为： + +- Python: `res.append(list(path))` 。 +- Java: `res.add(new LinkedList(path))` 。 +- C++: `res.push_back(path)` 。 + +三者的原理都是避免直接添加 `path` 对象，而是**拷贝**了一个 `path` 对象并加入到 `res` 。 + +```Python [] +class Solution: + def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]: + res, path = [], [] + def recur(root, tar): + if not root: return + path.append(root.val) + tar -= root.val + if tar == 0 and not root.left and not root.right: + res.append(list(path)) + recur(root.left, tar) + recur(root.right, tar) + path.pop() + + recur(root, targetSum) + return res +``` + +```Java [] +class Solution { + LinkedList> res = new LinkedList<>(); + LinkedList path = new LinkedList<>(); + public List> pathSum(TreeNode root, int targetSum) { + recur(root, targetSum); + return res; + } + void recur(TreeNode root, int tar) { + if (root == null) return; + path.add(root.val); + tar -= root.val; + if (tar == 0 && root.left == null && root.right == null) + res.add(new LinkedList(path)); + recur(root.left, tar); + recur(root.right, tar); + path.removeLast(); + } +} +``` + +```C++ [] +class Solution { +public: + vector> pathSum(TreeNode* root, int targetSum) { + recur(root, targetSum); + return res; + } +private: + vector> res; + vector path; + void recur(TreeNode* root, int tar) { + if (root == nullptr) return; + path.push_back(root->val); + tar -= root->val; + if (tar == 0 && root->left == nullptr && root->right == nullptr) + res.push_back(path); + recur(root->left, tar); + recur(root->right, tar); + path.pop_back(); + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数，先序遍历需要遍历所有节点。 +- **空间复杂度 $O(N)$ ：** 最差情况下，即树退化为链表时，`path` 存储所有树节点，使用 $O(N)$ 额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/121. \344\271\260\345\215\226\350\202\241\347\245\250\347\232\204\346\234\200\344\275\263\346\227\266\346\234\272.md" "b/selected_coding_interview/docs/121. \344\271\260\345\215\226\350\202\241\347\245\250\347\232\204\346\234\200\344\275\263\346\227\266\346\234\272.md" new file mode 100644 index 0000000..fc49258 --- /dev/null +++ "b/selected_coding_interview/docs/121. \344\271\260\345\215\226\350\202\241\347\245\250\347\232\204\346\234\200\344\275\263\346\227\266\346\234\272.md" @@ -0,0 +1,67 @@ +## 解题思路 + +先考虑最简单的「暴力遍历」，即枚举出所有情况，并从中选择最大利润。设数组 `prices` 的长度为 $n$ ，由于只能先买入后卖出，因此第 1 天买可在未来 $n - 1$ 天卖出，第 2 天买可在未来 $n - 2$ 天卖出……以此类推，共有 $(n - 1) + (n - 2) + \cdots + 0 = \frac{n (n - 1)}{2}$ 种情况，时间复杂度为 $O(N^2)$ 。考虑到题目给定的长度范围 $1 \leq prices.length \leq 10^5$ ，需要思考更优解法。 + +然而，暴力法会产生许多冗余计算。例如，若第 1 天价格低于第 2 天价格，即第 1 天成本更低，那么我们一定不会选择在第 2 天买入。进一步的，若在前 $i$ 天选择买入，若想达到最高利润，**则一定选择价格最低的交易日买入**。考虑根据此贪心思想，遍历价格列表 `prices` 并执行两步： + +> 由于初始值 $i = 0$ ，为了序号对应，本文设从第 0 天开始； + +1. 更新前 $i$ 天的最低价格，即最低买入成本 `cost`； +2. 更新前 $i$ 天的最高利润 `profit` ，即选择「前 $i-1$ 天最高利润 `profit` 」和「第 $i$ 天卖出的最高利润 `price - cost` 」中的最大值 ； + +![figures.gif](https://pic.leetcode-cn.com/1658590330-wivils-figures.gif) + +若感觉动图播放太快，可以一页页看 $\downarrow$ + + + +## 代码 + +```Python [] +class Solution: + def maxProfit(self, prices: List[int]) -> int: + cost, profit = float('+inf'), 0 + for price in prices: + cost = min(cost, price) + profit = max(profit, price - cost) + return profit +``` + +```Java [] +class Solution { + public int maxProfit(int[] prices) { + int cost = Integer.MAX_VALUE, profit = 0; + for (int price : prices) { + cost = Math.min(cost, price); + profit = Math.max(profit, price - cost); + } + return profit; + } +} +``` + +```C++ [] +class Solution { +public: + int maxProfit(vector& prices) { + int cost = INT_MAX, profit = 0; + for (int price : prices) { + cost = min(cost, price); + profit = max(profit, price - cost); + } + return profit; + } +}; +``` + +## 复杂度分析 + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为数组 `prices` 长度。遍历 `prices` 使用线性时间。 +- **空间复杂度 $O(1)$ ：** 变量 `cost` , `profit` 使用 $O(1)$ 空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/122. \344\271\260\345\215\226\350\202\241\347\245\250\347\232\204\346\234\200\344\275\263\346\227\266\346\234\272 II\302\240.md" "b/selected_coding_interview/docs/122. \344\271\260\345\215\226\350\202\241\347\245\250\347\232\204\346\234\200\344\275\263\346\227\266\346\234\272 II\302\240.md" new file mode 100644 index 0000000..a8dde6a --- /dev/null +++ "b/selected_coding_interview/docs/122. \344\271\260\345\215\226\350\202\241\347\245\250\347\232\204\346\234\200\344\275\263\346\227\266\346\234\272 II\302\240.md" @@ -0,0 +1,43 @@ +#### 解题思路： + +- **股票买卖策略：** + - **单独交易日：** 设今天价格 $p_1$、明天价格 $p_2$，则今天买入、明天卖出可赚取金额 $p_2 - p_1$ （负值代表亏损）。 + - **连续上涨交易日：** 设此上涨交易日股票价格分别为 $p_1, p_2, ... , p_n$，则第一天买最后一天卖收益最大，即 $p_n - p_1$；等价于每天都买卖，即 $p_n - p_1=(p_2 - p_1)+(p_3 - p_2)+...+(p_n - p_{n-1})$。 + - **连续下降交易日：** 则不买卖收益最大，即不会亏钱。 + +- **算法流程：** + - 遍历整个股票交易日价格列表 `price`，策略是所有上涨交易日都买卖（赚到所有利润），所有下降交易日都不买卖（永不亏钱）。 + 1. 设 `tmp` 为第 `i-1` 日买入与第 `i` 日卖出赚取的利润，即 `tmp = prices[i] - prices[i - 1]` ； + 2. 当该天利润为正 `tmp > 0`，则将利润加入总利润 `profit`；当利润为 $0$ 或为负，则直接跳过； + 3. 遍历完成后，返回总利润 `profit`。 + +- **复杂度分析：** + - **时间复杂度 $O(N)$ ：** 只需遍历一次`price`； + - **空间复杂度 $O(1)$ ：** 变量使用常数额外空间。 + + + +#### 代码： + +```Python [] +class Solution: + def maxProfit(self, prices: List[int]) -> int: + profit = 0 + for i in range(1, len(prices)): + tmp = prices[i] - prices[i - 1] + if tmp > 0: profit += tmp + return profit +``` + +```Java [] +class Solution { + public int maxProfit(int[] prices) { + int profit = 0; + for (int i = 1; i < prices.length; i++) { + int tmp = prices[i] - prices[i - 1]; + if (tmp > 0) profit += tmp; + } + return profit; + } +} +``` diff --git "a/selected_coding_interview/docs/135. \345\210\206\345\217\221\347\263\226\346\236\234.md" "b/selected_coding_interview/docs/135. \345\210\206\345\217\221\347\263\226\346\236\234.md" new file mode 100644 index 0000000..9fc3cbc --- /dev/null +++ "b/selected_coding_interview/docs/135. \345\210\206\345\217\221\347\263\226\346\236\234.md" @@ -0,0 +1,61 @@ +### 解题思路： + +- **规则定义：** 设学生 $A$ 和学生 $B$ 左右相邻，$A$ 在 $B$ 左边； + - **左规则：** 当 $ratings_B>ratings_A$时，$B$ 的糖比 $A$ 的糖数量多。 + - **右规则：** 当 $ratings_A>ratings_B$时，$A$ 的糖比 $B$ 的糖数量多。 + +> 相邻的学生中，评分高的学生必须获得更多的糖果 **等价于** 所有学生满足左规则且满足右规则。 + +- **算法流程：** + + 1. 先从左至右遍历学生成绩 `ratings`，按照以下规则给糖，并记录在 `left` 中： + + 1. 先给所有学生 $1$ 颗糖； + 2. 若 $ratings_i>ratings_{i-1}$，则第 $i$ 名学生糖比第 $i - 1$ 名学生多 $1$ 个。 + 3. 若 $ratings_i<=ratings_{i-1}$，则第 $i$ 名学生糖数量不变。（交由从右向左遍历时处理。） + + - 经过此规则分配后，可以保证所有学生糖数量 **满足左规则** 。 + + 2. 同理，在此规则下从右至左遍历学生成绩并记录在 `right` 中，可以保证所有学生糖数量 **满足右规则** 。 + 3. 最终，取以上 $2$ 轮遍历 `left` 和 `right` 对应学生糖果数的 **最大值** ，这样则 **同时满足左规则和右规则** ，即得到每个同学的最少糖果数量。 + +- **复杂度分析：** + - **时间复杂度 $O(N)$ ：** 遍历两遍数组即可得到结果； + - **空间复杂度 $O(N)$ ：** 需要借用 `left`，`right` 的线性额外空间。 + + + +#### 代码： + +```Python [] +class Solution: + def candy(self, ratings: List[int]) -> int: + left = [1 for _ in range(len(ratings))] + right = left[:] + for i in range(1, len(ratings)): + if ratings[i] > ratings[i - 1]: left[i] = left[i - 1] + 1 + count = left[-1] + for i in range(len(ratings) - 2, -1, -1): + if ratings[i] > ratings[i + 1]: right[i] = right[i + 1] + 1 + count += max(left[i], right[i]) + return count +``` + +```Java [] +class Solution { + public int candy(int[] ratings) { + int[] left = new int[ratings.length]; + int[] right = new int[ratings.length]; + Arrays.fill(left, 1); + Arrays.fill(right, 1); + for(int i = 1; i < ratings.length; i++) + if(ratings[i] > ratings[i - 1]) left[i] = left[i - 1] + 1; + int count = left[ratings.length - 1]; + for(int i = ratings.length - 2; i >= 0; i--) { + if(ratings[i] > ratings[i + 1]) right[i] = right[i + 1] + 1; + count += Math.max(left[i], right[i]); + } + return count; + } +} +``` diff --git "a/selected_coding_interview/docs/136. \345\217\252\345\207\272\347\216\260\344\270\200\346\254\241\347\232\204\346\225\260\345\255\227.md" "b/selected_coding_interview/docs/136. \345\217\252\345\207\272\347\216\260\344\270\200\346\254\241\347\232\204\346\225\260\345\255\227.md" new file mode 100644 index 0000000..cf14b76 --- /dev/null +++ "b/selected_coding_interview/docs/136. \345\217\252\345\207\272\347\216\260\344\270\200\346\254\241\347\232\204\346\225\260\345\255\227.md" @@ -0,0 +1,69 @@ +## 解题思路： + +题目要求时间复杂度 $O(N)$ ，空间复杂度 $O(1)$ ，因此首先排除 **暴力法** 和 **辅助哈希表法** 。 + +设整型数组 $nums$ 中出现一次的数字为 $x$ ，出现两次的数字为 $a, a, b, b, ...$ ，即： + +$$ +nums = [a, a, b, b, ..., x] +$$ + +异或运算有个重要的性质，两个相同数字异或为 $0$ ，即对于任意整数 $a$ 有 $a \oplus a = 0$ 。因此，若将 $nums$ 中所有数字执行异或运算，留下的结果则为 **出现一次的数字 $x$** ，即： + +$$ +\begin{aligned} +& \ \ a \oplus a \oplus b \oplus b \oplus ... \oplus x \\ += & \ \ 0 \oplus 0 \oplus ... \oplus x \\ += & \ \ x +\end{aligned} +$$ + +![Picture1.png](https://pic.leetcode-cn.com/1611393960-EnUIaQ-Picture1.png){:width=500} + +## 代码： + +异或运算满足交换律 $a \oplus b = b \oplus a$ ，即以上运算结果与 $nums$ 的元素顺序无关。代码如下： + +```Python [] +class Solution: + def singleNumber(self, nums: List[int]) -> List[int]: + x = 0 + for num in nums: # 1. 遍历 nums 执行异或运算 + x ^= num + return x; # 2. 返回出现一次的数字 x +``` + +```Java [] +class Solution { + public int singleNumber(int[] nums) { + int x = 0; + for (int num : nums) // 1. 遍历 nums 执行异或运算 + x ^= num; + return x; // 2. 返回出现一次的数字 x + } +} +``` + +```C++ [] +class Solution { +public: + int singleNumber(vector& nums) { + int x = 0; + for (int num : nums) // 1. 遍历 nums 执行异或运算 + x ^= num; + return x; // 2. 返回出现一次的数字 x + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 线性遍历 $nums$ 使用 $O(N)$ 时间，异或操作使用 $O(1)$ 时间。 +- **空间复杂度 $O(1)$ ：** 辅助变量 $a$ , $b$ , $x$ , $y$ 使用常数大小额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/137. \345\217\252\345\207\272\347\216\260\344\270\200\346\254\241\347\232\204\346\225\260\345\255\227 II.md" "b/selected_coding_interview/docs/137. \345\217\252\345\207\272\347\216\260\344\270\200\346\254\241\347\232\204\346\225\260\345\255\227 II.md" new file mode 100644 index 0000000..c8953b9 --- /dev/null +++ "b/selected_coding_interview/docs/137. \345\217\252\345\207\272\347\216\260\344\270\200\346\254\241\347\232\204\346\225\260\345\255\227 II.md" @@ -0,0 +1,206 @@ +#### 解题思路： + +如下图所示，考虑数字的二进制形式，对于出现三次的数字，各 **二进制位** 出现的次数都是 $3$ 的倍数。 +因此，统计所有数字的各二进制位中 $1$ 的出现次数，并对 $3$ 求余，结果则为只出现一次的数字。 + +![Picture1.png](https://pic.leetcode-cn.com/28f2379be5beccb877c8f1586d8673a256594e0fc45422b03773b8d4c8418825-Picture1.png){:width=450} + +#### 方法一：有限状态自动机 + +各二进制位的 **位运算规则相同** ，因此只需考虑一位即可。如下图所示，对于所有数字中的某二进制位 $1$ 的个数，存在 3 种状态，即对 3 余数为 $0, 1, 2$ 。 + +- 若输入二进制位 $1$ ，则状态按照以下顺序转换； +- 若输入二进制位 $0$ ，则状态不变。 + +$$ +0 \rightarrow 1 \rightarrow 2 \rightarrow 0 \rightarrow \cdots +$$ + +![Picture2.png](https://pic.leetcode-cn.com/ab00d4d1ad961a3cd4fc1840e34866992571162096000325e7ce10ff75fda770-Picture2.png){:width=400} + +如下图所示，由于二进制只能表示 $0, 1$ ，因此需要使用两个二进制位来表示 $3$ 个状态。设此两位分别为 $two$ , $one$ ，则状态转换变为： + +$$ +00 \rightarrow 01 \rightarrow 10 \rightarrow 00 \rightarrow \cdots +$$ + +![Picture3.png](https://pic.leetcode-cn.com/0a7ea5bca055b095673620d8bb4c98ef6c610a22f999294ed11ae35d43621e93-Picture3.png){:width=400} + +接下来，需要通过 **状态转换表** 导出 **状态转换的计算公式** 。首先回忆一下位运算特点，对于任意二进制位 $x$ ，有： + +- 异或运算：`x ^ 0 = x`​ ， `x ^ 1 = ~x` +- 与运算：`x & 0 = 0` ， `x & 1 = x` + +**计算 $one$ 方法：** + +设当前状态为 $two$ $one$ ，此时输入二进制位 $n$ 。如下图所示，通过对状态表的情况拆分，可推出 $one$ 的计算方法为： + +```python +if two == 0: + if n == 0: + one = one + if n == 1: + one = ~one +if two == 1: + one = 0 +``` + +引入 **异或运算** ，可将以上拆分简化为： + +```python +if two == 0: + one = one ^ n +if two == 1: + one = 0 +``` + +引入 **与运算** ，可继续简化为： + +```python +one = one ^ n & ~two +``` + +![Picture4.png](https://pic.leetcode-cn.com/f75d89219ad93c69757b187c64784b4c7a57dce7911884fe82f14073d654d32f-Picture4.png){:width=550} + +**计算 $two$ 方法：** + +由于是先计算 $one$ ，因此应在新 $one$ 的基础上计算 $two$ 。 +如下图所示，修改为新 $one$ 后，得到了新的状态图。观察发现，可以使用同样的方法计算 $two$ ，即： + +```python +two = two ^ n & ~one +``` + +![Picture5.png](https://pic.leetcode-cn.com/6ba76dba1ac98ee2bb982e011fdffd1df9a6963f157b2780461dbce453f0ded3-Picture5.png){:width=450} + +**返回值：** + +以上是对数字的二进制中 “一位” 的分析，而 `int` 类型的其他 31 位具有相同的运算规则，因此可将以上公式直接套用在 32 位数上。 + +遍历完所有数字后，各二进制位都处于状态 $00$ 和状态 $01$ （取决于 “只出现一次的数字” 的各二进制位是 $1$ 还是 $0$ ），而此两状态是由 $one$ 来记录的（此两状态下 $twos$ 恒为 $0$ ），因此返回 $ones$ 即可。 + +##### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 位数组 $nums$ 的长度；遍历数组占用 $O(N)$ ，每轮中的常数个位运算操作占用 $O(32 \times3 \times 2) = O(1)$ 。 +- **空间复杂度 $O(1)$ ：** 变量 $ones$ , $twos$ 使用常数大小的额外空间。 + + + +##### 代码： + +```Python [] +class Solution: + def singleNumber(self, nums: List[int]) -> int: + ones, twos = 0, 0 + for num in nums: + ones = ones ^ num & ~twos + twos = twos ^ num & ~ones + return ones +``` + +```Java [] +class Solution { + public int singleNumber(int[] nums) { + int ones = 0, twos = 0; + for(int num : nums){ + ones = ones ^ num & ~twos; + twos = twos ^ num & ~ones; + } + return ones; + } +} +``` + +#### 方法二：遍历统计 + +> 此方法相对容易理解，但效率较低，总体推荐方法一。 + +使用 **与运算** ，可获取二进制数字 $num$ 的最右一位 $n_1$ ： + +$$ +n_1 = num \& i +$$ + +配合 **无符号右移操作** ，可获取 $num$ 所有位的值（即 $n_1$ ~ $n_{32}$ ）： + +$$ +num = num >>> 1 +$$ + +建立一个长度为 32 的数组 $counts$ ，通过以上方法可记录所有数字的各二进制位的 $1$ 的出现次数。 + +```java +int[] counts = new int[32]; +for(int i = 0; i < nums.length; i++) { + for(int j = 0; j < 32; j++) { + counts[j] += nums[i] & 1; // 更新第 j 位 + nums[i] >>>= 1; // 第 j 位 --> 第 j + 1 位 + } +} +``` + +将 $counts$ 各元素对 $3$ 求余，则结果为 “只出现一次的数字” 的各二进制位。 + +```java +for(int i = 0; i < 32; i++) { + counts[i] %= 3; // 得到 只出现一次的数字 的第 (31 - i) 位 +} +``` + +利用 **左移操作** 和 **或运算** ，可将 $counts$ 数组中各二进位的值恢复到数字 $res$ 上（循环区间是 $i \in [0, 31]$ ）。 + +```java +for(int i = 0; i < counts.length; i++) { + res <<= 1; // 左移 1 位 + res |= counts[31 - i]; // 恢复第 i 位的值到 res +} +``` + +最终返回 $res$ 即可。 + +> 由于 Python 的存储负数的特殊性，需要先将 $0$ - $32$ 位取反（即 `res ^ 0xffffffff` ），再将所有位取反（即 `~` ）。 +> 两个组合操作实质上是将数字 $32$ 以上位取反， $0$ - $32$ 位不变。 + +##### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 位数组 $nums$ 的长度；遍历数组占用 $O(N)$ ，每轮中的常数个位运算操作占用 $O(1)$ 。 +- **空间复杂度 $O(1)$ ：** 数组 $counts$ 长度恒为 $32$ ，占用常数大小的额外空间。 + +##### 代码： + +实际上，只需要修改求余数值 $m$ ，即可实现解决 **除了一个数字以外，其余数字都出现 $m$ 次** 的通用问题。 + +```Python [] +class Solution: + def singleNumber(self, nums: List[int]) -> int: + counts = [0] * 32 + for num in nums: + for j in range(32): + counts[j] += num & 1 + num >>= 1 + res, m = 0, 3 + for i in range(32): + res <<= 1 + res |= counts[31 - i] % m + return res if counts[31] % m == 0 else ~(res ^ 0xffffffff) +``` + +```Java [] +class Solution { + public int singleNumber(int[] nums) { + int[] counts = new int[32]; + for(int num : nums) { + for(int j = 0; j < 32; j++) { + counts[j] += num & 1; + num >>>= 1; + } + } + int res = 0, m = 3; + for(int i = 0; i < 32; i++) { + res <<= 1; + res |= counts[31 - i] % m; + } + return res; + } +} +``` diff --git "a/selected_coding_interview/docs/138. \345\244\215\345\210\266\345\270\246\351\232\217\346\234\272\346\214\207\351\222\210\347\232\204\351\223\276\350\241\250.md" "b/selected_coding_interview/docs/138. \345\244\215\345\210\266\345\270\246\351\232\217\346\234\272\346\214\207\351\222\210\347\232\204\351\223\276\350\241\250.md" new file mode 100644 index 0000000..7def256 --- /dev/null +++ "b/selected_coding_interview/docs/138. \345\244\215\345\210\266\345\270\246\351\232\217\346\234\272\346\214\207\351\222\210\347\232\204\351\223\276\350\241\250.md" @@ -0,0 +1,360 @@ +## 解题思路： + +普通链表的节点定义如下： + +```Python [] +# Definition for a Node. +class Node: + def __init__(self, x: int, next: 'Node' = None): + self.val = int(x) + self.next = next +``` + +```Java [] +// Definition for a Node. +class Node { + int val; + Node next; + public Node(int val) { + this.val = val; + this.next = null; + } +} +``` + +```C++ [] +// Definition for a Node. +class Node { +public: + int val; + Node* next; + Node(int _val) { + val = _val; + next = NULL; + } +}; +``` + +本题链表的节点定义如下： + +```Python [] +# Definition for a Node. +class Node: + def __init__(self, x: int, next: 'Node' = None, random: 'Node' = None): + self.val = int(x) + self.next = next + self.random = random +``` + +```Java [] +// Definition for a Node. +class Node { + int val; + Node next, random; + public Node(int val) { + this.val = val; + this.next = null; + this.random = null; + } +} +``` + +```C++ [] +// Definition for a Node. +class Node { +public: + int val; + Node* next; + Node* random; + Node(int _val) { + val = _val; + next = NULL; + random = NULL; + } +}; +``` + +给定链表的头节点 `head` ，复制普通链表很简单，只需遍历链表，每轮建立新节点 + 构建前驱节点 `pre` 和当前节点 `node` 的引用指向即可。 + +本题链表的节点新增了 `random` 指针，指向链表中的 **任意节点** 或者 $null$ 。这个 `random` 指针意味着在复制过程中，除了构建前驱节点和当前节点的引用指向 `pre.next` ，还要构建前驱节点和其随机节点的引用指向 `pre.random` 。 + +**本题难点：** 在复制链表的过程中构建新链表各节点的 `random` 引用指向。 + +![Picture1.png](https://pic.leetcode-cn.com/1604747285-ELUgCd-Picture1.png){:width=450} + +```Python [] +class Solution: + def copyRandomList(self, head: 'Node') -> 'Node': + cur = head + dum = pre = Node(0) + while cur: + node = Node(cur.val) # 复制节点 cur + pre.next = node # 新链表的 前驱节点 -> 当前节点 + # pre.random = '???' # 新链表的 「 前驱节点 -> 当前节点 」 无法确定 + cur = cur.next # 遍历下一节点 + pre = node # 保存当前新节点 + return dum.next +``` + +```Java [] +class Solution { + public Node copyRandomList(Node head) { + Node cur = head; + Node dum = new Node(0), pre = dum; + while(cur != null) { + Node node = new Node(cur.val); // 复制节点 cur + pre.next = node; // 新链表的 前驱节点 -> 当前节点 + // pre.random = "???"; // 新链表的 「 前驱节点 -> 当前节点 」 无法确定 + cur = cur.next; // 遍历下一节点 + pre = node; // 保存当前新节点 + } + return dum.next; + } +} +``` + +```C++ [] +class Solution { +public: + Node* copyRandomList(Node* head) { + Node* cur = head; + Node* dum = new Node(0), *pre = dum; + while(cur != nullptr) { + Node* node = new Node(cur->val); // 复制节点 cur + pre->next = node; // 新链表的 前驱节点 -> 当前节点 + // pre->random = "???"; // 新链表的 「 前驱节点 -> 当前节点 」 无法确定 + cur = cur->next; // 遍历下一节点 + pre = node; // 保存当前新节点 + } + return dum->next; + } +}; +``` + +> 本文介绍 「哈希表」 ，「拼接 + 拆分」 两种方法。哈希表方法比较直观；拼接 + 拆分方法的空间复杂度更低。 + +## 方法一：哈希表 + +利用哈希表的查询特点，考虑构建 **原链表节点** 和 **新链表对应节点** 的键值对映射关系，再遍历构建新链表各节点的 `next` 和 `random` 引用指向即可。 + +### 算法流程： + +1. 若头节点 `head` 为空节点，直接返回 $null$ 。 +2. **初始化：** 哈希表 `dic` ， 节点 `cur` 指向头节点。 +3. **复制链表：** + 1. 建立新节点，并向 `dic` 添加键值对 `(原 cur 节点, 新 cur 节点）` 。 + 2. `cur` 遍历至原链表下一节点。 +4. **构建新链表的引用指向：** + 1. 构建新节点的 `next` 和 `random` 引用指向。 + 2. `cur` 遍历至原链表下一节点。 +5. **返回值：** 新链表的头节点 `dic[cur]` 。 + + + +### 代码： + +```Python [] +class Solution: + def copyRandomList(self, head: 'Node') -> 'Node': + if not head: return + dic = {} + # 3. 复制各节点，并建立 “原节点 -> 新节点” 的 Map 映射 + cur = head + while cur: + dic[cur] = Node(cur.val) + cur = cur.next + cur = head + # 4. 构建新节点的 next 和 random 指向 + while cur: + dic[cur].next = dic.get(cur.next) + dic[cur].random = dic.get(cur.random) + cur = cur.next + # 5. 返回新链表的头节点 + return dic[head] +``` + +```Java [] +class Solution { + public Node copyRandomList(Node head) { + if(head == null) return null; + Node cur = head; + Map map = new HashMap<>(); + // 3. 复制各节点，并建立 “原节点 -> 新节点” 的 Map 映射 + while(cur != null) { + map.put(cur, new Node(cur.val)); + cur = cur.next; + } + cur = head; + // 4. 构建新链表的 next 和 random 指向 + while(cur != null) { + map.get(cur).next = map.get(cur.next); + map.get(cur).random = map.get(cur.random); + cur = cur.next; + } + // 5. 返回新链表的头节点 + return map.get(head); + } +} +``` + +```C++ [] +class Solution { +public: + Node* copyRandomList(Node* head) { + if(head == nullptr) return nullptr; + Node* cur = head; + unordered_map map; + // 3. 复制各节点，并建立 “原节点 -> 新节点” 的 Map 映射 + while(cur != nullptr) { + map[cur] = new Node(cur->val); + cur = cur->next; + } + cur = head; + // 4. 构建新链表的 next 和 random 指向 + while(cur != nullptr) { + map[cur]->next = map[cur->next]; + map[cur]->random = map[cur->random]; + cur = cur->next; + } + // 5. 返回新链表的头节点 + return map[head]; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 两轮遍历链表，使用 $O(N)$ 时间。 +- **空间复杂度 $O(N)$ ：** 哈希表 `dic` 使用线性大小的额外空间。 + +## 方法二：拼接 + 拆分 + +考虑构建 `原节点 1 -> 新节点 1 -> 原节点 2 -> 新节点 2 -> ……` 的拼接链表，如此便可在访问原节点的 `random` 指向节点的同时找到新对应新节点的 `random` 指向节点。 + +### 算法流程： + +1. **复制各节点，构建拼接链表**：设原链表为 $node1 \rightarrow node2 \rightarrow \cdots$ ，构建的拼接链表如下所示： + +$$ +node1 \rightarrow node1_{new} \rightarrow node2 \rightarrow node2_{new} \rightarrow \cdots +$$ + +1. **构建新链表各节点的 `random` 指向**：当访问原节点 `cur` 的随机指向节点 `cur.random` 时，对应新节点 `cur.next` 的随机指向节点为 `cur.random.next` 。 + +2. **拆分原 / 新链表**：设置 `pre` / `cur` 分别指向原 / 新链表头节点，遍历执行 `pre.next = pre.next.next` 和 `cur.next = cur.next.next` 将两链表拆分开。 + +3. 返回新链表的头节点 `res` 即可。 + + + +### 代码： + +```Python [] +class Solution: + def copyRandomList(self, head: 'Node') -> 'Node': + if not head: return + cur = head + # 1. 复制各节点，并构建拼接链表 + while cur: + tmp = Node(cur.val) + tmp.next = cur.next + cur.next = tmp + cur = tmp.next + # 2. 构建各新节点的 random 指向 + cur = head + while cur: + if cur.random: + cur.next.random = cur.random.next + cur = cur.next.next + # 3. 拆分两链表 + cur = res = head.next + pre = head + while cur.next: + pre.next = pre.next.next + cur.next = cur.next.next + pre = pre.next + cur = cur.next + pre.next = None # 单独处理原链表尾节点 + return res # 返回新链表头节点 +``` + +```Java [] +class Solution { + public Node copyRandomList(Node head) { + if(head == null) return null; + Node cur = head; + // 1. 复制各节点，并构建拼接链表 + while(cur != null) { + Node tmp = new Node(cur.val); + tmp.next = cur.next; + cur.next = tmp; + cur = tmp.next; + } + // 2. 构建各新节点的 random 指向 + cur = head; + while(cur != null) { + if(cur.random != null) + cur.next.random = cur.random.next; + cur = cur.next.next; + } + // 3. 拆分两链表 + cur = head.next; + Node pre = head, res = head.next; + while(cur.next != null) { + pre.next = pre.next.next; + cur.next = cur.next.next; + pre = pre.next; + cur = cur.next; + } + pre.next = null; // 单独处理原链表尾节点 + return res; // 返回新链表头节点 + } +} +``` + +```C++ [] +class Solution { +public: + Node* copyRandomList(Node* head) { + if(head == nullptr) return nullptr; + Node* cur = head; + // 1. 复制各节点，并构建拼接链表 + while(cur != nullptr) { + Node* tmp = new Node(cur->val); + tmp->next = cur->next; + cur->next = tmp; + cur = tmp->next; + } + // 2. 构建各新节点的 random 指向 + cur = head; + while(cur != nullptr) { + if(cur->random != nullptr) + cur->next->random = cur->random->next; + cur = cur->next->next; + } + // 3. 拆分两链表 + cur = head->next; + Node* pre = head, *res = head->next; + while(cur->next != nullptr) { + pre->next = pre->next->next; + cur->next = cur->next->next; + pre = pre->next; + cur = cur->next; + } + pre->next = nullptr; // 单独处理原链表尾节点 + return res; // 返回新链表头节点 + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 三轮遍历链表，使用 $O(N)$ 时间。 +- **空间复杂度 $O(1)$ ：** 节点引用变量使用常数大小的额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/142. \347\216\257\345\275\242\351\223\276\350\241\250 II.md" "b/selected_coding_interview/docs/142. \347\216\257\345\275\242\351\223\276\350\241\250 II.md" new file mode 100644 index 0000000..acaf003 --- /dev/null +++ "b/selected_coding_interview/docs/142. \347\216\257\345\275\242\351\223\276\350\241\250 II.md" @@ -0,0 +1,78 @@ +### 解题思路： + +- 这类链表题目一般都是使用双指针法解决的，例如寻找距离尾部第 `K` 个节点、寻找环入口、寻找公共尾部入口等。 + +##### 算法流程： + +1. **双指针第一次相遇：** 设两指针 `fast`，`slow` 指向链表头部 `head`，`fast` 每轮走 $2$ 步，`slow` 每轮走 $1$ 步； + + 1. **第一种结果：** `fast` 指针走过链表末端，说明链表无环，直接返回 `null`； + + - TIPS: 若有环，两指针一定会相遇。因为每走 $1$ 轮，`fast` 与 `slow` 的间距 $+1$，`fast` 终会追上 `slow`； + + 2. **第二种结果：** 当`fast == slow`时， 两指针在环中 **第一次相遇** 。下面分析此时`fast` 与 `slow`走过的 **步数关系** ： + + - 设链表共有 $a+b$ 个节点，其中 **链表头部到链表入口** 有 $a$ 个节点（不计链表入口节点）， **链表环** 有 $b$ 个节点（这里需要注意，$a$ 和 $b$ 是未知数，例如图解上链表 $a=4$ , $b=5$）；设两指针分别走了 $f$，$s$ 步，则有： + + 1. `fast` 走的步数是`slow`步数的 $2$ 倍，即 $f = 2s$；（**解析：** `fast` 每轮走 $2$ 步） + 2. `fast` 比 `slow`多走了 $n$ 个环的长度，即 $f = s + nb$；（ **解析：** 双指针都走过 $a$ 步，然后在环内绕圈直到重合，重合时 `fast` 比 `slow` 多走 **环的长度整数倍** ）； + + - 以上两式相减得：$f = 2nb$，$s = nb$，即`fast`和`slow` 指针分别走了 $2n$，$n$ 个 **环的周长** （注意： $n$ 是未知数，不同链表的情况不同）。 + +2. **目前情况分析：** + + - 如果让指针从链表头部一直向前走并统计步数`k`，那么所有 **走到链表入口节点时的步数** 是：`k=a+nb`（先走 $a$ 步到入口节点，之后每绕 $1$ 圈环（ $b$ 步）都会再次到入口节点）。 + - 而目前，`slow` 指针走过的步数为 $nb$ 步。因此，我们只要想办法让 `slow` 再走 $a$ 步停下来，就可以到环的入口。 + - 但是我们不知道 $a$ 的值，该怎么办？依然是使用双指针法。我们构建一个指针，此指针需要有以下性质：此指针和`slow` 一起向前走 `a` 步后，两者在入口节点重合。那么从哪里走到入口节点需要 $a$ 步？答案是链表头部`head`。 + +3. **双指针第二次相遇：** + + - `slow`指针 **位置不变** ，将`fast`指针重新 **指向链表头部节点** ；`slow`和`fast`同时每轮向前走 $1$ 步； + - TIPS：此时 $f = 0$，$s = nb$ ； + - 当 `fast` 指针走到$f = a$ 步时，`slow` 指针走到步$s = a+nb$，此时 **两指针重合，并同时指向链表环入口** 。 + +4. **返回`slow`指针指向的节点。** + +##### **复杂度分析：** + +- **时间复杂度** $O(N)$ ：第二次相遇中，慢指针须走步数 $a < a + b$；第一次相遇中，慢指针须走步数 $a + b - x < a + b$，其中 $x$ 为双指针重合点与环入口距离；因此总体为线性复杂度； +- **空间复杂度** $O(1)$ ：双指针使用常数大小的额外空间。 + +{:width=500} +{:align=center} + +### 代码： + +```Python [] +class Solution(object): + def detectCycle(self, head): + fast, slow = head, head + while True: + if not (fast and fast.next): return + fast, slow = fast.next.next, slow.next + if fast == slow: break + fast = head + while fast != slow: + fast, slow = fast.next, slow.next + return fast +``` + +```Java [] +public class Solution { + public ListNode detectCycle(ListNode head) { + ListNode fast = head, slow = head; + while (true) { + if (fast == null || fast.next == null) return null; + fast = fast.next.next; + slow = slow.next; + if (fast == slow) break; + } + fast = head; + while (slow != fast) { + slow = slow.next; + fast = fast.next; + } + return fast; + } +} +``` diff --git "a/selected_coding_interview/docs/1480. \344\270\200\347\273\264\346\225\260\347\273\204\347\232\204\345\212\250\346\200\201\345\222\214.md" "b/selected_coding_interview/docs/1480. \344\270\200\347\273\264\346\225\260\347\273\204\347\232\204\345\212\250\346\200\201\345\222\214.md" new file mode 100644 index 0000000..e6d66b3 --- /dev/null +++ "b/selected_coding_interview/docs/1480. \344\270\200\347\273\264\346\225\260\347\273\204\347\232\204\345\212\250\346\200\201\345\222\214.md" @@ -0,0 +1,67 @@ +## 解题思路 + +此题使用求和公式暴力求解的效率较低，因为包含大量重复计算。考虑借助「前一个动态和 $f(i-1)$ 」来计算得到「当前动态和 $f(i)$ 」，此题被约化为一个简单动态规划问题。 + +- **状态定义：** 设前 $i + 1$ 个数字的和为 $f(i)$ ； +- **初始状态：** $f(0) = nums[0]$ ； +- **转移方程：** $f(i) = f(i - 1) + nums[i]$ ； +- **待求数值：** $f(n - 1)$ ，其中 $n$ 为数组 $nums$ 长度； + +![figures.gif](https://pic.leetcode-cn.com/1656947894-plQeLa-figures.gif) + +> 上为动态图，下为静态图，内容一致。 + + + +## 代码 + +细心的我们发现，如果原地修改 `nums` ，可以避免新建 `dp` 带来的内存开销。但通常情况下，不应改变输入变量，因此不建议原地修改 `nums` 数组。 + +```Python [] +class Solution: + def runningSum(self, nums: List[int]) -> List[int]: + dp = [0] * len(nums) + dp[0] = nums[0] + for i in range(1, len(nums)): + dp[i] = dp[i - 1] + nums[i] + return dp +``` + +```Java [] +class Solution { + public int[] runningSum(int[] nums) { + int[] dp = new int[nums.length]; + dp[0] = nums[0]; + for (int i = 1; i < nums.length; i++) { + dp[i] = dp[i - 1] + nums[i]; + } + return dp; + } +} +``` + +```C++ [] +class Solution { +public: + vector runningSum(vector& nums) { + vector dp(nums.size()); + dp[0] = nums[0]; + for (int i = 1; i < nums.size(); i++) { + dp[i] = dp[i - 1] + nums[i]; + } + return dp; + } +}; +``` + +## 复杂度分析 + +- **时间复杂度 $O(N)$ ：** 遍历 `nums` 使用线性时间。 +- **空间复杂度 $O(1)$ ：** 用于保存结果的 `dp` 是必须使用的空间，此处不计入。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/15. \344\270\211\346\225\260\344\271\213\345\222\214.md" "b/selected_coding_interview/docs/15. \344\270\211\346\225\260\344\271\213\345\222\214.md" new file mode 100644 index 0000000..037f2c8 --- /dev/null +++ "b/selected_coding_interview/docs/15. \344\270\211\346\225\260\344\271\213\345\222\214.md" @@ -0,0 +1,71 @@ +#### 解题思路： + +- 暴力法搜索为 $O(N^3)$ 时间复杂度，可通过双指针动态消去无效解来优化效率。 +- **双指针法铺垫：** 先将给定 `nums` 排序，复杂度为 $O(NlogN)$。 +- **双指针法思路：** 固定 $3$ 个指针中最左（最小）数字的指针 `k`，双指针 `i`，`j` 分设在数组索引 $(k, len(nums))$ 两端，通过双指针交替向中间移动，记录对于每个固定指针 `k` 的所有满足 `nums[k] + nums[i] + nums[j] == 0` 的 `i`,`j` 组合： + 1. 当 `nums[k] > 0` 时直接`break`跳出：因为 `nums[j] >= nums[i] >= nums[k] > 0`，即 $3$ 个数字都大于 $0$ ，在此固定指针 `k` 之后不可能再找到结果了。 + 2. 当 `k > 0`且`nums[k] == nums[k - 1]`时即跳过此元素`nums[k]`：因为已经将 `nums[k - 1]` 的所有组合加入到结果中，本次双指针搜索只会得到重复组合。 + 3. `i`，`j` 分设在数组索引 $(k, len(nums))$ 两端，当`i < j`时循环计算`s = nums[k] + nums[i] + nums[j]`，并按照以下规则执行双指针移动： + - 当`s < 0`时，`i += 1`并跳过所有重复的`nums[i]`； + - 当`s > 0`时，`j -= 1`并跳过所有重复的`nums[j]`； + - 当`s == 0`时，记录组合`[k, i, j]`至`res`，执行`i += 1`和`j -= 1`并跳过所有重复的`nums[i]`和`nums[j]`，防止记录到重复组合。 +- **复杂度分析：** + - 时间复杂度 $O(N^2)$：其中固定指针`k`循环复杂度 $O(N)$，双指针 `i`，`j` 复杂度 $O(N)$。 + - 空间复杂度 $O(1)$：指针使用常数大小的额外空间。 + + + +#### 代码： + +```Python [] +class Solution: + def threeSum(self, nums: [int]) -> [[int]]: + nums.sort() + res, k = [], 0 + for k in range(len(nums) - 2): + if nums[k] > 0: break # 1. because of j > i > k. + if k > 0 and nums[k] == nums[k - 1]: continue # 2. skip the same `nums[k]`. + i, j = k + 1, len(nums) - 1 + while i < j: # 3. double pointer + s = nums[k] + nums[i] + nums[j] + if s < 0: + i += 1 + while i < j and nums[i] == nums[i - 1]: i += 1 + elif s > 0: + j -= 1 + while i < j and nums[j] == nums[j + 1]: j -= 1 + else: + res.append([nums[k], nums[i], nums[j]]) + i += 1 + j -= 1 + while i < j and nums[i] == nums[i - 1]: i += 1 + while i < j and nums[j] == nums[j + 1]: j -= 1 + return res +``` + +```Java [] +class Solution { + public List> threeSum(int[] nums) { + Arrays.sort(nums); + List> res = new ArrayList<>(); + for(int k = 0; k < nums.length - 2; k++){ + if(nums[k] > 0) break; + if(k > 0 && nums[k] == nums[k - 1]) continue; + int i = k + 1, j = nums.length - 1; + while(i < j){ + int sum = nums[k] + nums[i] + nums[j]; + if(sum < 0){ + while(i < j && nums[i] == nums[++i]); + } else if (sum > 0) { + while(i < j && nums[j] == nums[--j]); + } else { + res.add(new ArrayList(Arrays.asList(nums[k], nums[i], nums[j]))); + while(i < j && nums[i] == nums[++i]); + while(i < j && nums[j] == nums[--j]); + } + } + } + return res; + } +} +``` diff --git "a/selected_coding_interview/docs/151. \345\217\215\350\275\254\345\255\227\347\254\246\344\270\262\344\270\255\347\232\204\345\215\225\350\257\215.md" "b/selected_coding_interview/docs/151. \345\217\215\350\275\254\345\255\227\347\254\246\344\270\262\344\270\255\347\232\204\345\215\225\350\257\215.md" new file mode 100644 index 0000000..de14f07 --- /dev/null +++ "b/selected_coding_interview/docs/151. \345\217\215\350\275\254\345\255\227\347\254\246\344\270\262\344\270\255\347\232\204\345\215\225\350\257\215.md" @@ -0,0 +1,110 @@ +## 方法一：双指针 + +### 算法解析： + +- 倒序遍历字符串 $s$ ，记录单词左右索引边界 $i$ , $j$ 。 +- 每确定一个单词的边界，则将其添加至单词列表 $res$ 。 +- 最终，将单词列表拼接为字符串，并返回即可。 + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为字符串 $s$ 的长度，线性遍历字符串。 +- **空间复杂度 $O(N)$ ：** 新建的 list(Python) 或 StringBuilder(Java) 中的字符串总长度 $\leq N$ ，占用 $O(N)$ 大小的额外空间。 + + + +### 代码： + +```Python [] +class Solution: + def reverseWords(self, s: str) -> str: + s = s.strip() # 删除首尾空格 + i = j = len(s) - 1 + res = [] + while i >= 0: + while i >= 0 and s[i] != ' ': i -= 1 # 搜索首个空格 + res.append(s[i + 1: j + 1]) # 添加单词 + while i >= 0 and s[i] == ' ': i -= 1 # 跳过单词间空格 + j = i # j 指向下个单词的尾字符 + return ' '.join(res) # 拼接并返回 +``` + +```Java [] +class Solution { + public String reverseWords(String s) { + s = s.trim(); // 删除首尾空格 + int j = s.length() - 1, i = j; + StringBuilder res = new StringBuilder(); + while (i >= 0) { + while (i >= 0 && s.charAt(i) != ' ') i--; // 搜索首个空格 + res.append(s.substring(i + 1, j + 1) + " "); // 添加单词 + while (i >= 0 && s.charAt(i) == ' ') i--; // 跳过单词间空格 + j = i; // j 指向下个单词的尾字符 + } + return res.toString().trim(); // 转化为字符串并返回 + } +} +``` + +## 方法二：分割 + 倒序 + +利用 “字符串分割”、“列表倒序” 的内置函数 *（面试时不建议使用）* ，可简便地实现本题的字符串翻转要求。 + +### 算法解析： + +- **Python ：** 由于 $split()$ 方法将单词间的 “多个空格看作一个空格” （参考自 [split()和split(' ')的区别](https://www.cnblogs.com/python-coder/p/10073329.html) ），因此不会出现多余的 “空单词” 。因此，直接利用 $reverse()$ 方法翻转单词列表 $strs$ ，拼接为字符串并返回即可。 + +![Picture1.png](https://pic.leetcode-cn.com/1600795186-pzCvCm-Picture1.png){:width=500} + +- **Java ：** 以空格为分割符完成字符串分割后，若两单词间有 $x > 1$ 个空格，则在单词列表 $strs$ 中，此两单词间会多出 $x - 1$ 个 “空单词” （即 `""` ）。解决方法：倒序遍历单词列表，并将单词逐个添加至 StringBuilder ，遇到空单词时跳过。 + +![Picture2.png](https://pic.leetcode-cn.com/1600795186-RmKJXL-Picture2.png){:width=500} + +### 代码： + +```Python [] +class Solution: + def reverseWords(self, s: str) -> str: + s = s.strip() # 删除首尾空格 + strs = s.split() # 分割字符串 + strs.reverse() # 翻转单词列表 + return ' '.join(strs) # 拼接为字符串并返回 +``` + +```Java [] +class Solution { + public String reverseWords(String s) { + String[] strs = s.trim().split(" "); // 删除首尾空格，分割字符串 + StringBuilder res = new StringBuilder(); + for (int i = strs.length - 1; i >= 0; i--) { // 倒序遍历单词列表 + if (strs[i].equals("")) continue; // 遇到空单词则跳过 + res.append(strs[i] + " "); // 将单词拼接至 StringBuilder + } + return res.toString().trim(); // 转化为字符串，删除尾部空格，并返回 + } +} +``` + +Python 可一行实现： + +```Python [] +class Solution: + def reverseWords(self, s: str) -> str: + return ' '.join(s.strip().split()[::-1]) +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 总体为线性时间复杂度，各函数时间复杂度和参考资料链接如下。 + - [`split()` 方法：](https://softwareengineering.stackexchange.com/questions/331909/whats-the-complexity-of-javas-string-split-function) 为 $O(N)$ 。 + - [`trim()` 和 `strip()` 方法：](https://stackoverflow.com/questions/51110114/is-string-trim-faster-than-string-replace) 最差情况下（当字符串全为空格时），为 $O(N)$ 。 + - [`join()` 方法：](https://stackoverflow.com/questions/37133547/time-complexity-of-string-concatenation-in-python) 为 $O(N)$ 。 + - [`reverse()` 方法：](https://stackoverflow.com/questions/37606159/what-is-the-time-complexity-of-python-list-reverse) 为 $O(N)$ 。 +- **空间复杂度 $O(N)$ ：** 单词列表 $strs$ 占用线性大小的额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/154. \345\257\273\346\211\276\346\227\213\350\275\254\346\216\222\345\272\217\346\225\260\347\273\204\344\270\255\347\232\204\346\234\200\345\260\217\345\200\274 II.md" "b/selected_coding_interview/docs/154. \345\257\273\346\211\276\346\227\213\350\275\254\346\216\222\345\272\217\346\225\260\347\273\204\344\270\255\347\232\204\346\234\200\345\260\217\345\200\274 II.md" new file mode 100644 index 0000000..ccd5e78 --- /dev/null +++ "b/selected_coding_interview/docs/154. \345\257\273\346\211\276\346\227\213\350\275\254\346\216\222\345\272\217\346\225\260\347\273\204\344\270\255\347\232\204\346\234\200\345\260\217\345\200\274 II.md" @@ -0,0 +1,166 @@ +## 解题思路： + +如下图所示，寻找旋转数组的最小元素即为寻找 **右排序数组** 的首个元素 $nums[x]$ ，称 $x$ 为 **旋转点** 。 + +![Picture1.png](https://pic.leetcode-cn.com/1599404042-JMvjtL-Picture1.png){:width=450} + +排序数组的查找问题首先考虑使用 **二分法** 解决，其可将 **遍历法** 的 **线性级别** 时间复杂度降低至 **对数级别** 。 + +### 算法流程： + +1. **初始化：** 声明 $i$, $j$ 双指针分别指向 $nums$ 数组左右两端。 +2. **循环二分：** 设 $m = (i + j) / 2$ 为每次二分的中点（ "`/`" 代表向下取整除法，因此恒有 $i \leq m < j$ ），可分为以下三种情况： + 1. **当 $nums[m] > nums[j]$ 时：** $m$ 一定在 左排序数组 中，即旋转点 $x$ 一定在 $[m + 1, j]$ 闭区间内，因此执行 $i = m + 1$。 + 2. **当 $nums[m] < nums[j]$ 时：** $m$ 一定在 右排序数组 中，即旋转点 $x$ 一定在$[i, m]$ 闭区间内，因此执行 $j = m$。 + 3. **当 $nums[m] = nums[j]$ 时：** 无法判断 $m$ 在哪个排序数组中，即无法判断旋转点 $x$ 在 $[i, m]$ 还是 $[m + 1, j]$ 区间中。**解决方案：** 执行 $j = j - 1$ 缩小判断范围，分析见下文。 +3. **返回值：** 当 $i = j$ 时跳出二分循环，并返回 **旋转点的值** $nums[i]$ 即可。 + +### 正确性证明： + +当 $nums[m] = nums[j]$ 时，无法判定 $m$ 在左（右）排序数组，自然也无法通过二分法安全地缩小区间，因为其会导致旋转点 $x$ 不在区间 $[i, j]$ 内。举例如下： + +> 设以下两个旋转点值为 $0$ 的示例数组，则当 $i = 0$, $j = 4$ 时 $m = 2$ ，两示例结果不同。 +> 示例一 $[1, 0, 1, 1, 1]$ ：旋转点 $x = 1$ ，因此 $m = 2$ 在 **右排序数组** 中。 +> 示例二 $[1, 1, 1, 0, 1]$ ：旋转点 $x = 3$ ，因此 $m = 2$ 在 **左排序数组** 中。 + +而证明 $j = j - 1$ 正确（缩小区间安全性），需分为两种情况： + +1. **当 $x < j$ 时：** 易得执行 $j = j - 1$ 后，旋转点 $x$ 仍在区间 $[i, j]$ 内。 +2. **当 $x = j$ 时：** 执行 $j = j - 1$ 后越过（丢失）了旋转点 $x$ ，但最终返回的元素值 $nums[i]$ 仍等于旋转点值 $nums[x]$ 。 + + 1. 由于 $x = j$ ，因此 $nums[x] = nums[j] = nums[m] \leq number[i]$ ; + 2. 又由于 $i \leq m 综上所述，此方法可以保证返回值 $nums[i]$ 等于旋转点值 $nums[x]$ ，但在少数特例下 $i \ne x$ ；而本题目只要求返回 “旋转点的值” ，因此本方法正确。 + +**补充思考：** 为什么本题二分法不用 $nums[m]$ 和 $nums[i]$ 作比较？ + +二分目的是判断 $m$ 在哪个排序数组中，从而缩小区间。而在 $nums[m] > nums[i]$情况下，无法判断 $m$ 在哪个排序数组中。本质上是由于 $j$ 初始值肯定在右排序数组中； $i$ 初始值无法确定在哪个排序数组中。举例如下： + +> 对于以下两示例，当 $i = 0, j = 4, m = 2$ 时，有 `nums[m] > nums[i]` ，而结果不同。 +> $[1, 2, 3, 4 ,5]$ 旋转点 $x = 0$ ： $m$ 在右排序数组（此示例只有右排序数组）。 +> $[3, 4, 5, 1 ,2]$ 旋转点 $x = 3$ ： $m$ 在左排序数组。 + + + +## 代码： + +```Python [] +class Solution: + def findMin(self, nums: [int]) -> int: + i, j = 0, len(nums) - 1 + while i < j: + m = (i + j) // 2 + if nums[m] > nums[j]: i = m + 1 + elif nums[m] < nums[j]: j = m + else: j -= 1 + return nums[i] +``` + +```Java [] +class Solution { + public int findMin(int[] nums) { + int i = 0, j = nums.length - 1; + while (i < j) { + int m = (i + j) / 2; + if (nums[m] > nums[j]) i = m + 1; + else if (nums[m] < nums[j]) j = m; + else j--; + } + return nums[i]; + } +} +``` + +```C++ [] +class Solution { +public: + int findMin(vector& nums) { + int i = 0, j = nums.size() - 1; + while (i < j) { + int m = (i + j) / 2; + if (nums[m] > nums[j]) i = m + 1; + else if (nums[m] < nums[j]) j = m; + else j--; + } + return nums[i]; + } +}; +``` + +实际上，当出现 $nums[m] = nums[j]$ 时，一定有区间 $[i, m]$ 内所有元素相等 或 区间 $[m, j]$ 内所有元素相等（或两者皆满足）。对于寻找此类数组的最小值问题，可直接放弃二分查找，而使用线性查找替代。 + +```Python [] +class Solution: + def findMin(self, nums: [int]) -> int: + i, j = 0, len(nums) - 1 + while i < j: + m = (i + j) // 2 + if nums[m] > nums[j]: i = m + 1 + elif nums[m] < nums[j]: j = m + else: return min(nums[i:j]) + return nums[i] +``` + +```Java [] +class Solution { + public int findMin(int[] nums) { + int i = 0, j = nums.length - 1; + while (i < j) { + int m = (i + j) / 2; + if (nums[m] > nums[j]) i = m + 1; + else if (nums[m] < nums[j]) j = m; + else { + int x = i; + for(int k = i + 1; k < j; k++) { + if(nums[k] < nums[x]) x = k; + } + return nums[x]; + } + } + return nums[i]; + } +} +``` + +```C++ [] +class Solution { +public: + int findMin(vector& nums) { + int i = 0, j = nums.size() - 1; + while (i < j) { + int m = (i + j) / 2; + if (nums[m] > nums[j]) i = m + 1; + else if (nums[m] < nums[j]) j = m; + else { + int x = i; + for(int k = i + 1; k < j; k++) { + if(nums[k] < nums[x]) x = k; + } + return nums[x]; + } + } + return nums[i]; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(\log_2 N)$ ：** 在特例情况下（例如 $[1, 1, 1, 1]$），会退化到 $O(N)$。 +- **空间复杂度 $O(1)$ ：** $i$ , $j$ , $m$ 变量使用常数大小的额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/155. \346\234\200\345\260\217\346\240\210\302\240.md" "b/selected_coding_interview/docs/155. \346\234\200\345\260\217\346\240\210\302\240.md" new file mode 100644 index 0000000..83f89a6 --- /dev/null +++ "b/selected_coding_interview/docs/155. \346\234\200\345\260\217\346\240\210\302\240.md" @@ -0,0 +1,64 @@ +#### 解题思路： + +- 借用一个辅助栈 `min_stack`，用于存获取 `stack` 中最小值。 + +- **算法流程：** + - **`push()` 方法：** 每当`push()`新值进来时，如果 **小于等于** `min_stack` 栈顶值，则一起 `push()` 到 `min_stack`，即更新了栈顶最小值； + - **`pop()` 方法：** 判断将 `pop()` 出去的元素值是否是 `min_stack` 栈顶元素值（即最小值），如果是则将 `min_stack` 栈顶元素一起 `pop()`，这样可以保证 `min_stack` 栈顶元素始终是 `stack` 中的最小值。 + - **`getMin()`方法：** 返回 `min_stack` 栈顶即可。 + +- **`min_stack` 作用分析：** + - `min_stack` 等价于遍历 `stack`所有元素，把升序的数字都删除掉，留下一个从栈底到栈顶降序的栈。 + - 相当于给 `stack` 中的降序元素做了标记，每当 `pop()` 这些降序元素，`min_stack` 会将相应的栈顶元素 `pop()` 出去，保证其栈顶元素始终是 `stack` 中的最小元素。 + +- **复杂度分析：** + - **时间复杂度 $O(1)$** ：压栈，出栈，获取最小值的时间复杂度都为 $O(1)$ 。 + - **空间复杂度 $O(N)$** ：包含 $N$ 个元素辅助栈占用线性大小的额外空间。 + +![155.gif](https://pic.leetcode-cn.com/28724fa9f92b6952f7fdaf8760edd1dea850b137c22df28751f1cdd4d2680992-155.gif) + +#### 代码： + +```Python [] +class MinStack: + def __init__(self): + self.stack = [] + self.min_stack = [] + def push(self, x: int) -> None: + self.stack.append(x) + if not self.min_stack or x <= self.min_stack[-1]: + self.min_stack.append(x) + def pop(self) -> None: + if self.stack.pop() == self.min_stack[-1]: + self.min_stack.pop() + def top(self) -> int: + return self.stack[-1] + def getMin(self) -> int: + return self.min_stack[-1] +``` + +```Java [] +class MinStack { + private Stack stack; + private Stack min_stack; + public MinStack() { + stack = new Stack<>(); + min_stack = new Stack<>(); + } + public void push(int x) { + stack.push(x); + if(min_stack.isEmpty() || x <= min_stack.peek()) + min_stack.push(x); + } + public void pop() { + if(stack.pop().equals(min_stack.peek())) + min_stack.pop(); + } + public int top() { + return stack.peek(); + } + public int getMin() { + return min_stack.peek(); + } +} +``` diff --git "a/selected_coding_interview/docs/160. \347\233\270\344\272\244\351\223\276\350\241\250.md" "b/selected_coding_interview/docs/160. \347\233\270\344\272\244\351\223\276\350\241\250.md" new file mode 100644 index 0000000..5893431 --- /dev/null +++ "b/selected_coding_interview/docs/160. \347\233\270\344\272\244\351\223\276\350\241\250.md" @@ -0,0 +1,81 @@ +#### 解题思路： + +设「第一个公共节点」为 `node` ，「链表 `headA`」的节点数量为 $a$ ，「链表 `headB`」的节点数量为 $b$ ，「两链表的公共尾部」的节点数量为 $c$ ，则有： + +- 头节点 `headA` 到 `node` 前，共有 $a - c$ 个节点； +- 头节点 `headB` 到 `node` 前，共有 $b - c$ 个节点； + +![Picture1.png](https://pic.leetcode-cn.com/1615224578-EBRtwv-Picture1.png){:width=500} + +考虑构建两个节点指针 `A​` , `B` 分别指向两链表头节点 `headA` , `headB` ，做如下操作： + +- 指针 `A` 先遍历完链表 `headA` ，再开始遍历链表 `headB` ，当走到 `node` 时，共走步数为： + +$$ +a + (b - c) +$$ + +- 指针 `B` 先遍历完链表 `headB` ，再开始遍历链表 `headA` ，当走到 `node` 时，共走步数为： + +$$ +b + (a - c) +$$ + +如下式所示，此时指针 `A` , `B` 重合，并有两种情况： + +$$ +a + (b - c) = b + (a - c) +$$ + +1. 若两链表 **有** 公共尾部 (即 $c > 0$ ) ：指针 `A` , `B` 同时指向「第一个公共节点」`node` 。 +2. 若两链表 **无** 公共尾部 (即 $c = 0$ ) ：指针 `A` , `B` 同时指向 $null$ 。 + +因此返回 `A` 即可。 + +> 如下图所示，为 $a = 5$ , $b = 3$ , $c = 2$ 示例的算法执行过程。 + + + +##### 复杂度分析： + +- **时间复杂度 $O(a + b)$ ：** 最差情况下（即 $|a - b| = 1$ , $c = 0$ ），此时需遍历 $a + b$ 个节点。 +- **空间复杂度 $O(1)$ ：** 节点指针 `A` , `B` 使用常数大小的额外空间。 + +#### 代码： + +```Python [] +class Solution: + def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode: + A, B = headA, headB + while A != B: + A = A.next if A else headB + B = B.next if B else headA + return A +``` + +```Java [] +public class Solution { + public ListNode getIntersectionNode(ListNode headA, ListNode headB) { + ListNode A = headA, B = headB; + while (A != B) { + A = A != null ? A.next : headB; + B = B != null ? B.next : headA; + } + return A; + } +} +``` + +```C++ [] +class Solution { +public: + ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { + ListNode *A = headA, *B = headB; + while (A != B) { + A = A != nullptr ? A->next : headB; + B = B != nullptr ? B->next : headA; + } + return A; + } +}; +``` diff --git "a/selected_coding_interview/docs/167. \344\270\244\346\225\260\344\271\213\345\222\214 II.md" "b/selected_coding_interview/docs/167. \344\270\244\346\225\260\344\271\213\345\222\214 II.md" new file mode 100644 index 0000000..764a781 --- /dev/null +++ "b/selected_coding_interview/docs/167. \344\270\244\346\225\260\344\271\213\345\222\214 II.md" @@ -0,0 +1,85 @@ +## 解题思路： + +利用 HashMap 可以通过遍历数组找到数字组合，时间和空间复杂度均为 $O(N)$ 。 +注意本题的 $numbers$ 是 **排序数组** ，因此可使用 **双指针法** 将空间复杂度降低至 $O(1)$ 。 + +### 算法流程： + +1. **初始化：** 双指针 $i$ , $j$ 分别指向数组 $numbers$ 的左右两端 *（俗称对撞双指针）*。 +2. **循环搜索：** 当双指针相遇时跳出。 + 1. 计算和 $s = numbers[i] + numbers[j]$ 。 + 2. 若 $s > target$ ，则指针 $j$ 向左移动，即执行 $j = j - 1$ 。 + 3. 若 $s < target$ ，则指针 $i$ 向右移动，即执行 $i = i + 1$ 。 + 4. 若 $s = target$ ，由于题目要求索引从 $1$ 开始，因此返回数组 $[i + 1, j + 1]$ 。 +3. 若循环结束，则返回空数组，代表无和为 $target$ 的数字组合。 + + + +### 正确性证明： + +> 记每个状态为 $S(i, j)$ ，即 $S(i, j) = numbers[i] + numbers[j]$ 。假设 $S(i, j) < target$ ，则执行 $i = i + 1$ ，即状态切换至 $S(i + 1, j)$ 。 + +状态 $S(i, j)$ 切换至 $S(i + 1, j)$ ，则会消去一行元素，相当于 **消去了状态集合** {$S(i, i + 1), S(i, i + 2), ..., S(i, j - 2), S(i, j - 1), S(i, j)$ } 。（由于双指针都是向中间收缩，因此这些状态之后不可能再遇到）。 + +由于 $numbers$ 是排序数组，因此这些 **消去的状态** 都一定满足 $S(i, j) < target$ ，即这些状态都 **不是解** 。 + +**结论：** 以上分析已证明 “每次指针 $i$ 的移动操作，都不会导致解的丢失” ，即指针 $i$ 的移动操作是安全的；同理，对于指针 $j$ 可得出同样推论；因此，此双指针法是正确的。 + +![Picture1.png](https://pic.leetcode-cn.com/1600794717-VSmNyQ-Picture1.png){:width=550} + +## 代码： + +```Python [] +class Solution: + def twoSum(self, numbers: List[int], target: int) -> List[int]: + i, j = 0, len(numbers) - 1 + while i < j: + s = numbers[i] + numbers[j] + if s > target: j -= 1 + elif s < target: i += 1 + else: return i + 1, j + 1 + return [] +``` + +```Java [] +class Solution { + public int[] twoSum(int[] numbers, int target) { + int i = 0, j = numbers.length - 1; + while (i < j) { + int s = numbers[i] + numbers[j]; + if (s < target) i++; + else if (s > target) j--; + else return new int[] { i + 1, j + 1 }; + } + return new int[0]; + } +} +``` + +```C++ [] +class Solution { +public: + vector twoSum(vector& numbers, int target) { + int i = 0, j = numbers.size() - 1; + while (i < j) { + int s = numbers[i] + numbers[j]; + if (s < target) i++; + else if (s > target) j--; + else return { i + 1, j + 1 }; + } + return {}; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为数组 $numbers$ 的长度；双指针共同线性遍历整个数组。 +- **空间复杂度 $O(1)$ ：** 变量 $i$, $j$ 使用常数大小的额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/169. \345\244\232\346\225\260\345\205\203\347\264\240.md" "b/selected_coding_interview/docs/169. \345\244\232\346\225\260\345\205\203\347\264\240.md" new file mode 100644 index 0000000..c639d38 --- /dev/null +++ "b/selected_coding_interview/docs/169. \345\244\232\346\225\260\345\205\203\347\264\240.md" @@ -0,0 +1,144 @@ +## 解题思路： + +> 在本文中，“数组中出现次数超过一半的数字” 被称为 **“众数”** 。 +> +> 需要注意的是，数学中众数的定义为 “数组中出现次数最多的数字” ，与本文定义不同。 + +本题常见的三种解法： + +1. **哈希表统计法：** 遍历数组 `nums` ，用 HashMap 统计各数字的数量，即可找出 众数 。此方法时间和空间复杂度均为 $O(N)$ 。 +2. **数组排序法：** 将数组 `nums` 排序，**数组中点的元素** 一定为众数。 +3. **摩尔投票法：** 核心理念为 **票数正负抵消** 。此方法时间和空间复杂度分别为 $O(N)$ 和 $O(1)$ ，为本题的最佳解法。 + +### 摩尔投票： + +> 设输入数组 `nums` 的众数为 $x$ ，数组长度为 $n$ 。 + +**推论一：** 若记 **众数** 的票数为 $+1$ ，**非众数** 的票数为 $-1$ ，则一定有所有数字的 **票数和 $> 0$** 。 + +**推论二：** 若数组的前 $a$ 个数字的 **票数和 $= 0$** ，则 数组剩余 $(n-a)$ 个数字的 **票数和一定仍 $>0$** ，即后 $(n-a)$ 个数字的 **众数仍为 $x$** 。 + +![Picture1.png](https://pic.leetcode-cn.com/1603612327-bOQxzq-Picture1.png){:width=500} + +根据以上推论，记数组首个元素为 $n_1$ ，众数为 $x$ ，遍历并统计票数。当发生 **票数和 $= 0$** 时，**剩余数组的众数一定不变** ，这是由于： + +- **当 $n_1 = x$ ：** 抵消的所有数字中，有一半是众数 $x$ 。 +- **当 $n_1 \neq x$ ：** 抵消的所有数字中，众数 $x$ 的数量最少为 0 个，最多为一半。 + +利用此特性，每轮假设发生 **票数和 $= 0$** 都可以 **缩小剩余数组区间** 。当遍历完成时，最后一轮假设的数字即为众数。 + +### 算法流程: + +1. **初始化：** 票数统计 `votes = 0` ， 众数 `x`。 +2. **循环：** 遍历数组 `nums` 中的每个数字 `num` 。 + 1. 当 票数 `votes` 等于 0 ，则假设当前数字 `num` 是众数。 + 2. 当 `num = x` 时，票数 `votes` 自增 1 ；当 `num != x` 时，票数 `votes` 自减 1 。 +3. **返回值：** 返回 `x` 即可。 + + + +## 代码： + +```Python [] +class Solution: + def majorityElement(self, nums: List[int]) -> int: + votes = 0 + for num in nums: + if votes == 0: x = num + votes += 1 if num == x else -1 + return x +``` + +```Java [] +class Solution { + public int majorityElement(int[] nums) { + int x = 0, votes = 0; + for (int num : nums){ + if (votes == 0) x = num; + votes += num == x ? 1 : -1; + } + return x; + } +} +``` + +```C++ [] +class Solution { +public: + int majorityElement(vector& nums) { + int x = 0, votes = 0; + for (int num : nums){ + if (votes == 0) x = num; + votes += num == x ? 1 : -1; + } + return x; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为数组 `nums` 长度。 +- **空间复杂度 $O(1)$ ：** `votes` 变量使用常数大小的额外空间。 + +### 拓展： + +由于题目说明“给定的数组总是存在多数元素”，因此本题不用考虑 **数组不存在众数** 的情况。若考虑，需要加入一个 “验证环节” ，遍历数组 `nums` 统计 `x` 的数量。 + +- 若 `x` 的数量超过数组长度一半，则返回 `x` 。 +- 否则，返回未找到众数。 + +```Python [] +class Solution: + def majorityElement(self, nums: List[int]) -> int: + votes, count = 0, 0 + for num in nums: + if votes == 0: x = num + votes += 1 if num == x else -1 + # 验证 x 是否为众数 + for num in nums: + if num == x: count += 1 + return x if count > len(nums) // 2 else 0 # 当无众数时返回 0 +``` + +```Java [] +class Solution { + public int majorityElement(int[] nums) { + int x = 0, votes = 0, count = 0; + for (int num : nums){ + if (votes == 0) x = num; + votes += num == x ? 1 : -1; + } + // 验证 x 是否为众数 + for (int num : nums) + if (num == x) count++; + return count > nums.length / 2 ? x : 0; // 当无众数时返回 0 + } +} +``` + +```C++ [] +class Solution { +public: + int majorityElement(vector& nums) { + int x = 0, votes = 0, count = 0; + for (int num : nums){ + if (votes == 0) x = num; + votes += num == x ? 1 : -1; + } + // 验证 x 是否为众数 + for (int num : nums) + if (num == x) count++; + return count > nums.size() / 2 ? x : 0; // 当无众数时返回 0 + } +}; +``` + +时间复杂度和空间复杂度都不变，仍为 $O(N)$ 和 $O(1)$ 。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/179. \346\234\200\345\244\247\346\225\260.md" "b/selected_coding_interview/docs/179. \346\234\200\345\244\247\346\225\260.md" new file mode 100644 index 0000000..e8be5d8 --- /dev/null +++ "b/selected_coding_interview/docs/179. \346\234\200\345\244\247\346\225\260.md" @@ -0,0 +1,221 @@ +## 解题思路： + +此题求拼接起来的最大数字。设数组 $nums$ 中任意两数字的字符串为 $x$ 和 $y$ ，则规定 **贪心策略**： + +- 若拼接字符串 $x + y > y + x$ ，则 $x$ “大于” $y$ 。 +- 反之，若 $x + y < y + x$ ，则 $x$ “小于” $y$ 。 + +> $x$ “小于” $y$ 代表：排序完成后，数组中 $x$ 应在 $y$ 左边；“大于” 则反之。 + +根据以上规则，套用任何排序方法对 $nums$ 执行排序即可。 + +![Picture1.png](https://pic.leetcode.cn/1690468266-PzVUaU-Picture1.png){:width=500} + +### 算法流程： + +1. **初始化：** 字符串列表 $strs$ ，保存各数字的字符串格式。 +2. **列表排序：** 根据贪心策略对 $strs$ 进行从大到小排序。 +3. **返回值：** 拼接 $strs$ 中的所有字符串，并返回。 + +### 正确性证明： + +本题的贪心策略的正确性证明包括以下两个命题： + +- **反身性**：对于任意的数字 $x$，有 $xx = xx$。 +- **传递性**：假设对于任意的数字 $x$ , $y$ , $z$ ，如果 $xy < yx$ , $yz < zy$ ，那么 $xz < zx$ 一定成立。 + +反身性是显然成立的，因为任意数字和自己拼接总是相等。下面证明传递性： + +设十进制数 $x$ , $y$ , $z$ 分别有 $a$ , $b$ , $c$ 位，则有（左边是字符串拼接，右边是十进制数计算，两者等价）： + +$$ +xy = x \times 10^b + y \\ +yx = y \times 10^a + x +$$ + +则 $xy < yx$ 可转化为： + +$$ +x \times 10^b + y < y \times 10^a + x \\ +x (10^b - 1) < y (10^a - 1) \\ +x / (10^a - 1) < y / (10^b - 1) +$$ + +同理， 可将 $yz < zy$ 转化为： + +$$ +y / (10^b - 1) < z / (10^c - 1) +$$ + +将以上两式合并，整理得： + +$$ +x / (10^a - 1) < y / (10^b - 1) < z / (10^c - 1) \\ +x / (10^a - 1) < z / (10^c - 1) \\ +x (10^c - 1) < z (10^a - 1) \\ +x \times 10^c + z < z \times 10^a + x +$$ + +可推出 $xz$ < $zx$ ，传递性证毕。 + +因此贪心策略是正确的，所得排序结果是全局最优的。 + +## 代码（内置函数）： + +可使用编程语言的内置排序函数实现，需自定义判断规则。 + +```Python [] +class Solution: + def largestNumber(self, nums: List[int]) -> str: + def sort_rule(x, y): + a, b = x + y, y + x + if a < b: return 1 + elif a > b: return -1 + else: return 0 + + strs = [str(num) for num in nums] + strs.sort(key = cmp_to_key(sort_rule)) + if strs[0] == "0": + return "0" + return ''.join(strs) +``` + +```Java [] +class Solution { + public String largestNumber(int[] nums) { + String[] strs = new String[nums.length]; + for (int i = 0; i < nums.length; i++) + strs[i] = String.valueOf(nums[i]); + Arrays.sort(strs, (x, y) -> (y + x).compareTo(x + y)); + if (strs[0].equals("0")) + return "0"; + StringBuilder res = new StringBuilder(); + for (String s : strs) + res.append(s); + return res.toString(); + } +} +``` + +```C++ [] +class Solution { +public: + string largestNumber(vector& nums) { + vector strs; + string res; + for (int i = 0; i < nums.size(); i++) + strs.push_back(to_string(nums[i])); + sort(strs.begin(), strs.end(), [](string& x, string& y){ return y + x < x + y; }); + if (strs[0] == "0") + return "0"; + for (int i = 0; i < strs.size(); i++) + res.append(strs[i]); + return res; + } +}; +``` + +## 代码（快速排序）： + +需修改快速排序函数中的排序判断规则。字符串大小（字典序）对比的实现方法： + +- Python/C++ 中可直接用 `<` , `>`。 +- Java 中使用函数 `A.compareTo(B)`。 + +```Python [] +class Solution: + def largestNumber(self, nums: List[int]) -> str: + def quick_sort(l , r): + if l >= r: return + i, j = l, r + while i < j: + while strs[j] + strs[l] >= strs[l] + strs[j] and i < j: j -= 1 + while strs[i] + strs[l] <= strs[l] + strs[i] and i < j: i += 1 + strs[i], strs[j] = strs[j], strs[i] + strs[i], strs[l] = strs[l], strs[i] + quick_sort(l, i - 1) + quick_sort(i + 1, r) + + strs = [str(num) for num in nums] + quick_sort(0, len(strs) - 1) + if strs[-1] == "0": + return "0" + return ''.join(strs[::-1]) +``` + +```Java [] +class Solution { + void quickSort(String[] strs, int l, int r) { + if (l >= r) return; + int i = l, j = r; + String tmp = strs[i]; + while (i < j) { + while ((strs[j] + strs[l]).compareTo(strs[l] + strs[j]) >= 0 && i < j) j--; + while ((strs[i] + strs[l]).compareTo(strs[l] + strs[i]) <= 0 && i < j) i++; + tmp = strs[i]; + strs[i] = strs[j]; + strs[j] = tmp; + } + strs[i] = strs[l]; + strs[l] = tmp; + quickSort(strs, l, i - 1); + quickSort(strs, i + 1, r); + } + + public String largestNumber(int[] nums) { + String[] strs = new String[nums.length]; + for(int i = 0; i < nums.length; i++) + strs[i] = String.valueOf(nums[i]); + quickSort(strs, 0, strs.length - 1); + StringBuilder res = new StringBuilder(); + if (strs[strs.length - 1].equals("0")) + return "0"; + for(int i = strs.length - 1; i >=0; i--) + res.append(strs[i]); + return res.toString(); + } +} +``` + +```C++ [] +class Solution { +public: + string largestNumber(vector& nums) { + vector strs; + for (int i = 0; i < nums.size(); i++) + strs.push_back(to_string(nums[i])); + quickSort(strs, 0, strs.size() - 1); + if (strs[strs.size() - 1] == "0") + return "0"; + string res; + for (int i = nums.size() - 1; i >=0; i--) + res.append(strs[i]); + return res; + } +private: + void quickSort(vector& strs, int l, int r) { + if (l >= r) return; + int i = l, j = r; + while (i < j) { + while (strs[j] + strs[l] >= strs[l] + strs[j] && i < j) j--; + while (strs[i] + strs[l] <= strs[l] + strs[i] && i < j) i++; + swap(strs[i], strs[j]); + } + swap(strs[i], strs[l]); + quickSort(strs, l, i - 1); + quickSort(strs, i + 1, r); + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N \log N)$ ：** $N$ 为最终返回值的字符数量（ $strs$ 列表的长度 $\leq N$ ）；使用快排或内置函数的平均时间复杂度为 $O(N \log N)$ ，最差为 $O(N^2)$ 。 +- **空间复杂度 $O(N)$ ：** 字符串列表 $strs$ 占用线性大小的额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/1823. \346\211\276\345\207\272\346\270\270\346\210\217\347\232\204\350\216\267\350\203\234\350\200\205.md" "b/selected_coding_interview/docs/1823. \346\211\276\345\207\272\346\270\270\346\210\217\347\232\204\350\216\267\350\203\234\350\200\205.md" new file mode 100644 index 0000000..fc129fb --- /dev/null +++ "b/selected_coding_interview/docs/1823. \346\211\276\345\207\272\346\270\270\346\210\217\347\232\204\350\216\267\350\203\234\350\200\205.md" @@ -0,0 +1,128 @@ +## 解题思路： + +模拟整个删除过程最直观，即构建一个长度为 $n$ 的链表，各节点值为对应的顺序索引；每轮删除第 $k$ 个节点，直至链表长度为 1 时结束，返回最后剩余节点的值即可。 + +模拟法需要循环删除 $n - 1$ 轮，每轮在链表中寻找删除节点需要 $k$ 次访问操作（链表线性遍历），因此总体时间复杂度为 $O(nm)$ 。题目给定的 $k, n$ 取值范围如下所示，观察可知此时间复杂度是不可接受的。 + +$$ +1 \leq n \leq 10^5 \\ +1 \leq k \leq 10^6 +$$ + +> 实际上，本题是著名的 “约瑟夫环” 问题，可使用 **动态规划** 解决。 + +输入 $n, k$ ，记此约瑟夫环问题为 「$n, k$ 问题」 ，设解（即最后留下的数字）为 $f(n)$ ，则有： + +- 「$n, k$ 问题」：数字环为 $0, 1, 2, ..., n - 1$ ，解为 $f(n)$ 。 +- 「$n-1, k$ 问题」：数字环为 $0, 1, 2, ..., n - 2$ ，解为 $f(n-1)$ 。 +- 以此类推…… + +> 请注意，数字环是 **首尾相接** 的，为方便行文，本文使用列表形式表示。 + +对于「$n, k$ 问题」，首轮删除环中第 $k$ 个数字后，得到一个长度为 $n - 1$ 的数字环。由于有可能 $k > n$ ，因此删除的数字为 $(k - 1) \% n$ ，删除后的数字环从下个数字（即 $k \% n$ ）开始，设 $t = k \% n$ ，可得数字环： + +$$ +t, t + 1, t + 2, ..., 0, 1, ..., t - 3, t - 2 +$$ + +删除一轮后的数字环也变为一个「$n-1, k$ 问题」，观察以下数字编号对应关系： + +$$ +\begin{aligned} +「n-1, k 问题」 && \rightarrow && 「n, k 问题」删除后 \\ +0 && \rightarrow && t + 0 \\ +1 && \rightarrow && t + 1 \\ +... && \rightarrow && ... \\ +n - 2 && \rightarrow && t - 2 \\ +\end{aligned} +$$ + +设「$n-1, k$ 问题」某数字为 $x$ ，则可得递推关系： + +$$ +x \rightarrow (x + t) \% n \\ +$$ + +换而言之，若已知「$n-1, k$ 问题」的解 $f(n - 1)$ ，则可通过以上公式计算得到「$n, k$ 问题」的解 $f(n)$ ，即： + +$$ +\begin{aligned} +f(n) & = (f(n - 1) + t) \% n \\ +& = (f(n - 1) + k \% n) \% n \\ +& = (f(n - 1) + k) \% n +\end{aligned} +$$ + +> 以 $n = 5$ , $k = 3$ 的示例如下图所示。 + +![Picture1.png](https://pic.leetcode-cn.com/1615096532-kUoKUe-Picture1.png){:width=550} + +$f(n)$ 可由 $f(n - 1)$ 得到，$f(n - 1)$ 可由 $f(n - 2)$ 得到，……，$f(2)$ 可由 $f(1)$ 得到；因此，若给定 $f(1)$ 的值，就可以递推至任意 $f(n)$ 。而「$1, k$ 问题」的解 $f(1) = 0$ 恒成立，即无论 $k$ 为何值，长度为 1 的数字环留下的是一定是数字 $0$ 。 + +> 以上数学推导本质是得出动态规划的 转移方程 和 初始状态 。 + +### 算法流程： + +1. **状态定义：** 设「$i, k$ 问题」的解为 $dp[i]$ 。 +2. **转移方程：** 通过以下公式可从 $dp[i - 1]$ 递推得到 $dp[i]$ 。 + +$$ +dp[i] = (dp[i - 1] + k) \% i +$$ + +3. **初始状态：**「$1, k$ 问题」的解恒为 $0$ ，即 $dp[1] = 0$ 。 +4. **返回值：** 返回「$n, k$ 问题」的解 $dp[n]$ 。 + +> 如下图所示，为 $n = 5$ , $k = 3$ 时的状态转移和对应的模拟删除过程。 + +![Picture2.png](https://pic.leetcode-cn.com/1613584667-AQpTlK-Picture2.png) + +## 代码： + +根据状态转移方程的递推特性，无需建立状态列表 $dp$ ，而使用一个变量 $x$ 执行状态转移即可。 + +```Python [] +class Solution: + def findTheWinner(self, n: int, k: int) -> int: + x = 0 + for i in range(2, n + 1): + x = (x + k) % i + return x + 1 +``` + +```Java [] +class Solution { + public int findTheWinner(int n, int k) { + int x = 0; + for (int i = 2; i <= n; i++) { + x = (x + k) % i; + } + return x + 1; + } +} +``` + +```C++ [] +class Solution { +public: + int findTheWinner(int n, int k) { + int x = 0; + for (int i = 2; i <= n; i++) { + x = (x + k) % i; + } + return x + 1; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(n)$ ：** 状态转移循环 $n - 1$ 次使用 $O(n)$ 时间，状态转移方程计算使用 $O(1)$ 时间。 +- **空间复杂度 $O(1)$ ：** 使用常数大小的额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/191. \344\275\2151\347\232\204\344\270\252\346\225\260.md" "b/selected_coding_interview/docs/191. \344\275\2151\347\232\204\344\270\252\346\225\260.md" new file mode 100644 index 0000000..3f55d9d --- /dev/null +++ "b/selected_coding_interview/docs/191. \344\275\2151\347\232\204\344\270\252\346\225\260.md" @@ -0,0 +1,134 @@ +## 方法一：逐位判断 + +根据 **与运算** 定义，设二进制数字 $n$ ，则有： + +- 若 $n \& 1 = 0$ ，则 $n$ 二进制 **最右一位** 为 $0$ 。 +- 若 $n \& 1 = 1$ ，则 $n$ 二进制 **最右一位** 为 $1$ 。 + +根据以上特点，考虑以下 **循环判断** ： + +1. 判断 $n$ 最右一位是否为 $1$ ，根据结果计数。 +2. 将 $n$ 右移一位（本题要求把数字 $n$ 看作无符号数，因此使用 **无符号右移** 操作）。 + +### 算法流程： + +1. 初始化数量统计变量 $res = 0$ 。 +2. 循环逐位判断： 当 $n = 0$ 时跳出。 + 1. **`res += n & 1` ：** 若 $n \& 1 = 1$ ，则统计数 $res$ 加一。 + 2. **`n >>= 1` ：** 将二进制数字 $n$ 无符号右移一位（ Java 中无符号右移为 "$>>>$" ） 。 +3. 返回统计数量 $res$ 。 + + + +### 代码： + +```Python [] +class Solution: + def hammingWeight(self, n: int) -> int: + res = 0 + while n: + res += n & 1 + n >>= 1 + return res +``` + +```Java [] +public class Solution { + public int hammingWeight(int n) { + int res = 0; + while (n != 0) { + res += n & 1; + n >>>= 1; + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int hammingWeight(uint32_t n) { + unsigned int res = 0; // c++ 使用无符号数 + while (n != 0) { + res += n & 1; + n >>= 1; + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(\log n)$ ：** 此算法循环内部仅有 **移位、与、加** 等基本运算，占用 $O(1)$ ；逐位判断需循环 $log_2 n$ 次，其中 $\log_2 n$ 代表数字 $n$ 最高位 $1$ 的所在位数（例如 $\log_2 4 = 2$, $\log_2 16 = 4$）。 +- **空间复杂度 $O(1)$ ：** 变量 $res$ 使用常数大小额外空间。 + +## 方法二：巧用 $n \& (n - 1)$ + +**$(n - 1)$ 作用：** 二进制数字 $n$ 最右边的 $1$ 变成 $0$ ，此 $1$ 右边的 $0$ 都变成 $1$ 。 + +**$n \& (n - 1)$ 作用：** 二进制数字 $n$ 最右边的 $1$ 变成 $0$ ，其余不变。 + +![Picture1.png](https://pic.leetcode-cn.com/f23d9ef4fcfd65d7fbe29e477cbf36110b2f34558020e8cff09a1e13c0275c43-Picture1.png){:width=400} + +### 算法流程： + +1. 初始化数量统计变量 $res$ 。 +2. 循环消去最右边的 $1$ ：当 $n = 0$ 时跳出。 + 1. **`res += 1` ：** 统计变量加 $1$ 。 + 2. **`n &= n - 1` ：** 消去数字 $n$ 最右边的 $1$ 。 +3. 返回统计数量 $res$ 。 + + + +### 代码： + +```Python [] +class Solution: + def hammingWeight(self, n: int) -> int: + res = 0 + while n: + res += 1 + n &= n - 1 + return res +``` + +```Java [] +public class Solution { + public int hammingWeight(int n) { + int res = 0; + while (n != 0) { + res++; + n &= n - 1; + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int hammingWeight(uint32_t n) { + int res = 0; + while (n != 0) { + res++; + n &= n - 1; + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(M)$ ：** $n \& (n - 1)$ 操作仅有减法和与运算，占用 $O(1)$ ；设 $M$ 为二进制数字 $n$ 中 $1$ 的个数，则需循环 $M$ 次（每轮消去一个 $1$ ），占用 $O(M)$ 。 +- **空间复杂度 $O(1)$ ：** 变量 $res$ 使用常数大小额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/198. \346\211\223\345\256\266\345\212\253\350\210\215.md" "b/selected_coding_interview/docs/198. \346\211\223\345\256\266\345\212\253\350\210\215.md" new file mode 100644 index 0000000..fcacfec --- /dev/null +++ "b/selected_coding_interview/docs/198. \346\211\223\345\256\266\345\212\253\350\210\215.md" @@ -0,0 +1,64 @@ +#### 解题思路： + +> 典型的动态规划，以下按照标准流程解题。 + +- **状态定义：** + + - 设动态规划列表 $dp$ ，$dp[i]$ 代表前 $i$ 个房子在满足条件下的能偷窃到的最高金额。 + +- **转移方程：** + + - **设：** 有 $n$ 个房子，前 $n$ 间能偷窃到的最高金额是 $dp[n]$ ，前 $n-1$ 间能偷窃到的最高金额是 $dp[n-1]$ ，此时向这些房子后加一间房，此房间价值为 $num$ ； + - **加一间房间后：** 由于不能抢相邻的房子，意味着抢第 $n+1$ 间就不能抢第 $n$ 间；那么前 $n+1$ 间房能偷取到的最高金额 $dp[n+1]$ 一定是以下两种情况的 **较大值** ： + 1. 不抢第 $n+1$ 个房间，因此等于前 $n$ 个房子的最高金额，即 $dp[n+1] = dp[n]$ ； + 2. 抢第 $n+1$ 个房间，此时不能抢第 $n$ 个房间；因此等于前 $n-1$ 个房子的最高金额加上当前房间价值，即 $dp[n+1] = dp[n-1] + num$ ； + + - **细心的我们发现：** 难道在前 $n$ 间的最高金额 $dp[n]$ 情况下，第 $n$ 间一定被偷了吗？假设没有被偷，那 $n+1$ 间的最大值应该也可能是 $dp[n+1] = dp[n] + num$ 吧？其实这种假设的情况可以被省略，这是因为： + 1. 假设第 $n$ 间没有被偷，那么此时 $dp[n] = dp[n-1]$ ，此时 $dp[n+1] = dp[n] + num = dp[n-1] + num$ ，即两种情况可以 **合并为一种情况** 考虑； + 2. 假设第 $n$ 间被偷，那么此时 $dp[n+1] = dp[n] + num$ **不可取** ，因为偷了第 $n$ 间就不能偷第 $n+1$ 间。 + + - **最终的转移方程：** $dp[n+1] = max(dp[n],dp[n-1]+num)$ + +- **初始状态：** + + - 前 $0$ 间房子的最大偷窃价值为 $0$ ，即 $dp[0] = 0$ 。 + +- **返回值：** + + - 返回 $dp$ 列表最后一个元素值，即所有房间的最大偷窃价值。 + +- **简化空间复杂度：** + + - 我们发现 $dp[n]$ 只与 $dp[n-1]$ 和 $dp[n-2]$ 有关系，因此我们可以设两个变量 `cur`和 `pre` 交替记录，将空间复杂度降到 $O(1)$ 。 + +#### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 遍历 `nums` 需要线性时间； +- **空间复杂度 $O(1)$ ：** `cur`和 `pre` 使用常数大小的额外空间。 + + + +#### 代码： + +```Python [] +class Solution: + def rob(self, nums: List[int]) -> int: + cur, pre = 0, 0 + for num in nums: + cur, pre = max(pre + num, cur), cur + return cur +``` + +```Java [] +class Solution { + public int rob(int[] nums) { + int pre = 0, cur = 0, tmp; + for(int num : nums) { + tmp = cur; + cur = Math.max(pre + num, cur); + pre = tmp; + } + return cur; + } +} +``` diff --git "a/selected_coding_interview/docs/20. \346\234\211\346\225\210\347\232\204\346\213\254\345\217\267.md" "b/selected_coding_interview/docs/20. \346\234\211\346\225\210\347\232\204\346\213\254\345\217\267.md" new file mode 100644 index 0000000..54e09ac --- /dev/null +++ "b/selected_coding_interview/docs/20. \346\234\211\346\225\210\347\232\204\346\213\254\345\217\267.md" @@ -0,0 +1,48 @@ +#### 解题思路： + +- **算法原理** + - 栈先入后出特点恰好与本题括号排序特点一致，即若遇到左括号入栈，遇到右括号时将对应栈顶左括号出栈，则遍历完所有括号后 `stack` 仍然为空； + - 建立哈希表 `dic` 构建左右括号对应关系：$key$ 左括号，$value$ 右括号；这样查询 $2$ 个括号是否对应只需 $O(1)$ 时间复杂度；建立栈 `stack`，遍历字符串 `s` 并按照算法流程一一判断。 +- **算法流程** + 1. 如果 `c` 是左括号，则入栈 $push$； + 2. 否则通过哈希表判断括号对应关系，若 `stack` 栈顶出栈括号 `stack.pop()` 与当前遍历括号 `c` 不对应，则提前返回 $false$。 +- **提前返回** $false$ + - **提前返回优点：** 在迭代过程中，提前发现不符合的括号并且返回，提升算法效率。 + - **解决边界问题：** + - **栈** `stack` **为空：** 此时 `stack.pop()` 操作会报错；因此，我们采用一个取巧方法，给 `stack` 赋初值 $?$ ，并在哈希表 `dic` 中建立 $key: '?'，value:'?'$ 的对应关系予以配合。此时当 `stack` 为空且 `c` 为右括号时，可以正常提前返回 $false$； + - **字符串** `s` **以左括号结尾：** 此情况下可以正常遍历完整个 `s`，但 `stack` 中遗留未出栈的左括号；因此，最后需返回 `len(stack) == 1`，以判断是否是有效的括号组合。 +- **复杂度分析** + - 时间复杂度 $O(N)$：正确的括号组合需要遍历 $1$ 遍 `s`； + - 空间复杂度 $O(N)$：哈希表和栈使用线性的空间大小。 + + + +#### 代码： + +```Python [] +class Solution: + def isValid(self, s: str) -> bool: + dic = {'{': '}', '[': ']', '(': ')', '?': '?'} + stack = ['?'] + for c in s: + if c in dic: stack.append(c) + elif dic[stack.pop()] != c: return False + return len(stack) == 1 +``` + +```Java [] +class Solution { + private static final Map map = new HashMap(){{ + put('{','}'); put('[',']'); put('(',')'); put('?','?'); + }}; + public boolean isValid(String s) { + if(s.length() > 0 && !map.containsKey(s.charAt(0))) return false; + LinkedList stack = new LinkedList() {{ add('?'); }}; + for(Character c : s.toCharArray()){ + if(map.containsKey(c)) stack.addLast(c); + else if(map.get(stack.removeLast()) != c) return false; + } + return stack.size() == 1; + } +} +``` diff --git "a/selected_coding_interview/docs/205. \345\220\214\346\236\204\345\255\227\347\254\246\344\270\262.md" "b/selected_coding_interview/docs/205. \345\220\214\346\236\204\345\255\227\347\254\246\344\270\262.md" new file mode 100644 index 0000000..1f3f1d4 --- /dev/null +++ "b/selected_coding_interview/docs/205. \345\220\214\346\236\204\345\255\227\347\254\246\344\270\262.md" @@ -0,0 +1,90 @@ +## 解题思路 + +首先复习一下数学中映射的相关概念定义。设集合 `s` , `t` 中的某字符为 `x` , `y` ， + +- 单射：对于任意 `x` ，都有唯一的 `y` 与之对应。 +- 满射：对于任意 `y` ，至少存在一个 `x` 与之对应。 +- 双射：既是单射又是满射，又称为一一对应。 + +![Slide1.png](https://pic.leetcode-cn.com/1656945936-BsSBMu-Slide1.png){:width=600} + +接下来，抽象理解题目给定条件， + +- “每个出现的字符都应当映射到另一个字符”。代表字符集合 `s` , `t` 之间是「满射」。 +- “相同字符只能映射到同一个字符上，不同字符不能映射到同一个字符上”。代表字符集合 `s` , `t` 之间是「单射」。 + +因此， `s` 和 `t` 之间是「双射」，满足**一一对应**。考虑遍历字符串，使用哈希表 `s2t` , `t2s` 分别记录 $s \rightarrow t$ , $t \rightarrow s$ 的映射，当发现任意「一对多」的关系时返回 false 即可。 + +> 首页为动态图，其余为静态图。 + + + +## 代码 + +```Python [] +class Solution: + def isIsomorphic(self, s: str, t: str) -> bool: + s2t, t2s = {}, {} + for a, b in zip(s, t): + # 对于已有映射 a -> s2t[a]，若和当前字符映射 a -> b 不匹配， + # 说明有一对多的映射关系，则返回 false ； + # 对于映射 b -> a 也同理 + if a in s2t and s2t[a] != b or \ + b in t2s and t2s[b] != a: + return False + s2t[a], t2s[b] = b, a + return True +``` + +```Java [] +class Solution { + public boolean isIsomorphic(String s, String t) { + Map s2t = new HashMap<>(), t2s = new HashMap<>(); + for (int i = 0; i < s.length(); i++) { + char a = s.charAt(i), b = t.charAt(i); + // 对于已有映射 a -> s2t[a]，若和当前字符映射 a -> b 不匹配， + // 说明有一对多的映射关系，则返回 false ； + // 对于映射 b -> a 也同理 + if (s2t.containsKey(a) && s2t.get(a) != b || + t2s.containsKey(b) && t2s.get(b) != a) + return false; + s2t.put(a, b); + t2s.put(b, a); + } + return true; + } +} +``` + +```C++ [] +class Solution { +public: + bool isIsomorphic(string s, string t) { + unordered_map t2s, s2t; + for (int i = 0; i < s.size(); i++) { + char a = s[i], b = t[i]; + // 对于已有映射 a -> s2t[a]，若和当前字符映射 a -> b 不匹配， + // 说明有一对多的映射关系，则返回 false ； + // 对于映射 b -> a 也同理 + if (s2t.find(a) != s2t.end() && s2t[a] != b || + t2s.find(b) != t2s.end() && t2s[b] != a) + return false; + s2t[a] = b; + t2s[b] = a; + } + return true; + } +}; +``` + +## 复杂度分析 + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为字符串 `s` , `t` 的长度。遍历字符串 `s` , `t` 使用线性时间，hashmap 查询操作使用 $O(1)$ 时间。 +- **空间复杂度 $O(1)$ ：** 题目说明 `s` 和 `t` 由任意有效的 ASCII 字符组成。由于 ASCII 字符共 128 个，因此 hashmap `s2t` , `t2s` 使用 $O(128) = O(1)$ 空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/206. \345\217\215\350\275\254\351\223\276\350\241\250.md" "b/selected_coding_interview/docs/206. \345\217\215\350\275\254\351\223\276\350\241\250.md" new file mode 100644 index 0000000..57ea5f9 --- /dev/null +++ "b/selected_coding_interview/docs/206. \345\217\215\350\275\254\351\223\276\350\241\250.md" @@ -0,0 +1,145 @@ +## 解题思路： + +如下图所示，题目要求将链表反转。本文介绍迭代（双指针）、递归两种实现方法。 + +![Picture1.png](https://pic.leetcode-cn.com/1604779288-WXygqL-Picture1.png){:width=400} + +## 方法一：迭代（双指针） + +考虑遍历链表，并在访问各节点时修改 `next` 引用指向，算法流程见注释。 + + + +### 代码： + +```Python [] +class Solution: + def reverseList(self, head: ListNode) -> ListNode: + cur, pre = head, None + while cur: + tmp = cur.next # 暂存后继节点 cur.next + cur.next = pre # 修改 next 引用指向 + pre = cur # pre 暂存 cur + cur = tmp # cur 访问下一节点 + return pre +``` + +```Java [] +class Solution { + public ListNode reverseList(ListNode head) { + ListNode cur = head, pre = null; + while(cur != null) { + ListNode tmp = cur.next; // 暂存后继节点 cur.next + cur.next = pre; // 修改 next 引用指向 + pre = cur; // pre 暂存 cur + cur = tmp; // cur 访问下一节点 + } + return pre; + } +} +``` + +```C++ [] +class Solution { +public: + ListNode* reverseList(ListNode* head) { + ListNode *cur = head, *pre = nullptr; + while(cur != nullptr) { + ListNode* tmp = cur->next; // 暂存后继节点 cur.next + cur->next = pre; // 修改 next 引用指向 + pre = cur; // pre 暂存 cur + cur = tmp; // cur 访问下一节点 + } + return pre; + } +}; +``` + +利用 Python 语言的平行赋值语法，可以进一步简化代码（但可读性下降）： + +```Python [] +class Solution: + def reverseList(self, head: ListNode) -> ListNode: + cur, pre = head, None + while cur: + cur.next, pre, cur = pre, cur, cur.next + return pre +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 遍历链表使用线性大小时间。 +- **空间复杂度 $O(1)$ ：** 变量 `pre` 和 `cur` 使用常数大小额外空间。 + +## 方法二：递归 + +考虑使用递归法遍历链表，当越过尾节点后终止递归，在回溯时修改各节点的 `next` 引用指向。 + +### `recur(cur, pre)` 递归函数： + +1. 终止条件：当 `cur` 为空，则返回尾节点 `pre` （即反转链表的头节点）； +2. 递归后继节点，记录返回值（即反转链表的头节点）为 `res` ； +3. 修改当前节点 `cur` 引用指向前驱节点 `pre` ； +4. 返回反转链表的头节点 `res` ； + +### `reverseList(head)` 函数： + +调用并返回 `recur(head, null)` 。传入 `null` 是因为反转链表后， `head` 节点指向 `null` ； + + + +### 代码： + +```Python [] +class Solution: + def reverseList(self, head: ListNode) -> ListNode: + def recur(cur, pre): + if not cur: return pre # 终止条件 + res = recur(cur.next, cur) # 递归后继节点 + cur.next = pre # 修改节点引用指向 + return res # 返回反转链表的头节点 + + return recur(head, None) # 调用递归并返回 +``` + +```Java [] +class Solution { + public ListNode reverseList(ListNode head) { + return recur(head, null); // 调用递归并返回 + } + private ListNode recur(ListNode cur, ListNode pre) { + if (cur == null) return pre; // 终止条件 + ListNode res = recur(cur.next, cur); // 递归后继节点 + cur.next = pre; // 修改节点引用指向 + return res; // 返回反转链表的头节点 + } +} +``` + +```C++ [] +class Solution { +public: + ListNode* reverseList(ListNode* head) { + return recur(head, nullptr); // 调用递归并返回 + } +private: + ListNode* recur(ListNode* cur, ListNode* pre) { + if (cur == nullptr) return pre; // 终止条件 + ListNode* res = recur(cur->next, cur); // 递归后继节点 + cur->next = pre; // 修改节点引用指向 + return res; // 返回反转链表的头节点 + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 遍历链表使用线性大小时间。 +- **空间复杂度 $O(N)$ ：** 遍历链表的递归深度达到 $N$ ，系统使用 $O(N)$ 大小额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/207. \350\257\276\347\250\213\350\241\250.md" "b/selected_coding_interview/docs/207. \350\257\276\347\250\213\350\241\250.md" new file mode 100644 index 0000000..34ac999 --- /dev/null +++ "b/selected_coding_interview/docs/207. \350\257\276\347\250\213\350\241\250.md" @@ -0,0 +1,159 @@ +#### 解题思路： + +- **本题可约化为：** 课程安排图是否是 **有向无环图(DAG)**。即课程间规定了前置条件，但不能构成任何环路，否则课程前置条件将不成立。 +- 思路是通过 **拓扑排序** 判断此课程安排图是否是 **有向无环图(DAG)** 。 **拓扑排序原理：** 对 DAG 的顶点进行排序，使得对每一条有向边 $(u, v)$，均有 $u$（在排序记录中）比 $v$ 先出现。亦可理解为对某点 $v$ 而言，只有当 $v$ 的所有源点均出现了，$v$ 才能出现。 +- 通过课程前置条件列表 `prerequisites` 可以得到课程安排图的 **邻接表** `adjacency`，以降低算法时间复杂度，以下两种方法都会用到邻接表。 + +#### 方法一：入度表（广度优先遍历） + +##### 算法流程： + +1. 统计课程安排图中每个节点的入度，生成 **入度表** `indegrees`。 +2. 借助一个队列 `queue`，将所有入度为 $0$ 的节点入队。 +3. 当 `queue` 非空时，依次将队首节点出队，在课程安排图中删除此节点 `pre`： + - 并不是真正从邻接表中删除此节点 `pre`，而是将此节点对应所有邻接节点 `cur` 的入度 $-1$，即 `indegrees[cur] -= 1`。 + - 当入度 $-1$后邻接节点 `cur` 的入度为 $0$，说明 `cur` 所有的前驱节点已经被 “删除”，此时将 `cur` 入队。 +4. 在每次 `pre` 出队时，执行 `numCourses--`； + - 若整个课程安排图是有向无环图（即可以安排），则所有节点一定都入队并出队过，即完成拓扑排序。换个角度说，若课程安排图中存在环，一定有节点的入度始终不为 $0$。 + - 因此，拓扑排序出队次数等于课程个数，返回 `numCourses == 0` 判断课程是否可以成功安排。 + +##### 复杂度分析： + +- **时间复杂度 $O(N + M)$：** 遍历一个图需要访问所有节点和所有临边，$N$ 和 $M$ 分别为节点数量和临边数量； +- **空间复杂度 $O(N + M)$：** 为建立邻接表所需额外空间，`adjacency` 长度为 $N$ ，并存储 $M$ 条临边的数据。 + + + +#### 代码： + +```Python [] +from collections import deque + +class Solution: + def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool: + indegrees = [0 for _ in range(numCourses)] + adjacency = [[] for _ in range(numCourses)] + queue = deque() + # Get the indegree and adjacency of every course. + for cur, pre in prerequisites: + indegrees[cur] += 1 + adjacency[pre].append(cur) + # Get all the courses with the indegree of 0. + for i in range(len(indegrees)): + if not indegrees[i]: queue.append(i) + # BFS TopSort. + while queue: + pre = queue.popleft() + numCourses -= 1 + for cur in adjacency[pre]: + indegrees[cur] -= 1 + if not indegrees[cur]: queue.append(cur) + return not numCourses +``` + +```Java [] +class Solution { + public boolean canFinish(int numCourses, int[][] prerequisites) { + int[] indegrees = new int[numCourses]; + List> adjacency = new ArrayList<>(); + Queue queue = new LinkedList<>(); + for(int i = 0; i < numCourses; i++) + adjacency.add(new ArrayList<>()); + // Get the indegree and adjacency of every course. + for(int[] cp : prerequisites) { + indegrees[cp[0]]++; + adjacency.get(cp[1]).add(cp[0]); + } + // Get all the courses with the indegree of 0. + for(int i = 0; i < numCourses; i++) + if(indegrees[i] == 0) queue.add(i); + // BFS TopSort. + while(!queue.isEmpty()) { + int pre = queue.poll(); + numCourses--; + for(int cur : adjacency.get(pre)) + if(--indegrees[cur] == 0) queue.add(cur); + } + return numCourses == 0; + } +} +``` + +--- + +#### 方法二：深度优先遍历 + +原理是通过 DFS 判断图中是否有环。 + +##### 算法流程： + +1. 借助一个标志列表 `flags`，用于判断每个节点 `i` （课程）的状态： + 1. 未被 DFS 访问：`i == 0`； + 2. 已被**其他节点启动**的 DFS 访问：`i == -1`； + 3. 已被**当前节点启动**的 DFS 访问：`i == 1`。 +2. 对 `numCourses` 个节点依次执行 DFS，判断每个节点起步 DFS 是否存在环，若存在环直接返回 $False$。DFS 流程； + 1. 终止条件： + - 当 `flag[i] == -1`，说明当前访问节点已被其他节点启动的 DFS 访问，无需再重复搜索，直接返回 $True$。 + - 当 `flag[i] == 1`，说明在本轮 DFS 搜索中节点 `i` 被第 $2$ 次访问，即 **课程安排图有环** ，直接返回 $False$。 + 2. 将当前访问节点 `i` 对应 `flag[i]` 置 $1$，即标记其被本轮 DFS 访问过； + 3. 递归访问当前节点 `i` 的所有邻接节点 `j`，当发现环直接返回 $False$； + 4. 当前节点所有邻接节点已被遍历，并没有发现环，则将当前节点 `flag` 置为 $-1$ 并返回 $True$。 +3. 若整个图 DFS 结束并未发现环，返回 $True$。 + +##### 复杂度分析： + +- **时间复杂度 $O(N + M)$：** 遍历一个图需要访问所有节点和所有临边，$N$ 和 $M$ 分别为节点数量和临边数量； +- **空间复杂度 $O(N + M)$：** 为建立邻接表所需额外空间，`adjacency` 长度为 $N$ ，并存储 $M$ 条临边的数据。 + + + +#### 代码： + +```Python [] +class Solution: + def canFinish(self, numCourses: int, prerequisites: List[List[int]]) -> bool: + def dfs(i, adjacency, flags): + if flags[i] == -1: return True + if flags[i] == 1: return False + flags[i] = 1 + for j in adjacency[i]: + if not dfs(j, adjacency, flags): return False + flags[i] = -1 + return True + + adjacency = [[] for _ in range(numCourses)] + flags = [0 for _ in range(numCourses)] + for cur, pre in prerequisites: + adjacency[pre].append(cur) + for i in range(numCourses): + if not dfs(i, adjacency, flags): return False + return True +``` + +```Java [] +class Solution { + public boolean canFinish(int numCourses, int[][] prerequisites) { + List> adjacency = new ArrayList<>(); + for(int i = 0; i < numCourses; i++) + adjacency.add(new ArrayList<>()); + int[] flags = new int[numCourses]; + for(int[] cp : prerequisites) + adjacency.get(cp[1]).add(cp[0]); + for(int i = 0; i < numCourses; i++) + if(!dfs(adjacency, flags, i)) return false; + return true; + } + private boolean dfs(List> adjacency, int[] flags, int i) { + if(flags[i] == 1) return false; + if(flags[i] == -1) return true; + flags[i] = 1; + for(Integer j : adjacency.get(i)) + if(!dfs(adjacency, flags, j)) return false; + flags[i] = -1; + return true; + } +} +``` + +> 感谢评论区各位大佬 @马嘉利 @GSbeegnnord @mountaincode @kin @131xxxx8381 @dddong @chuwenli @JiangJian @番茄大大 @zjma 勘误。 +> 本篇初稿错误频出，实属汗颜 Orz ，现已一一修正。再次感谢！ diff --git "a/selected_coding_interview/docs/21. \345\220\210\345\271\266\344\270\244\344\270\252\346\234\211\345\272\217\351\223\276\350\241\250.md" "b/selected_coding_interview/docs/21. \345\220\210\345\271\266\344\270\244\344\270\252\346\234\211\345\272\217\351\223\276\350\241\250.md" new file mode 100644 index 0000000..658e418 --- /dev/null +++ "b/selected_coding_interview/docs/21. \345\220\210\345\271\266\344\270\244\344\270\252\346\234\211\345\272\217\351\223\276\350\241\250.md" @@ -0,0 +1,95 @@ +## 解题思路： + +根据题目描述， 链表 $l_1$ , $l_2$ 是 **递增** 的，因此容易想到使用双指针 $l_1$ 和 $l_2$ 遍历两链表，根据 $l_1.val$ 和 $l_2.val$ 的大小关系确定节点添加顺序，两节点指针交替前进，直至遍历完毕。 + +**引入伪头节点：** 由于初始状态合并链表中无节点，因此循环第一轮时无法将节点添加到合并链表中。解决方案：初始化一个辅助节点 $dum$ 作为合并链表的伪头节点，将各节点添加至 $dum$ 之后。 + +![Picture1.png](https://pic.leetcode-cn.com/e4c8c97883da50d81498fd1f1e6cdd575429bd65f9f2babb00dc2b709f7ad8b2-Picture1.png){:width=400} + +### 算法流程： + +1. **初始化：** 伪头节点 $dum$ ，节点 $cur$ 指向 $dum$ 。 +2. **循环合并：** 当 $l_1$ 或 $l_2$ 为空时跳出。 + 1. 当 $l_1.val < l_2.val$ 时： $cur$ 的后继节点指定为 $l_1$ ，并 $l_1$ 向前走一步。 + 2. 当 $l_1.val \geq l_2.val$ 时： $cur$ 的后继节点指定为 $l_2$ ，并 $l_2$ 向前走一步 。 + 3. 节点 $cur$ 向前走一步，即 $cur = cur.next$ 。 +3. **合并剩余尾部：** 跳出时有两种情况，即 $l_1$ 为空 **或** $l_2$ 为空。 + 1. 若 $l_1 \ne null$ ： 将 $l_1$ 添加至节点 $cur$ 之后。 + 2. 否则： 将 $l_2$ 添加至节点 $cur$ 之后。 +4. **返回值：** 合并链表在伪头节点 $dum$ 之后，因此返回 $dum.next$ 即可。 + + + +## 代码： + +Python 三元表达式写法 `A if x else B` ，代表当 $x = True$ 时执行 $A$ ，否则执行 $B$ 。 + +```Python [] +class Solution: + def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]: + cur = dum = ListNode(0) + while list1 and list2: + if list1.val < list2.val: + cur.next, list1 = list1, list1.next + else: + cur.next, list2 = list2, list2.next + cur = cur.next + cur.next = list1 if list1 else list2 + return dum.next +``` + +```Java [] +class Solution { + public ListNode mergeTwoLists(ListNode list1, ListNode list2) { + ListNode dum = new ListNode(0), cur = dum; + while (list1 != null && list2 != null) { + if (list1.val < list2.val) { + cur.next = list1; + list1 = list1.next; + } + else { + cur.next = list2; + list2 = list2.next; + } + cur = cur.next; + } + cur.next = list1 != null ? list1 : list2; + return dum.next; + } +} +``` + +```C++ [] +class Solution { +public: + ListNode* mergeTwoLists(ListNode* list1, ListNode* list2) { + ListNode* dum = new ListNode(0); + ListNode* cur = dum; + while (list1 != nullptr && list2 != nullptr) { + if (list1->val < list2->val) { + cur->next = list1; + list1 = list1->next; + } + else { + cur->next = list2; + list2 = list2->next; + } + cur = cur->next; + } + cur->next = list1 != nullptr ? list1 : list2; + return dum->next; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(M+N)$ ：** $M, N$ 分别为链表 $l_1$, $l_2$ 的长度，合并操作需遍历两链表。 +- **空间复杂度 $O(1)$ ：** 节点引用 $dum$ , $cur$ 使用常数大小的额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/213. \346\211\223\345\256\266\345\212\253\350\210\215 II.md" "b/selected_coding_interview/docs/213. \346\211\223\345\256\266\345\212\253\350\210\215 II.md" new file mode 100644 index 0000000..f5d420a --- /dev/null +++ "b/selected_coding_interview/docs/213. \346\211\223\345\256\266\345\212\253\350\210\215 II.md" @@ -0,0 +1,76 @@ +### 解题思路： + +#### 总体思路： + +- **此题是 [198. 打家劫舍](https://leetcode-cn.com/problems/house-robber/solution/da-jia-jie-she-dong-tai-gui-hua-jie-gou-hua-si-lu-/) 的拓展版：** 唯一的区别是此题中的房间是 **环状排列** 的（即首尾相接），而 $198.$ 题中的房间是 **单排排列** 的；而这也是此题的难点。 + +- **环状排列** 意味着第一个房子和最后一个房子中 **只能选择一个偷窃**，因此可以把此 **环状排列房间** 问题约化为两个 **单排排列房间** 子问题： + + 1. 在不偷窃第一个房子的情况下（即 $nums[1:]$），最大金额是 $p_1$ ； + 2. 在不偷窃最后一个房子的情况下（即 $nums[:n-1]$），最大金额是 $p_2$ 。 + + - **综合偷窃最大金额：** 为以上两种情况的较大值，即 $max(p1,p2)$ 。 + +- 下面的任务则是解决 **单排排列房间（即 [198. 打家劫舍](https://leetcode-cn.com/problems/house-robber/solution/da-jia-jie-she-dong-tai-gui-hua-jie-gou-hua-si-lu-/)）** 问题。推荐可以先把 $198.$ 做完再做这道题。 + +#### 198. 解题思路： + +典型的动态规划，以下按照标准流程解题。 + +- **状态定义：** + - 设动态规划列表 $dp$ ，$dp[i]$ 代表前 $i$ 个房子在满足条件下的能偷窃到的最高金额。 +- **转移方程：** + - **设：** 有 $n$ 个房子，前 $n$ 间能偷窃到的最高金额是 $dp[n]$ ，前 $n-1$ 间能偷窃到的最高金额是 $dp[n-1]$ ，此时向这些房子后加一间房，此房间价值为 $num$ ； + - **加一间房间后：** 由于不能抢相邻的房子，意味着抢第 $n+1$ 间就不能抢第 $n$ 间；那么前 $n+1$ 间房能偷取到的最高金额 $dp[n+1]$ 一定是以下两种情况的 **较大值** ： + 1. 不抢第 $n+1$ 个房间，因此等于前 $n$ 个房子的最高金额，即 $dp[n+1] = dp[n]$ ； + 2. 抢第 $n+1$ 个房间，此时不能抢第 $n$ 个房间；因此等于前 $n-1$ 个房子的最高金额加上当前房间价值，即 $dp[n+1] = dp[n-1] + num$ ； + - **细心的我们发现：** 难道在前 $n$ 间的最高金额 $dp[n]$ 情况下，第 $n$ 间一定被偷了吗？假设没有被偷，那 $n+1$ 间的最大值应该也可能是 $dp[n+1] = dp[n] + num$ 吧？其实这种假设的情况可以被省略，这是因为： + 1. 假设第 $n$ 间没有被偷，那么此时 $dp[n] = dp[n-1]$ ，此时 $dp[n+1] = dp[n] + num = dp[n-1] + num$ ，即可以将 **两种情况合并为一种情况** 考虑； + 2. 假设第 $n$ 间被偷，那么此时 $dp[n+1] = dp[n] + num$ **不可取** ，因为偷了第 $n$ 间就不能偷第 $n+1$ 间。 + - **最终的转移方程：** $dp[n+1] = max(dp[n],dp[n-1]+num)$ +- **初始状态：** + - 前 $0$ 间房子的最大偷窃价值为 $0$ ，即 $dp[0] = 0$ 。 +- **返回值：** + - 返回 $dp$ 列表最后一个元素值，即所有房间的最大偷窃价值。 +- **简化空间复杂度：** + - 我们发现 $dp[n]$ 只与 $dp[n-1]$ 和 $dp[n-2]$ 有关系，因此我们可以设两个变量 `cur`和 `pre` 交替记录，将空间复杂度降到 $O(1)$ 。 + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 两次遍历 `nums` 需要线性时间； +- **空间复杂度 $O(1)$ ：** `cur`和 `pre` 使用常数大小的额外空间。 + + + +### 代码： + +```Python [] +class Solution: + def rob(self, nums: [int]) -> int: + def my_rob(nums): + cur, pre = 0, 0 + for num in nums: + cur, pre = max(pre + num, cur), cur + return cur + return max(my_rob(nums[:-1]),my_rob(nums[1:])) if len(nums) != 1 else nums[0] +``` + +```Java [] +class Solution { + public int rob(int[] nums) { + if(nums.length == 0) return 0; + if(nums.length == 1) return nums[0]; + return Math.max(myRob(Arrays.copyOfRange(nums, 0, nums.length - 1)), + myRob(Arrays.copyOfRange(nums, 1, nums.length))); + } + private int myRob(int[] nums) { + int pre = 0, cur = 0, tmp; + for(int num : nums) { + tmp = cur; + cur = Math.max(pre + num, cur); + pre = tmp; + } + return cur; + } +} +``` diff --git "a/selected_coding_interview/docs/215. \346\225\260\347\273\204\344\270\255\347\232\204\347\254\254K\344\270\252\346\234\200\345\244\247\345\205\203\347\264\240.md" "b/selected_coding_interview/docs/215. \346\225\260\347\273\204\344\270\255\347\232\204\347\254\254K\344\270\252\346\234\200\345\244\247\345\205\203\347\264\240.md" new file mode 100644 index 0000000..2a70c69 --- /dev/null +++ "b/selected_coding_interview/docs/215. \346\225\260\347\273\204\344\270\255\347\232\204\347\254\254K\344\270\252\346\234\200\345\244\247\345\205\203\347\264\240.md" @@ -0,0 +1,190 @@ +## 方法一：快速排序 + +本题使用排序算法解决最直观：对数组 `nums` 执行排序，再返回倒数第 $k$ 个元素即可。使用任意排序算法皆可，本文采用并介绍 **快速排序** ，为下文 **方法二** 做铺垫。 + +快速排序算法有两个核心点，分别为 “哨兵划分” 和 “递归” 。 + +- **哨兵划分操作：** 以数组某个元素（一般选取首元素）为 **基准数** ，将所有小于基准数的元素移动至其左边，大于基准数的元素移动至其右边。 +- **递归：** 对 **左子数组** 和 **右子数组** 递归执行 **哨兵划分**，直至子数组长度为 1 时终止递归，即可完成对整个数组的排序。 + +> 如下图所示，为哨兵划分操作流程。通过一轮 **哨兵划分** ，可将数组排序问题拆分为 **两个较短数组的排序问题** （本文称之为左（右）子数组）。 + + + +如下图所示，为示例数组 `[2,4,1,0,3,5]` 的快速排序流程。观察发现，快速排序和**二分查找**的原理类似，都是以 $O(\log n)$ 时间复杂度实现搜索区间缩小。 + +![Picture1.png](https://pic.leetcode-cn.com/1612615552-rifQwI-Picture1.png){:width=550} + +### 代码： + +```Python [] +class Solution: + def findKthLargest(self, nums: List[int], k: int) -> int: + def quick_sort(nums, l, r): + # 子数组长度为 1 时终止递归 + if l >= r: return + # 哨兵划分操作（以 nums[l] 作为基准数） + i, j = l, r + while i < j: + while i < j and nums[j] >= nums[l]: j -= 1 + while i < j and nums[i] <= nums[l]: i += 1 + nums[i], nums[j] = nums[j], nums[i] + nums[l], nums[i] = nums[i], nums[l] + # 递归左（右）子数组执行哨兵划分 + quick_sort(nums, l, i - 1) + quick_sort(nums, i + 1, r) + + quick_sort(nums, 0, len(nums) - 1) + return nums[-k] +``` + +```Java [] +class Solution { + private void swap(int[] nums, int i, int j) { + int tmp = nums[i]; + nums[i] = nums[j]; + nums[j] = tmp; + } + + private void quickSort(int[] nums, int l, int r) { + // 子数组长度为 1 时终止递归 + if (l >= r) return; + // 哨兵划分操作（以 nums[l] 作为基准数） + int i = l, j = r; + while (i < j) { + while (i < j && nums[j] >= nums[l]) j--; + while (i < j && nums[i] <= nums[l]) i++; + swap(nums, i, j); + } + swap(nums, i, l); + // 递归左（右）子数组执行哨兵划分 + quickSort(nums, l, i - 1); + quickSort(nums, i + 1, r); + } + + public int findKthLargest(int[] nums, int k) { + quickSort(nums, 0, nums.length - 1); + return nums[nums.length - k]; + } +} +``` + +```C++ [] +class Solution { +public: + int findKthLargest(vector& nums, int k) { + quickSort(nums, 0, nums.size() - 1); + return nums[nums.size() - k]; + } +private: + void quickSort(vector& nums, int l, int r) { + // 子数组长度为 1 时终止递归 + if (l >= r) return; + // 哨兵划分操作（以 nums[l] 作为基准数） + int i = l, j = r; + while (i < j) { + while (i < j && nums[j] >= nums[l]) j--; + while (i < j && nums[i] <= nums[l]) i++; + swap(nums[i], nums[j]); + } + swap(nums[i], nums[l]); + // 递归左（右）子数组执行哨兵划分 + quickSort(nums, l, i - 1); + quickSort(nums, i + 1, r); + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N \log N)$ ：** 库函数、快排等排序算法的平均时间复杂度为 $O(N \log N)$ 。 +- **空间复杂度 $O(N)$ ：** 快速排序的递归深度最好（平均）为 $O(\log N)$ ，最差情况（即输入数组完全倒序）为 $O(N)$。 + +## 方法二： 基于快速排序的分治 + +设 $n$ 是数组长度。根据快速排序原理，如果某次哨兵划分后，**基准数的索引正好是 $n-k$ ，则意味着它就是第 $k$ 大的数字** ，那么就可以直接返回它，无需继续递归下去了。 + +### 代码： + +```Python [] +class Solution: + def findKthLargest(self, nums: List[int], k: int) -> int: + def quick_sort(l, r): + i, j = l, r + while i < j: + while i < j and nums[j] >= nums[l]: j -= 1 + while i < j and nums[i] <= nums[l]: i += 1 + nums[i], nums[j] = nums[j], nums[i] + nums[l], nums[i] = nums[i], nums[l] + if i > len(nums) - k: return quick_sort(l, i - 1) + if i < len(nums) - k: return quick_sort(i + 1, r) + # 若基准数索引为 n - k ，则直接返回该元素 + return nums[-k] + + return quick_sort(0, len(nums) - 1) +``` + +```Java [] +class Solution { + private void swap(int[] nums, int i, int j) { + int tmp = nums[i]; + nums[i] = nums[j]; + nums[j] = tmp; + } + + private int quickSort(int[] nums, int k, int l, int r) { + int i = l, j = r; + while (i < j) { + while (i < j && nums[j] >= nums[l]) j--; + while (i < j && nums[i] <= nums[l]) i++; + swap(nums, i, j); + } + swap(nums, i, l); + if (i > nums.length - k) return quickSort(nums, k, l, i - 1); + if (i < nums.length - k) return quickSort(nums, k, i + 1, r); + // 若基准数索引为 n - k ，则直接返回该元素 + return nums[nums.length - k]; + } + + public int findKthLargest(int[] nums, int k) { + return quickSort(nums, k, 0, nums.length - 1); + } +} +``` + +```C++ [] +class Solution { +public: + int findKthLargest(vector& nums, int k) { + return quickSort(nums, k, 0, nums.size() - 1); + } +private: + int quickSort(vector& nums, int k, int l, int r) { + int i = l, j = r; + while (i < j) { + while (i < j && nums[j] >= nums[l]) j--; + while (i < j && nums[i] <= nums[l]) i++; + swap(nums[i], nums[j]); + } + swap(nums[i], nums[l]); + if (i > nums.size() - k) return quickSort(nums, k, l, i - 1); + if (i < nums.size() - k) return quickSort(nums, k, i + 1, r); + // 若基准数索引为 n - k ，则直接返回该元素 + return nums[nums.size() - k]; + } +}; +``` + +### 复杂度分析： + +本方法优化时间复杂度的本质是通过判断舍去了不必要的哨兵划分的递归操作。 + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为数组元素数量；对于长度为 $N$ 的数组执行哨兵划分操作的时间复杂度为 $O(N)$ ；每轮哨兵划分后根据 $k$ 和 $i$ 的大小关系选择递归，由于 $i$ 分布的随机性，则向下递归子数组的平均长度为 $\frac{N}{2}$ ；因此平均情况下，哨兵划分操作一共有 $N + \frac{N}{2} + \frac{N}{4} + ... + \frac{N}{N} = \frac{N - \frac{1}{2}}{1 - \frac{1}{2}} = 2N - 1$ （等比数列求和），即总体时间复杂度为 $O(N)$ 。 +- **空间复杂度 $O(\log N)$ ：** 划分函数的平均递归深度为 $O(\log N)$ 。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/226. \347\277\273\350\275\254\344\272\214\345\217\211\346\240\221.md" "b/selected_coding_interview/docs/226. \347\277\273\350\275\254\344\272\214\345\217\211\346\240\221.md" new file mode 100644 index 0000000..98eba7c --- /dev/null +++ "b/selected_coding_interview/docs/226. \347\277\273\350\275\254\344\272\214\345\217\211\346\240\221.md" @@ -0,0 +1,160 @@ +## 解题思路： + +**二叉树镜像定义：** 对于二叉树中任意节点 $root$ ，设其左 / 右子节点分别为 $left, right$ ；则在二叉树的镜像中的对应 $root$ 节点，其左 / 右子节点分别为 $right, left$ 。 + +![Picture1.png](https://pic.leetcode-cn.com/20717714d97fa04d509e4f0525a3089efefc2ea02cc08ce92b77978e9b51f15f-Picture1.png){:width=450} + +## 方法一：递归法 + +根据二叉树镜像的定义，考虑递归遍历（dfs）二叉树，交换每个节点的左 / 右子节点，即可生成二叉树的镜像。 + +### 递归解析： + +1. **终止条件：** 当节点 $root$ 为空时（即越过叶节点），则返回 $null$ 。 +2. **递推工作：** + 1. 初始化节点 $tmp$ ，用于暂存 $root$ 的左子节点。 + 2. 开启递归 **右子节点** $invertTree(root.right)$ ，并将返回值作为 $root$ 的 **左子节点** 。 + 3. 开启递归 **左子节点** $invertTree(tmp)$ ，并将返回值作为 $root$ 的 **右子节点** 。 +3. **返回值：** 返回当前节点 $root$ 。 + +> **Q：** 为何需要暂存 $root$ 的左子节点？ +> **A：** 在递归右子节点 “$root.left = invertTree(root.right);$” 执行完毕后， $root.left$ 的值已经发生改变，此时递归左子节点 $invertTree(root.left)$ 则会出问题。 + + + +### 代码： + +Python 利用平行赋值的写法（即 $a, b = b, a$ ），可省略暂存操作。其原理是先将等号右侧打包成元组 $(b,a)$ ，再序列地分给等号左侧的 $a, b$ 序列。 + +```Python [] +class Solution: + def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]: + if not root: return + tmp = root.left + root.left = self.invertTree(root.right) + root.right = self.invertTree(tmp) + return root +``` + +```Python [] +class Solution: + def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]: + if not root: return + root.left, root.right = self.invertTree(root.right), self.invertTree(root.left) + return root +``` + +```Java [] +class Solution { + public TreeNode invertTree(TreeNode root) { + if (root == null) return null; + TreeNode tmp = root.left; + root.left = invertTree(root.right); + root.right = invertTree(tmp); + return root; + } +} +``` + +```C++ [] +class Solution { +public: + TreeNode* invertTree(TreeNode* root) { + if (root == nullptr) return nullptr; + TreeNode* tmp = root->left; + root->left = invertTree(root->right); + root->right = invertTree(tmp); + return root; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为二叉树的节点数量，建立二叉树镜像需要遍历树的所有节点，占用 $O(N)$ 时间。 +- **空间复杂度 $O(N)$ ：** 最差情况下（当二叉树退化为链表），递归时系统需使用 $O(N)$ 大小的栈空间。 + +## 方法二：辅助栈（或队列） + +利用栈（或队列）遍历树的所有节点 $node$ ，并交换每个 $node$ 的左 / 右子节点。 + +### 算法流程： + +1. **特例处理：** 当 $root$ 为空时，直接返回 $null$ 。 +2. **初始化：** 栈（或队列），本文用栈，并加入根节点 $root$ 。 +3. **循环交换：** 当栈 $stack$ 为空时跳出。 + 1. **出栈：** 记为 $node$ 。 + 2. **添加子节点：** 将 $node$ 左和右子节点入栈。 + 3. **交换：** 交换 $node$ 的左 / 右子节点。 +4. **返回值：** 返回根节点 $root$ 。 + + + +### 代码： + +```Python [] +class Solution: + def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]: + if not root: return + stack = [root] + while stack: + node = stack.pop() + if node.left: stack.append(node.left) + if node.right: stack.append(node.right) + node.left, node.right = node.right, node.left + return root +``` + +```Java [] +class Solution { + public TreeNode invertTree(TreeNode root) { + if (root == null) return null; + Stack stack = new Stack<>() {{ add(root); }}; + while (!stack.isEmpty()) { + TreeNode node = stack.pop(); + if (node.left != null) stack.add(node.left); + if (node.right != null) stack.add(node.right); + TreeNode tmp = node.left; + node.left = node.right; + node.right = tmp; + } + return root; + } +} +``` + +```C++ [] +class Solution { +public: + TreeNode* invertTree(TreeNode* root) { + if (root == nullptr) return nullptr; + stack stack; + stack.push(root); + while (!stack.empty()) + { + TreeNode* node = stack.top(); + stack.pop(); + if (node->left != nullptr) stack.push(node->left); + if (node->right != nullptr) stack.push(node->right); + TreeNode* tmp = node->left; + node->left = node->right; + node->right = tmp; + } + return root; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为二叉树的节点数量，建立二叉树镜像需要遍历树的所有节点，占用 $O(N)$ 时间。 +- **空间复杂度 $O(N)$ ：** 如下图所示，最差情况下，栈 $stack$ 最多同时存储 $\frac{N + 1}{2}$ 个节点，占用 $O(N)$ 额外空间。 + +![Picture0.png](https://pic.leetcode-cn.com/1614450330-bTAcyj-Picture0.png){:width=450} + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/230. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\344\270\255\347\254\254K\345\260\217\347\232\204\345\205\203\347\264\240.md" "b/selected_coding_interview/docs/230. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\344\270\255\347\254\254K\345\260\217\347\232\204\345\205\203\347\264\240.md" new file mode 100644 index 0000000..2cc5043 --- /dev/null +++ "b/selected_coding_interview/docs/230. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\344\270\255\347\254\254K\345\260\217\347\232\204\345\205\203\347\264\240.md" @@ -0,0 +1,84 @@ +## 解题思路： + +在二叉搜索树中，任意子节点都满足“左子节点 $<$ 根节点 $<$ 右子节点”的规则。因此二叉搜索树具有一个重要性质：**二叉搜索树的中序遍历为递增序列**。 + +也就是说，本题可被转化为求中序遍历的第 $k$ 个节点。 + +![Picture1.png](https://pic.leetcode.cn/1690460306-SMjxpo-Picture1.png){:width=500} + +为求第 $k$ 个节点，需要实现以下三项工作： + +1. 递归遍历时计数，统计当前节点的序号。 +2. 递归到第 $k$ 个节点时，应记录结果 $res$ 。 +3. 记录结果后，后续的遍历即失去意义，应提前返回。 + +## 代码： + +题目指出：$1 \leq k \leq N$ （二叉搜索树节点个数）；因此无需考虑 $k > N$ 的情况。 +若考虑，可以在中序遍历完成后判断 $k > 0$ 是否成立，若成立则说明 $k > N$ 。 + +```Python [] +class Solution: + def kthSmallest(self, root: Optional[TreeNode], k: int) -> int: + def dfs(root): + if not root: return + dfs(root.left) + if self.k == 0: return + self.k -= 1 + if self.k == 0: self.res = root.val + dfs(root.right) + + self.k = k + dfs(root) + return self.res +``` + +```Java [] +class Solution { + int res, k; + void dfs(TreeNode root) { + if (root == null) return; + dfs(root.left); + if (k == 0) return; + if (--k == 0) res = root.val; + dfs(root.right); + } + public int kthSmallest(TreeNode root, int k) { + this.k = k; + dfs(root); + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int kthSmallest(TreeNode* root, int k) { + this->k = k; + dfs(root); + return res; + } +private: + int res, k; + void dfs(TreeNode* root) { + if (root == nullptr) return; + dfs(root->left); + if (k == 0) return; + if (--k == 0) res = root->val; + dfs(root->right); + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 当树退化为链表，即全部为左子节点时，无论 $k$ 的值大小，递归深度都为 $N$ ，使用 $O(N)$ 时间。 +- **空间复杂度 $O(N)$ ：** 当树退化为链表时，系统使用 $O(N)$ 大小的栈空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/231. 2 \347\232\204\345\271\202.md" "b/selected_coding_interview/docs/231. 2 \347\232\204\345\271\202.md" new file mode 100644 index 0000000..4732689 --- /dev/null +++ "b/selected_coding_interview/docs/231. 2 \347\232\204\345\271\202.md" @@ -0,0 +1,32 @@ +#### 解题思路： + +- 若 $n = 2^x$ 且 $x$ 为自然数（即 $n$ 为 $2$ 的幂），则一定满足以下条件： + 1. 恒有 `n & (n - 1) == 0`，这是因为： + - $n$ 二进制最高位为 $1$，其余所有位为 $0$； + - $n - 1$ 二进制最高位为 $0$，其余所有位为 $1$； + 2. 一定满足 `n > 0`。 +- 因此，通过 `n > 0` 且 `n & (n - 1) == 0` 即可判定是否满足 $n = 2^x$。 + +| 2^x | n | n - 1 | n & (n - 1) | +| ----- | ------ | ------ | -------------------- | +| $2^0$ | $0001$ | $0000$ | (0001) & (0000) == 0 | +| $2^1$ | $0010$ | $0001$ | (0010) & (0001) == 0 | +| $2^2$ | $0100$ | $0011$ | (0100) & (0011) == 0 | +| $2^3$ | $1000$ | $0111$ | (1000) & (0111) == 0 | +| ... | ... | ... | ... | + +#### 代码： + +```Python [] +class Solution: + def isPowerOfTwo(self, n: int) -> bool: + return n > 0 and n & (n - 1) == 0 +``` + +```Java [] +class Solution { + public boolean isPowerOfTwo(int n) { + return n > 0 && (n & (n - 1)) == 0; + } +} +``` diff --git "a/selected_coding_interview/docs/232. \347\224\250\346\240\210\345\256\236\347\216\260\351\230\237\345\210\227.md" "b/selected_coding_interview/docs/232. \347\224\250\346\240\210\345\256\236\347\216\260\351\230\237\345\210\227.md" new file mode 100644 index 0000000..a574868 --- /dev/null +++ "b/selected_coding_interview/docs/232. \347\224\250\346\240\210\345\256\236\347\216\260\351\230\237\345\210\227.md" @@ -0,0 +1,135 @@ +## 解题思路： + +栈实现队列的出队操作效率低下：栈底元素（对应队首元素）无法直接删除，需要将上方所有元素出栈。 + +**两个栈可实现将列表倒序**：设有含三个元素的栈 `A = [1,2,3]` 和空栈 `B = []` 。若循环执行 `A` 元素出栈并添加入栈 `B` ，直到栈 `A` 为空，则 `A = []` , `B = [3,2,1]` ，即栈 `B` 元素为栈 `A` 元素倒序。 + +**利用栈 `B` 删除队首元素**：倒序后，`B` 执行出栈则相当于删除了 `A` 的栈底元素，即对应队首元素。 + +![Picture1.png](https://pic.leetcode-cn.com/1599286207-HnnMhX-Picture1.png){:width=500} + +因此，可以设计栈 `A` 用于加入队尾操作，栈 `B` 用于将元素倒序，从而实现删除队首元素。 + +### 函数设计： + +1. **加入队尾 `push()` ：** 将数字 `val` 加入栈 `A` 即可。 +2. **获取队首元素 `peek()` ：** + 1. **当栈 `B` 不为空：** `B`中仍有已完成倒序的元素，因此直接返回 `B` 的栈顶元素。 + 2. **否则，当 `A` 为空：** 即两个栈都为空，无元素，因此返回 -1 。 + 3. **否则：** 将栈 `A` 元素全部转移至栈 `B` 中，实现元素倒序，并返回栈 `B` 的栈顶元素。 +3. **弹出队首元素 `pop()` ：** + 1. 执行 `peek()` ，获取队首元素。 + 2. 弹出 `B` 的栈顶元素。 +4. **队列判空 `empty()` ：** 当栈 `A` 和 `B` 都为空时，队列为空。 + +## 代码： + +```Python [] +class MyQueue: + + def __init__(self): + self.A, self.B = [], [] + + def push(self, x: int) -> None: + self.A.append(x) + + def pop(self) -> int: + peek = self.peek() + self.B.pop() + return peek + + def peek(self) -> int: + if self.B: return self.B[-1] + if not self.A: return -1 + # 将栈 A 的元素依次移动至栈 B + while self.A: + self.B.append(self.A.pop()) + return self.B[-1] + + def empty(self) -> bool: + return not self.A and not self.B + +``` + +```Java [] +class MyQueue { + private Stack A; + private Stack B; + + public MyQueue() { + A = new Stack<>(); + B = new Stack<>(); + } + + public void push(int x) { + A.push(x); + } + + public int pop() { + int peek = peek(); + B.pop(); + return peek; + } + + public int peek() { + if (!B.isEmpty()) return B.peek(); + if (A.isEmpty()) return -1; + while (!A.isEmpty()){ + B.push(A.pop()); + } + return B.peek(); + } + + public boolean empty() { + return A.isEmpty() && B.isEmpty(); + } +} +``` + +```C++ [] +class MyQueue { +private: + std::stack A, B; + +public: + MyQueue() {} + + void push(int x) { + A.push(x); + } + + int pop() { + int peek = this->peek(); + B.pop(); + return peek; + } + + int peek() { + if (!B.empty()) return B.top(); + if (A.empty()) return -1; + while (!A.empty()){ + B.push(A.top()), A.pop(); + } + int res = B.top(); + return res; + } + + bool empty() { + return A.empty() && B.empty(); + } +}; +``` + +### 复杂度分析： + +以下分析仅满足添加 $N$ 个元素并删除 $N$ 个元素，即栈初始和结束状态下都为空的情况。 + +- **时间复杂度：** `push()`, `empty()` 函数的时间复杂度为 $O(1)$ ；`peek()` , `pop()` 函数在 $N$ 次队首元素删除操作中总共需完成 $N$ 个元素的倒序，均摊时间复杂度为 $O(1)$ 。 +- **空间复杂度 $O(N)$ ：** 最差情况下，栈 `A` 和 `B` 共保存 $N$ 个元素。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/233. \346\225\260\345\255\227 1 \347\232\204\344\270\252\346\225\260.md" "b/selected_coding_interview/docs/233. \346\225\260\345\255\227 1 \347\232\204\344\270\252\346\225\260.md" new file mode 100644 index 0000000..6ea21df --- /dev/null +++ "b/selected_coding_interview/docs/233. \346\225\260\345\255\227 1 \347\232\204\344\270\252\346\225\260.md" @@ -0,0 +1,168 @@ +## 解题思路： + +将 $1$ ~ $n$ 的个位、十位、百位、...的 $1$ 出现次数相加，即为 $1$ 出现的总次数。 + +设数字 $n$ 是个 $x$ 位数，记 $n$ 的第 $i$ 位为 $n_i$ ，则可将 $n$ 写为 $n_{x} n_{x-1} \cdots n_{2} n_{1}$ ；本文名词规定如下： + +- 称 「 $n_i$ 」称为 **当前位** ，记为 $cur$ 。 +- 将 「 $n_{i-1} n_{i-2} \cdots n_{2} n_{1}$ 」称为 **低位** ，记为 $low$ 。 +- 将 「 $n_{x} n_{x-1} \cdots n_{i+2} n_{i+1}$ 」称为 **高位** ，记为 $high$ 。 +- 将 「 $10^i$ 」称为 **位因子** ，记为 $digit$ 。 + +### 某位中 $1$ 出现次数的计算方法： + +根据当前位 $cur$ 值的不同，分为以下三种情况： + +1. 当 **$cur = 0$ 时：** 此位 $1$ 的出现次数只由高位 $high$ 决定，计算公式为： + +$$ +high \times digit +$$ + +> 如下图所示，以 $n = 2304$ 为例，求 $digit = 10$ （即十位）的 $1$ 出现次数。 + +![Picture1.png](https://pic.leetcode-cn.com/1599887431-cVmcVA-Picture1.png){:width=450} + +2. 当 **$cur = 1$ 时：** 此位 $1$ 的出现次数由高位 $high$ 和低位 $low$ 决定，计算公式为： + +$$ +high \times digit + low + 1 +$$ + +> 如下图所示，以 $n = 2314$ 为例，求 $digit = 10$ （即十位）的 $1$ 出现次数。 + +![Picture2.png](https://pic.leetcode-cn.com/1599887431-HAAvVp-Picture2.png){:width=450} + +3. 当 **$cur = 2, 3, \cdots, 9$ 时：** 此位 $1$ 的出现次数只由高位 $high$ 决定，计算公式为： + +$$ +(high + 1) \times digit +$$ + +> 如下图所示，以 $n = 2324$ 为例，求 $digit = 10$ （即十位）的 $1$ 出现次数。 + +![Picture3.png](https://pic.leetcode-cn.com/1599887431-djUZTe-Picture3.png){:width=450} + +### 变量递推公式： + +设计按照 “个位、十位、...” 的顺序计算，则 $high / cur / low / digit$ 应初始化为： + +```Python [] +high = n // 10 +cur = n % 10 +low = 0 +digit = 1 # 个位 +``` + +```Java [] +int high = n / 10; +int cur = n % 10; +int low = 0; +int digit = 1; // 个位 +``` + +```C++ [] +int high = n / 10; +int cur = n % 10; +int low = 0; +int digit = 1; // 个位 +``` + +因此，从个位到最高位的变量递推公式为： + +```Python [] +while high != 0 or cur != 0: # 当 high 和 cur 同时为 0 时，说明已经越过最高位，因此跳出 + low += cur * digit # 将 cur 加入 low ，组成下轮 low + cur = high % 10 # 下轮 cur 是本轮 high 的最低位 + high //= 10 # 将本轮 high 最低位删除，得到下轮 high + digit *= 10 # 位因子每轮 × 10 +``` + +```Java [] +while (high != 0 || cur != 0) { // 当 high 和 cur 同时为 0 时，说明已经越过最高位，因此跳出 + low += cur * digit; // 将 cur 加入 low ，组成下轮 low + cur = high % 10; // 下轮 cur 是本轮 high 的最低位 + high /= 10; // 将本轮 high 最低位删除，得到下轮 high + digit *= 10; // 位因子每轮 × 10 +} +``` + +```C++ [] +while (high != 0 || cur != 0) { // 当 high 和 cur 同时为 0 时，说明已经越过最高位，因此跳出 + low += cur * digit; // 将 cur 加入 low ，组成下轮 low + cur = high % 10; // 下轮 cur 是本轮 high 的最低位 + high /= 10; // 将本轮 high 最低位删除，得到下轮 high + digit *= 10; // 位因子每轮 × 10 +} +``` + + + +### 代码： + +```Python [] +class Solution: + def countDigitOne(self, n: int) -> int: + digit, res = 1, 0 + high, cur, low = n // 10, n % 10, 0 + while high != 0 or cur != 0: + if cur == 0: res += high * digit + elif cur == 1: res += high * digit + low + 1 + else: res += (high + 1) * digit + low += cur * digit + cur = high % 10 + high //= 10 + digit *= 10 + return res +``` + +```Java [] +class Solution { + public int countDigitOne(int n) { + int digit = 1, res = 0; + int high = n / 10, cur = n % 10, low = 0; + while (high != 0 || cur != 0) { + if (cur == 0) res += high * digit; + else if (cur == 1) res += high * digit + low + 1; + else res += (high + 1) * digit; + low += cur * digit; + cur = high % 10; + high /= 10; + digit *= 10; + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int countDigitOne(int n) { + long digit = 1; + int high = n / 10, cur = n % 10, low = 0, res = 0; + while (high != 0 || cur != 0) { + if (cur == 0) res += high * digit; + else if (cur == 1) res += high * digit + low + 1; + else res += (high + 1) * digit; + low += cur * digit; + cur = high % 10; + high /= 10; + digit *= 10; + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(\log n)$ ：** 循环内的计算操作使用 $O(1)$ 时间；循环次数为数字 $n$ 的位数，即 $\log_{10}{n}$ ，因此循环使用 $O(\log n)$ 时间。 +- **空间复杂度 $O(1)$ ：** 几个变量使用常数大小的额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/235. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\347\232\204\346\234\200\350\277\221\345\205\254\345\205\261\347\245\226\345\205\210.md" "b/selected_coding_interview/docs/235. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\347\232\204\346\234\200\350\277\221\345\205\254\345\205\261\347\245\226\345\205\210.md" new file mode 100644 index 0000000..de41558 --- /dev/null +++ "b/selected_coding_interview/docs/235. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\347\232\204\346\234\200\350\277\221\345\205\254\345\205\261\347\245\226\345\205\210.md" @@ -0,0 +1,190 @@ +## 解题思路： + +**祖先的定义：** 若节点 $p$ 在节点 $root$ 的左（右）子树中，或 $p = root$，则称 $root$ 是 $p$ 的祖先。 + +**最近公共祖先的定义：** 设节点 $root$ 为节点 $p,q$ 的某公共祖先，若其左子节点 $root.left$ 和右子节点 $root.right$ 都不是 $p,q$ 的公共祖先，则称 $root$ 是 “最近的公共祖先” 。 + +![Picture1.png](https://pic.leetcode-cn.com/1599885085-LbAmPd-Picture1.png){:width=400} + +根据以上定义，若 $root$ 是 $p,q$ 的 **最近公共祖先** ，则只可能为以下三种情况之一： + +1. $p$ 和 $q$ 在 $root$ 的子树中，且分列 $root$ 的 **异侧**（即分别在左、右子树中）。 +2. $p = root$ 且 $q$ 在 $root$ 的左或右子树中； +3. $q = root$ 且 $p$ 在 $root$ 的左或右子树中； + +![Picture2.png](https://pic.leetcode-cn.com/1599885085-mTpblH-Picture2.png){:width=400} + +本题给定了两个重要条件：① 树为 **二叉搜索树** ，② 树的所有节点的值都是 **唯一** 的。根据以上条件，可方便地判断 $p,q$ 与 $root$ 的子树关系，即： + +- 若 $root.val < p.val$ ，则 $p$ 在 $root$ **右子树** 中。 +- 若 $root.val > p.val$ ，则 $p$ 在 $root$ **左子树** 中。 +- 若 $root.val = p.val$ ，则 $p$ 和 $root$ 指向 **同一节点** 。 + +## 方法一：迭代 + +1. **循环搜索：** 当节点 $root$ 为空时跳出。 + 1. 当 $p, q$ 都在 $root$ 的 **右子树** 中，则遍历至 $root.right$ 。 + 2. 否则，当 $p, q$ 都在 $root$ 的 **左子树** 中，则遍历至 $root.left$ 。 + 3. 否则，说明找到了 **最近公共祖先** ，跳出。 +2. **返回值：** 最近公共祖先 $root$ 。 + + + +### 代码： + +```Python [] +class Solution: + def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode': + while root: + if root.val < p.val and root.val < q.val: # p,q 都在 root 的右子树中 + root = root.right # 遍历至右子节点 + elif root.val > p.val and root.val > q.val: # p,q 都在 root 的左子树中 + root = root.left # 遍历至左子节点 + else: break + return root +``` + +```Java [] +class Solution { + public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { + while (root != null) { + if (root.val < p.val && root.val < q.val) // p,q 都在 root 的右子树中 + root = root.right; // 遍历至右子节点 + else if (root.val > p.val && root.val > q.val) // p,q 都在 root 的左子树中 + root = root.left; // 遍历至左子节点 + else break; + } + return root; + } +} +``` + +```C++ [] +class Solution { +public: + TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { + while (root != nullptr) { + if (root->val < p->val && root->val < q->val) // p,q 都在 root 的右子树中 + root = root->right; // 遍历至右子节点 + else if (root->val > p->val && root->val > q->val) // p,q 都在 root 的左子树中 + root = root->left; // 遍历至左子节点 + else break; + } + return root; + } +}; +``` + +代码优化：若可保证 $p.val < q.val$ ，则在循环中可减少判断条件，提升计算效率。 + +```Python [] +class Solution: + def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode': + if p.val > q.val: p, q = q, p # 保证 p.val < q.val + while root: + if root.val < p.val: # p,q 都在 root 的右子树中 + root = root.right # 遍历至右子节点 + elif root.val > q.val: # p,q 都在 root 的左子树中 + root = root.left # 遍历至左子节点 + else: break + return root +``` + +```Java [] +class Solution { + public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { + if (p.val > q.val) { // 保证 p.val < q.val + TreeNode tmp = p; + p = q; + q = tmp; + } + while (root != null) { + if (root.val < p.val) // p,q 都在 root 的右子树中 + root = root.right; // 遍历至右子节点 + else if (root.val > q.val) // p,q 都在 root 的左子树中 + root = root.left; // 遍历至左子节点 + else break; + } + return root; + } +} +``` + +```C++ [] +class Solution { +public: + TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { + if (p->val > q->val) + swap(p, q); + while (root != nullptr) { + if (root->val < p->val) // p,q 都在 root 的右子树中 + root = root->right; // 遍历至右子节点 + else if (root->val > q->val) // p,q 都在 root 的左子树中 + root = root->left; // 遍历至左子节点 + else break; + } + return root; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为二叉树节点数；每循环一轮排除一层，二叉搜索树的层数最小为 $\log N$ （满二叉树），最大为 $N$ （退化为链表）。 +- **空间复杂度 $O(1)$ ：** 使用常数大小的额外空间。 + +## 方法二：递归 + +1. **递推工作：** + 1. 当 $p, q$ 都在 $root$ 的 **右子树** 中，则开启递归 $root.right$ 并返回。 + 2. 否则，当 $p, q$ 都在 $root$ 的 **左子树** 中，则开启递归 $root.left$ 并返回。 +2. **返回值：** 最近公共祖先 $root$ 。 + +### 代码： + +```Python [] +class Solution: + def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode': + if root.val < p.val and root.val < q.val: + return self.lowestCommonAncestor(root.right, p, q) + if root.val > p.val and root.val > q.val: + return self.lowestCommonAncestor(root.left, p, q) + return root +``` + +```Java [] +class Solution { + public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { + if (root.val < p.val && root.val < q.val) + return lowestCommonAncestor(root.right, p, q); + if (root.val > p.val && root.val > q.val) + return lowestCommonAncestor(root.left, p, q); + return root; + } +} +``` + +```C++ [] +class Solution { +public: + TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { + if (root->val < p->val && root->val < q->val) + return lowestCommonAncestor(root->right, p, q); + if (root->val > p->val && root->val > q->val) + return lowestCommonAncestor(root->left, p, q); + return root; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为二叉树节点数；每循环一轮排除一层，二叉搜索树的层数最小为 $\log N$ （满二叉树），最大为 $N$ （退化为链表）。 +- **空间复杂度 $O(N)$ ：** 最差情况下，即树退化为链表时，递归深度达到树的层数 $N$ 。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/236. \344\272\214\345\217\211\346\240\221\347\232\204\346\234\200\350\277\221\345\205\254\345\205\261\347\245\226\345\205\210.md" "b/selected_coding_interview/docs/236. \344\272\214\345\217\211\346\240\221\347\232\204\346\234\200\350\277\221\345\205\254\345\205\261\347\245\226\345\205\210.md" new file mode 100644 index 0000000..3e3d129 --- /dev/null +++ "b/selected_coding_interview/docs/236. \344\272\214\345\217\211\346\240\221\347\232\204\346\234\200\350\277\221\345\205\254\345\205\261\347\245\226\345\205\210.md" @@ -0,0 +1,125 @@ +#### 解题思路： + +**祖先的定义：** 若节点 $p$ 在节点 $root$ 的左（右）子树中，或 $p = root$ ，则称 $root$ 是 $p$ 的祖先。 + +**最近公共祖先的定义：** 设节点 $root$ 为节点 $p, q$ 的某公共祖先，若其左子节点 $root.left$ 和右子节点 $root.right$ 都不是 $p,q$ 的公共祖先，则称 $root$ 是 “最近的公共祖先” 。 + +![Picture1.png](https://pic.leetcode-cn.com/1599885247-rxcHcZ-Picture1.png){:width=450} + +根据以上定义，若 $root$ 是 $p, q$ 的 **最近公共祖先** ，则只可能为以下情况之一： + +1. $p$ 和 $q$ 在 $root$ 的子树中，且分列 $root$ 的 **异侧**（即分别在左、右子树中）； +2. $p = root$ ，且 $q$ 在 $root$ 的左或右子树中； +3. $q = root$ ，且 $p$ 在 $root$ 的左或右子树中； + +![Picture2.png](https://pic.leetcode-cn.com/1599885247-mgYjRv-Picture2.png){:width=450} + +考虑通过递归对二叉树进行先序遍历，当遇到节点 $p$ 或 $q$ 时返回。从底至顶回溯，当节点 $p, q$ 在节点 $root$ 的异侧时，节点 $root$ 即为最近公共祖先，则向上返回 $root$ 。 + +##### 递归解析： + +1. **终止条件：** + 1. 当越过叶节点，则直接返回 $null$ ； + 2. 当 $root$ 等于 $p, q$ ，则直接返回 $root$ ； +2. **递推工作：** + 1. 开启递归左子节点，返回值记为 $left$ ； + 2. 开启递归右子节点，返回值记为 $right$ ； +3. **返回值：** 根据 $left$ 和 $right$ ，可展开为四种情况； + 1. 当 $left$ 和 $right$ **同时为空** ：说明 $root$ 的左 / 右子树中都不包含 $p,q$ ，返回 $null$ ； + 2. 当 $left$ 和 $right$ **同时不为空** ：说明 $p, q$ 分列在 $root$ 的 **异侧** （分别在 左 / 右子树），因此 $root$ 为最近公共祖先，返回 $root$ ； + 3. 当 $left$ **为空** ，$right$ **不为空** ：$p,q$ 都不在 $root$ 的左子树中，直接返回 $right$ 。具体可分为两种情况： + 1. $p,q$ 其中一个在 $root$ 的 **右子树** 中，此时 $right$ 指向 $p$（假设为 $p$ ）； + 2. $p,q$ 两节点都在 $root$ 的 **右子树** 中，此时的 $right$ 指向 **最近公共祖先节点** ； + 4. 当 $left$ **不为空** ， $right$ **为空** ：与情况 `3.` 同理； + +> 观察发现， 情况 `1.` 可合并至 `3.` 和 `4.` 内，详见文章末尾代码。 + + + +##### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为二叉树节点数；最差情况下，需要递归遍历树的所有节点。 +- **空间复杂度 $O(N)$ ：** 最差情况下，递归深度达到 $N$ ，系统使用 $O(N)$ 大小的额外空间。 + +##### 代码： + +```Python [] +class Solution: + def lowestCommonAncestor(self, root: TreeNode, p: TreeNode, q: TreeNode) -> TreeNode: + if not root or root == p or root == q: return root + left = self.lowestCommonAncestor(root.left, p, q) + right = self.lowestCommonAncestor(root.right, p, q) + if not left: return right + if not right: return left + return root +``` + +```Java [] +class Solution { + public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { + if(root == null || root == p || root == q) return root; + TreeNode left = lowestCommonAncestor(root.left, p, q); + TreeNode right = lowestCommonAncestor(root.right, p, q); + if(left == null) return right; + if(right == null) return left; + return root; + } +} +``` + +```C++ [] +class Solution { +public: + TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { + if(root == nullptr || root == p || root == q) return root; + TreeNode *left = lowestCommonAncestor(root->left, p, q); + TreeNode *right = lowestCommonAncestor(root->right, p, q); + if(left == nullptr) return right; + if(right == nullptr) return left; + return root; + } +}; +``` + +情况 `1.` , `2.` , `3.` , `4.` 的展开写法如下。 + +```Python [] +class Solution: + def lowestCommonAncestor(self, root: TreeNode, p: TreeNode, q: TreeNode) -> TreeNode: + if not root or root == p or root == q: return root + left = self.lowestCommonAncestor(root.left, p, q) + right = self.lowestCommonAncestor(root.right, p, q) + if not left and not right: return # 1. + if not left: return right # 3. + if not right: return left # 4. + return root # 2. if left and right: +``` + +```Java [] +class Solution { + public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { + if(root == null || root == p || root == q) return root; + TreeNode left = lowestCommonAncestor(root.left, p, q); + TreeNode right = lowestCommonAncestor(root.right, p, q); + if(left == null && right == null) return null; // 1. + if(left == null) return right; // 3. + if(right == null) return left; // 4. + return root; // 2. if(left != null and right != null) + } +} +``` + +```C++ [] +class Solution { +public: + TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { + if(root == nullptr || root == p || root == q) return root; + TreeNode *left = lowestCommonAncestor(root->left, p, q); + TreeNode *right = lowestCommonAncestor(root->right, p, q); + if(left == nullptr && right == nullptr) return nullptr; // 1. + if(left == nullptr) return right; // 3. + if(right == nullptr) return left; // 4. + return root; // 2. if(left != null and right != null) + } +}; +``` diff --git "a/selected_coding_interview/docs/237. \345\210\240\351\231\244\351\223\276\350\241\250\344\270\255\347\232\204\350\212\202\347\202\271.md" "b/selected_coding_interview/docs/237. \345\210\240\351\231\244\351\223\276\350\241\250\344\270\255\347\232\204\350\212\202\347\202\271.md" new file mode 100644 index 0000000..999a3ce --- /dev/null +++ "b/selected_coding_interview/docs/237. \345\210\240\351\231\244\351\223\276\350\241\250\344\270\255\347\232\204\350\212\202\347\202\271.md" @@ -0,0 +1,91 @@ +## 解题思路： + +普通情况：设前驱节点 `pre` 、当前节点 `cur` 、后继节点 `cur.next` ，则执行 `pre.next = cur.next` 可**将节点 `cur` 从链表中删除**。 + +![ccw-02-03.001.png](https://pic.leetcode-cn.com/1642318280-EhKkoi-ccw-02-03.001.png) + +本题仅传入「待删除节点 `node` 」，由于普通链表只有「单向指针」，因此无法访问到 `node` 的「前驱节点」，进而无法使用以上方法删除节点 `node` 。 + +为了删除节点 `node` ，可使用以下方法： + +1. 复制后继节点 `node.next` 的「节点值」至节点 `node` ； +2. 使用上述方法将 `node.next` 从链表中删除即可； + +> 如下图所示，示例链表 $4 \rightarrow 5 \rightarrow 1 \rightarrow 9$ 和待删除节点 $5$ ，完成删除后链表变为 $4 \rightarrow 1 \rightarrow 9$ 。 + + + +## 代码： + +> 后三个 Tab 为「代码注释解析」。 + +```Python [] +class Solution: + def deleteNode(self, node): + node.val = node.next.val + node.next = node.next.next +``` + +```Java [] +class Solution { + public void deleteNode(ListNode node) { + node.val = node.next.val; + node.next = node.next.next; + } +} +``` + +```C++ [] +class Solution { +public: + void deleteNode(ListNode* node) { + node->val = node->next->val; + node->next = node->next->next; + } +}; +``` + +```Python [] +class Solution: + def deleteNode(self, node): + # 复制 node.next 到 node + node.val = node.next.val + # 从链表中删除 node.next + node.next = node.next.next +``` + +```Java [] +class Solution { + public void deleteNode(ListNode node) { + // 复制 node.next 到 node + node.val = node.next.val; + // 从链表中删除 node.next + node.next = node.next.next; + } +} +``` + +```C++ [] +class Solution { +public: + void deleteNode(ListNode* node) { + // 复制 node.next 到 node + node->val = node->next->val; + // 从链表中删除 node.next + node->next = node->next->next; + } +}; +``` + +### 复杂度分析： + +**时间复杂度 $O(1)$ ：** 使用常数时间。 + +**空间复杂度 $O(1)$ ：** 使用常数大小额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/238. \351\231\244\350\207\252\350\272\253\344\273\245\345\244\226\346\225\260\347\273\204\347\232\204\344\271\230\347\247\257.md" "b/selected_coding_interview/docs/238. \351\231\244\350\207\252\350\272\253\344\273\245\345\244\226\346\225\260\347\273\204\347\232\204\344\271\230\347\247\257.md" new file mode 100644 index 0000000..3f58f1b --- /dev/null +++ "b/selected_coding_interview/docs/238. \351\231\244\350\207\252\350\272\253\344\273\245\345\244\226\346\225\260\347\273\204\347\232\204\344\271\230\347\247\257.md" @@ -0,0 +1,85 @@ +## 解题思路： + +本题的难点在于 **不能使用除法** ，即需要 **只用乘法** 生成数组 $ans$ 。根据题目对 $ans[i]$ 的定义，可列出下图所示的表格。 + +根据表格的主对角线（全为 $1$ ），可将表格分为 **上三角** 和 **下三角** 两部分。分别迭代计算下三角和上三角两部分的乘积，即可 **不使用除法** 就获得结果。 + +> 下图中 $A = nums$ , $B = ans$ + +![Picture1.png](https://pic.leetcode-cn.com/1624619180-vpyyqh-Picture1.png){:width=500} + +### 算法流程： + +1. 初始化：数组 $ans$ ，其中 $ans[0] = 1$ ；辅助变量 $tmp = 1$ 。 +2. 计算 $ans[i]$ 的 **下三角** 各元素的乘积，直接乘入 $ans[i]$ 。 +3. 计算 $ans[i]$ 的 **上三角** 各元素的乘积，记为 $tmp$ ，并乘入 $ans[i]$ 。 +4. 返回 $ans$ 。 + + + +## 代码： + +```Python [] +class Solution: + def productExceptSelf(self, nums: List[int]) -> List[int]: + ans, tmp = [1] * len(nums), 1 + for i in range(1, len(nums)): + ans[i] = ans[i - 1] * nums[i - 1] # 下三角 + for i in range(len(nums) - 2, -1, -1): + tmp *= nums[i + 1] # 上三角 + ans[i] *= tmp # 下三角 * 上三角 + return ans +``` + +```Java [] +class Solution { + public int[] productExceptSelf(int[] nums) { + int len = nums.length; + if (len == 0) return new int[0]; + int[] ans = new int[len]; + ans[0] = 1; + int tmp = 1; + for (int i = 1; i < len; i++) { + ans[i] = ans[i - 1] * nums[i - 1]; + } + for (int i = len - 2; i >= 0; i--) { + tmp *= nums[i + 1]; + ans[i] *= tmp; + } + return ans; + } +} +``` + +```C++ [] +class Solution { +public: + vector productExceptSelf(vector& nums) { + int len = nums.size(); + if (len == 0) return {}; + vector ans(len, 1); + ans[0] = 1; + int tmp = 1; + for (int i = 1; i < len; i++) { + ans[i] = ans[i - 1] * nums[i - 1]; + } + for (int i = len - 2; i >= 0; i--) { + tmp *= nums[i + 1]; + ans[i] *= tmp; + } + return ans; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为数组长度，两轮遍历数组 $nums$ ，使用 $O(N)$ 时间。 +- **空间复杂度 $O(1)$ ：** 变量 $tmp$ 使用常数大小额外空间（数组 $ans$ 作为返回值，不计入复杂度考虑）。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/239. \346\273\221\345\212\250\347\252\227\345\217\243\346\234\200\345\244\247\345\200\274.md" "b/selected_coding_interview/docs/239. \346\273\221\345\212\250\347\252\227\345\217\243\346\234\200\345\244\247\345\200\274.md" new file mode 100644 index 0000000..fa28595 --- /dev/null +++ "b/selected_coding_interview/docs/239. \346\273\221\345\212\250\347\252\227\345\217\243\346\234\200\345\244\247\345\200\274.md" @@ -0,0 +1,233 @@ +# 方法一：辅助矩阵 + +如下图所示，矩阵顺时针旋转 90º 后，可找到以下规律： + +- 「第 $i$ 行」元素旋转到「第 $n - 1 - i$ 列」元素； +- 「第 $j$ 列」元素旋转到「第 $j$ 行」元素； + +因此，对于矩阵任意第 $i$ 行、第 $j$ 列元素 $matrix[i][j]$ ，矩阵旋转 90º 后「元素位置旋转公式」为： + +$$ +\begin{aligned} +matrix[i][j] & \rightarrow matrix[j][n - 1 - i] \\ +原索引位置 & \rightarrow 旋转后索引位置 +\end{aligned} +$$ + +![ccw-01-07.001.png](https://pic.leetcode-cn.com/1638557961-AVzCQb-ccw-01-07.001.png) + +根据以上「元素旋转公式」，考虑遍历矩阵，将各元素依次写入到旋转后的索引位置。但仍**存在问题**：在写入一个元素 $matrix[i][j] \rightarrow matrix[j][n - 1 - i]$ 后，原矩阵元素 $matrix[j][n - 1 - i]$ 就会**被覆盖（即丢失）**，而此丢失的元素就无法被写入到旋转后的索引位置了。 + +为解决此问题，考虑借助一个「辅助矩阵」暂存原矩阵，通过遍历辅助矩阵所有元素，将各元素填入「原矩阵」旋转后的新索引位置即可。 + +```Python [] +class Solution: + def rotate(self, matrix: List[List[int]]) -> None: + n = len(matrix) + # 深拷贝 matrix -> tmp + tmp = copy.deepcopy(matrix) + # 根据元素旋转公式，遍历修改原矩阵 matrix 的各元素 + for i in range(n): + for j in range(n): + matrix[j][n - 1 - i] = tmp[i][j] +``` + +```Java [] +class Solution { + public void rotate(int[][] matrix) { + int n = matrix.length; + // 深拷贝 matrix -> tmp + int[][] tmp = new int[n][]; + for (int i = 0; i < n; i++) + tmp[i] = matrix[i].clone(); + // 根据元素旋转公式，遍历修改原矩阵 matrix 的各元素 + for (int i = 0; i < n; i++) { + for (int j = 0; j < n; j++) { + matrix[j][n - 1 - i] = tmp[i][j]; + } + } + } +} +``` + +```C++ [] +class Solution { +public: + void rotate(vector>& matrix) { + int n = matrix.size(); + // 深拷贝 matrix -> tmp + vector> tmp = matrix; + // 根据元素旋转公式，遍历修改原矩阵 matrix 的各元素 + for (int i = 0; i < n; i++) { + for (int j = 0; j < n; j++) { + matrix[j][n - 1 - i] = tmp[i][j]; + } + } + } +}; +``` + +如以上代码所示，遍历矩阵所有元素的时间复杂度为 $O(N^2)$ ；由于借助了一个辅助矩阵，**空间复杂度**为 $O(N^2)$ 。 + +# 方法二：原地修改 + +考虑不借助辅助矩阵，通过在原矩阵中直接「原地修改」，实现空间复杂度 $O(1)$ 的解法。 + +以位于矩阵四个角点的元素为例，设矩阵左上角元素 $A$ 、右上角元素 $B$ 、右下角元素 $C$ 、左下角元素 $D$ 。矩阵旋转 90º 后，相当于依次先后执行 $D \rightarrow A$ , $C \rightarrow D$ , $B \rightarrow C$ , $A \rightarrow B$ 修改元素，即如下「首尾相接」的元素旋转操作： + +$$ +A \leftarrow D \leftarrow C \leftarrow B \leftarrow A +$$ + +![ccw-01-07.002.png](https://pic.leetcode-cn.com/1638557961-BSxFQQ-ccw-01-07.002.png) + +如上图所示，由于第 $1$ 步 $D \rightarrow A$ 已经将 $A$ 覆盖（导致 $A$ 丢失），此丢失导致最后第 $4$ 步 $A \rightarrow B$ 无法赋值。为解决此问题，考虑借助一个「辅助变量 $tmp$ 」预先存储 $A$ ，此时的旋转操作变为： + +$$ +暂存 \ tmp = A \\ +A \leftarrow D \leftarrow C \leftarrow B \leftarrow tmp +$$ + +![ccw-01-07.003.png](https://pic.leetcode-cn.com/1638557961-hYpOoH-ccw-01-07.003.png) + +如上图所示，一轮可以完成矩阵 4 个元素的旋转。因而，只要分别以矩阵左上角 $1/4$ 的各元素为起始点执行以上旋转操作，即可完整实现矩阵旋转。 + +具体来看，当矩阵大小 $n$ 为偶数时，取前 $\frac{n}{2}$ 行、前 $\frac{n}{2}$ 列的元素为起始点；当矩阵大小 $n$ 为奇数时，取前 $\frac{n}{2}$ 行、前 $\frac{n + 1}{2}$ 列的元素为起始点。 + +令 $matrix[i][j] = A$ ，根据文章开头的元素旋转公式，可推导得适用于任意起始点的元素旋转操作： + +$$ +暂存 tmp = matrix[i][j] \\ +matrix[i][j] \leftarrow matrix[n - 1 - j][i] \leftarrow matrix[n - 1 - i][n - 1 - j] \leftarrow matrix[j][n - 1 - i] \leftarrow tmp +$$ + +> 如下图所示，为示例矩阵的算法执行流程。 + + + +## 代码 + +> 后三个 Tab 为「代码注释解析」。 + +```Python [] +class Solution: + def rotate(self, matrix: List[List[int]]) -> None: + n = len(matrix) + for i in range(n // 2): + for j in range((n + 1) // 2): + tmp = matrix[i][j] + matrix[i][j] = matrix[n - 1 - j][i] + matrix[n - 1 - j][i] = matrix[n - 1 - i][n - 1 - j] + matrix[n - 1 - i][n - 1 - j] = matrix[j][n - 1 - i] + matrix[j][n - 1 - i] = tmp +``` + +```Java [] +class Solution { + public void rotate(int[][] matrix) { + int n = matrix.length; + for (int i = 0; i < n / 2; i++) { + for (int j = 0; j < (n + 1) / 2; j++) { + int tmp = matrix[i][j]; + matrix[i][j] = matrix[n - 1 - j][i]; + matrix[n - 1 - j][i] = matrix[n - 1 - i][n - 1 - j]; + matrix[n - 1 - i][n - 1 - j] = matrix[j][n - 1 - i]; + matrix[j][n - 1 - i] = tmp; + } + } + } +} +``` + +```C++ [] +class Solution { +public: + void rotate(vector>& matrix) { + int n = matrix.size(); + for (int i = 0; i < n / 2; i++) { + for (int j = 0; j < (n + 1) / 2; j++) { + int tmp = matrix[i][j]; + matrix[i][j] = matrix[n - 1 - j][i]; + matrix[n - 1 - j][i] = matrix[n - 1 - i][n - 1 - j]; + matrix[n - 1 - i][n - 1 - j] = matrix[j][n - 1 - i]; + matrix[j][n - 1 - i] = tmp; + } + } + } +}; +``` + +```Python [] +class Solution: + def rotate(self, matrix: List[List[int]]) -> None: + # 设矩阵行列数为 n + n = len(matrix) + # 起始点范围为 0 <= i < n // 2 , 0 <= j < (n + 1) // 2 + # 其中 '//' 为整数除法 + for i in range(n // 2): + for j in range((n + 1) // 2): + # 暂存 A 至 tmp + tmp = matrix[i][j] + # 元素旋转操作 A <- D <- C <- B <- tmp + matrix[i][j] = matrix[n - 1 - j][i] + matrix[n - 1 - j][i] = matrix[n - 1 - i][n - 1 - j] + matrix[n - 1 - i][n - 1 - j] = matrix[j][n - 1 - i] + matrix[j][n - 1 - i] = tmp +``` + +```Java [] +class Solution { + public void rotate(int[][] matrix) { + // 设矩阵行列数为 n + int n = matrix.length; + // 起始点范围为 0 <= i < n / 2 , 0 <= j < (n + 1) / 2 + // 其中 '/' 为整数除法 + for (int i = 0; i < n / 2; i++) { + for (int j = 0; j < (n + 1) / 2; j++) { + // 暂存 A 至 tmp + int tmp = matrix[i][j]; + // 元素旋转操作 A <- D <- C <- B <- tmp + matrix[i][j] = matrix[n - 1 - j][i]; + matrix[n - 1 - j][i] = matrix[n - 1 - i][n - 1 - j]; + matrix[n - 1 - i][n - 1 - j] = matrix[j][n - 1 - i]; + matrix[j][n - 1 - i] = tmp; + } + } + } +} +``` + +```C++ [] +class Solution { +public: + void rotate(vector>& matrix) { + // 设矩阵行列数为 n + int n = matrix.size(); + // 起始点范围为 0 <= i < n / 2 , 0 <= j < (n + 1) / 2 + // 其中 '/' 为整数除法 + for (int i = 0; i < n / 2; i++) { + for (int j = 0; j < (n + 1) / 2; j++) { + // 暂存 A 至 tmp + int tmp = matrix[i][j]; + // 元素旋转操作 A <- D <- C <- B <- tmp + matrix[i][j] = matrix[n - 1 - j][i]; + matrix[n - 1 - j][i] = matrix[n - 1 - i][n - 1 - j]; + matrix[n - 1 - i][n - 1 - j] = matrix[j][n - 1 - i]; + matrix[j][n - 1 - i] = tmp; + } + } + } +}; +``` + +## 复杂度分析 + +- **时间复杂度 $O(N^2)$ ：** 其中 $N$ 为输入矩阵的行（列）数。需要将矩阵中每个元素旋转到新的位置，即对矩阵所有元素操作一次，使用 $O(N^2)$ 时间。 +- **空间复杂度 $O(1)$ ：** 临时变量 $tmp$ 使用常数大小的额外空间。值得注意，当循环中进入下轮迭代，上轮迭代初始化的 $tmp$ 占用的内存就会被自动释放，因此无累计使用空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/240. \346\220\234\347\264\242\344\272\214\347\273\264\347\237\251\351\230\265 II.md" "b/selected_coding_interview/docs/240. \346\220\234\347\264\242\344\272\214\347\273\264\347\237\251\351\230\265 II.md" new file mode 100644 index 0000000..aa81196 --- /dev/null +++ "b/selected_coding_interview/docs/240. \346\220\234\347\264\242\344\272\214\347\273\264\347\237\251\351\230\265 II.md" @@ -0,0 +1,80 @@ +## 解题思路： + +> 若使用暴力法遍历矩阵 `matrix` ，则时间复杂度为 $O(NM)$ 。暴力法未利用矩阵 **“从上到下递增、从左到右递增”** 的特点，显然不是最优解法。 + +如下图所示，我们将矩阵逆时针旋转 45° ，并将其转化为图形式，发现其类似于 **二叉搜索树** ，即对于每个元素，其左分支元素更小、右分支元素更大。因此，通过从 “根节点” 开始搜索，遇到比 `target` 大的元素就向左，反之向右，即可找到目标值 `target` 。 + +![Picture1.png](https://pic.leetcode-cn.com/6584ea93812d27112043d203ea90e4b0950117d45e0452d0c630fcb247fbc4af-Picture1.png){:width=450} + +“根节点” 对应的是矩阵的 “左下角” 和 “右上角” 元素，本文称之为 **标志数** ，以 `matrix` 中的 **左下角元素** 为标志数 `flag` ，则有: + +1. 若 `flag > target` ，则 `target` 一定在 `flag` 所在 **行的上方** ，即 `flag` 所在行可被消去。 +2. 若 `flag < target` ，则 `target` 一定在 `flag` 所在 **列的右方** ，即 `flag` 所在列可被消去。 + +### **算法流程：** + +1. 从矩阵 `matrix` 左下角元素（索引设为 `(i, j)` ）开始遍历，并与目标值对比： + - 当 `matrix[i][j] > target` 时，执行 `i--` ，即消去第 `i` 行元素。 + - 当 `matrix[i][j] < target` 时，执行 `j++` ，即消去第 `j` 列元素。 + - 当 `matrix[i][j] = target` 时，返回 $true$ ，代表找到目标值。 +2. 若行索引或列索引越界，则代表矩阵中无目标值，返回 $false$ 。 + +> 每轮 `i` 或 `j` 移动后，相当于生成了“消去一行（列）的新矩阵”， 索引`(i,j)` 指向新矩阵的左下角元素（标志数），因此可重复使用以上性质消去行（列）。 + + + +## 代码： + +```Python [] +class Solution: + def searchMatrix(self, matrix: List[List[int]], target: int) -> bool: + i, j = len(matrix) - 1, 0 + while i >= 0 and j < len(matrix[0]): + if matrix[i][j] > target: i -= 1 + elif matrix[i][j] < target: j += 1 + else: return True + return False +``` + +```Java [] +class Solution { + public boolean searchMatrix(int[][] matrix, int target) { + int i = matrix.length - 1, j = 0; + while(i >= 0 && j < matrix[0].length) + { + if(matrix[i][j] > target) i--; + else if(matrix[i][j] < target) j++; + else return true; + } + return false; + } +} +``` + +```C++ [] +class Solution { +public: + bool searchMatrix(vector>& matrix, int target) { + int i = matrix.size() - 1, j = 0; + while(i >= 0 && j < matrix[0].size()) + { + if(matrix[i][j] > target) i--; + else if(matrix[i][j] < target) j++; + else return true; + } + return false; + } +}; +``` + +### 复杂度分析： + +- 时间复杂度 $O(M+N)$ ：其中，$N$ 和 $M$ 分别为矩阵行数和列数，此算法最多循环 $M+N$ 次。 +- 空间复杂度 $O(1)$ : `i`, `j` 指针使用常数大小额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/242. \346\234\211\346\225\210\347\232\204\345\255\227\346\257\215\345\274\202\344\275\215\350\257\215.md" "b/selected_coding_interview/docs/242. \346\234\211\346\225\210\347\232\204\345\255\227\346\257\215\345\274\202\344\275\215\350\257\215.md" new file mode 100644 index 0000000..d480c82 --- /dev/null +++ "b/selected_coding_interview/docs/242. \346\234\211\346\225\210\347\232\204\345\255\227\346\257\215\345\274\202\344\275\215\350\257\215.md" @@ -0,0 +1,178 @@ +## 解题思路： + +设两字符串 $s_1$ , $s_2$ ，则两者互为重排的「充要条件」为：两字符串 $s_1$ , $s_2$ 包含的字符是一致的，即 $s_1$ , $s_2$ 所有对应字符数量都相同，仅排列顺序不同。 + +![ccw-01-02.001.png](https://pic.leetcode-cn.com/1637859370-KkSeGV-ccw-01-02.001.png) + +根据以上分析，可借助「哈希表」分别统计 $s_1$ , $s_2$ 中各字符数量 `key: 字符, value: 数量` ，分为以下情况： + +- 若 $s_1$ , $s_2$ 字符串长度不相等，则「不互为重排」； +- 若 $s_1$ , $s_2$ 某对应字符数量不同，则「不互为重排」； +- 否则，若 $s_1$ , $s_2$ 所有对应字符数量都相同，则「互为重排」； + +具体上看，我们可以统计 $s_1$ 各字符时执行 $+1$ ，统计 $s_2$ 各字符时 $-1$ 。若两字符串互为重排，则最终哈希表中所有字符统计数值都应为 0 。 + +> 如下图所示，为 $s_1 = "abc"$ , $s_2 = "bad"$ 的算法执行过程。 + + + +## 代码： + +Python 代码使用 `collections.defaultdict()` 类，可指定所有 `key` 对应的默认 `value` 。 + +> 后三个 Tab 为「代码注释解析」。 + +```Python [] +class Solution: + def isAnagram(self, s: str, t: str) -> bool: + if len(s) != len(t): + return False + dic = defaultdict(int) + for c in s: + dic[c] += 1 + for c in t: + dic[c] -= 1 + for val in dic.values(): + if val != 0: + return False + return True +``` + +```Java [] +class Solution { + public boolean isAnagram(String s, String t) { + int len1 = s.length(), len2 = t.length(); + if (len1 != len2) + return false; + HashMap dic = new HashMap<>(); + for (int i = 0; i < len1; i++) { + dic.put(s.charAt(i) , dic.getOrDefault(s.charAt(i), 0) + 1); + } + for (int i = 0; i < len2; i++) { + dic.put(t.charAt(i) , dic.getOrDefault(t.charAt(i), 0) - 1); + } + for (int val : dic.values()) { + if (val != 0) + return false; + } + return true; + } +} +``` + +```C++ [] +class Solution { +public: + bool isAnagram(string s, string t) { + if (s.length() != t.length()) + return false; + unordered_map dic; + for (char c : s) { + dic[c] += 1; + } + for (char c : t) { + dic[c] -= 1; + } + for (auto kv : dic) { + if (kv.second != 0) + return false; + } + return true; + } +}; +``` + +```Python [] +class Solution: + def isAnagram(self, s: str, t: str) -> bool: + # 若 s, t 长度不同，则不互为重排 + if len(s) != len(t): + return False + # 初始化字典 dic ，且所有 key 的初始 value 都为 0 + dic = defaultdict(int) + # 统计字符串 s 各字符数量，遇到 +1 + for c in s: + dic[c] += 1 + # 统计字符串 t 各字符数量，遇到 -1 + for c in t: + dic[c] -= 1 + # 遍历 s, t 中各字符的数量差 + for val in dic.values(): + # 若 s, t 中某字符的数量不一致，则不互为重排 + if val != 0: + return False + # 所有字符数量都一致，因此互为重排 + return True +``` + +```Java [] +class Solution { + public boolean isAnagram(String s, String t) { + int len1 = s.length(), len2 = t.length(); + // 若 s, t 长度不同，则不互为重排 + if (len1 != len2) + return false; + // 初始化哈希表 dic + HashMap dic = new HashMap<>(); + // 统计字符串 s 各字符数量，遇到 +1 + for (int i = 0; i < len1; i++) { + // dic.getOrDefault(key, default) 函数：在 key 存在时返回对应 value ，在 key 不存在时默认返回 default 值； + dic.put(s.charAt(i) , dic.getOrDefault(s.charAt(i), 0) + 1); + } + // 统计字符串 t 各字符数量，遇到 -1 + for (int i = 0; i < len2; i++) { + dic.put(t.charAt(i) , dic.getOrDefault(t.charAt(i), 0) - 1); + } + // 遍历 s, t 中各字符的数量差 + for (int val : dic.values()) { + // 若 s, t 中某字符的数量不一致，则不互为重排 + if (val != 0) + return false; + } + // 所有字符数量都一致，因此互为重排 + return true; + } +} +``` + +```C++ [] +class Solution { +public: + bool isAnagram(string s, string t) { + // 若 s, t 长度不同，则不互为重排 + if (s.length() != t.length()) + return false; + // 初始化哈希表 dic + unordered_map dic; + // 统计字符串 s 各字符数量，遇到 +1 + for (char c : s) { + dic[c] += 1; + } + // 统计字符串 t 各字符数量，遇到 -1 + for (char c : t) { + dic[c] -= 1; + } + // 遍历 s, t 中各字符的数量差 + for (auto kv : dic) { + // 若 s, t 中某字符的数量不一致，则不互为重排 + if (kv.second != 0) + return false; + } + // 所有字符数量都一致，因此互为重排 + return true; + } +}; +``` + +### 复杂度分析： + +**时间复杂度 $O(M + N)$ ：** 其 $M$ , $N$ 分别为字符串 $s_1$ , $s_2$ 长度。当 $s_1$ , $s_2$ 无相同字符时，三轮循环的总迭代次数最多为 $2M + 2N$ ，使用 $O(M + N)$ 线性时间。 + +**空间复杂度 $O(1)$ ：** 由于字符种类是有限的（常量），一般 ASCII 码共包含 128 个字符，因此可假设使用 $O(1)$ 大小的额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/264. \344\270\221\346\225\260 II\302\240.md" "b/selected_coding_interview/docs/264. \344\270\221\346\225\260 II\302\240.md" new file mode 100644 index 0000000..d6f1d72 --- /dev/null +++ "b/selected_coding_interview/docs/264. \344\270\221\346\225\260 II\302\240.md" @@ -0,0 +1,89 @@ +## 解题思路： + +根据题意，每个丑数都可以由其他较小的丑数通过乘以 $2$ 或 $3$ 或 $5$ 得到。 + +所以，可以考虑使用一个优先队列保存所有的丑数，每次取出最小的那个，然后乘以 $2$ , $3$ , $5$ 后放回队列。然而，**这样做会出现重复的丑数**。例如： + +```shell +初始化丑数列表 [1] +第一轮： 1 -> 2, 3, 5 ，丑数列表变为 [1, 2, 3, 5] +第二轮： 2 -> 4, 6, 10 ，丑数列表变为 [1, 2, 3, 4, 6, 10] +第三轮： 3 -> 6, 9, 15 ，出现重复的丑数 6 +``` + +为了避免重复，我们可以用三个指针 $a$ , $b$, $c$ ，分别表示下一个丑数是当前指针指向的丑数乘以 $2$ , $3$ , $5$ 。 + +利用三个指针生成丑数的算法流程： + +1. 初始化丑数列表 $res$ ，首个丑数为 $1$ ，三个指针 $a$ , $b$, $c$ 都指向首个丑数。 +2. 开启循环生成丑数： + 1. 计算下一个丑数的候选集 $res[a] \cdot 2$ , $res[b] \cdot 3$ , $res[c] \cdot 5$ 。 + 2. 选择丑数候选集中最小的那个作为下一个丑数，填入 $res$ 。 + 3. 将被选中的丑数对应的指针向右移动一格。 +3. 返回 $res$ 的最后一个元素即可。 + + + +## 代码： + +```Python [] +class Solution: + def nthUglyNumber(self, n: int) -> int: + res, a, b, c = [1] * n, 0, 0, 0 + for i in range(1, n): + n2, n3, n5 = res[a] * 2, res[b] * 3, res[c] * 5 + res[i] = min(n2, n3, n5) + if res[i] == n2: a += 1 + if res[i] == n3: b += 1 + if res[i] == n5: c += 1 + return res[-1] +``` + +```Java [] +class Solution { + public int nthUglyNumber(int n) { + int a = 0, b = 0, c = 0; + int[] res = new int[n]; + res[0] = 1; + for(int i = 1; i < n; i++) { + int n2 = res[a] * 2, n3 = res[b] * 3, n5 = res[c] * 5; + res[i] = Math.min(Math.min(n2, n3), n5); + if (res[i] == n2) a++; + if (res[i] == n3) b++; + if (res[i] == n5) c++; + } + return res[n - 1]; + } +} +``` + +```C++ [] +class Solution { +public: + int nthUglyNumber(int n) { + int a = 0, b = 0, c = 0; + int res[n]; + res[0] = 1; + for(int i = 1; i < n; i++) { + int n2 = res[a] * 2, n3 = res[b] * 3, n5 = res[c] * 5; + res[i] = min(min(n2, n3), n5); + if (res[i] == n2) a++; + if (res[i] == n3) b++; + if (res[i] == n5) c++; + } + return res[n - 1]; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(n)$ ：** 计算 $res$ 列表需遍历 $n-1$ 轮。 +- **空间复杂度 $O(n)$ ：** 长度为 $n$ 的 $res$ 列表使用 $O(n)$ 的额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/266. \345\233\236\346\226\207\346\216\222\345\210\227.md" "b/selected_coding_interview/docs/266. \345\233\236\346\226\207\346\216\222\345\210\227.md" new file mode 100644 index 0000000..9116451 --- /dev/null +++ "b/selected_coding_interview/docs/266. \345\233\236\346\226\207\346\216\222\345\210\227.md" @@ -0,0 +1,158 @@ +## 解题思路： + +根据字符串长度，「回文串」可分为两种情况： + +- 「回文串」长度为偶数：所有不同字符的出现次数都为「偶数」； +- 「回文串」长度为奇数：位于中点的字符出现「奇数」次，其余字符出现「偶数」次； + +因此，某字符串是回文串排列之一的「充要条件」为：此字符串中，**最多只有一种字符**的出现次数为「奇数」，**其余所有字符**的出现次数都为「偶数」。 + +![ccw-01-04.001.png](https://pic.leetcode-cn.com/1638093879-zybntU-ccw-01-04.001.png) + +考虑使用「哈希表」统计给定字符串中各字符的数量，再根据以上规则判断字符串是否为回文串排列之一。 + +> 如下图所示，为判断示例字符串 `"tactcoa"` 是否为回文串排列之一的算法执行流程。 + + + +## 代码： + +Python 代码使用 `collections.defaultdict()` 类，可指定所有 `key` 对应的默认 `value` 。 + +> 后三个 Tab 的代码包括注释解析。 + +```Python [] +class Solution: + def canPermutePalindrome(self, s: str) -> bool: + dic = defaultdict(int) + for c in s: + dic[c] += 1 + odd = 0 + for val in dic.values(): + if val % 2 == 1: + odd += 1 + if odd > 1: + return False + return True +``` + +```Java [] +class Solution { + public boolean canPermutePalindrome(String s) { + HashMap dic = new HashMap<>(); + for (int i = 0; i < s.length(); i++) { + dic.put(s.charAt(i), dic.getOrDefault(s.charAt(i), 0) + 1); + } + int odd = 0; + for (int val : dic.values()) { + if (val % 2 == 1) { + if (++odd > 1) + return false; + } + } + return true; + } +} +``` + +```C++ [] +class Solution { +public: + bool canPermutePalindrome(string s) { + unordered_map dic; + for (char c : s) { + dic[c] += 1; + } + int odd = 0; + for (auto kv : dic) { + if (kv.second % 2 == 1) { + if (++odd > 1) + return false; + } + } + return true; + } +}; +``` + +```Python [] +class Solution: + def canPermutePalindrome(self, s: str) -> bool: + # 初始化哈希表 + dic = defaultdict(int) + # 统计字符串中各字符的数量 + for c in s: + dic[c] += 1 + odd = 0 + for val in dic.values(): + # 统计“数量为奇数”字符的个数 + if val % 2 == 1: + odd += 1 + # 若“数量为奇数”的字符个数 > 1 ，则不是回文串排列 + if odd > 1: + return False + # 若“数量为奇数”的字符个数 <= 1 ，则是回文串排列 + return True +``` + +```Java [] +class Solution { + public boolean canPermutePalindrome(String s) { + // 初始化哈希表 + HashMap dic = new HashMap<>(); + // 统计字符串中各字符的数量 + for (int i = 0; i < s.length(); i++) { + dic.put(s.charAt(i), dic.getOrDefault(s.charAt(i), 0) + 1); + } + int odd = 0; + for (int val : dic.values()) { + // 统计“数量为奇数”字符的个数 + if (val % 2 == 1) { + // 若“数量为奇数”的字符个数 > 1 ，则不是回文串排列 + if (++odd > 1) // 注意 ++odd > 1 是先执行 odd 自增，再执行逻辑判断； odd++ 的顺序反之 + return false; + } + } + // 若“数量为奇数”的字符个数 <= 1 ，则是回文串排列 + return true; + } +} +``` + +```C++ [] +class Solution { +public: + bool canPermutePalindrome(string s) { + // 初始化哈希表 + unordered_map dic; + // 统计字符串中各字符的数量 + for (char c : s) { + dic[c] += 1; + } + int odd = 0; + for (auto kv : dic) { + // 统计“数量为奇数”字符的个数 + if (kv.second % 2 == 1) { + // 若“数量为奇数”的字符个数 > 1 ，则不是回文串排列 + if (++odd > 1) // 注意 ++odd > 1 是先执行 odd 自增，再执行逻辑判断； odd++ 的顺序反之 + return false; + } + } + // 若“数量为奇数”的字符个数 <= 1 ，则是回文串排列 + return true; + } +}; +``` + +### 复杂度分析： + +**时间复杂度 $O(N)$ ：** 其中 $N$ 为字符串长度；哈希表统计字符数量迭代 $N$ 次，判断是否为回文串最多迭代 $N$ 次，总体使用 $O(N + N) = O(N)$ 时间。 + +**空间复杂度 $O(N)$ ：** 哈希表 `dic` 使用 $O(N)$ 大小的额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/278. \347\254\254\344\270\200\344\270\252\351\224\231\350\257\257\347\232\204\347\211\210\346\234\254.md" "b/selected_coding_interview/docs/278. \347\254\254\344\270\200\344\270\252\351\224\231\350\257\257\347\232\204\347\211\210\346\234\254.md" new file mode 100644 index 0000000..ce9efc9 --- /dev/null +++ "b/selected_coding_interview/docs/278. \347\254\254\344\270\200\344\270\252\351\224\231\350\257\257\347\232\204\347\211\210\346\234\254.md" @@ -0,0 +1,77 @@ +## 解题思路 + +根据题目描述 “错误的版本之后的所有版本都是错的” ，说明给定的版本正确性列表是「有序的」，即以某个版本为分界点，左边（右边）都是正确（错误）版本。因此，考虑使用「二分查找」来查找首个错误版本。 + +本文使用与 [704. 二分查找](https://leetcode.cn/problems/binary-search/solution/by-jyd-i7xr/) 相同的写法，二分查找缩窄区间的含义请参考代码注释。 + +![figures.gif](https://pic.leetcode-cn.com/1658594283-NxBxSc-figures.gif) + +若感觉动图播放太快，可以一页页看 $\downarrow$ + + + +## 代码 + +```Python [] +class Solution: + def firstBadVersion(self, n: int) -> int: + i, j = 1, n + while i <= j: + # 向下取整除法计算中点 m + m = (i + j) // 2 + # 若 m 是错误版本，则最后一个正确版本一定在闭区间 [i, m - 1] + if isBadVersion(m): j = m - 1 + # 若 m 是正确版本，则首个错误版本一定在闭区间 [m + 1, j] + else: i = m + 1 + # i 指向首个错误版本，j 指向最后一个正确版本 + return i +``` + +```Java [] +public class Solution extends VersionControl { + public int firstBadVersion(int n) { + int i = 1, j = n; + while (i <= j) { + // 向下取整除法计算中点 m + int m = i + (j - i) / 2; + // 若 m 是错误版本，则最后一个正确版本一定在闭区间 [i, m - 1] + if (isBadVersion(m)) j = m - 1; + // 若 m 是正确版本，则首个错误版本一定在闭区间 [m + 1, j] + else i = m + 1; + } + // i 指向首个错误版本，j 指向最后一个正确版本 + return i; + } +} +``` + +```C++ [] +class Solution { +public: + int firstBadVersion(int n) { + int i = 1, j = n; + while (i <= j) { + // 向下取整除法计算中点 m + int m = i + (j - i) / 2; + // 若 m 是错误版本，则最后一个正确版本一定在闭区间 [i, m - 1] + if (isBadVersion(m)) j = m - 1; + // 若 m 是正确版本，则首个错误版本一定在闭区间 [m + 1, j] + else i = m + 1; + } + // i 指向首个错误版本，j 指向最后一个正确版本 + return i; + } +}; +``` + +## 复杂度分析 + +- **时间复杂度 $O(\log n)$ ：** 其中 $n$ 为版本数；二分查找使用对数级别时间。 +- **空间复杂度 $O(1)$ ：** 变量 $i$ , $j$ 使用常数大小空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/287. \345\257\273\346\211\276\351\207\215\345\244\215\346\225\260.md" "b/selected_coding_interview/docs/287. \345\257\273\346\211\276\351\207\215\345\244\215\346\225\260.md" new file mode 100644 index 0000000..5d0010d --- /dev/null +++ "b/selected_coding_interview/docs/287. \345\257\273\346\211\276\351\207\215\345\244\215\346\225\260.md" @@ -0,0 +1,145 @@ +## 方法一：哈希表 + +利用数据结构特点，容易想到使用哈希表（Set）记录数组的各个数字，当查找到重复数字则直接返回。 + +### 算法流程： + +1. 初始化： 新建 HashSet ，记为 $hmap$ 。 +2. 遍历数组 $nums$ 中的每个数字 $num$ ： + 1. 当 $num$ 在 $hmap$ 中，说明重复，直接返回 $num$ 。 + 2. 将 $num$ 添加至 $hmap$ 中。 +3. 返回 $-1$ 。本题中一定有重复数字，因此这里返回多少都可以。 + + + +### 代码： + +```Python [] +class Solution: + def findDuplicate(self, nums: List[int]) -> int: + hmap = set() + for num in nums: + if num in hmap: return num + hmap.add(num) + return -1 +``` + +```Java [] +class Solution { + public int findDuplicate(int[] nums) { + Set hmap = new HashSet<>(); + for(int num : nums) { + if(hmap.contains(num)) return num; + hmap.add(num); + } + return -1; + } +} +``` + +```C++ [] +class Solution { +public: + int findDuplicate(vector& nums) { + unordered_map map; + for(int num : nums) { + if(map[num]) return num; + map[num] = true; + } + return -1; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 遍历数组使用 $O(N)$ ，HashSet 添加与查找元素皆为 $O(1)$ 。 +- **空间复杂度 $O(N)$ ：** HashSet 占用 $O(N)$ 大小的额外空间。 + +## 方法二：原地交换 + +题目说明尚未被充分使用，即 `在一个长度为 n 的数组 nums 里的所有数字都在 0 ~ n-1 的范围内` 。 此说明含义：数组元素的 **索引** 和 **值** 是 **一对多** 的关系。 +因此，可遍历数组并通过交换操作，使元素的 **索引** 与 **值** 一一对应（即 $nums[i] = i$ ）。因而，就能通过索引映射对应的值，起到与字典等价的作用。 + +![Picture0.png](https://pic.leetcode-cn.com/1618146573-bOieFQ-Picture0.png){:width=500} + +遍历中，第一次遇到数字 $x$ 时，将其交换至索引 $x$ 处；而当第二次遇到数字 $x$ 时，一定有 $nums[x] = x$ ，此时即可得到一组重复数字。 + +### 算法流程： + +1. 遍历数组 $nums$ ，设索引初始值为 $i = 0$ : + 1. **若 $nums[i] = i$ ：** 说明此数字已在对应索引位置，无需交换，因此跳过。 + 2. **若 $nums[nums[i]] = nums[i]$ ：** 代表索引 $nums[i]$ 处和索引 $i$ 处的元素值都为 $nums[i]$ ，即找到一组重复值，返回此值 $nums[i]$ 。 + 3. **否则：** 交换索引为 $i$ 和 $nums[i]$ 的元素值，将此数字交换至对应索引位置。 + +2. 若遍历完毕尚未返回，则返回 $-1$ 。 + + + +### 代码： + +Python 中， $a, b = c, d$ 操作的原理是先暂存元组 $(c, d)$ ，然后 “按左右顺序” 赋值给 a 和 b 。 +因此，若写为 $nums[i], nums[nums[i]] = nums[nums[i]], nums[i]$ ，则 $nums[i]$ 会先被赋值，之后 $nums[nums[i]]$ 指向的元素则会出错。 + +```Python [] +class Solution: + def findDuplicate(self, nums: List[int]) -> int: + i = 0 + while i < len(nums): + if nums[i] == i: + i += 1 + continue + if nums[nums[i]] == nums[i]: return nums[i] + nums[nums[i]], nums[i] = nums[i], nums[nums[i]] + return -1 +``` + +```Java [] +class Solution { + public int findDuplicate(int[] nums) { + int i = 0; + while(i < nums.length) { + if(nums[i] == i) { + i++; + continue; + } + if(nums[nums[i]] == nums[i]) return nums[i]; + int tmp = nums[i]; + nums[i] = nums[tmp]; + nums[tmp] = tmp; + } + return -1; + } +} +``` + +```C++ [] +class Solution { +public: + int findDuplicate(vector& nums) { + int i = 0; + while(i < nums.size()) { + if(nums[i] == i) { + i++; + continue; + } + if(nums[nums[i]] == nums[i]) + return nums[i]; + swap(nums[i],nums[nums[i]]); + } + return -1; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 遍历数组使用 $O(N)$ ，每轮遍历的判断和交换操作使用 $O(1)$ 。 +- **空间复杂度 $O(1)$ ：** 使用常数复杂度的额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/295. \346\225\260\346\215\256\346\265\201\347\232\204\344\270\255\344\275\215\346\225\260.md" "b/selected_coding_interview/docs/295. \346\225\260\346\215\256\346\265\201\347\232\204\344\270\255\344\275\215\346\225\260.md" new file mode 100644 index 0000000..5b1ed19 --- /dev/null +++ "b/selected_coding_interview/docs/295. \346\225\260\346\215\256\346\265\201\347\232\204\344\270\255\344\275\215\346\225\260.md" @@ -0,0 +1,141 @@ +## 解题思路： + +> 给定一长度为 $N$ 的无序数组，其中位数的计算方法：首先对数组执行排序（使用 $O(N \log N)$ 时间），然后返回中间元素即可（使用 $O(1)$ 时间）。 + +根据以上思路，可以将数据流保存在一个列表中，并在添加元素时 **保持数组有序** 。此方法的时间复杂度为 $O(N)$ ，其中包括： 查找元素插入位置 $O(\log N)$ （二分查找）、向数组某位置插入元素 $O(N)$ （插入位置之后的元素都需要向后移动一位）。 + +> 借助 **堆** 可进一步优化时间复杂度。 + +建立一个 **小顶堆** $A$ 和 **大顶堆** $B$ ，各保存列表的一半元素，且规定： + +- $A$ 保存 **较大** 的一半，长度为 $\frac{N}{2}$（ $N$ 为偶数）或 $\frac{N+1}{2}$（ $N$ 为奇数）。 +- $B$ 保存 **较小** 的一半，长度为 $\frac{N}{2}$（ $N$ 为偶数）或 $\frac{N-1}{2}$（ $N$ 为奇数）。 + +随后，中位数可仅根据 $A, B$ 的堆顶元素计算得到。 + +![Picture1.png](https://pic.leetcode-cn.com/bcfaca2b1920d2dd6bbb01aeff990698eb36d53830c38ed499ea3239a15296b3-Picture1.png){:width=500} + +### 算法流程： + +> 设元素总数为 $N = m + n$ ，其中 $m$ 和 $n$ 分别为 $A$ 和 $B$ 中的元素个数。 + +**函数 `addNum(num)` ：** + +1. 当 $m = n$（即 $N$ 为 **偶数**）：需向 $A$ 添加一个元素。实现方法：将新元素 $num$ 插入至 $B$ ，再将 $B$ 堆顶元素插入至 $A$ 。 +2. 当 $m \ne n$（即 $N$ 为 **奇数**）：需向 $B$ 添加一个元素。实现方法：将新元素 $num$ 插入至 $A$ ，再将 $A$ 堆顶元素插入至 $B$ 。 + +> 假设插入数字 $num$ 遇到情况 `1.` 。由于 $num$ 可能属于 “较小的一半” （即属于 $B$ ），因此不能将 $nums$ 直接插入至 $A$ 。而应先将 $num$ 插入至 $B$ ，再将 $B$ 堆顶元素插入至 $A$ 。这样就可以始终保持 $A$ 保存较大一半、 $B$ 保存较小一半。 + +**函数 `findMedian()` ：** + +1. 当 $m = n$（ $N$ 为 **偶数**）：则中位数为 $($ $A$ 的堆顶元素 + $B$ 的堆顶元素 $)/2$。 +2. 当 $m \ne n$（ $N$ 为 **奇数**）：则中位数为 $A$ 的堆顶元素。 + + + +## 代码： + +- Python 中 heapq 模块是小顶堆。实现 **大顶堆** 方法： 小顶堆的插入和弹出操作均将元素 **取反** 即可。 +- Java 使用 `PriorityQueue<>((x, y) -> (y - x))` 可方便实现大顶堆。 +- C++ 中 `greater` 为小顶堆， `less` 为大顶堆。 + +```Python [] +from heapq import * + +class MedianFinder: + def __init__(self): + self.A = [] # 小顶堆，保存较大的一半 + self.B = [] # 大顶堆，保存较小的一半 + + def addNum(self, num: int) -> None: + if len(self.A) != len(self.B): + heappush(self.A, num) + heappush(self.B, -heappop(self.A)) + else: + heappush(self.B, -num) + heappush(self.A, -heappop(self.B)) + + def findMedian(self) -> float: + return self.A[0] if len(self.A) != len(self.B) else (self.A[0] - self.B[0]) / 2.0 +``` + +```Java [] +class MedianFinder { + Queue A, B; + public MedianFinder() { + A = new PriorityQueue<>(); // 小顶堆，保存较大的一半 + B = new PriorityQueue<>((x, y) -> (y - x)); // 大顶堆，保存较小的一半 + } + public void addNum(int num) { + if (A.size() != B.size()) { + A.add(num); + B.add(A.poll()); + } else { + B.add(num); + A.add(B.poll()); + } + } + public double findMedian() { + return A.size() != B.size() ? A.peek() : (A.peek() + B.peek()) / 2.0; + } +} +``` + +```C++ [] +class MedianFinder { +public: + priority_queue, greater> A; // 小顶堆，保存较大的一半 + priority_queue, less> B; // 大顶堆，保存较小的一半 + MedianFinder() { } + void addNum(int num) { + if (A.size() != B.size()) { + A.push(num); + B.push(A.top()); + A.pop(); + } else { + B.push(num); + A.push(B.top()); + B.pop(); + } + } + double findMedian() { + return A.size() != B.size() ? A.top() : (A.top() + B.top()) / 2.0; + } +}; +``` + +> Push item on the heap, then pop and return the smallest item from the heap. The combined action runs more efficiently than heappush() followed by a separate call to heappop(). + +根据以上文档，可将 Python 代码优化为： + +```Python [] +from heapq import * + +class MedianFinder: + def __init__(self): + self.A = [] # 小顶堆，保存较大的一半 + self.B = [] # 大顶堆，保存较小的一半 + + def addNum(self, num: int) -> None: + if len(self.A) != len(self.B): + heappush(self.B, -heappushpop(self.A, num)) + else: + heappush(self.A, -heappushpop(self.B, -num)) + + def findMedian(self) -> float: + return self.A[0] if len(self.A) != len(self.B) else (self.A[0] - self.B[0]) / 2.0 +``` + +### 复杂度分析： + +- **时间复杂度 $O(\log N)$ ：** + - **查找中位数 $O(1)$ ：** 获取堆顶元素使用 $O(1)$ 时间。 + - **添加数字 $O(\log N)$ ：** 堆的插入和弹出操作使用 $O(\log N)$ 时间。 +- **空间复杂度 $O(N)$ ：** 其中 $N$ 为数据流中的元素数量，小顶堆 $A$ 和大顶堆 $B$ 最多同时保存 $N$ 个元素。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/297. \344\272\214\345\217\211\346\240\221\347\232\204\345\272\217\345\210\227\345\214\226\344\270\216\345\217\215\345\272\217\345\210\227\345\214\226.md" "b/selected_coding_interview/docs/297. \344\272\214\345\217\211\346\240\221\347\232\204\345\272\217\345\210\227\345\214\226\344\270\216\345\217\215\345\272\217\345\210\227\345\214\226.md" new file mode 100644 index 0000000..92bee38 --- /dev/null +++ "b/selected_coding_interview/docs/297. \344\272\214\345\217\211\346\240\221\347\232\204\345\272\217\345\210\227\345\214\226\344\270\216\345\217\215\345\272\217\345\210\227\345\214\226.md" @@ -0,0 +1,164 @@ +## 解题思路： + +通常使用的前序、中序、后序、层序遍历记录的二叉树的信息不完整，即唯一的输出序列可能对应着多种二叉树可能性。题目要求的 序列化 和 反序列化 是 **可逆操作** 。因此，序列化的字符串应携带 **完整的二叉树信息** 。 + +> 观察题目示例，序列化的字符串实际上是二叉树的 “层序遍历”（BFS）结果，本文也采用层序遍历。 + +为完整表示二叉树，考虑将叶节点下的 `null` 也记录。在此基础上，对于列表中任意某节点 `node` ，其左子节点 `node.left` 和右子节点 `node.right` 在序列中的位置都是 **唯一确定** 的。如下图所示： + +![Picture1.png](https://pic.leetcode-cn.com/1603117385-ehAGsP-Picture1.png){:width=550} + +上图规律可总结为下表： + +| `node.val` | `node` 的索引 | `node.left` 的索引 | `node.right` 的索引 | +| :--------: | :-----------: | :----------------: | :-----------------: | +| $1$ | $0$ | $1$ | $2$ | +| $2$ | $1$ | $3$ | $4$ | +| $3$ | $2$ | $5$ | $6$ | +| $4$ | $5$ | $7$ | $8$ | +| $5$ | $6$ | $9$ | $10$ | + +设 $m$ 为列表区间 $[0, n]$ 中的 `null` 节点个数，则可总结出根节点、左子节点、右子节点的列表索引的递推公式： + +| `node.val` | `node` 的列表索引 | `node.left` 的列表索引 | `node.right` 的列表索引 | +| :----------: | :---------------: | :--------------------: | :---------------------: | +| $\ne$ `null` | $n$ | $2(n-m) + 1$ | $2(n-m) + 2$ | +| $=$ `null` | $n$ | 无 | 无 | + +**序列化** 使用层序遍历实现。**反序列化** 通过以上递推公式反推各节点在序列中的索引，进而实现。 + +## 序列化 Serialize ： + +借助队列，对二叉树做层序遍历，并将越过叶节点的 `null` 也打印出来。 + +### 算法流程： + +1. **特例处理：** 若 `root` 为空，则直接返回空列表 `"[]"` 。 +2. **初始化：** 队列 `queue` （包含根节点 `root` ）；序列化列表 `res` 。 +3. **层序遍历：** 当 `queue` 为空时跳出。 + 1. 节点出队，记为 `node` 。 + 2. 若 `node` 不为空：① 打印字符串 `node.val` ，② 将左、右子节点加入 `queue` 。 + 3. 否则（若 `node` 为空）：打印字符串 `"null"` 。 +4. **返回值：** 拼接列表，用 `','` 隔开，首尾添加中括号。 + + + +## 反序列化 Deserialize ： + +基于本文开始推出的 `node` , `node.left` , `node.right` 在序列化列表中的位置关系，可实现反序列化。 + +利用队列按层构建二叉树，借助一个指针 `i` 指向节点 `node` 的左、右子节点，每构建一个 `node` 的左、右子节点，指针 `i` 就向右移动 $1$ 位。 + +### 算法流程： + +1. **特例处理：** 若 `data` 为空，直接返回 `null` 。 +2. **初始化：** 序列化列表 `vals` （先去掉首尾中括号，再用逗号隔开），指针 `i = 1` ，根节点 `root` （值为 `vals[0]` ），队列 `queue`（包含 `root` ）。 +3. **按层构建：** 当 `queue` 为空时跳出。 + 1. 节点出队，记为 `node` 。 + 2. 构建 `node` 的左子节点：`node.left` 的值为 `vals[i]` ，并将 `node.left` 入队。 + 3. 执行 `i += 1` 。 + 4. 构建 `node` 的右子节点：`node.right` 的值为 `vals[i]` ，并将 `node.right` 入队。 + 5. 执行 `i += 1` 。 +4. **返回值：** 返回根节点 `root` 即可。 + + + +## 代码： + +```Python [] +class Codec: + def serialize(self, root): + if not root: return "[]" + queue = collections.deque() + queue.append(root) + res = [] + while queue: + node = queue.popleft() + if node: + res.append(str(node.val)) + queue.append(node.left) + queue.append(node.right) + else: res.append("null") + return '[' + ','.join(res) + ']' + + def deserialize(self, data): + if data == "[]": return + vals, i = data[1:-1].split(','), 1 + root = TreeNode(int(vals[0])) + queue = collections.deque() + queue.append(root) + while queue: + node = queue.popleft() + if vals[i] != "null": + node.left = TreeNode(int(vals[i])) + queue.append(node.left) + i += 1 + if vals[i] != "null": + node.right = TreeNode(int(vals[i])) + queue.append(node.right) + i += 1 + return root +``` + +```Java [] +public class Codec { + public String serialize(TreeNode root) { + if (root == null) return "[]"; + StringBuilder res = new StringBuilder("["); + Queue queue = new LinkedList<>() {{ add(root); }}; + while (!queue.isEmpty()) { + TreeNode node = queue.poll(); + if (node != null) { + res.append(node.val + ","); + queue.add(node.left); + queue.add(node.right); + } + else res.append("null,"); + } + res.deleteCharAt(res.length() - 1); + res.append("]"); + return res.toString(); + } + + public TreeNode deserialize(String data) { + if (data.equals("[]")) return null; + String[] vals = data.substring(1, data.length() - 1).split(","); + TreeNode root = new TreeNode(Integer.parseInt(vals[0])); + Queue queue = new LinkedList<>() {{ add(root); }}; + int i = 1; + while (!queue.isEmpty()) { + TreeNode node = queue.poll(); + if (!vals[i].equals("null")) { + node.left = new TreeNode(Integer.parseInt(vals[i])); + queue.add(node.left); + } + i++; + if (!vals[i].equals("null")) { + node.right = new TreeNode(Integer.parseInt(vals[i])); + queue.add(node.right); + } + i++; + } + return root; + } +} +``` + +### 复杂度分析： + +**序列化：** + +- **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数，层序遍历需要访问所有节点，最差情况下需要访问 $N + 1$ 个 `null` ，总体复杂度为 $O(2N + 1) = O(N)$ 。 +- **空间复杂度 $O(N)$ ：** 最差情况下，队列 `queue` 同时存储 $\frac{N + 1}{2}$ 个节点（或 $N+1$ 个 `null` ），使用 $O(N)$ ；列表 `res` 使用 $O(N)$ 。 + +**反序列化：** + +- **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数，按层构建二叉树需要遍历整个 $vals$ ，其长度最大为 $2N+1$ 。 +- **空间复杂度 $O(N)$ ：** 最差情况下，队列 `queue` 同时存储 $\frac{N + 1}{2}$ 个节点，因此使用 $O(N)$ 额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/3. \346\227\240\351\207\215\345\244\215\345\255\227\347\254\246\347\232\204\346\234\200\351\225\277\345\255\220\344\270\262.md" "b/selected_coding_interview/docs/3. \346\227\240\351\207\215\345\244\215\345\255\227\347\254\246\347\232\204\346\234\200\351\225\277\345\255\220\344\270\262.md" new file mode 100644 index 0000000..b843447 --- /dev/null +++ "b/selected_coding_interview/docs/3. \346\227\240\351\207\215\345\244\215\345\255\227\347\254\246\347\232\204\346\234\200\351\225\277\345\255\220\344\270\262.md" @@ -0,0 +1,174 @@ +## 解题思路： + +长度为 $N$ 的字符串共有 $\frac{(1 + N)N}{2}$ 个子字符串（复杂度为 $O(N^2)$ ），判断长度为 $N$ 的字符串是否有重复字符的复杂度为 $O(N)$ ，因此本题使用暴力法解决的复杂度为 $O(N^3)$ 。 + +本题有滑动窗口和动态规划两种解法。 + +## 方法一：滑动窗口 + 哈希表 + +**哈希表 $dic$ 统计：** 指针 $j$ 遍历字符 $s$ ，哈希表统计字符 $s[j]$ **最后一次出现的索引** 。 + +**更新左指针 $i$ ：** 根据上轮左指针 $i$ 和 $dic[s[j]]$ ，每轮更新左边界 $i$ ，保证区间 $[i + 1, j]$ 内无重复字符且最大。 + +$$ +i = \max(dic[s[j]], i) +$$ + +**更新结果 $res$ ：** 取上轮 $res$ 和本轮双指针区间 $[i + 1,j]$ 的宽度（即 $j - i$ ）中的最大值。 + +$$ +res = \max(res, j - i) +$$ + + + +### 代码： + +```Python [] +class Solution: + def lengthOfLongestSubstring(self, s: str) -> int: + dic, res, i = {}, 0, -1 + for j in range(len(s)): + if s[j] in dic: + i = max(dic[s[j]], i) # 更新左指针 i + dic[s[j]] = j # 哈希表记录 + res = max(res, j - i) # 更新结果 + return res +``` + +```Java [] +class Solution { + public int lengthOfLongestSubstring(String s) { + Map dic = new HashMap<>(); + int i = -1, res = 0, len = s.length(); + for(int j = 0; j < len; j++) { + if (dic.containsKey(s.charAt(j))) + i = Math.max(i, dic.get(s.charAt(j))); // 更新左指针 i + dic.put(s.charAt(j), j); // 哈希表记录 + res = Math.max(res, j - i); // 更新结果 + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int lengthOfLongestSubstring(string s) { + unordered_map dic; + int i = -1, res = 0, len = s.size(); + for(int j = 0; j < len; j++) { + if (dic.find(s[j]) != dic.end()) + i = max(i, dic.find(s[j])->second); // 更新左指针 + dic[s[j]] = j; // 哈希表记录 + res = max(res, j - i); // 更新结果 + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为字符串长度，动态规划需遍历计算 $dp$ 列表。 +- **空间复杂度 $O(1)$ ：** 字符的 ASCII 码范围为 $0$ ~ $127$ ，哈希表 $dic$ 最多使用 $O(128) = O(1)$ 大小的额外空间。 + +## 方法二：动态规划 + 哈希表 + +- **状态定义：** 设动态规划列表 $dp$ ，$dp[j]$ 代表以字符 $s[j]$ 为结尾的 “最长不重复子字符串” 的长度。 +- **转移方程：** 固定右边界 $j$ ，设字符 $s[j]$ 左边距离最近的相同字符为 $s[i]$ ，即 $s[i] = s[j]$ 。 + 1. 当 $i < 0$ ，即 $s[j]$ 左边无相同字符，则 $dp[j] = dp[j-1] + 1$ 。 + 2. 当 $dp[j - 1] < j - i$ ，说明字符 $s[i]$ 在子字符串 $dp[j-1]$ **区间之外** ，则 $dp[j] = dp[j - 1] + 1$ 。 + 3. 当 $dp[j - 1] \geq j - i$ ，说明字符 $s[i]$ 在子字符串 $dp[j-1]$ **区间之中** ，则 $dp[j]$ 的左边界由 $s[i]$ 决定，即 $dp[j] = j - i$ 。 + + > 当 $i < 0$ 时，由于 $dp[j - 1] \leq j$ 恒成立，因而 $dp[j - 1] < j - i$ 恒成立，因此分支 `1.` 和 `2.` 可被合并。 + +$$ +dp[j] = +\begin{cases} +dp[j - 1] + 1 & , dp[j-1] < j - i \\ +j - i & , dp[j-1] \geq j - i +\end{cases} +$$ + +- **返回值：** $\max(dp)$ ，即全局的 “最长不重复子字符串” 的长度。 + +![Picture1.png](https://pic.leetcode-cn.com/1599287290-mTdFye-Picture1.png){:width=500} + +### 状态压缩： + +- 由于返回值是取 $dp$ 列表最大值，因此可借助变量 $tmp$ 存储 $dp[j]$ ，变量 $res$ 每轮更新最大值即可。 +- 此优化可节省 $dp$ 列表使用的 $O(N)$ 大小的额外空间。 + +### 哈希表记录： + +观察转移方程，可知关键问题：每轮遍历字符 $s[j]$ 时，如何计算索引 $i$ ？ + +- **哈希表统计：** 遍历字符串 $s$ 时，使用哈希表（记为 $dic$ ）统计 **各字符最后一次出现的索引位置** 。 +- **左边界 $i$ 获取方式：** 遍历到 $s[j]$ 时，可通过访问哈希表 $dic[s[j]]$ 获取最近的相同字符的索引 $i$ 。 + + + +### 代码： + +Python 的 `get(key, default)` 方法和 Java 的 `getOrDefault(key, default)` ，代表当哈希表包含键 `key` 时返回对应 `value` ，不包含时返回默认值 `default` 。 + +```Python [] +class Solution: + def lengthOfLongestSubstring(self, s: str) -> int: + dic = {} + res = tmp = 0 + for j in range(len(s)): + i = dic.get(s[j], -1) # 获取索引 i + dic[s[j]] = j # 更新哈希表 + tmp = tmp + 1 if tmp < j - i else j - i # dp[j - 1] -> dp[j] + res = max(res, tmp) # max(dp[j - 1], dp[j]) + return res +``` + +```Java [] +class Solution { + public int lengthOfLongestSubstring(String s) { + Map dic = new HashMap<>(); + int res = 0, tmp = 0, len = s.length(); + for(int j = 0; j < len; j++) { + int i = dic.getOrDefault(s.charAt(j), -1); // 获取索引 i + dic.put(s.charAt(j), j); // 更新哈希表 + tmp = tmp < j - i ? tmp + 1 : j - i; // dp[j - 1] -> dp[j] + res = Math.max(res, tmp); // max(dp[j - 1], dp[j]) + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int lengthOfLongestSubstring(string s) { + unordered_map dic; + int res = 0, tmp = 0, len = s.size(), i; + for(int j = 0; j < len; j++) { + if (dic.find(s[j]) == dic.end()) i = - 1; + else i = dic.find(s[j])->second; // 获取索引 i + dic[s[j]] = j; // 更新哈希表 + tmp = tmp < j - i ? tmp + 1 : j - i; // dp[j - 1] -> dp[j] + res = max(res, tmp); // max(dp[j - 1], dp[j]) + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为字符串长度，动态规划需遍历计算 $dp$ 列表。 +- **空间复杂度 $O(1)$ ：** 字符的 ASCII 码范围为 $0$ ~ $127$ ，哈希表 $dic$ 最多使用 $O(128) = O(1)$ 大小的额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/300. \346\234\200\351\225\277\351\200\222\345\242\236\345\255\220\345\272\217\345\210\227.md" "b/selected_coding_interview/docs/300. \346\234\200\351\225\277\351\200\222\345\242\236\345\255\220\345\272\217\345\210\227.md" new file mode 100644 index 0000000..d0c146c --- /dev/null +++ "b/selected_coding_interview/docs/300. \346\234\200\351\225\277\351\200\222\345\242\236\345\255\220\345\272\217\345\210\227.md" @@ -0,0 +1,140 @@ +### 解法一：动态规划 + +#### 解题思路： + +- **状态定义：** + - $dp[i]$ 的值代表 `nums` 以 $nums[i]$ 结尾的最长子序列长度。 + +- **转移方程：** 设 $j∈[0,i)$，考虑每轮计算新 $dp[i]$ 时，遍历 $[0,i)$ 列表区间，做以下判断： + 1. **当 $nums[i] > nums[j]$ 时：** $nums[i]$ 可以接在 $nums[j]$ 之后（此题要求严格递增），此情况下最长上升子序列长度为 $dp[j] + 1$ ； + 2. **当 $nums[i] <= nums[j]$ 时：** $nums[i]$ 无法接在 $nums[j]$ 之后，此情况上升子序列不成立，跳过。 + - 上述所有 **`1.` 情况** 下计算出的 $dp[j] + 1$ 的最大值，为直到 $i$ 的最长上升子序列长度（即 $dp[i]$ ）。实现方式为遍历 $j$ 时，每轮执行 $dp[i] = max(dp[i], dp[j] + 1)$。 + - **转移方程：** `dp[i] = max(dp[i], dp[j] + 1) for j in [0, i)`。 + +- **初始状态：** + - $dp[i]$ 所有元素置 $1$，含义是每个元素都至少可以单独成为子序列，此时长度都为 $1$。 + +- **返回值：** + - 返回 $dp$ 列表最大值，即可得到全局最长上升子序列长度。 + +#### 复杂度分析： + +- **时间复杂度 $O(N^2)$ ：** 遍历计算 $dp$ 列表需 $O(N)$，计算每个 $dp[i]$ 需 $O(N)$。 +- **空间复杂度 $O(N)$ ：** $dp$ 列表占用线性大小额外空间。 + + + +#### 代码： + +```Python [] +# Dynamic programming. +class Solution: + def lengthOfLIS(self, nums: List[int]) -> int: + if not nums: return 0 + dp = [1] * len(nums) + for i in range(len(nums)): + for j in range(i): + if nums[j] < nums[i]: # 如果要求非严格递增，将此行 '<' 改为 '<=' 即可。 + dp[i] = max(dp[i], dp[j] + 1) + return max(dp) +``` + +```Java [] +// Dynamic programming. +class Solution { + public int lengthOfLIS(int[] nums) { + if(nums.length == 0) return 0; + int[] dp = new int[nums.length]; + int res = 0; + Arrays.fill(dp, 1); + for(int i = 0; i < nums.length; i++) { + for(int j = 0; j < i; j++) { + if(nums[j] < nums[i]) dp[i] = Math.max(dp[i], dp[j] + 1); + } + res = Math.max(res, dp[i]); + } + return res; + } +} +``` + +--- + +### 解法二：动态规划 + 二分查找 + +#### 解题思路： + +- **降低复杂度切入点：** 解法一中，遍历计算 $dp$ 列表需 $O(N)$，计算每个 $dp[k]$ 需 $O(N)$。 + 1. 动态规划中，通过线性遍历来计算 $dp$ 的复杂度无法降低； + 2. 每轮计算中，需要通过线性遍历 $[0,k)$ 区间元素来得到 $dp[k]$ 。我们考虑：是否可以通过重新设计**状态定义**，使整个 $dp$ 为一个**排序列表**；这样在计算每个 $dp[k]$ 时，就可以通过二分法遍历 $[0,k)$ 区间元素，将此部分复杂度由 $O(N)$ 降至 $O(logN)$。 + +- **设计思路：** + - **新的状态定义：** + - 我们考虑维护一个列表 $tails$，其中每个元素 $tails[k]$ 的值代表 **长度为 $k+1$ 的子序列尾部元素的值**。 + - 如 $[1,4,6]$ 序列，长度为 $1,2,3$ 的子序列尾部元素值分别为 $tails = [1,4,6]$。 + - **状态转移设计：** + - 设常量数字 $N$，和随机数字 $x$，我们可以容易推出：当 $N$ 越小时，$N= tails[i]$，代表较短子序列的尾部元素的值 $>$ 较长子序列的尾部元素的值。这是不可能的，因为从长度为 $i$ 的子序列尾部倒序删除 $i-1$ 个元素，剩下的为长度为 $k$ 的子序列，设此序列尾部元素值为 $v$，则一定有 $v=tails[i]$ 矛盾。 + - 既然严格递增，每轮计算 $tails[k]$ 时就可以使用二分法查找需要更新的尾部元素值的对应索引 $i$。 + +- **算法流程：** + - **状态定义：** + - $tails[k]$ 的值代表 长度为 $k+1$ 子序列 的尾部元素值。 + + - **转移方程：** 设 $res$ 为 $tails$ 当前长度，代表直到当前的最长上升子序列长度。设 $j∈[0,res)$，考虑每轮遍历 $nums[k]$ 时，通过二分法遍历 $[0,res)$ 列表区间，找出 $nums[k]$ 的大小分界点，会出现两种情况： + - **区间中存在 $tails[i] > nums[k]$ ：** 将第一个满足 $tails[i] > nums[k]$ 执行 $tails[i] = nums[k]$ ；因为更小的 $nums[k]$ 后更可能接一个比它大的数字（前面分析过）。 + - **区间中不存在 $tails[i] > nums[k]$ ：** 意味着 $nums[k]$ 可以接在前面所有长度的子序列之后，因此肯定是接到最长的后面（长度为 $res$ ），新子序列长度为 $res + 1$。 + + - **初始状态：** + - 令 $tails$ 列表所有值 $=0$。 + + - **返回值：** + - 返回 $res$ ，即最长上升子子序列长度。 + +#### 复杂度分析： + +- **时间复杂度 $O(NlogN)$ ：** 遍历 $nums$ 列表需 $O(N)$，在每个 $nums[i]$ 二分法需 $O(logN)$。 +- **空间复杂度 $O(N)$ ：** $tails$ 列表占用线性大小额外空间。 + + + +#### 代码： + +```Python [] +# Dynamic programming + Dichotomy. +class Solution: + def lengthOfLIS(self, nums: [int]) -> int: + tails, res = [0] * len(nums), 0 + for num in nums: + i, j = 0, res + while i < j: + m = (i + j) // 2 + if tails[m] < num: i = m + 1 # 如果要求非严格递增，将此行 '<' 改为 '<=' 即可。 + else: j = m + tails[i] = num + if j == res: res += 1 + return res +``` + +```Java [] +// Dynamic programming + Dichotomy. +class Solution { + public int lengthOfLIS(int[] nums) { + int[] tails = new int[nums.length]; + int res = 0; + for(int num : nums) { + int i = 0, j = res; + while(i < j) { + int m = (i + j) / 2; + if(tails[m] < num) i = m + 1; + else j = m; + } + tails[i] = num; + if(res == j) res++; + } + return res; + } +} +``` diff --git "a/selected_coding_interview/docs/343. \346\225\264\346\225\260\346\213\206\345\210\206.md" "b/selected_coding_interview/docs/343. \346\225\264\346\225\260\346\213\206\345\210\206.md" new file mode 100644 index 0000000..a2ebd5e --- /dev/null +++ "b/selected_coding_interview/docs/343. \346\225\264\346\225\260\346\213\206\345\210\206.md" @@ -0,0 +1,108 @@ +#### 解题思路： + +- 设将整数 $n$ 拆分为 $a$ 个小数字： + +$$ +n = n_1 + n_2 + ... + n_a +$$ + +- 本题等价于求解： + +$$ +\max(n_1 \times n_2 \times ... \times n_a) +$$ + +> 以下数学推导总体分为两步：① 当所有拆分出的数字相等时，乘积最大。② 最优拆分数字为 $3$ 。 + +##### 数学推导： + +- 以下公式为“算术几何均值不等式” ，等号当且仅当 $n_1 = n_2 = ... = n_a$ 时成立。 + +$$ +\frac{n_1 + n_2 + ... + n_a}{a} \geq \sqrt[a]{n_1 n_2 ... n_a} +$$ + +> **推论一：** 若拆分的数量 $a$ 确定， 则 **各拆分数字相等时** ，乘积最大。 + +- 设将数字以因子 $x$ 等分为 $a$ 个，即 $n = ax$ ，则乘积为 $x^a$ 。观察以下公式，由于 $n$ 为常数，因此当 $x^{\frac{1}{x}}$ 取最大值时， 乘积达到最大值。 + +$$ +x^a = x^{\frac{n}{x}} = (x^{\frac{1}{x}})^n +$$ + +- 根据分析，可将问题转化为求 $y = x^{\frac{1}{x}}$ 的极大值，因此对 $x$ 求导数。 + +$$ +\begin{aligned} \ln y & = \frac{1}{x} \ln x & \text{取对数} \\ \frac{1}{y} \dot {y} & = \frac{1}{x^2} - \frac{1}{x^2} \ln x & \text{对 $x$ 求导} \\ & = \frac{1 - \ln x}{x^2} \\ \dot {y} & = \frac{1 - \ln x}{x^2} x^{\frac{1}{x}} & \text{整理得}\end{aligned} +$$ + +- 令 $\dot {y} = 0$ ，则 $1 - \ln x = 0$ ，易得驻点为 $x_0 = e \approx 2.7$ ；根据以下公式，可知 $x_0$ 为极大值点。 + +$$ +\dot {y}\begin{cases} > 0 & , x \in [- \infty, e) \\ < 0 & , x \in (e, \infty] \\\end{cases} +$$ + +- 由于因子 $x$ 必须为整数，最接近 $e$ 的整数为 $2$ 或 $3$ 。如下式所示，代入 $x = 2$ 和 $x = 3$ ，得出 $x = 3$ 时，乘积达到最大。 + +$$ +y(3) = 3^{1/3} \approx 1.44 \\ +y(2) = 2^{1/2} \approx 1.41 +$$ + +- 口算对比方法：给两数字同时取 $6$ 次方，再对比。 + +$$ +[y(3)]^6 = (3^{1/3})^6 = 9 \\ +[y(2)]^6 = (2^{1/2})^6 = 8 +$$ + +> **推论二：** 将数字 $n$ 尽可能以因子 $3$ 等分时，乘积最大。 + +##### 拆分规则： + +1. **最优：** $3$ 。把数字 $n$ 可能拆为多个因子 $3$ ，余数可能为 $0,1,2$ 三种情况。 +2. **次优：** $2$ 。若余数为 $2$ ；则保留，不再拆为 $1+1$ 。 +3. **最差：** $1$ 。若余数为 $1$ ；则应把一份 $3 + 1$ 替换为 $2 + 2$，因为 $2 \times 2 > 3 \times 1$。 + +##### 算法流程： + +1. 当 $n \leq 3$ 时，按照规则应不拆分，但由于题目要求必须拆分，因此必须拆出一个因子 $1$ ，即返回 $n - 1$ 。 +2. 当 $n>3$ 时，求 $n$ 除以 $3$ 的 整数部分 $a$ 和 余数部分 $b$ （即 $n = 3a + b$ ），并分为以下三种情况： + - 当 $b = 0$ 时，直接返回 $3^a$； + - 当 $b = 1$ 时，要将一个 $1 + 3$ 转换为 $2+2$，因此返回 $3^{a-1} \times 4$； + - 当 $b = 2$ 时，返回 $3^a \times 2$。 + +![Picture2.png](https://pic.leetcode-cn.com/1d32896766463a7a74ffafe47e7f57008e563b8fe7a8e4d52525732ac8d34275-Picture2.png){:width=600} + +##### 复杂度分析： + +- **时间复杂度 $O(1)$ ：** 仅有求整、求余、次方运算。 + - [求整和求余运算](https://stackoverflow.com/questions/35189851/time-complexity-of-modulo-operator-in-python)：查阅资料，提到不超过机器数的整数可以看作是 $O(1)$ ； + - [幂运算](https://stackoverflow.com/questions/32418731/java-math-powa-b-time-complexity)：查阅资料，提到浮点取幂为 $O(1)$ 。 +- **空间复杂度 $O(1)$ ：** `a` 和 `b` 使用常数大小额外空间。 + +#### 代码： + +> Python 中常见有三种幂计算函数： **`*`** 和 **`pow()`** 的时间复杂度均为 $O(\log a)$ ；而 **`math.pow()`** 始终调用 C 库的 `pow()` 函数，其执行浮点取幂，时间复杂度为 $O(1)$ 。 + +```Python [] +class Solution: + def integerBreak(self, n: int) -> int: + if n <= 3: return n - 1 + a, b = n // 3, n % 3 + if b == 0: return int(math.pow(3, a)) + if b == 1: return int(math.pow(3, a - 1) * 4) + return int(math.pow(3, a) * 2) +``` + +```Java [] +class Solution { + public int integerBreak(int n) { + if(n <= 3) return n - 1; + int a = n / 3, b = n % 3; + if(b == 0) return (int)Math.pow(3, a); + if(b == 1) return (int)Math.pow(3, a - 1) * 4; + return (int)Math.pow(3, a) * 2; + } +} +``` diff --git "a/selected_coding_interview/docs/371. \344\270\244\346\225\264\346\225\260\344\271\213\345\222\214.md" "b/selected_coding_interview/docs/371. \344\270\244\346\225\264\346\225\260\344\271\213\345\222\214.md" new file mode 100644 index 0000000..8d215ee --- /dev/null +++ "b/selected_coding_interview/docs/371. \344\270\244\346\225\264\346\225\260\344\271\213\345\222\214.md" @@ -0,0 +1,110 @@ +## 解题思路： + +本题考察对位运算的灵活使用，即使用位运算实现加法。 + +设两数字的二进制形式 $a, b$ ，其求和 $s = a + b$ ，$a(i)$ 代表 $a$ 的二进制第 $i$ 位，则分为以下四种情况： + +| $a(i)$ | $b(i)$ | 无进位和 $n(i)$ | 进位 $c(i+1)$ | +| :----: | :----: | :-------------: | :-----------: | +| $0$ | $0$ | $0$ | $0$ | +| $0$ | $1$ | $1$ | $0$ | +| $1$ | $0$ | $1$ | $0$ | +| $1$ | $1$ | $0$ | $1$ | + +观察发现，**无进位和** 与 **异或运算** 规律相同，**进位** 和 **与运算** 规律相同（并需左移一位）。因此，无进位和 $n$ 与进位 $c$ 的计算公式如下。 + +$$ +\begin{cases} +n = a \oplus b & 非进位和：异或运算 \\ +c = a \& b << 1 & 进位：与运算 + 左移一位 +\end{cases} +$$ + +（和 $s$ ）$=$（非进位和 $n$ ）$+$（进位 $c$ ）。即可将 $s = a + b$ 转化为： + +$$ +s = a + b \Rightarrow s = n + c +$$ + +循环求 $n$ 和 $c$ ，直至进位 $c = 0$ ；此时 $s = n$ ，返回 $n$ 即可。 + +![Picture1.png](https://pic.leetcode-cn.com/9716b1a1ead21824b8216c7d54910bee4d838c011581f4e3d82a14f71cb392a1-Picture1.png){:width=500} + +值得注意的是，在计算机系统中，数值一律用**补码**来表示和存储。**补码的优势：** 加法、减法可以统一处理，从而使得 CPU 只需加法器。这意味着以上方法 **同时适用于正数和负数的加法** 。 + + + +## 代码： + +```Java [] +class Solution { + public int getSum(int a, int b) { + // 循环，当进位为 0 时跳出 + while (b != 0) { + int c = (a & b) << 1; // c = 进位 + a ^= b; // a = 非进位和 + b = c; // b = 进位 + } + return a; + } +} +``` + +```C++ [] +class Solution { +public: + int getSum(int a, int b) { + // 循环，当进位为 0 时跳出 + while (b != 0) { + int c = (unsigned int)(a & b) << 1; // c = 进位 + a ^= b; // a = 非进位和 + b = c; // b = 进位 + } + return a; + } +}; +``` + +```Python [] +class Solution: + def getSum(self, a: int, b: int) -> int: + x = 0xffffffff + a, b = a & x, b & x + # 循环，当进位为 0 时跳出 + while b != 0: + # a, b = 非进位和, 进位 + a, b = (a ^ b), (a & b) << 1 & x + return a if a <= 0x7fffffff else ~(a ^ x) +``` + +### 复杂度分析： + +- **时间复杂度 $O(1)$ ：** 最差情况下（例如 $a = \text{0x7fffffff}$ , $b = 1$ 时），需循环 32 次，使用 $O(1)$ 时间；每轮中的常数次位操作使用 $O(1)$ 时间。 +- **空间复杂度 $O(1)$ ：** 使用常数大小的额外空间。 + +### 负数的存储： + +> 由于 Python 的数字存储特点，需要做特殊考虑。 + +Python，Java, C++ 等语言中的数字都是以 **补码** 形式存储的。但 Python 没有 `int` , `long` 等不同长度变量，即在编程时无位数的概念。 + +**获取负数的补码：** 需要将数字与十六进制数 $\text{0xffffffff}$ 相与。可理解为舍去此数字 32 位以上的数字（将 32 位以上都变为 $0$ ），从无限长度变为一个 32 位整数。 + +**返回前数字还原：** 若补码 $a$ 为负数（ $\text{0x7fffffff}$ 是最大的正数的补码 ），需执行 `~(a ^ x)` 操作，将补码还原至 Python 的存储格式。 `a ^ x` 运算将 1 至 32 位按位取反； `~` 运算是将整个数字取反；因此， `~(a ^ x)` 是将 32 位以上的位取反，1 至 32 位不变。 + +```Python +print(hex(1)) # = 0x1 补码 +print(hex(-1)) # = -0x1 负号 + 原码 （ Python 特色，Java 会直接输出补码） + +print(hex(1 & 0xffffffff)) # = 0x1 正数补码 +print(hex(-1 & 0xffffffff)) # = 0xffffffff 负数补码 + +print(-1 & 0xffffffff) # = 4294967295 （ Python 将其认为正数） +``` + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/387. \345\255\227\347\254\246\344\270\262\344\270\255\347\232\204\347\254\254\344\270\200\344\270\252\345\224\257\344\270\200\345\255\227\347\254\246.md" "b/selected_coding_interview/docs/387. \345\255\227\347\254\246\344\270\262\344\270\255\347\232\204\347\254\254\344\270\200\344\270\252\345\224\257\344\270\200\345\255\227\347\254\246.md" new file mode 100644 index 0000000..bb24a9b --- /dev/null +++ "b/selected_coding_interview/docs/387. \345\255\227\347\254\246\344\270\262\344\270\255\347\232\204\347\254\254\344\270\200\344\270\252\345\224\257\344\270\200\345\255\227\347\254\246.md" @@ -0,0 +1,73 @@ +## 解题思路 + +1. 遍历字符串 `s` ，使用哈希表统计 “各字符数量是否 $> 1$ ”。 +2. 再遍历字符串 `s` ，在哈希表中找到首个 “数量为 $1$ 的字符”，并返回。 + +![Picture1.png](https://pic.leetcode-cn.com/ed093aabc9195caff6d088454eaebe3cad875e8ca4a643c004ef25e4e5e9e174-Picture1.png){:width=450} + +### 算法流程： + +1. **初始化：** 字典 (Python)、HashMap(Java)、map(C++)，记为 `dic` 。 +2. **字符统计：** 遍历字符串 `s` 中的每个字符 `c` 。 + 1. 若 `dic` 中 **不包含** 键(key) `c` ：则向 `dic` 中添加键值对 `(c, True)` ，代表字符 `c` 的数量为 $1$ 。 + 2. 若 `dic` 中 **包含** 键(key) `c` ：则修改键 `c` 的键值对为 `(c, False)` ，代表字符 `c` 的数量 $> 1$ 。 +3. **查找数量为 $1$ 的字符：** 遍历字符串 `s` 中的每个字符 `c` 。 + 1. 若 `dic`中键 `c` 对应的值为 `True` ：，则返回其索引。 + 2. 否则，返回 `-1` ，代表不存在数量为 $1$ 的字符。 + + + +### 代码： + +Python 代码中的 `not c in dic` 整体为一个布尔值； `c in dic` 为判断字典中是否含有键 `c` 。 + +```Python [] +class Solution: + def firstUniqChar(self, s: str) -> int: + dic = {} + for c in s: + dic[c] = not c in dic + for i, c in enumerate(s): + if dic[c]: return i + return -1 +``` + +```Java [] +class Solution { + public int firstUniqChar(String s) { + HashMap dic = new HashMap<>(); + char[] sc = s.toCharArray(); + for(char c : sc) + dic.put(c, !dic.containsKey(c)); + for(int i = 0; i < sc.length; i++) + if(dic.get(sc[i])) return i; + return -1; + } +} +``` + +```C++ [] +class Solution { +public: + int firstUniqChar(string s) { + unordered_map dic; + for(char c : s) + dic[c] = dic.find(c) == dic.end(); + for(int i = 0; i < s.size(); i++) + if(dic[s[i]]) return i; + return -1; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为字符串 `s` 的长度；需遍历 `s` 两轮，使用 $O(N)$ ；HashMap 查找操作的复杂度为 $O(1)$ 。 +- **空间复杂度 $O(1)$ ：** 由于题目指出 `s` 只包含小写字母，因此最多有 26 个不同字符，HashMap 存储需占用 $O(26) = O(1)$ 的额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/39. \347\273\204\345\220\210\346\200\273\345\222\214.md" "b/selected_coding_interview/docs/39. \347\273\204\345\220\210\346\200\273\345\222\214.md" new file mode 100644 index 0000000..ee4af2d --- /dev/null +++ "b/selected_coding_interview/docs/39. \347\273\204\345\220\210\346\200\273\345\222\214.md" @@ -0,0 +1,161 @@ +## 解题思路： + +例如，输入集合 $\{3, 4, 5\}$ 和目标整数 $9$ ，解为 $\{3, 3, 3\}, \{4, 5\}$ 。需要注意两点： + +- 输入集合中的元素可以被无限次重复选取。 +- 子集是不区分元素顺序的，比如 $\{4, 5\}$ 和 $\{5, 4\}$ 是同一个子集。 + +向「[全排列](https://leetcode.cn/problems/permutations/solutions/2363882/46-quan-pai-lie-hui-su-qing-xi-tu-jie-by-6o7h/)」代码输入数组 $[3, 4, 5]$ 和目标元素 $9$ ，输出结果为 $[3, 3, 3], [4, 5], [5, 4]$ 。**虽然成功找出了所有和为 $9$ 的子集，但其中存在重复的子集 $[4, 5]$ 和 $[5, 4]$** 。 + +这是因为搜索过程是区分选择顺序的，然而子集不区分选择顺序。如下图所示，先选 $4$ 后选 $5$ 与先选 $5$ 后选 $4$ 是两个不同的分支，但两者对应同一个子集。 + +![subset_sum_i_naive.png](https://pic.leetcode.cn/1690624999-SRITpI-subset_sum_i_naive.png) + +为了去除重复子集，**一种直接的思路是对结果列表进行去重**。但这个方法效率很低，因为： + +- 当数组元素较多，尤其是当 `target` 较大时，搜索过程会产生大量的重复子集。 +- 比较子集（数组）的异同非常耗时，需要先排序数组，再比较数组中每个元素的异同。 + +### 重复子集剪枝： + +**我们考虑在搜索过程中通过剪枝进行去重**。观察下图，重复子集是在以不同顺序选择数组元素时产生的，具体来看： + +1. 第一轮和第二轮分别选择 $3$ , $4$ ，会生成包含这两个元素的所有子集，记为 $[3, 4, \cdots]$ 。 +2. 若第一轮选择 $4$ ，**则第二轮应该跳过 $3$** ，因为该选择产生的子集 $[4, 3, \cdots]$ 和 `1.` 中生成的子集完全重复。 + +分支越靠右，需要排除的分支也越多，例如： + +1. 前两轮选择 $3$ , $5$ ，生成子集 $[3, 5, \cdots]$ 。 +2. 前两轮选择 $4$ , $5$ ，生成子集 $[4, 5, \cdots]$ 。 +3. 若第一轮选择 $5$ ，**则第二轮应该跳过 $3$ 和 $4$** ，因为子集 $[5, 3, \cdots]$ 和子集 $[5, 4, \cdots]$ 和 `1.` , `2.` 中生成的子集完全重复。 + +![subset_sum_i_pruning.png](https://pic.leetcode.cn/1690625058-WYmZtD-subset_sum_i_pruning.png) + +总结来看，给定输入数组 $[x_1, x_2, \cdots, x_n]$ ，设搜索过程中的选择序列为 $[x_{i_1}, x_{i_2}, \cdots , x_{i_m}]$ ，则该选择序列需要满足 $i_1 \leq i_2 \leq \cdots \leq i_m$ ，**不满足该条件的选择序列都会造成重复，应当剪枝**。 + +## 代码： + +为实现该剪枝，我们初始化变量 `start` ，用于指示遍历起点。**当做出选择 $x_{i}$ 后，设定下一轮从索引 $i$ 开始遍历**。这样做就可以让选择序列满足 $i_1 \leq i_2 \leq \cdots \leq i_m$ ，从而保证子集唯一。 + +除此之外，我们还对代码进行了两项优化： + +- 在开启搜索前，先将数组 `nums` 排序。在遍历所有选择时，**当子集和超过 `target` 时直接结束循环**，因为后边的元素更大，其子集和都一定会超过 `target` 。 +- 省去元素和变量 `total`，**通过在 `target` 上执行减法来统计元素和**，当 `target` 等于 $0$ 时记录解。 + +```java [] +class Solution { + void backtrack(List state, int target, int[] choices, int start, List> res) { + // 子集和等于 target 时，记录解 + if (target == 0) { + res.add(new ArrayList<>(state)); + return; + } + // 遍历所有选择 + // 剪枝二：从 start 开始遍历，避免生成重复子集 + for (int i = start; i < choices.length; i++) { + // 剪枝一：若子集和超过 target ，则直接结束循环 + // 这是因为数组已排序，后边元素更大，子集和一定超过 target + if (target - choices[i] < 0) { + break; + } + // 尝试：做出选择，更新 target, start + state.add(choices[i]); + // 进行下一轮选择 + backtrack(state, target - choices[i], choices, i, res); + // 回退：撤销选择，恢复到之前的状态 + state.remove(state.size() - 1); + } + } + + public List> combinationSum(int[] candidates, int target) { + List state = new ArrayList<>(); // 状态（子集） + Arrays.sort(candidates); // 对 candidates 进行排序 + int start = 0; // 遍历起始点 + List> res = new ArrayList<>(); // 结果列表（子集列表） + backtrack(state, target, candidates, start, res); + return res; + } +} +``` + +```cpp [] +class Solution { +public: + vector> combinationSum(vector& candidates, int target) { + vector state; // 状态（子集） + sort(candidates.begin(), candidates.end()); // 对 candidates 进行排序 + int start = 0; // 遍历起始点 + vector> res; // 结果列表（子集列表） + backtrack(state, target, candidates, start, res); + return res; + } +private: + void backtrack(vector &state, int target, vector &choices, int start, vector> &res) { + // 子集和等于 target 时，记录解 + if (target == 0) { + res.push_back(state); + return; + } + // 遍历所有选择 + // 剪枝二：从 start 开始遍历，避免生成重复子集 + for (int i = start; i < choices.size(); i++) { + // 剪枝一：若子集和超过 target ，则直接结束循环 + // 这是因为数组已排序，后边元素更大，子集和一定超过 target + if (target - choices[i] < 0) { + break; + } + // 尝试：做出选择，更新 target, start + state.push_back(choices[i]); + // 进行下一轮选择 + backtrack(state, target - choices[i], choices, i, res); + // 回退：撤销选择，恢复到之前的状态 + state.pop_back(); + } + } +}; +``` + +```python [] +class Solution: + def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]: + def backtrack( + state: list[int], target: int, choices: list[int], start: int, res: list[list[int]] + ): + """回溯算法：子集和 I""" + # 子集和等于 target 时，记录解 + if target == 0: + res.append(list(state)) + return + # 遍历所有选择 + # 剪枝二：从 start 开始遍历，避免生成重复子集 + for i in range(start, len(choices)): + # 剪枝一：若子集和超过 target ，则直接结束循环 + # 这是因为数组已排序，后边元素更大，子集和一定超过 target + if target - choices[i] < 0: + break + # 尝试：做出选择，更新 target, start + state.append(choices[i]) + # 进行下一轮选择 + backtrack(state, target - choices[i], choices, i, res) + # 回退：撤销选择，恢复到之前的状态 + state.pop() + + state = [] # 状态（子集） + candidates.sort() # 对 candidates 进行排序 + start = 0 # 遍历起始点 + res = [] # 结果列表（子集列表） + backtrack(state, target, candidates, start, res) + return res + +``` + +如下图所示，为将数组 $[3, 4, 5]$ 和目标元素 $9$ 输入到以上代码后的整体回溯过程。 + +![subset_sum_i.png](https://pic.leetcode.cn/1690624990-TxtFOY-subset_sum_i.png) + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/392. \345\210\244\346\226\255\345\255\220\345\272\217\345\210\227.md" "b/selected_coding_interview/docs/392. \345\210\244\346\226\255\345\255\220\345\272\217\345\210\227.md" new file mode 100644 index 0000000..cdb4556 --- /dev/null +++ "b/selected_coding_interview/docs/392. \345\210\244\346\226\255\345\255\220\345\272\217\345\210\227.md" @@ -0,0 +1,76 @@ +## 解题思路 + +设置双指针 `i` , `j` 分别指向字符串 `s` , `t` 的首个字符，遍历字符串 `t` ： + +- 当 `s[i] == t[j] ` 时，代表匹配成功，此时同时 `i++` , `j++` ； + - 进而，若 `i` 已走过 `s` 尾部，代表 `s` 是 `t` 的子序列，此时应提前返回 true ； +- 当 `s[i] != t[j] ` 时，代表匹配失败，此时仅 `j++` ； + +若遍历完字符串 `t` 后，字符串 `s` 仍未遍历完，代表 `s` 不是 `t` 的子序列，此时返回 false 。 + + + +> 上为静态图，下为动态图，内容一致。 + +![figures.gif](https://pic.leetcode-cn.com/1656946529-bhpXbL-figures.gif) + +## 代码 + +```Python [] +class Solution: + def isSubsequence(self, s: str, t: str) -> bool: + if not s: return True + i = 0 + for c in t: + if s[i] == c: + i += 1 + # 若已经遍历完 s ，则提前返回 true + if i == len(s): + return True + return False +``` + +```Java [] +class Solution { + public boolean isSubsequence(String s, String t) { + if (s.length() == 0) return true; + for (int i = 0, j = 0; j < t.length(); j++) { + if (s.charAt(i) == t.charAt(j)) { + // 若已经遍历完 s ，则提前返回 true + if (++i == s.length()) + return true; + } + } + return false; + } +} +``` + +```C++ [] +class Solution { +public: + bool isSubsequence(string s, string t) { + if (s.size() == 0) return true; + for (int i = 0, j = 0; j < t.size(); j++) { + if (s[i] == t[j]) { + // 若已经遍历完 s ，则提前返回 true + if (++i == s.size()) + return true; + } + } + return false; + } +}; +``` + +## 复杂度分析 + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为字符串 `t` 的长度。最差情况下需完整遍历 `t` 。 +- **空间复杂度 $O(1)$ ：** `i` , `j` 变量使用常数大小空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/394. \345\255\227\347\254\246\344\270\262\350\247\243\347\240\201.md" "b/selected_coding_interview/docs/394. \345\255\227\347\254\246\344\270\262\350\247\243\347\240\201.md" new file mode 100644 index 0000000..9b96071 --- /dev/null +++ "b/selected_coding_interview/docs/394. \345\255\227\347\254\246\344\270\262\350\247\243\347\240\201.md" @@ -0,0 +1,131 @@ +#### 解法一：辅助栈法 + +- 本题难点在于括号内嵌套括号，需要**从内向外**生成与拼接字符串，这与栈的**先入后出**特性对应。 +- **算法流程：** + 1. 构建辅助栈 `stack`， 遍历字符串 `s` 中每个字符 `c`； + - 当 `c` 为数字时，将数字字符转化为数字 `multi`，用于后续倍数计算； + - 当 `c` 为字母时，在 `res` 尾部添加 `c`； + - 当 `c` 为 `[` 时，将当前 `multi` 和 `res` 入栈，并分别置空置 $0$： + - 记录此 `[` 前的临时结果 `res` 至栈，用于发现对应 `]` 后的拼接操作； + - 记录此 `[` 前的倍数 `multi` 至栈，用于发现对应 `]` 后，获取 `multi × [...]` 字符串。 + - 进入到新 `[` 后，`res` 和 `multi` 重新记录。 + - 当 `c` 为 `]` 时，`stack` 出栈，拼接字符串 `res = last_res + cur_multi * res`，其中: + - `last_res`是上个 `[` 到当前 `[` 的字符串，例如 `"3[a2[c]]"` 中的 `a`； + - `cur_multi`是当前 `[` 到 `]` 内字符串的重复倍数，例如 `"3[a2[c]]"` 中的 `2`。 + 2. 返回字符串 `res`。 + +- **复杂度分析：** + - 时间复杂度 $O(N)$，一次遍历 `s`； + - 空间复杂度 $O(N)$，辅助栈在极端情况下需要线性空间，例如 `2[2[2[a]]]`。 + + + +```Python [] +class Solution: + def decodeString(self, s: str) -> str: + stack, res, multi = [], "", 0 + for c in s: + if c == '[': + stack.append([multi, res]) + res, multi = "", 0 + elif c == ']': + cur_multi, last_res = stack.pop() + res = last_res + cur_multi * res + elif '0' <= c <= '9': + multi = multi * 10 + int(c) + else: + res += c + return res +``` + +```Java [] +class Solution { + public String decodeString(String s) { + StringBuilder res = new StringBuilder(); + int multi = 0; + LinkedList stack_multi = new LinkedList<>(); + LinkedList stack_res = new LinkedList<>(); + for(Character c : s.toCharArray()) { + if(c == '[') { + stack_multi.addLast(multi); + stack_res.addLast(res.toString()); + multi = 0; + res = new StringBuilder(); + } + else if(c == ']') { + StringBuilder tmp = new StringBuilder(); + int cur_multi = stack_multi.removeLast(); + for(int i = 0; i < cur_multi; i++) tmp.append(res); + res = new StringBuilder(stack_res.removeLast() + tmp); + } + else if(c >= '0' && c <= '9') multi = multi * 10 + Integer.parseInt(c + ""); + else res.append(c); + } + return res.toString(); + } +} +``` + +--- + +#### 解法二：递归法 + +- 总体思路与辅助栈法一致，不同点在于将 `[` 和 `]` 分别作为递归的开启与终止条件： + - 当 `s[i] == ']'` 时，返回当前括号内记录的 `res` 字符串与 `]` 的索引 `i` （更新上层递归指针位置）； + - 当 `s[i] == '['` 时，开启新一层递归，记录此 `[...]` 内字符串 `tmp` 和递归后的最新索引 `i`，并执行 `res + multi * tmp` 拼接字符串。 + - 遍历完毕后返回 `res`。 + +- **复杂度分析：** + - 时间复杂度 $O(N)$，递归会更新索引，因此实际上还是一次遍历 `s`； + - 空间复杂度 $O(N)$，极端情况下递归深度将会达到线性级别。 + +```Python [] +class Solution: + def decodeString(self, s: str) -> str: + def dfs(s, i): + res, multi = "", 0 + while i < len(s): + if '0' <= s[i] <= '9': + multi = multi * 10 + int(s[i]) + elif s[i] == '[': + i, tmp = dfs(s, i + 1) + res += multi * tmp + multi = 0 + elif s[i] == ']': + return i, res + else: + res += s[i] + i += 1 + return res + return dfs(s,0) +``` + +```Java [] +class Solution { + public String decodeString(String s) { + return dfs(s, 0)[0]; + } + private String[] dfs(String s, int i) { + StringBuilder res = new StringBuilder(); + int multi = 0; + while(i < s.length()) { + if(s.charAt(i) >= '0' && s.charAt(i) <= '9') + multi = multi * 10 + Integer.parseInt(String.valueOf(s.charAt(i))); + else if(s.charAt(i) == '[') { + String[] tmp = dfs(s, i + 1); + i = Integer.parseInt(tmp[0]); + while(multi > 0) { + res.append(tmp[1]); + multi--; + } + } + else if(s.charAt(i) == ']') + return new String[] { String.valueOf(i), res.toString() }; + else + res.append(String.valueOf(s.charAt(i))); + i++; + } + return new String[] { res.toString() }; + } +} +``` diff --git "a/selected_coding_interview/docs/40. \347\273\204\345\220\210\346\200\273\345\222\214 II.md" "b/selected_coding_interview/docs/40. \347\273\204\345\220\210\346\200\273\345\222\214 II.md" new file mode 100644 index 0000000..36592a7 --- /dev/null +++ "b/selected_coding_interview/docs/40. \347\273\204\345\220\210\346\200\273\345\222\214 II.md" @@ -0,0 +1,148 @@ +## 解题思路： + +请先做前置题目「[39. 组合总和](https://leetcode.cn/problems/combination-sum/solutions/2363929/39-zu-he-zong-he-hui-su-qing-xi-tu-jie-b-9zx7/)」。 + +相比于上题，**本题的输入数组可能包含重复元素**，这引入了新的问题。例如，给定数组 $[4, \hat{4}, 5]$ 和目标元素 $9$ ，则现有代码的输出结果为 $[4, 5], [\hat{4}, 5]$ ，出现了重复子集。 + +**造成这种重复的原因是相等元素在某轮中被多次选择**。如下图所示，第一轮共有三个选择，其中两个都为 $4$ ，会产生两个重复的搜索分支，从而输出重复子集；同理，第二轮的两个 $4$ 也会产生重复子集。 + +![subset_sum_ii_repeat.png](https://pic.leetcode.cn/1690625348-eGjYFi-subset_sum_ii_repeat.png) + +### 相等元素剪枝： + +为解决此问题，**我们需要限制相等元素在每一轮中只被选择一次**。实现方式比较巧妙：由于数组是已排序的，因此相等元素都是相邻的。这意味着在某轮选择中，若当前元素与其左边元素相等，则说明它已经被选择过，因此直接跳过当前元素。 + +与此同时，**本题规定中的每个数组元素只能被选择一次**。幸运的是，我们也可以利用变量 `start` 来满足该约束：当做出选择 $x_{i}$ 后，设定下一轮从索引 $i + 1$ 开始向后遍历。这样即能去除重复子集，也能避免重复选择元素。 + +## 代码： + +```python [] +class Solution: + def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]: + def backtrack( + state: list[int], target: int, choices: list[int], start: int, res: list[list[int]] + ): + """回溯算法：子集和 II""" + # 子集和等于 target 时，记录解 + if target == 0: + res.append(list(state)) + return + # 遍历所有选择 + # 剪枝二：从 start 开始遍历，避免生成重复子集 + # 剪枝三：从 start 开始遍历，避免重复选择同一元素 + for i in range(start, len(choices)): + # 剪枝一：若子集和超过 target ，则直接结束循环 + # 这是因为数组已排序，后边元素更大，子集和一定超过 target + if target - choices[i] < 0: + break + # 剪枝四：如果该元素与左边元素相等，说明该搜索分支重复，直接跳过 + if i > start and choices[i] == choices[i - 1]: + continue + # 尝试：做出选择，更新 target, start + state.append(choices[i]) + # 进行下一轮选择 + backtrack(state, target - choices[i], choices, i + 1, res) + # 回退：撤销选择，恢复到之前的状态 + state.pop() + + state = [] # 状态（子集） + candidates.sort() # 对 candidates 进行排序 + start = 0 # 遍历起始点 + res = [] # 结果列表（子集列表） + backtrack(state, target, candidates, start, res) + return res +``` + +```java [] +class Solution { + void backtrack(List state, int target, int[] choices, int start, List> res) { + // 子集和等于 target 时，记录解 + if (target == 0) { + res.add(new ArrayList<>(state)); + return; + } + // 遍历所有选择 + // 剪枝二：从 start 开始遍历，避免生成重复子集 + // 剪枝三：从 start 开始遍历，避免重复选择同一元素 + for (int i = start; i < choices.length; i++) { + // 剪枝一：若子集和超过 target ，则直接结束循环 + // 这是因为数组已排序，后边元素更大，子集和一定超过 target + if (target - choices[i] < 0) { + break; + } + // 剪枝四：如果该元素与左边元素相等，说明该搜索分支重复，直接跳过 + if (i > start && choices[i] == choices[i - 1]) { + continue; + } + // 尝试：做出选择，更新 target, start + state.add(choices[i]); + // 进行下一轮选择 + backtrack(state, target - choices[i], choices, i + 1, res); + // 回退：撤销选择，恢复到之前的状态 + state.remove(state.size() - 1); + } + } + + public List> combinationSum2(int[] candidates, int target) { + List state = new ArrayList<>(); // 状态（子集） + Arrays.sort(candidates); // 对 candidates 进行排序 + int start = 0; // 遍历起始点 + List> res = new ArrayList<>(); // 结果列表（子集列表） + backtrack(state, target, candidates, start, res); + return res; + } +} +``` + +```cpp [] +class Solution { +public: + vector> combinationSum2(vector& candidates, int target) { + vector state; // 状态（子集） + sort(candidates.begin(), candidates.end()); // 对 candidates 进行排序 + int start = 0; // 遍历起始点 + vector> res; // 结果列表（子集列表） + backtrack(state, target, candidates, start, res); + return res; + } +private: + void backtrack(vector &state, int target, vector &choices, int start, vector> &res) { + // 子集和等于 target 时，记录解 + if (target == 0) { + res.push_back(state); + return; + } + // 遍历所有选择 + // 剪枝二：从 start 开始遍历，避免生成重复子集 + // 剪枝三：从 start 开始遍历，避免重复选择同一元素 + for (int i = start; i < choices.size(); i++) { + // 剪枝一：若子集和超过 target ，则直接结束循环 + // 这是因为数组已排序，后边元素更大，子集和一定超过 target + if (target - choices[i] < 0) { + break; + } + // 剪枝四：如果该元素与左边元素相等，说明该搜索分支重复，直接跳过 + if (i > start && choices[i] == choices[i - 1]) { + continue; + } + // 尝试：做出选择，更新 target, start + state.push_back(choices[i]); + // 进行下一轮选择 + backtrack(state, target - choices[i], choices, i + 1, res); + // 回退：撤销选择，恢复到之前的状态 + state.pop_back(); + } + } +}; +``` + +下图展示了数组 $[4, 4, 5]$ 和目标元素 $9$ 的回溯过程，共包含四种剪枝操作。请你将图示与代码注释相结合，理解整个搜索过程，以及每种剪枝操作是如何工作的。 + +![subset_sum_ii.png](https://pic.leetcode.cn/1690625346-mkEVHR-subset_sum_ii.png) + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/400. \347\254\254 N \344\275\215\346\225\260\345\255\227.md" "b/selected_coding_interview/docs/400. \347\254\254 N \344\275\215\346\225\260\345\255\227.md" new file mode 100644 index 0000000..a0c511a --- /dev/null +++ "b/selected_coding_interview/docs/400. \347\254\254 N \344\275\215\346\225\260\345\255\227.md" @@ -0,0 +1,176 @@ +## 解题思路： + +本文名词规定如下： + +1. 将 $101112 \cdots$ 中的每一位称为 **数位** ，记为 $n$ 。 +2. 将 $10, 11, 12, \cdots$ 称为 **数字** ，记为 $num$ 。 +3. 数字 $10$ 是一个两位数，称此数字的 **位数** 为 $2$ ，记为 $digit$ 。 +4. 每 $digit$ 位数的起始数字（即：$1, 10, 100, \cdots$），记为 $start$ 。 + +![Picture1.png](https://pic.leetcode-cn.com/1599888213-CYhLfm-Picture1.png){:width=400} + +观察上表，可推出各 $digit$ 下的数位数量 $count$ 的计算公式： + +$$ +count = 9 \times start \times digit +$$ + +根据以上分析，可将求解分为三步： + +1. 确定 $n$ 所在 **数字** 的 **位数** ，记为 $digit$ 。 +2. 确定 $n$ 所在的 **数字** ，记为 $num$ 。 +3. 确定 $n$ 是 $num$ 中的哪一数位，并返回结果。 + +### 第一步：确定所求数位的所在数字的位数 + +如下图所示，循环执行 $n$ 减去 一位数、两位数、... 的数位数量 $count$ ，直至 $n \leq count$ 时跳出。 + +由于 $n$ 已经减去了一位数、两位数、...、$(digit-1)$ 位数的 **数位数量** $count$ ，因而此时的 $n$ 是从起始数字 $start$ 开始计数的。 + +```Python [] +digit, start, count = 1, 1, 9 +while n > count: + n -= count + start *= 10 # 1, 10, 100, ... + digit += 1 # 1, 2, 3, ... + count = 9 * start * digit # 9, 180, 2700, ... +``` + +```Java [] +int digit = 1; +long start = 1; +long count = 9; +while (n > count) { + n -= count; + start *= 10; // 1, 10, 100, ... + digit += 1; // 1, 2, 3, ... + count = digit * start * 9; // 9, 180, 2700, ... +} +``` + +```C++ [] +int digit = 1; +long start = 1; +long count = 9; +while (n > count) { // 1. + n -= count; + start *= 10; // 1, 10, 100, ... + digit += 1; // 1, 2, 3, ... + count = digit * start * 9; // 9, 180, 2700, ... +} +``` + +**结论：** 所求数位 ① 在某个 $digit$ 位数中； ② 为从数字 $start$ 开始的第 $n$ 个数位。 + +![Picture2.png](https://pic.leetcode-cn.com/1599888496-HivJvS-Picture2.png){:width=500} + +### 第二步：确定所求数位所在的数字 + +如下图所示，所求数位 在从数字 $start$ 开始的第 $[(n - 1) / digit]$ 个 **数字** 中（ $start$ 为第 0 个数字）。 + +```Python [] +num = start + (n - 1) // digit +``` + +```Java [] +long num = start + (n - 1) / digit; +``` + +```C++ [] +long num = start + (n - 1) / digit; +``` + +**结论：** 所求数位在数字 $num$ 中。 + +![Picture3.png](https://pic.leetcode-cn.com/1599888213-gCcnEA-Picture3.png){:width=500} + +### 第三步：确定所求数位在 $num$ 的哪一数位 + +如下图所示，所求数位为数字 $num$ 的第 $(n - 1) \% digit$ 位（ 数字的首个数位为第 0 位）。 + +```Python [] +s = str(num) # 转化为 string +res = int(s[(n - 1) % digit]) # 获得 num 的 第 (n - 1) % digit 个数位，并转化为 int +``` + +```Java [] +String s = Long.toString(num); // 转化为 string +int res = s.charAt((n - 1) % digit) - '0'; // 获得 num 的 第 (n - 1) % digit 个数位，并转化为 int +``` + +```C++ [] +string s = to_string(num); // 转化为 string +int res = s[(n - 1) % digit] - '0'; // 获得 num 的 第 (n - 1) % digit 个数位，并转化为 int +``` + +**结论：** 所求数位是 $res$ 。 + +![Picture4.png](https://pic.leetcode-cn.com/1599888395-oeWGAH-Picture4.png){:width=500} + +完整流程如下图所示。 + + + +## 代码： + +```Python [] +class Solution: + def findNthDigit(self, n: int) -> int: + digit, start, count = 1, 1, 9 + while n > count: # 1. + n -= count + start *= 10 + digit += 1 + count = 9 * start * digit + num = start + (n - 1) // digit # 2. + return int(str(num)[(n - 1) % digit]) # 3. +``` + +```Java [] +class Solution { + public int findNthDigit(int n) { + int digit = 1; + long start = 1; + long count = 9; + while (n > count) { // 1. + n -= count; + start *= 10; + digit += 1; + count = digit * start * 9; + } + long num = start + (n - 1) / digit; // 2. + return Long.toString(num).charAt((n - 1) % digit) - '0'; // 3. + } +} +``` + +```C++ [] +class Solution { +public: + int findNthDigit(int n) { + int digit = 1; + long start = 1; + long count = 9; + while (n > count) { // 1. + n -= count; + start *= 10; + digit += 1; + count = digit * start * 9; + } + long num = start + (n - 1) / digit; // 2. + return to_string(num)[(n - 1) % digit] - '0'; // 3. + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(\log n)$ ：** 所求数位 $n$ 对应数字 $num$ 的位数 $digit$ 最大为 $O(\log n)$ ；第一步最多循环 $O(\log n)$ 次；第三步中将 $num$ 转化为字符串使用 $O(\log n)$ 时间；因此总体为 $O(\log n)$ 。 +- **空间复杂度 $O(\log n)$ ：** 将数字 $num$ 转化为字符串 `str(num)` ，占用 $O(\log n)$ 的额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/409. \346\234\200\351\225\277\345\233\236\346\226\207\344\270\262.md" "b/selected_coding_interview/docs/409. \346\234\200\351\225\277\345\233\236\346\226\207\344\270\262.md" new file mode 100644 index 0000000..15ed9a2 --- /dev/null +++ "b/selected_coding_interview/docs/409. \346\234\200\351\225\277\345\233\236\346\226\207\344\270\262.md" @@ -0,0 +1,93 @@ +## 解题思路 + +「回文串」是指倒序后和自身完全相同的字符串，即具有**关于中心轴对称**的性质。观察发现， + +- 当回文串长度为偶数时，则所有字符都出现了偶数次； +- 当回文串长度为奇数时，则位于中心的字符出现了奇数次，其余所有字符出现偶数次； + +根据以上分析，字符串能被构造成回文串的**充要条件**为：除了一种字符出现**奇数**次外，其余所有字符出现**偶数**次。判别流程如下： + +1. 借助一个 HashMap ，统计字符串 `s` 中各字符的出现次数； +2. 遍历 HashMap ，统计构造回文串的最大长度， + 1. 将当前字符的出现次数**向下取偶数**（即若为偶数则不变，若为奇数则减 1 ），**出现偶数次则都可组成回文串**，因此计入 `res` ； + 2. 若此字符出现次数为奇数，则**可将此字符放到回文串中心**，因此将 `odd` 置 1 ； +3. 返回 `res + odd` 即可。 + + + +## 代码 + +```Python [] +class Solution: + def longestPalindrome(self, s: str) -> int: + # 统计各字符数量 + counter = collections.defaultdict(int) + for c in s: + counter[c] += 1 + res, odd = 0, 0 + # 统计构造回文串的最大长度 + for count in counter.values(): + # 将当前字符出现次数向下取偶数，并计入 res + rem = count % 2 + res += count - rem + # 若当前字符出现次数为奇数，则将 odd 置 1 + if rem == 1: odd = 1 + return res + odd +``` + +```Java [] +class Solution { + public int longestPalindrome(String s) { + // 统计各字符数量 + HashMap counter = new HashMap<>(); + for (int i = 0; i < s.length(); i++) + counter.merge(s.charAt(i), 1, (a, b) -> a + b); + // 统计构造回文串的最大长度 + int res = 0, odd = 0; + for (Map.Entry kv : counter.entrySet()) { + // 将当前字符出现次数向下取偶数，并计入 res + int count = kv.getValue(); + int rem = count % 2; + res += count - rem; + // 若当前字符出现次数为奇数，则将 odd 置 1 + if (rem == 1) odd = 1; + } + return res + odd; + } +} +``` + +```C++ [] +class Solution { +public: + int longestPalindrome(string s) { + // 统计各字符数量 + unordered_map counter; + for (char c : s) + counter[c]++; + // 统计构造回文串的最大长度 + int res = 0, odd = 0; + for (auto kv : counter) { + // 将当前字符出现次数向下取偶数，并计入 res + int count = kv.second; + int rem = count % 2; + res += count - rem; + // 若当前字符出现次数为奇数，则将 odd 置 1 + if (rem == 1) odd = 1; + } + return res + odd; + } +}; +``` + +## 复杂度分析 + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为字符串 `s` 长度。遍历字符串 `s` 和哈希表 `counter` 皆使用线性时间。 +- **空间复杂度 $O(1)$ ：** 由于 ASCII 字符数量为 128 ，哈希表 `counter` 最多使用 $O(128) = O(1)$ 空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/415. \345\255\227\347\254\246\344\270\262\347\233\270\345\212\240.md" "b/selected_coding_interview/docs/415. \345\255\227\347\254\246\344\270\262\347\233\270\345\212\240.md" new file mode 100644 index 0000000..65b1022 --- /dev/null +++ "b/selected_coding_interview/docs/415. \345\255\227\347\254\246\344\270\262\347\233\270\345\212\240.md" @@ -0,0 +1,49 @@ +#### 解题思路： + +- **算法流程：** 设定 `i`，`j` 两指针分别指向 `num1`，`num2` 尾部，模拟人工加法； + - **计算进位：** 计算 `carry = tmp // 10`，代表当前位相加是否产生进位； + - **添加当前位：** 计算 `tmp = n1 + n2 + carry`，并将当前位 `tmp % 10` 添加至 `res` 头部； + - **索引溢出处理：** 当指针 `i`或`j` 走过数字首部后，给 `n1`，`n2` 赋值为 $0$，相当于给 `num1`，`num2` 中长度较短的数字前面填 $0$，以便后续计算。 + - 当遍历完 `num1`，`num2` 后跳出循环，并根据 `carry` 值决定是否在头部添加进位 $1$，最终返回 `res` 即可。 + +- **复杂度分析：** + - 时间复杂度 $O(max(M,N))$：其中 $M$，$N$ 为 $2$ 数字长度，按位遍历一遍数字（以较长的数字为准）； + - 空间复杂度 $O(1)$：指针与变量使用常数大小空间。 + + + +#### 代码： + +```Python [] +class Solution: + def addStrings(self, num1: str, num2: str) -> str: + res = "" + i, j, carry = len(num1) - 1, len(num2) - 1, 0 + while i >= 0 or j >= 0: + n1 = int(num1[i]) if i >= 0 else 0 + n2 = int(num2[j]) if j >= 0 else 0 + tmp = n1 + n2 + carry + carry = tmp // 10 + res = str(tmp % 10) + res + i, j = i - 1, j - 1 + return "1" + res if carry else res +``` + +```Java [] +class Solution { + public String addStrings(String num1, String num2) { + StringBuilder res = new StringBuilder(""); + int i = num1.length() - 1, j = num2.length() - 1, carry = 0; + while(i >= 0 || j >= 0){ + int n1 = i >= 0 ? num1.charAt(i) - '0' : 0; + int n2 = j >= 0 ? num2.charAt(j) - '0' : 0; + int tmp = n1 + n2 + carry; + carry = tmp / 10; + res.append(tmp % 10); + i--; j--; + } + if(carry == 1) res.append(1); + return res.reverse().toString(); + } +} +``` diff --git "a/selected_coding_interview/docs/426. \345\260\206\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\350\275\254\345\214\226\344\270\272\346\216\222\345\272\217\347\232\204\345\217\214\345\220\221\351\223\276\350\241\250.md" "b/selected_coding_interview/docs/426. \345\260\206\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\350\275\254\345\214\226\344\270\272\346\216\222\345\272\217\347\232\204\345\217\214\345\220\221\351\223\276\350\241\250.md" new file mode 100644 index 0000000..32486f8 --- /dev/null +++ "b/selected_coding_interview/docs/426. \345\260\206\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\350\275\254\345\214\226\344\270\272\346\216\222\345\272\217\347\232\204\345\217\214\345\220\221\351\223\276\350\241\250.md" @@ -0,0 +1,149 @@ +## 解题思路： + +本文解法基于性质：二叉搜索树的中序遍历为 **递增序列** 。 + +将 二叉搜索树 转换成一个 “排序的循环双向链表” ，其中包含三个要素： + +1. **排序链表：** 节点应从小到大排序，因此应使用 **中序遍历** “从小到大”访问树的节点。 +2. **双向链表：** 在构建相邻节点的引用关系时，设前驱节点 `pre` 和当前节点 `cur` ，不仅应构建 `pre.right = cur` ，也应构建 `cur.left = pre` 。 +3. **循环链表：** 设链表头节点 `head` 和尾节点 `tail` ，则应构建 `head.left = tail` 和 `tail.right = head` 。 + +![Picture1.png](https://pic.leetcode-cn.com/1599401091-PKIjds-Picture1.png){:width=500} + +**中序遍历** 为对二叉树作 “左、根、右” 顺序遍历，递归实现如下： + +```Python [] +# 打印中序遍历 +def dfs(root): + if not root: return + dfs(root.left) # 左 + print(root.val) # 根 + dfs(root.right) # 右 +``` + +```Java [] +// 打印中序遍历 +void dfs(Node root) { + if (root == null) return; + dfs(root.left); // 左 + System.out.println(root.val); // 根 + dfs(root.right); // 右 +} +``` + +```C++ [] +// 打印中序遍历 +void dfs(Node* root) { + if (root == nullptr) return; + dfs(root->left); // 左 + cout << root->val << endl; // 根 + dfs(root->right); // 右 +} +``` + +根据以上分析，考虑使用中序遍历访问树的各节点 `cur` ，并在访问每个节点时构建 `cur` 和前驱节点 `pre` 的引用指向。在中序遍历完成后，最后构建头节点和尾节点的引用指向即可。 + +### 算法流程： + +**函数 `dfs(cur)` ：** 中序遍历。 + +1. **终止条件：** 当节点 `cur` 为空，代表越过叶节点，直接返回。 +2. 递归左子树，即 `dfs(cur.left)` 。 +3. **构建链表：** + 1. **当 `pre` 为空时：** 代表正在访问链表头节点，记为 `head` 。 + 2. **当 `pre` 不为空时：** 修改双向节点引用，即 `pre.right = cur` ， `cur.left = pre` 。 + 3. **保存 `cur` ：** 更新 `pre = cur` ，即节点 `cur` 是后继节点的 `pre` 。 +4. 递归右子树，即 `dfs(cur.right)` 。 + +**函数 `treeToDoublyList(root)` ：** + +1. **特例处理：** 若节点 `root` 为空，则直接返回。 +2. **初始化：** 空节点 `pre` 。 +3. **转化为双向链表：** 调用 `dfs(root)` 。 +4. **构建循环链表：** 中序遍历完成后，`head` 指向头节点， `pre` 指向尾节点，因此修改 `head` 和 `pre` 的双向节点引用即可。 +5. **返回值：** 返回链表的头节点 `head` 即可。 + + + +## 代码： + +```Python [] +class Solution: + def treeToDoublyList(self, root: 'Optional[Node]') -> 'Optional[Node]': + def dfs(cur): + if not cur: return + dfs(cur.left) # 递归左子树 + if self.pre: # 修改节点引用 + self.pre.right, cur.left = cur, self.pre + else: # 记录头节点 + self.head = cur + self.pre = cur # 保存 cur + dfs(cur.right) # 递归右子树 + + if not root: return + self.pre = None + dfs(root) + self.head.left, self.pre.right = self.pre, self.head + return self.head +``` + +```Java [] +class Solution { + Node pre, head; + + void dfs(Node cur) { + if (cur == null) return; + dfs(cur.left); + if (pre != null) pre.right = cur; + else head = cur; + cur.left = pre; + pre = cur; + dfs(cur.right); + } + + public Node treeToDoublyList(Node root) { + if (root == null) return null; + dfs(root); + head.left = pre; + pre.right = head; + return head; + } +} +``` + +```C++ [] +class Solution { +public: + Node* treeToDoublyList(Node* root) { + if (root == nullptr) return nullptr; + dfs(root); + head->left = pre; + pre->right = head; + return head; + } + +private: + Node *pre, *head; + void dfs(Node* cur) { + if (cur == nullptr) return; + dfs(cur->left); + if (pre != nullptr) pre->right = cur; + else head = cur; + cur->left = pre; + pre = cur; + dfs(cur->right); + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数，中序遍历需要访问所有节点。 +- **空间复杂度 $O(N)$ ：** 最差情况下，即树退化为链表时，递归深度达到 $N$，系统使用 $O(N)$ 栈空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/46. \345\205\250\346\216\222\345\210\227.md" "b/selected_coding_interview/docs/46. \345\205\250\346\216\222\345\210\227.md" new file mode 100644 index 0000000..a7fa6e3 --- /dev/null +++ "b/selected_coding_interview/docs/46. \345\205\250\346\216\222\345\210\227.md" @@ -0,0 +1,115 @@ +## 解题思路： + +对于一个长度为 $n$ 的数组（假设元素互不重复），其排列方案数共有： + +$$ +n \times (n-1) \times (n-2) … \times 2 \times 1 +$$ + +**排列方案的生成：** + +根据数组排列的特点，考虑深度优先搜索所有排列方案。即通过元素交换，先固定第 $1$ 位元素（ $n$ 种情况）、再固定第 $2$ 位元素（ $n-1$ 种情况）、... 、最后固定第 $n$ 位元素（ $1$ 种情况）。 + +![Picture1.png](https://pic.leetcode.cn/1690622875-QNAmSJ-Picture1.png) +{:width=500} + +### 递归解析： + +1. **终止条件：** 当 `x = len(nums) - 1` 时，代表所有位已固定（最后一位只有 $1$ 种情况），则将当前组合 `nums` 转化为数组并加入 `res` ，并返回。 +2. **递推参数：** 当前固定位 `x` 。 +3. **递推工作：** 将第 `x` 位元素与 `i` $\in$ `[x, len(nums)]` 元素分别交换，并进入下层递归。 + 1. **固定元素：** 将元素 `nums[i]` 和 `nums[x]` 交换，即固定 `nums[i]` 为当前位元素。 + 2. **开启下层递归：** 调用 `dfs(x + 1)` ，即开始固定第 `x + 1` 个元素。 + 3. **还原交换：** 将元素 `nums[i]` 和 `nums[x]` 交换（还原之前的交换）。 + +> 下图中 `list` 对应文中的列表 `nums` ，`"abc"` 对应 `123` 。 + + + +## 代码： + +```Python [] +class Solution: + def permute(self, nums: List[int]) -> List[List[int]]: + def dfs(x): + if x == len(nums) - 1: + res.append(list(nums)) # 添加排列方案 + return + for i in range(x, len(nums)): + nums[i], nums[x] = nums[x], nums[i] # 交换，将 nums[i] 固定在第 x 位 + dfs(x + 1) # 开启固定第 x + 1 位元素 + nums[i], nums[x] = nums[x], nums[i] # 恢复交换 + res = [] + dfs(0) + return res +``` + +```Java [] +class Solution { + List nums; + List> res; + + void swap(int a, int b) { + int tmp = nums.get(a); + nums.set(a, nums.get(b)); + nums.set(b, tmp); + } + + void dfs(int x) { + if (x == nums.size() - 1) { + res.add(new ArrayList<>(nums)); // 添加排列方案 + return; + } + for (int i = x; i < nums.size(); i++) { + swap(i, x); // 交换，将 nums[i] 固定在第 x 位 + dfs(x + 1); // 开启固定第 x + 1 位元素 + swap(i, x); // 恢复交换 + } + } + + public List> permute(int[] nums) { + this.res = new ArrayList<>(); + this.nums = new ArrayList<>(); + for (int num : nums) { + this.nums.add(num); + } + dfs(0); + return res; + } +} +``` + +```C++ [] +class Solution { +public: + vector> permute(vector& nums) { + dfs(nums, 0); + return res; + } +private: + vector> res; + void dfs(vector nums, int x) { + if (x == nums.size() - 1) { + res.push_back(nums); // 添加排列方案 + return; + } + for (int i = x; i < nums.size(); i++) { + swap(nums[i], nums[x]); // 交换，将 nums[i] 固定在第 x 位 + dfs(nums, x + 1); // 开启固定第 x + 1 位元素 + swap(nums[i], nums[x]); // 恢复交换 + } + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N!N)$ ：** $N$ 为数组 `nums` 的长度；时间复杂度和数组排列的方案数成线性关系，方案数为 $N \times (N-1) \times (N-2) … \times 2 \times 1$ ，即复杂度为 $O(N!)$ ；数组拼接操作 `join()` 使用 $O(N)$ ；因此总体时间复杂度为 $O(N!N)$ 。 +- **空间复杂度 $O(N)$ ：** 全排列的递归深度为 $N$ ，系统累计使用栈空间大小为 $O(N)$ 。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/47. \345\205\250\346\216\222\345\210\227 II.md" "b/selected_coding_interview/docs/47. \345\205\250\346\216\222\345\210\227 II.md" new file mode 100644 index 0000000..cba1780 --- /dev/null +++ "b/selected_coding_interview/docs/47. \345\205\250\346\216\222\345\210\227 II.md" @@ -0,0 +1,124 @@ +## 解题思路： + +请先做前置题目「[46. 全排列](https://leetcode.cn/problems/permutations/solutions/2363882/46-quan-pai-lie-hui-su-qing-xi-tu-jie-by-6o7h/)」。 + +本题和上一题的区别是数组中“存在重复元素”。当数组存在重复元素时，排列方案中也存在重复的排列方案。 + +为了排除这些重复方案，需在固定某位元素时，保证“每种元素只在此位固定一次”，**即遇到重复元素时不交换，直接跳过**，从而将生成重复排列的搜索分支进行“剪枝” 。 + +![Picture2.png](https://pic.leetcode.cn/1690622903-LENLFB-Picture2.png) +{:width=500} + +### 递归解析： + +1. **终止条件：** 当 `x = len(nums) - 1` 时，代表所有位已固定（最后一位只有 $1$ 种情况），则将当前组合 `nums` 转化为数组并加入 `res` ，并返回。 +2. **递推参数：** 当前固定位 `x` 。 +3. **递推工作：** 初始化一个 Set ，用于排除重复的元素；将第 `x` 位元素与 `i` $\in$ `[x, len(nums)]` 元素分别交换，并进入下层递归。 + 1. **剪枝：** 若 `nums[i]` 在 Set​ 中，代表其是重复元素，因此 “剪枝” 。 + 2. 将 `nums[i]` 加入 Set​ ，以便之后遇到重复元素时剪枝。 + 3. **固定元素：** 将元素 `nums[i]` 和 `nums[x]` 交换，即固定 `nums[i]` 为当前位元素。 + 4. **开启下层递归：** 调用 `dfs(x + 1)` ，即开始固定第 `x + 1` 个元素。 + 5. **还原交换：** 将元素 `nums[i]` 和 `nums[x]` 交换（还原之前的交换）。 + +> 下图中 `list` 对应文中的列表 `nums` ，`"abc"` 对应 `123` 。 + + + +## 代码： + +```Python [] +class Solution: + def permuteUnique(self, nums: List[int]) -> List[List[int]]: + def dfs(x): + if x == len(nums) - 1: + res.append(list(nums)) # 添加排列方案 + return + dic = set() + for i in range(x, len(nums)): + if nums[i] in dic: continue # 重复，因此剪枝 + dic.add(nums[i]) + nums[i], nums[x] = nums[x], nums[i] # 交换，将 nums[i] 固定在第 x 位 + dfs(x + 1) # 开启固定第 x + 1 位元素 + nums[i], nums[x] = nums[x], nums[i] # 恢复交换 + res = [] + dfs(0) + return res +``` + +```Java [] +class Solution { + List nums; + List> res; + + void swap(int a, int b) { + int tmp = nums.get(a); + nums.set(a, nums.get(b)); + nums.set(b, tmp); + } + + void dfs(int x) { + if (x == nums.size() - 1) { + res.add(new ArrayList<>(nums)); // 添加排列方案 + return; + } + HashSet set = new HashSet<>(); + for (int i = x; i < nums.size(); i++) { + if (set.contains(nums.get(i))) + continue; // 重复，因此剪枝 + set.add(nums.get(i)); + swap(i, x); // 交换，将 nums[i] 固定在第 x 位 + dfs(x + 1); // 开启固定第 x + 1 位元素 + swap(i, x); // 恢复交换 + } + } + + public List> permuteUnique(int[] nums) { + this.res = new ArrayList<>(); + this.nums = new ArrayList<>(); + for (int num : nums) { + this.nums.add(num); + } + dfs(0); + return res; + } +} +``` + +```C++ [] +class Solution { +public: + vector> permuteUnique(vector& nums) { + dfs(nums, 0); + return res; + } +private: + vector> res; + void dfs(vector nums, int x) { + if (x == nums.size() - 1) { + res.push_back(nums); // 添加排列方案 + return; + } + set st; + for (int i = x; i < nums.size(); i++) { + if (st.find(nums[i]) != st.end()) + continue; // 重复，因此剪枝 + st.insert(nums[i]); + swap(nums[i], nums[x]); // 交换，将 nums[i] 固定在第 x 位 + dfs(nums, x + 1); // 开启固定第 x + 1 位元素 + swap(nums[i], nums[x]); // 恢复交换 + } + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N!N)$ ：** 与上一题相同 。 +- **空间复杂度 $O(N^2)$ ：** 递归中辅助 Set 累计存储的元素数量最多为 $N + (N-1) + ... + 2 + 1 = (N+1)N/2$ ，因此占用 $O(N^2)$ 的额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/48. \346\227\213\350\275\254\345\233\276\345\203\217.md" "b/selected_coding_interview/docs/48. \346\227\213\350\275\254\345\233\276\345\203\217.md" new file mode 100644 index 0000000..ae20ed3 --- /dev/null +++ "b/selected_coding_interview/docs/48. \346\227\213\350\275\254\345\233\276\345\203\217.md" @@ -0,0 +1,233 @@ +# 方法一：辅助矩阵 + +如下图所示，矩阵顺时针旋转 90º 后，可找到以下规律： + +- 「第 $i$ 行」元素旋转到「第 $n - 1 - i$ 列」元素； +- 「第 $j$ 列」元素旋转到「第 $j$ 行」元素； + +因此，对于矩阵任意第 $i$ 行、第 $j$ 列元素 $matrix[i][j]$ ，矩阵旋转 90º 后「元素位置旋转公式」为： + +$$ +\begin{aligned} +matrix[i][j] & \rightarrow matrix[j][n - 1 - i] \\ +原索引位置 & \rightarrow 旋转后索引位置 +\end{aligned} +$$ + +![ccw-01-07.001.png](https://pic.leetcode-cn.com/1638557961-AVzCQb-ccw-01-07.001.png) + +根据以上「元素旋转公式」，考虑遍历矩阵，将各元素依次写入到旋转后的索引位置。但仍**存在问题**：在写入一个元素 $matrix[i][j] \rightarrow matrix[j][n - 1 - i]$ 后，原矩阵元素 $matrix[j][n - 1 - i]$ 就会**被覆盖（即丢失）**，而此丢失的元素就无法被写入到旋转后的索引位置了。 + +为解决此问题，考虑借助一个「辅助矩阵」暂存原矩阵，通过遍历辅助矩阵所有元素，将各元素填入「原矩阵」旋转后的新索引位置即可。 + +```Python [] +class Solution: + def rotate(self, matrix: List[List[int]]) -> None: + n = len(matrix) + # 深拷贝 matrix -> tmp + tmp = copy.deepcopy(matrix) + # 根据元素旋转公式，遍历修改原矩阵 matrix 的各元素 + for i in range(n): + for j in range(n): + matrix[j][n - 1 - i] = tmp[i][j] +``` + +```Java [] +class Solution { + public void rotate(int[][] matrix) { + int n = matrix.length; + // 深拷贝 matrix -> tmp + int[][] tmp = new int[n][]; + for (int i = 0; i < n; i++) + tmp[i] = matrix[i].clone(); + // 根据元素旋转公式，遍历修改原矩阵 matrix 的各元素 + for (int i = 0; i < n; i++) { + for (int j = 0; j < n; j++) { + matrix[j][n - 1 - i] = tmp[i][j]; + } + } + } +} +``` + +```C++ [] +class Solution { +public: + void rotate(vector>& matrix) { + int n = matrix.size(); + // 深拷贝 matrix -> tmp + vector> tmp = matrix; + // 根据元素旋转公式，遍历修改原矩阵 matrix 的各元素 + for (int i = 0; i < n; i++) { + for (int j = 0; j < n; j++) { + matrix[j][n - 1 - i] = tmp[i][j]; + } + } + } +}; +``` + +如以上代码所示，遍历矩阵所有元素的时间复杂度为 $O(N^2)$ ；由于借助了一个辅助矩阵，**空间复杂度**为 $O(N^2)$ 。 + +# 方法二：原地修改 + +考虑不借助辅助矩阵，通过在原矩阵中直接「原地修改」，实现空间复杂度 $O(1)$ 的解法。 + +以位于矩阵四个角点的元素为例，设矩阵左上角元素 $A$ 、右上角元素 $B$ 、右下角元素 $C$ 、左下角元素 $D$ 。矩阵旋转 90º 后，相当于依次先后执行 $D \rightarrow A$ , $C \rightarrow D$ , $B \rightarrow C$ , $A \rightarrow B$ 修改元素，即如下「首尾相接」的元素旋转操作： + +$$ +A \leftarrow D \leftarrow C \leftarrow B \leftarrow A +$$ + +![ccw-01-07.002.png](https://pic.leetcode-cn.com/1638557961-BSxFQQ-ccw-01-07.002.png) + +如上图所示，由于第 $1$ 步 $D \rightarrow A$ 已经将 $A$ 覆盖（导致 $A$ 丢失），此丢失导致最后第 $4$ 步 $A \rightarrow B$ 无法赋值。为解决此问题，考虑借助一个「辅助变量 $tmp$ 」预先存储 $A$ ，此时的旋转操作变为： + +$$ +暂存 \ tmp = A \\ +A \leftarrow D \leftarrow C \leftarrow B \leftarrow tmp +$$ + +![ccw-01-07.003.png](https://pic.leetcode-cn.com/1638557961-hYpOoH-ccw-01-07.003.png) + +如上图所示，一轮可以完成矩阵 4 个元素的旋转。因而，只要分别以矩阵左上角 $1/4$ 的各元素为起始点执行以上旋转操作，即可完整实现矩阵旋转。 + +具体来看，当矩阵大小 $n$ 为偶数时，取前 $\frac{n}{2}$ 行、前 $\frac{n}{2}$ 列的元素为起始点；当矩阵大小 $n$ 为奇数时，取前 $\frac{n}{2}$ 行、前 $\frac{n + 1}{2}$ 列的元素为起始点。 + +令 $matrix[i][j] = A$ ，根据文章开头的元素旋转公式，可推导得适用于任意起始点的元素旋转操作： + +$$ +暂存 tmp = matrix[i][j] \\ +matrix[i][j] \leftarrow matrix[n - 1 - j][i] \leftarrow matrix[n - 1 - i][n - 1 - j] \leftarrow matrix[j][n - 1 - i] \leftarrow tmp +$$ + +> 如下图所示，为示例矩阵的算法执行流程。 + + + +## 代码 + +> 后三个 Tab 为「代码注释解析」。 + +```Python [] +class Solution: + def rotate(self, matrix: List[List[int]]) -> None: + n = len(matrix) + for i in range(n // 2): + for j in range((n + 1) // 2): + tmp = matrix[i][j] + matrix[i][j] = matrix[n - 1 - j][i] + matrix[n - 1 - j][i] = matrix[n - 1 - i][n - 1 - j] + matrix[n - 1 - i][n - 1 - j] = matrix[j][n - 1 - i] + matrix[j][n - 1 - i] = tmp +``` + +```Java [] +class Solution { + public void rotate(int[][] matrix) { + int n = matrix.length; + for (int i = 0; i < n / 2; i++) { + for (int j = 0; j < (n + 1) / 2; j++) { + int tmp = matrix[i][j]; + matrix[i][j] = matrix[n - 1 - j][i]; + matrix[n - 1 - j][i] = matrix[n - 1 - i][n - 1 - j]; + matrix[n - 1 - i][n - 1 - j] = matrix[j][n - 1 - i]; + matrix[j][n - 1 - i] = tmp; + } + } + } +} +``` + +```C++ [] +class Solution { +public: + void rotate(vector>& matrix) { + int n = matrix.size(); + for (int i = 0; i < n / 2; i++) { + for (int j = 0; j < (n + 1) / 2; j++) { + int tmp = matrix[i][j]; + matrix[i][j] = matrix[n - 1 - j][i]; + matrix[n - 1 - j][i] = matrix[n - 1 - i][n - 1 - j]; + matrix[n - 1 - i][n - 1 - j] = matrix[j][n - 1 - i]; + matrix[j][n - 1 - i] = tmp; + } + } + } +}; +``` + +```Python [] +class Solution: + def rotate(self, matrix: List[List[int]]) -> None: + # 设矩阵行列数为 n + n = len(matrix) + # 起始点范围为 0 <= i < n // 2 , 0 <= j < (n + 1) // 2 + # 其中 '//' 为整数除法 + for i in range(n // 2): + for j in range((n + 1) // 2): + # 暂存 A 至 tmp + tmp = matrix[i][j] + # 元素旋转操作 A <- D <- C <- B <- tmp + matrix[i][j] = matrix[n - 1 - j][i] + matrix[n - 1 - j][i] = matrix[n - 1 - i][n - 1 - j] + matrix[n - 1 - i][n - 1 - j] = matrix[j][n - 1 - i] + matrix[j][n - 1 - i] = tmp +``` + +```Java [] +class Solution { + public void rotate(int[][] matrix) { + // 设矩阵行列数为 n + int n = matrix.length; + // 起始点范围为 0 <= i < n / 2 , 0 <= j < (n + 1) / 2 + // 其中 '/' 为整数除法 + for (int i = 0; i < n / 2; i++) { + for (int j = 0; j < (n + 1) / 2; j++) { + // 暂存 A 至 tmp + int tmp = matrix[i][j]; + // 元素旋转操作 A <- D <- C <- B <- tmp + matrix[i][j] = matrix[n - 1 - j][i]; + matrix[n - 1 - j][i] = matrix[n - 1 - i][n - 1 - j]; + matrix[n - 1 - i][n - 1 - j] = matrix[j][n - 1 - i]; + matrix[j][n - 1 - i] = tmp; + } + } + } +} +``` + +```C++ [] +class Solution { +public: + void rotate(vector>& matrix) { + // 设矩阵行列数为 n + int n = matrix.size(); + // 起始点范围为 0 <= i < n / 2 , 0 <= j < (n + 1) / 2 + // 其中 '/' 为整数除法 + for (int i = 0; i < n / 2; i++) { + for (int j = 0; j < (n + 1) / 2; j++) { + // 暂存 A 至 tmp + int tmp = matrix[i][j]; + // 元素旋转操作 A <- D <- C <- B <- tmp + matrix[i][j] = matrix[n - 1 - j][i]; + matrix[n - 1 - j][i] = matrix[n - 1 - i][n - 1 - j]; + matrix[n - 1 - i][n - 1 - j] = matrix[j][n - 1 - i]; + matrix[j][n - 1 - i] = tmp; + } + } + } +}; +``` + +## 复杂度分析 + +- **时间复杂度 $O(N^2)$ ：** 其中 $N$ 为输入矩阵的行（列）数。需要将矩阵中每个元素旋转到新的位置，即对矩阵所有元素操作一次，使用 $O(N^2)$ 时间。 +- **空间复杂度 $O(1)$ ：** 临时变量 $tmp$ 使用常数大小的额外空间。值得注意，当循环中进入下轮迭代，上轮迭代初始化的 $tmp$ 占用的内存就会被自动释放，因此无累计使用空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git a/selected_coding_interview/docs/50. Pow(x, n).md b/selected_coding_interview/docs/50. Pow(x, n).md new file mode 100644 index 0000000..effed17 --- /dev/null +++ b/selected_coding_interview/docs/50. Pow(x, n).md @@ -0,0 +1,107 @@ +#### 解题思路： + +求 $x^n$ 最简单的方法是通过循环将 $n$ 个 $x$ 乘起来，依次求 $x^1, x^2, ..., x^{n-1}, x^n$ ，时间复杂度为 $O(n)$ 。 +**快速幂法** 可将时间复杂度降低至 $O(\log n)$ ，以下从 **“二分法”** 和 **“二进制”** 两个角度解析快速幂法。 + +##### 快速幂解析（二进制角度）： + +> 利用十进制数字 $n$ 的二进制表示，可对快速幂进行数学化解释。 + +- 对于任何十进制正整数 $n$ ，设其二进制为 "$b_m...b_3b_2b_1$"（ $b_i$ 为二进制某位值，$i \in [1,m]$ ），则有： + - **二进制转十进制：** $n = 1b_1 + 2b_2 + 4b_3 + ... + 2^{m-1}b_m$ *（即二进制转十进制公式）* ； + - **幂的二进制展开：** $x^n = x^{1b_1 + 2b_2 + 4b_3 + ... + 2^{m-1}b_m} = x^{1b_1}x^{2b_2}x^{4b_3}...x^{2^{m-1}b_m}$ ； +- 根据以上推导，可把计算 $x^n$ 转化为解决以下两个问题： + - **计算 $x^1, x^2, x^4, ..., x^{2^{m-1}}$ 的值：** 循环赋值操作 $x = x^2$ 即可； + - **获取二进制各位 $b_1, b_2, b_3, ..., b_m$ 的值：** 循环执行以下操作即可。 + 1. **$n \& 1$ （与操作）：** 判断 $n$ 二进制最右一位是否为 $1$ ； + 2. **$n>>1$ （移位操作）：** $n$ 右移一位（可理解为删除最后一位）。 +- 因此，应用以上操作，可在循环中依次计算 $x^{2^{0}b_1}, x^{2^{1}b_2}, ..., x^{2^{m-1}b_m}$ 的值，并将所有 $x^{2^{i-1}b_i}$ 累计相乘即可，其中： + +$$ +x^{2^{i-1}b_i}= +\begin{cases} + 1 & , b_i = 0 \\ + x^{2^{i-1}} & , b_i = 1 \\ +\end{cases} +$$ + +![Picture1.png](https://pic.leetcode-cn.com/40a7a874523e26cacae9c502a6e8cf8b58dba878739f17e6bb3ed6be76e97569-Picture1.png){:width=450} + +##### 快速幂解析（分治法角度）： + +> 快速幂实际上是分治思想的一种应用。 + +- $x^n = x^{n/2} \times x^{n/2} = (x^2)^{n/2}$ ，令 $n/2$ 为整数，则需要分为奇偶两种情况（设向下取整除法符号为 "$//$" ）： + +$$ +x^n = +\begin{cases} + (x^2)^{n//2} & , n 为偶数 \\ + x(x^2)^{n//2} & , n 为奇数 \\ +\end{cases} +$$ + +> 观察发现，当 $n$ 为奇数时，二分后会多出一项 $x$ 。 + +- **幂结果获取：** + - 根据推导，可通过循环 $x = x^2$ 操作，每次把幂从 $n$ 降至 $n//2$ ，直至将幂降为 $0$ ； + - 设 $res=1$ ，则初始状态 $x^n = x^n \times res$ 。在循环二分时，每当 $n$ 为奇数时，将多出的一项 $x$ 乘入 $res$ ，则最终可化至 $x^n = x^0 \times res = res$ ，返回 $res$ 即可。 + +![Picture2.png](https://pic.leetcode-cn.com/379a042b9d8df3a96d1ac0f27346718033bf3bfce69731bab52bf6f372b4c8f4-Picture2.png){:width=450} + +- **转化为位运算：** + - 向下整除 $n // 2$ **等价于** 右移一位 $n >> 1$ ； + - 取余数 $n \% 2$ **等价于** 判断二进制最右位 $n \& 1$ ； + +##### 算法流程： + +1. 当 $x = 0.0$ 时：直接返回 $0.0$ ，以避免后续 $1$ 除以 $0$ 操作报错。**分析：** 数字 $0$ 的正数次幂恒为 $0$ ； $0$ 的 $0$ 次幂和负数次幂没有意义，因此直接返回 $0.0$ 即可。 +2. 初始化 $res = 1$ 。 +3. 当 $n < 0$ 时：把问题转化至 $n \geq 0$ 的范围内，即执行 $x = 1/x$ ，$n = - n$ 。 +4. 循环计算：当 $n = 0$ 时跳出。 + 1. 当 $n \& 1 = 1$ 时：将当前 $x$ 乘入 $res$ （即 $res *= x$ ）。 + 2. 执行 $x = x^2$ （即 $x *= x$ ）。 + 3. 执行 $n$ 右移一位（即 $n >>= 1$）。 +5. 返回 $res$ 。 + +##### 复杂度分析： + +- **时间复杂度 $O(\log n)$ ：** 二分的时间复杂度为对数级别。 +- **空间复杂度 $O(1)$ ：** $res$, $b$ 等变量占用常数大小额外空间。 + +#### 代码： + +> Java 代码中 `int32` 变量 $n \in [-2147483648, 2147483647]$ ，因此当 $n = -2147483648$ 时执行 $n = -n$ 会因越界而赋值出错。解决方法是先将 $n$ 存入 `long` 变量 $b$ ，后面用 $b$ 操作即可。 + +```Python [] +class Solution: + def myPow(self, x: float, n: int) -> float: + if x == 0.0: return 0.0 + res = 1 + if n < 0: x, n = 1 / x, -n + while n: + if n & 1: res *= x + x *= x + n >>= 1 + return res +``` + +```Java [] +class Solution { + public double myPow(double x, int n) { + if(x == 0.0f) return 0.0d; + long b = n; + double res = 1.0; + if(b < 0) { + x = 1 / x; + b = -b; + } + while(b > 0) { + if((b & 1) == 1) res *= x; + x *= x; + b >>= 1; + } + return res; + } +} +``` diff --git "a/selected_coding_interview/docs/509. \346\226\220\346\263\242\351\202\243\345\245\221\346\225\260.md" "b/selected_coding_interview/docs/509. \346\226\220\346\263\242\351\202\243\345\245\221\346\225\260.md" new file mode 100644 index 0000000..b6a88e8 --- /dev/null +++ "b/selected_coding_interview/docs/509. \346\226\220\346\263\242\351\202\243\345\245\221\346\225\260.md" @@ -0,0 +1,83 @@ +## 解题思路： + +斐波那契数列的定义是 $f(n + 1) = f(n) + f(n - 1)$ ，生成第 $n$ 项的做法有以下几种： + +1. **暴力搜索：** + - **原理：** 把 $f(n)$ 问题的计算拆分成 $f(n-1)$ 和 $f(n-2)$ 两个子问题的计算，并递归，以 $f(0)$ 和 $f(1)$ 为终止条件。 + - **缺点：** 大量重复的递归计算，例如 $f(n)$ 和 $f(n - 1)$ 两者向下递归需要 **各自计算** $f(n - 2)$ 的值。 +2. **记忆化递归：** + - **原理：** 在递归法的基础上，新建一个长度为 $n$ 的数组，用于在递归时存储 $f(0)$ 至 $f(n)$ 的数字值，重复遇到某数字则直接从数组取用，避免了重复的递归计算。 + - **缺点：** 记忆化存储需要使用 $O(N)$ 的额外空间。 +3. **动态规划：** + - **原理：** 以斐波那契数列性质 $f(n + 1) = f(n) + f(n - 1)$ 为转移方程。 + - 从计算效率、空间复杂度上看，动态规划是本题的最佳解法。 + +下图帮助理解暴力搜索中的“重叠子问题”概念。 + +![Picture1.png](https://pic.leetcode-cn.com/1599882883-mtYecf-Picture1.png){:width=500} + +### 动态规划解析： + +- **状态定义：** 设 $dp$ 为一维数组，其中 $dp[i]$ 的值代表 斐波那契数列第 $i$ 个数字 。 +- **转移方程：** $dp[i + 1] = dp[i] + dp[i - 1]$ ，即对应数列定义 $f(n + 1) = f(n) + f(n - 1)$ 。 +- **初始状态：** $dp[0] = 0$, $dp[1] = 1$ ，即初始化前两个数字。 +- **返回值：** $dp[n]$ ，即斐波那契数列的第 $n$ 个数字。 + +### 状态压缩： + +> 若新建长度为 $n$ 的 $dp$ 列表，则空间复杂度为 $O(N)$ 。 + +- 由于 $dp$ 列表第 $i$ 项只与第 $i-1$ 和第 $i-2$ 项有关，因此只需要初始化三个整形变量 `sum`, `a`, `b` ，利用辅助变量 $sum$ 使 $a, b$ 两数字交替前进即可 *（具体实现见代码）* 。 +- 节省了 $dp$ 列表空间，因此空间复杂度降至 $O(1)$ 。 + +## 代码： + +```Python [] +class Solution: + def fib(self, n: int) -> int: + a, b = 0, 1 + for _ in range(n): + a, b = b, a + b + return a +``` + +```Java [] +class Solution { + public int fib(int n) { + int a = 0, b = 1, sum; + for(int i = 0; i < n; i++){ + sum = a + b; + a = b; + b = sum; + } + return a; + } +} +``` + +```C++ [] +class Solution { +public: + int fib(int n) { + int a = 0, b = 1, sum; + for(int i = 0; i < n; i++){ + sum = a + b; + a = b; + b = sum; + } + return a; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(n)$ ：** 计算 $f(n)$ 需循环 $n$ 次，每轮循环内计算操作使用 $O(1)$ 。 +- **空间复杂度 $O(1)$ ：** 几个标志变量使用常数大小的额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/53. \346\234\200\345\244\247\345\255\220\346\225\260\347\273\204\345\222\214.md" "b/selected_coding_interview/docs/53. \346\234\200\345\244\247\345\255\220\346\225\260\347\273\204\345\222\214.md" new file mode 100644 index 0000000..fe42a24 --- /dev/null +++ "b/selected_coding_interview/docs/53. \346\234\200\345\244\247\345\255\220\346\225\260\347\273\204\345\222\214.md" @@ -0,0 +1,86 @@ +## 解题思路： + +动态规划是本题的最优解法，以下按照标准流程解题。 + +| 常见解法 | 时间复杂度 | 空间复杂度 | +| -------- | ------------- | ----------- | +| 暴力搜索 | $O(N^2)$ | $O(1)$ | +| 分治思想 | $O(N \log N)$ | $O(\log N)$ | +| 动态规划 | $O(N)$ | $O(1)$ | + +### 动态规划解析： + +- **状态定义：** 设动态规划列表 $dp$ ，$dp[i]$ 代表以元素 $nums[i]$ 为结尾的连续子数组最大和。 + - 为何定义最大和 $dp[i]$ 中必须包含元素 $nums[i]$ ：保证 $dp[i]$ 递推到 $dp[i+1]$ 的正确性；如果不包含 $nums[i]$ ，递推时则不满足题目的 **连续子数组** 要求。 + +- **转移方程：** 若 $dp[i-1] \leq 0$ ，说明 $dp[i - 1]$ 对 $dp[i]$ 产生负贡献，即 $dp[i-1] + nums[i]$ 还不如 $nums[i]$ 本身大。 + +$$ +dp[i] = +\begin{cases} + dp[i-1] + nums[i] & , dp[i - 1] > 0 \\ + nums[i] & , dp[i - 1] \leq 0 \\ +\end{cases} +$$ + +- **初始状态：** $dp[0] = nums[0]$，即以 $nums[0]$ 结尾的连续子数组最大和为 $nums[0]$ 。 + +- **返回值：** 返回 $dp$ 列表中的最大值，代表全局最大值。 + +![Picture1.png](https://pic.leetcode-cn.com/77d1aa6a444743d3c8606ac951cd7fc38faf68a62064fd2639df517cd666a4d0-Picture1.png){:width=500} + +### 状态压缩： + +- 由于 $dp[i]$ 只与 $dp[i-1]$ 和 $nums[i]$ 有关系，因此可以将原数组 $nums$ 用作 $dp$ 列表，即直接在 $nums$ 上修改即可。 +- 由于省去 $dp$ 列表使用的额外空间，因此空间复杂度从 $O(N)$ 降至 $O(1)$ 。 + + + +## 代码： + +```Python [] +class Solution: + def maxSubArray(self, nums: List[int]) -> int: + for i in range(1, len(nums)): + nums[i] += max(nums[i - 1], 0) + return max(nums) +``` + +```Java [] +class Solution { + public int maxSubArray(int[] nums) { + int res = nums[0]; + for(int i = 1; i < nums.length; i++) { + nums[i] += Math.max(nums[i - 1], 0); + res = Math.max(res, nums[i]); + } + return res; + } +} +``` + +```C++ [] +class Solution { +public: + int maxSubArray(vector& nums) { + int res = nums[0]; + for(int i = 1; i < nums.size(); i++) { + if (nums[i - 1] > 0) nums[i] += nums[i - 1]; + if (nums[i] > res) res = nums[i]; + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 线性遍历数组 $nums$ 即可获得结果，使用 $O(N)$ 时间。 +- **空间复杂度 $O(1)$ ：** 使用常数大小的额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/54. \350\236\272\346\227\213\347\237\251\351\230\265.md" "b/selected_coding_interview/docs/54. \350\236\272\346\227\213\347\237\251\351\230\265.md" new file mode 100644 index 0000000..ef91bda --- /dev/null +++ "b/selected_coding_interview/docs/54. \350\236\272\346\227\213\347\237\251\351\230\265.md" @@ -0,0 +1,112 @@ +## 解题思路： + +根据题目示例 `matrix = [[1,2,3],[4,5,6],[7,8,9]]` 的对应输出 `[1,2,3,6,9,8,7,4,5]` 可以发现，顺时针打印矩阵的顺序是 **“从左向右、从上向下、从右向左、从下向上”** 循环。 + +因此，考虑设定矩阵的 “左、上、右、下” 四个边界，模拟以上矩阵遍历顺序。 + +![Picture1.png](https://pic.leetcode-cn.com/7605d807782923e4ad3c7995dc2485f538f202ac326bb330fe997f449123a548-Picture1.png){:width=400} + +### 算法流程： + +1. **空值处理：** 当 `matrix` 为空时，直接返回空列表 `[]` 即可。 +2. **初始化：** 矩阵 左、右、上、下 四个边界 `l` , `r` , `t` , `b` ，用于打印的结果列表 `res` 。 +3. **循环打印：** “从左向右、从上向下、从右向左、从下向上” 四个方向循环打印。 + 1. 根据边界打印，即将元素按顺序添加至列表 `res` 尾部。 + 2. 边界向内收缩 1 （代表已被打印）。 + 3. 判断边界是否相遇（是否打印完毕），若打印完毕则跳出。 +4. **返回值：** 返回 `res` 即可。 + +| 打印方向 | 1. 根据边界打印 | 2. 边界向内收缩 | 3. 是否打印完毕 | +| -------- | ---------------------- | ----------------- | --------------- | +| 从左向右 | 左边界`l` ，右边界 `r` | 上边界 `t` 加 $1$ | 是否 `t > b` | +| 从上向下 | 上边界 `t` ，下边界`b` | 右边界 `r` 减 $1$ | 是否 `l > r` | +| 从右向左 | 右边界 `r` ，左边界`l` | 下边界 `b` 减 $1$ | 是否 `t > b` | +| 从下向上 | 下边界 `b` ，上边界`t` | 左边界 `l` 加 $1$ | 是否 `l > r` | + + + +## 代码： + +Java, C++ 代码利用了 `++` 操作的便利性，详情可见 [++i 和 i++ 的区别](https://www.jianshu.com/p/b62eac216499) 。 + +- `res[x++]` 等价于先给 `res[x]` 赋值，再给 `x` 自增 $1$ 。 +- `++t > b` 等价于先给 `t` 自增 $1$ ，再判断 `t > b` 逻辑表达式。 + +> TIPS： 请注意区分数字 `1` 和字母 `l` 。 + +```Python [] +class Solution: + def spiralOrder(self, matrix: List[List[int]]) -> List[int]: + if not matrix: return [] + l, r, t, b, res = 0, len(matrix[0]) - 1, 0, len(matrix) - 1, [] + while True: + for i in range(l, r + 1): res.append(matrix[t][i]) # left to right + t += 1 + if t > b: break + for i in range(t, b + 1): res.append(matrix[i][r]) # top to bottom + r -= 1 + if l > r: break + for i in range(r, l - 1, -1): res.append(matrix[b][i]) # right to left + b -= 1 + if t > b: break + for i in range(b, t - 1, -1): res.append(matrix[i][l]) # bottom to top + l += 1 + if l > r: break + return res +``` + +```Java [] +class Solution { + public List spiralOrder(int[][] matrix) { + if (matrix.length == 0) + return new ArrayList(); + int l = 0, r = matrix[0].length - 1, t = 0, b = matrix.length - 1, x = 0; + Integer[] res = new Integer[(r + 1) * (b + 1)]; + while (true) { + for (int i = l; i <= r; i++) res[x++] = matrix[t][i]; // left to right + if (++t > b) break; + for (int i = t; i <= b; i++) res[x++] = matrix[i][r]; // top to bottom + if (l > --r) break; + for (int i = r; i >= l; i--) res[x++] = matrix[b][i]; // right to left + if (t > --b) break; + for (int i = b; i >= t; i--) res[x++] = matrix[i][l]; // bottom to top + if (++l > r) break; + } + return Arrays.asList(res); + } +} +``` + +```C++ [] +class Solution { +public: + vector spiralOrder(vector>& matrix) { + if (matrix.empty()) return {}; + int l = 0, r = matrix[0].size() - 1, t = 0, b = matrix.size() - 1; + vector res; + while (true) { + for (int i = l; i <= r; i++) res.push_back(matrix[t][i]); // left to right + if (++t > b) break; + for (int i = t; i <= b; i++) res.push_back(matrix[i][r]); // top to bottom + if (l > --r) break; + for (int i = r; i >= l; i--) res.push_back(matrix[b][i]); // right to left + if (t > --b) break; + for (int i = b; i >= t; i--) res.push_back(matrix[i][l]); // bottom to top + if (++l > r) break; + } + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(MN)$ ：** $M, N$ 分别为矩阵行数和列数。 +- **空间复杂度 $O(1)$ ：** 四个边界 `l` , `r` , `t` , `b` 使用常数大小的额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/59. \350\236\272\346\227\213\347\237\251\351\230\265 II.md" "b/selected_coding_interview/docs/59. \350\236\272\346\227\213\347\237\251\351\230\265 II.md" new file mode 100644 index 0000000..6dc49ee --- /dev/null +++ "b/selected_coding_interview/docs/59. \350\236\272\346\227\213\347\237\251\351\230\265 II.md" @@ -0,0 +1,60 @@ +### 思路： + +- 生成一个 `n×n` 空矩阵 `mat`，随后模拟整个向内环绕的填入过程： + - 定义当前左右上下边界 `l,r,t,b`，初始值 `num = 1`，迭代终止值 `tar = n * n`； + - 当 `num <= tar` 时，始终按照 `从左到右` `从上到下` `从右到左` `从下到上` 填入顺序循环，每次填入后： + - 执行 `num += 1`：得到下一个需要填入的数字； + - 更新边界：例如从左到右填完后，上边界` t += 1`，相当于上边界向内缩 1。 + - 使用`num <= tar`而不是`l < r || t < b`作为迭代条件，是为了解决当`n`为奇数时，矩阵中心数字无法在迭代过程中被填充的问题。 +- 最终返回 `mat` 即可。 + +![Picture1.png](https://pic.leetcode-cn.com/ccff416fa39887c938d36fec8e490e1861813d3bba7836eda941426f13420759-Picture1.png){:width=500} + +### 代码： + +```Java [] +class Solution { + public int[][] generateMatrix(int n) { + int l = 0, r = n - 1, t = 0, b = n - 1; + int[][] mat = new int[n][n]; + int num = 1, tar = n * n; + while(num <= tar){ + for(int i = l; i <= r; i++) mat[t][i] = num++; // left to right. + t++; + for(int i = t; i <= b; i++) mat[i][r] = num++; // top to bottom. + r--; + for(int i = r; i >= l; i--) mat[b][i] = num++; // right to left. + b--; + for(int i = b; i >= t; i--) mat[i][l] = num++; // bottom to top. + l++; + } + return mat; + } +} +``` + +```Python [] +class Solution: + def generateMatrix(self, n: int) -> [[int]]: + l, r, t, b = 0, n - 1, 0, n - 1 + mat = [[0 for _ in range(n)] for _ in range(n)] + num, tar = 1, n * n + while num <= tar: + for i in range(l, r + 1): # left to right + mat[t][i] = num + num += 1 + t += 1 + for i in range(t, b + 1): # top to bottom + mat[i][r] = num + num += 1 + r -= 1 + for i in range(r, l - 1, -1): # right to left + mat[b][i] = num + num += 1 + b -= 1 + for i in range(b, t - 1, -1): # bottom to top + mat[i][l] = num + num += 1 + l += 1 + return mat +``` diff --git "a/selected_coding_interview/docs/6. N \345\255\227\345\275\242\345\217\230\346\215\242.md" "b/selected_coding_interview/docs/6. N \345\255\227\345\275\242\345\217\230\346\215\242.md" new file mode 100644 index 0000000..b586706 --- /dev/null +++ "b/selected_coding_interview/docs/6. N \345\255\227\345\275\242\345\217\230\346\215\242.md" @@ -0,0 +1,49 @@ +#### 解题思路： + +- **题目理解：** + - 字符串 `s` 是以 $Z$ 字形为顺序存储的字符串，目标是按行打印。 + - 设 `numRows` 行字符串分别为 $s_1$ , $s_2$ ,..., $s_n$，则容易发现：按顺序遍历字符串 `s` 时，每个字符 `c` 在 $Z$ 字形中对应的 **行索引** 先从 $s_1$ 增大至 $s_n$，再从 $s_n$ 减小至 $s_1$ …… 如此反复。 + - 因此，解决方案为：模拟这个行索引的变化，在遍历 `s` 中把每个字符填到正确的行 `res[i]` 。 +- **算法流程：** 按顺序遍历字符串 `s`； + 1. **`res[i] += c`：** 把每个字符 `c` 填入对应行 $s_i$； + 2. **`i += flag`：** 更新当前字符 `c` 对应的行索引； + 3. **`flag = - flag`：** 在达到 $Z$ 字形转折点时，执行反向。 +- **复杂度分析：** + - 时间复杂度 $O(N)$ ：遍历一遍字符串 `s`； + - 空间复杂度 $O(N)$ ：各行字符串共占用 $O(N)$ 额外空间。 + + + +#### 代码： + +```Python [] +class Solution: + def convert(self, s: str, numRows: int) -> str: + if numRows < 2: return s + res = ["" for _ in range(numRows)] + i, flag = 0, -1 + for c in s: + res[i] += c + if i == 0 or i == numRows - 1: flag = -flag + i += flag + return "".join(res) +``` + +```Java [] +class Solution { + public String convert(String s, int numRows) { + if(numRows < 2) return s; + List rows = new ArrayList(); + for(int i = 0; i < numRows; i++) rows.add(new StringBuilder()); + int i = 0, flag = -1; + for(char c : s.toCharArray()) { + rows.get(i).append(c); + if(i == 0 || i == numRows -1) flag = - flag; + i += flag; + } + StringBuilder res = new StringBuilder(); + for(StringBuilder row : rows) res.append(row); + return res.toString(); + } +} +``` diff --git "a/selected_coding_interview/docs/64. \346\234\200\345\260\217\350\267\257\345\276\204\345\222\214.md" "b/selected_coding_interview/docs/64. \346\234\200\345\260\217\350\267\257\345\276\204\345\222\214.md" new file mode 100644 index 0000000..72563e8 --- /dev/null +++ "b/selected_coding_interview/docs/64. \346\234\200\345\260\217\350\267\257\345\276\204\345\222\214.md" @@ -0,0 +1,59 @@ +### 解题思路： + +此题是典型的动态规划题目。 + +- **状态定义：** + - 设 $dp$ 为大小 $m \times n$ 矩阵，其中 $dp[i][j]$ 的值代表直到走到 $(i,j)$ 的最小路径和。 + +- **转移方程：** + > 题目要求，只能向右或向下走，换句话说，当前单元格 $(i,j)$ 只能从左方单元格 $(i-1,j)$ 或上方单元格 $(i,j-1)$ 走到，因此只需要考虑矩阵左边界和上边界。 + - 走到当前单元格 $(i,j)$ 的最小路径和 $=$ “从左方单元格 $(i-1,j)$ 与 从上方单元格 $(i,j-1)$ 走来的 **两个最小路径和中较小的** ” $+$ 当前单元格值 $grid[i][j]$ 。具体分为以下 $4$ 种情况： + 1. **当左边和上边都不是矩阵边界时：** 即当$i \not= 0$, $j \not= 0$时，$dp[i][j] = min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j]$ ； + 2. **当只有左边是矩阵边界时：** 只能从上面来，即当$i = 0, j \not= 0$时， $dp[i][j] = dp[i][j - 1] + grid[i][j]$ ； + 3. **当只有上边是矩阵边界时：** 只能从左面来，即当$i \not= 0, j = 0$时， $dp[i][j] = dp[i - 1][j] + grid[i][j]$ ； + 4. **当左边和上边都是矩阵边界时：** 即当$i = 0, j = 0$时，其实就是起点， $dp[i][j] = grid[i][j]$； + +- **初始状态：** + - $dp$ 初始化即可，不需要修改初始 $0$ 值。 + +- **返回值：** + - 返回 $dp$ 矩阵右下角值，即走到终点的最小路径和。 + +其实我们完全不需要建立 $dp$ 矩阵浪费额外空间，直接遍历 $grid[i][j]$ 修改即可。这是因为：`grid[i][j] = min(grid[i - 1][j], grid[i][j - 1]) + grid[i][j]` ；原 $grid$ 矩阵元素中被覆盖为 $dp$ 元素后（都处于当前遍历点的左上方），不会再被使用到。 + +#### 复杂度分析： + +- **时间复杂度 $O(M \times N)$ ：** 遍历整个 $grid$ 矩阵元素。 +- **空间复杂度 $O(1)$ ：** 直接修改原矩阵，不使用额外空间。 + + + +#### 代码： + +```Python [] +class Solution: + def minPathSum(self, grid: [[int]]) -> int: + for i in range(len(grid)): + for j in range(len(grid[0])): + if i == j == 0: continue + elif i == 0: grid[i][j] = grid[i][j - 1] + grid[i][j] + elif j == 0: grid[i][j] = grid[i - 1][j] + grid[i][j] + else: grid[i][j] = min(grid[i - 1][j], grid[i][j - 1]) + grid[i][j] + return grid[-1][-1] +``` + +```Java [] +class Solution { + public int minPathSum(int[][] grid) { + for(int i = 0; i < grid.length; i++) { + for(int j = 0; j < grid[0].length; j++) { + if(i == 0 && j == 0) continue; + else if(i == 0) grid[i][j] = grid[i][j - 1] + grid[i][j]; + else if(j == 0) grid[i][j] = grid[i - 1][j] + grid[i][j]; + else grid[i][j] = Math.min(grid[i - 1][j], grid[i][j - 1]) + grid[i][j]; + } + } + return grid[grid.length - 1][grid[0].length - 1]; + } +} +``` diff --git "a/selected_coding_interview/docs/65. \346\234\211\346\225\210\346\225\260\345\255\227.md" "b/selected_coding_interview/docs/65. \346\234\211\346\225\210\346\225\260\345\255\227.md" new file mode 100644 index 0000000..7cfeaa2 --- /dev/null +++ "b/selected_coding_interview/docs/65. \346\234\211\346\225\210\346\225\260\345\255\227.md" @@ -0,0 +1,146 @@ +## 解题思路： + +本题使用有限状态自动机。根据字符类型和合法数值的特点，先定义状态，再画出状态转移图，最后编写代码即可。 + +**字符类型**：空格 「 」、数字「 $0—9$ 」 、正负号 「 $+$, $-$ 」 、小数点 「 $.$ 」 、幂符号 「 $e$, $E$ 」 。 + +**状态定义**：按照字符串从左到右的顺序，定义以下 9 种状态。 + +0. 开始的空格 +1. 幂符号前的正负号 +2. 小数点前的数字 +3. 小数点、小数点后的数字 +4. 当小数点前为空格时，小数点、小数点后的数字 +5. 幂符号 +6. 幂符号后的正负号 +7. 幂符号后的数字 +8. 结尾的空格 + +**结束状态**：合法的结束状态有 2, 3, 7, 8 。 + +![Picture1.png](https://pic.leetcode-cn.com/1599283151-YmPMis-Picture1.png){:width=650} + +### 算法流程： + +1. **初始化：** + 1. **状态转移表 `states` ：** 设 `states[i]` ，其中 `i` 为所处状态， `states[i]` 使用哈希表存储可转移至的状态。键值对 `(key, value)` 含义：输入字符 `key` ，则从状态 `i` 转移至状态 `value` 。 + 2. **当前状态 `p` ：** 起始状态初始化为 `p = 0` 。 + +2. **状态转移循环：** 遍历字符串 `s` 的每个字符 `c` 。 + 1. **记录字符类型 `t` ：** 分为四种情况。 + - 当 `c` 为正负号时，执行 `t = 's'` ; + - 当 `c` 为数字时，执行 `t = 'd'` ; + - 当 `c` 为 `e` 或 `E` 时，执行 `t = 'e'` ; + - 当 `c` 为 `.` 或 `空格` 时，执行 `t = c` （即用字符本身表示字符类型）; + - 否则，执行 `t = '?'` ，代表为不属于判断范围的非法字符，后续直接返回 $false$ 。 + 2. **终止条件：** 若字符类型 `t` 不在哈希表 `states[p]` 中，说明无法转移至下一状态，因此直接返回 $false$ 。 + 3. **状态转移：** 状态 `p` 转移至 `states[p][t]` 。 + +3. **返回值：** 跳出循环后，若状态 `p` $\in {2, 3, 7, 8}$ ，说明结尾合法，返回 $true$ ，否则返回 $false$ 。 + + + +## 代码： + +Java 的状态转移表 `states` 使用 Map[] 数组存储。 + +```Python [] +class Solution: + def isNumber(self, s: str) -> bool: + states = [ + { ' ': 0, 's': 1, 'd': 2, '.': 4 }, # 0. start with 'blank' + { 'd': 2, '.': 4 } , # 1. 'sign' before 'e' + { 'd': 2, '.': 3, 'e': 5, ' ': 8 }, # 2. 'digit' before 'dot' + { 'd': 3, 'e': 5, ' ': 8 }, # 3. 'digit' after 'dot' + { 'd': 3 }, # 4. 'digit' after 'dot' (‘blank’ before 'dot') + { 's': 6, 'd': 7 }, # 5. 'e' + { 'd': 7 }, # 6. 'sign' after 'e' + { 'd': 7, ' ': 8 }, # 7. 'digit' after 'e' + { ' ': 8 } # 8. end with 'blank' + ] + p = 0 # start with state 0 + for c in s: + if '0' <= c <= '9': t = 'd' # digit + elif c in "+-": t = 's' # sign + elif c in "eE": t = 'e' # e or E + elif c in ". ": t = c # dot, blank + else: t = '?' # unknown + if t not in states[p]: return False + p = states[p][t] + return p in (2, 3, 7, 8) +``` + +```Java [] +class Solution { + public boolean isNumber(String s) { + Map[] states = { + new HashMap<>() {{ put(' ', 0); put('s', 1); put('d', 2); put('.', 4); }}, // 0. + new HashMap<>() {{ put('d', 2); put('.', 4); }}, // 1. + new HashMap<>() {{ put('d', 2); put('.', 3); put('e', 5); put(' ', 8); }}, // 2. + new HashMap<>() {{ put('d', 3); put('e', 5); put(' ', 8); }}, // 3. + new HashMap<>() {{ put('d', 3); }}, // 4. + new HashMap<>() {{ put('s', 6); put('d', 7); }}, // 5. + new HashMap<>() {{ put('d', 7); }}, // 6. + new HashMap<>() {{ put('d', 7); put(' ', 8); }}, // 7. + new HashMap<>() {{ put(' ', 8); }} // 8. + }; + int p = 0; + char t; + for(char c : s.toCharArray()) { + if (c >= '0' && c <= '9') t = 'd'; + else if (c == '+' || c == '-') t = 's'; + else if (c == 'e' || c == 'E') t = 'e'; + else if (c == '.' || c == ' ') t = c; + else t = '?'; + if (!states[p].containsKey(t)) return false; + p = (int)states[p].get(t); + } + return p == 2 || p == 3 || p == 7 || p == 8; + } +} +``` + +```C++ [] +class Solution { +public: + bool isNumber(string s) { + std::vector> states = { + { {' ', 0}, {'s', 1}, {'d', 2}, {'.', 4} }, // 0. start with 'blank' + { {'d', 2}, {'.', 4} }, // 1. 'sign' before 'e' + { {'d', 2}, {'.', 3}, {'e', 5}, {' ', 8} }, // 2. 'digit' before 'dot' + { {'d', 3}, {'e', 5}, {' ', 8} }, // 3. 'digit' after 'dot' + { {'d', 3} }, // 4. 'digit' after 'dot' (‘blank’ before 'dot') + { {'s', 6}, {'d', 7} }, // 5. 'e' + { {'d', 7} }, // 6. 'sign' after 'e' + { {'d', 7}, {' ', 8} }, // 7. 'digit' after 'e' + { {' ', 8} } // 8. end with 'blank' + }; + + int p = 0; // start with state 0 + for (char c : s) { + char t; + if (c >= '0' && c <= '9') t = 'd'; // digit + else if (c == '+' || c == '-') t = 's'; // sign + else if (c == 'e' || c == 'E') t = 'e'; // e or E + else if (c == '.' || c == ' ') t = c; // dot, blank + else t = '?'; // unknown + + if (states[p].count(t) == 0) return false; + p = states[p][t]; + } + return p == 2 || p == 3 || p == 7 || p == 8; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为字符串 `s` 的长度，判断需遍历字符串，每轮状态转移的使用 $O(1)$ 时间。 +- **空间复杂度 $O(1)$ ：** `states` 和 `p` 使用常数大小的额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/70. \347\210\254\346\245\274\346\242\257.md" "b/selected_coding_interview/docs/70. \347\210\254\346\245\274\346\242\257.md" new file mode 100644 index 0000000..8d2c3e5 --- /dev/null +++ "b/selected_coding_interview/docs/70. \347\210\254\346\245\274\346\242\257.md" @@ -0,0 +1,78 @@ +## 解题思路： + +设跳上 $n$ 级台阶有 $f(n)$ 种跳法。在所有跳法中，青蛙的最后一步只有两种情况： **跳上 $1$ 级或 $2$ 级台阶**。 + +1. **当为 $1$ 级台阶：** 剩 $n-1$ 个台阶，此情况共有 $f(n-1)$ 种跳法。 +2. **当为 $2$ 级台阶：** 剩 $n-2$ 个台阶，此情况共有 $f(n-2)$ 种跳法。 + +即 $f(n)$ 为以上两种情况之和，即 $f(n)=f(n-1)+f(n-2)$ ，以上递推性质为斐波那契数列。因此，本题可转化为 **求斐波那契数列的第 $n$ 项**，区别仅在于初始值不同： + +- 青蛙跳台阶问题： $f(0)=1$ , $f(1)=1$ , $f(2)=2$ 。 +- 斐波那契数列问题： $f(0)=0$ , $f(1)=1$ , $f(2)=1$ 。 + +![Picture1.png](https://pic.leetcode-cn.com/1599883153-UckfTw-Picture1.png){:width=500} + +### 动态规划解析： + +- **状态定义：** 设 $dp$ 为一维数组，其中 $dp[i]$ 的值代表斐波那契数列的第 $i$ 个数字。 +- **转移方程：** $dp[i + 1] = dp[i] + dp[i - 1]$ ，即对应数列定义 $f(n + 1) = f(n) + f(n - 1)$ 。 +- **初始状态：** $dp[0] = 1$, $dp[1] = 1$ ，即初始化前两个数字。 +- **返回值：** $dp[n]$ ，即斐波那契数列的第 $n$ 个数字。 + +### 状态压缩： + +若新建长度为 $n$ 的 $dp$ 列表，则空间复杂度为 $O(N)$ 。 + +由于 $dp$ 列表第 $i$ 项只与第 $i-1$ 和第 $i-2$ 项有关，因此只需要初始化三个整形变量 `sum`, `a`, `b` ，利用辅助变量 $sum$ 使 $a, b$ 两数字交替前进即可 *（具体实现见代码）* 。由于省去了 $dp$ 列表空间，因此空间复杂度降至 $O(1)$ 。 + +## 代码： + +```Python [] +class Solution: + def climbStairs(self, n: int) -> int: + a, b = 1, 1 + for _ in range(n - 1): + a, b = b, a + b + return b +``` + +```Java [] +class Solution { + public int climbStairs(int n) { + int a = 1, b = 1, sum; + for(int i = 0; i < n - 1; i++){ + sum = a + b; + a = b; + b = sum; + } + return b; + } +} +``` + +```C++ [] +class Solution { +public: + int climbStairs(int n) { + int a = 1, b = 1, sum; + for(int i = 0; i < n - 1; i++){ + sum = a + b; + a = b; + b = sum; + } + return b; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(n)$ ：** 计算 $f(n)$ 需循环 $n$ 次，每轮循环内计算操作使用 $O(1)$ 。 +- **空间复杂度 $O(1)$ ：** 几个标志变量使用常数大小的额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/704. \344\272\214\345\210\206\346\237\245\346\211\276.md" "b/selected_coding_interview/docs/704. \344\272\214\345\210\206\346\237\245\346\211\276.md" new file mode 100644 index 0000000..8fcc61b --- /dev/null +++ "b/selected_coding_interview/docs/704. \344\272\214\345\210\206\346\237\245\346\211\276.md" @@ -0,0 +1,86 @@ +## 解题思路 + +「二分查找」是利用数组的有序性，每轮缩窄一半的查找区间（即排除一半元素），直到找到目标值或查找区间为空时返回。 + +由于每轮可以排除一半元素，因此查找最多循环 $\log_2 N$ 次，时间复杂度 $O(\log N)$ 。在数据量较大时，「二分查找 $O(\log N)$ 」效率大幅高于「线性查找 $O(N)$ 」。 + +理解二分查找的关键是理解**缩窄区间的含义**。给定升序数组 $nums$ 和目标值 $target$ ，二分查找流程如下： + +1. **定义查找区间：** 初始化双指针 $i$ , $j$ 分别指向数组首、尾元素，代表查找区间为**闭区间** $[i, j]$ ； + +2. **循环二分，缩窄查找区间：** + + - 使用向下取整除法，计算区间 $[i, j]$ 的中点 $m$ ； + + - 若 $nums[m] < target$ ，根据数组有序性，易得 $target$ **一定不在**闭区间 $[i, m]$ 中，因此执行 $i = m + 1$ ，即将查找区间缩窄至 $[m + 1, j]$ ； + + - 若 $nums[m] > target$ ， 根据数组有序性，易得 $target$ **一定不在**闭区间 $[m, j]$ 中，因此执行 $j = m - 1$ ，即将查找区间缩窄至 $[i, m - 1]$ ； + + - 若 $nums[m] = target$ ，说明找到 $target$ ，返回索引 $m$ 即可； + +3. 不满足 $i \leq j$ 时跳出循环，此时代表无法在数组中找到 $target$ ，因此返回 $-1$ ； + +![figures.gif](https://pic.leetcode-cn.com/1658569201-hBJKhZ-figures.gif) + +动图播放较快，可以在 $\downarrow$ 一页页看 + + + +## 代码 + +需注意，若数组长度取值范围较大，计算中点操作 $m = \frac{i + j}{2}$ 中的 $i + j$ 可能超出 int 类型的取值范围，从而导致计算错误。此时，需使用 $m = i + \frac{j - i}{2}$ 计算中点，避免此问题。 + +```Python [] +class Solution: + def search(self, nums: List[int], target: int) -> int: + i, j = 0, len(nums) - 1 + while i <= j: + m = (i + j) // 2 + if nums[m] < target: i = m + 1 + elif nums[m] > target: j = m - 1 + else: return m + return -1 +``` + +```Java [] +class Solution { + public int search(int[] nums, int target) { + int i = 0, j = nums.length - 1; + while (i <= j) { + int m = (i + j) / 2; + if (nums[m] < target) i = m + 1; + else if (nums[m] > target) j = m - 1; + else return m; + } + return -1; + } +} +``` + +```C++ [] +class Solution { +public: + int search(vector& nums, int target) { + int i = 0, j = nums.size() - 1; + while (i <= j) { + int m = (i + j) / 2; + if (nums[m] < target) i = m + 1; + else if (nums[m] > target) j = m - 1; + else return m; + } + return -1; + } +}; +``` + +## 复杂度分析 + +- **时间复杂度 $O(\log N)$ ：** 其中 $N$ 为数组 $nums$ 长度。二分查找使用对数级别时间。 +- **空间复杂度 $O(1)$ ：** 变量 $i$ , $j$ 使用常数大小空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/724. \345\257\273\346\211\276\346\225\260\347\273\204\347\232\204\344\270\255\345\277\203\344\270\213\346\240\207.md" "b/selected_coding_interview/docs/724. \345\257\273\346\211\276\346\225\260\347\273\204\347\232\204\344\270\255\345\277\203\344\270\213\346\240\207.md" new file mode 100644 index 0000000..1ff9265 --- /dev/null +++ "b/selected_coding_interview/docs/724. \345\257\273\346\211\276\346\225\260\347\273\204\347\232\204\344\270\255\345\277\203\344\270\213\346\240\207.md" @@ -0,0 +1,83 @@ +## 解题思路 + +题目仅说明是整数数组，无其他已知条件，因此考虑直接遍历数组。 + +- 设索引 $i$ 对应变量「左侧元素相加和 `sum_left` 」和「右侧元素相加和 `sum_right` 」。 +- 遍历数组 `nums` ，每轮更新 `sum_left` 和 `sum_right` 。 +- 遍历中，遇到满足 `sum_left == sum_right` 时，说明当前索引为中心下标，返回即可。 +- 若遍历完成，仍未找到「中心下标」，则返回 -1 。 + +初始化时，相当于索引 $i = - 1$ ，此时 `sum_left = 0` , `sum_right = 所有元素的和` 。 + +> 首页为动态图，其余页为静态图，方便一步步看。 + + + +## 代码 + +需要考虑大数越界问题。题目给定整数数组 `nums` ，并给定取值范围。 + +$$ +1 \leq nums.length \leq 10^4 \\ +-1000 \leq nums[i] \leq 1000 +$$ + +易得「元素相加和」的取值范围为 $[-10^7, 10^7]$ ，在 `int` 类型的取值范围内，因此 `sum_left` 和 `sum_right` 使用 int 类型即可。 + +```Python [] +class Solution: + def pivotIndex(self, nums: List[int]) -> int: + sum_left, sum_right = 0, sum(nums) + for i in range(len(nums)): + sum_right -= nums[i] + # 若左侧元素和等于右侧元素和，返回中心下标 i + if sum_left == sum_right: + return i + sum_left += nums[i] + return -1 +``` + +```Java [] +class Solution { + public int pivotIndex(int[] nums) { + int sumLeft = 0, sumRight = Arrays.stream(nums).sum(); + for (int i = 0; i < nums.length; i++) { + sumRight -= nums[i]; + // 若左侧元素和等于右侧元素和，返回中心下标 i + if (sumLeft == sumRight) + return i; + sumLeft += nums[i]; + } + return -1; + } +} +``` + +```C++ [] +class Solution { +public: + int pivotIndex(vector& nums) { + int sumLeft = 0, sumRight = accumulate(nums.begin(), nums.end(), 0); + for (int i = 0; i < nums.size(); i++) { + sumRight -= nums[i]; + // 若左侧元素和等于右侧元素和，返回中心下标 i + if (sumLeft == sumRight) + return i; + sumLeft += nums[i]; + } + return -1; + } +}; +``` + +## 复杂度分析 + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为数组 `nums` 长度。求和操作使用 $O(N)$ 线性时间，遍历 `nums` 最差使用 $O(N)$ 线性时间。 +- **空间复杂度 $O(1)$ ：** 变量 `sum_left` , `sum_right` 使用常数大小空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/768. \346\234\200\345\244\232\350\203\275\345\256\214\346\210\220\346\216\222\345\272\217\347\232\204\345\235\227 II.md" "b/selected_coding_interview/docs/768. \346\234\200\345\244\232\350\203\275\345\256\214\346\210\220\346\216\222\345\272\217\347\232\204\345\235\227 II.md" new file mode 100644 index 0000000..11710e2 --- /dev/null +++ "b/selected_coding_interview/docs/768. \346\234\200\345\244\232\350\203\275\345\256\214\346\210\220\346\216\222\345\272\217\347\232\204\345\235\227 II.md" @@ -0,0 +1,67 @@ +#### 解题思路： + +- **排序块定义：** + - **排序块** 充分条件： 设此块中最大数字为 $head$ , 若此块后面的所有数字都 $>= head$ ，则此块为排序块。 + - **排序块** 最短长度为 $1$，即单个元素可以独立看作一个排序块。 + +- **贪心法则** （划分出尽可能多的排序块）： + - **思路一：** + - 设定双指针指向数组头部，判断双指针内数字集合形成的块是否满足排序块条件，并尽量使窗口最小（贪心）。 + - 每次形成排序块时计数，并越过此排序块重新指定双指针位置，重复以上步骤直到划分完整个数组。 + - 此思路容易理解，但每次确定 $1$ 个块都需要遍历整个数组，在某些极端情况（例如如 $[1,2,3,4,5]$ ）时间复杂度达到 $O(N^2)$ 。 + - **思路二（本题解采用）：** + - 判断是否是排序块只需要用到该块的 **元素最大值** $head$ 。我们联想到，是否可以遍历一遍数组 $arr$ ，动态判断到目前数字 $num$ 为止最多能分出多少排序块，并保存每个排序块的最大值 $head$ 。每遍历到下个数字 $num$ ，动态判断前面所有的排序块是否成立，并更新所有排序块： + - 当某排序块 $num < head$ ：将此排序块`[A]`与 `num` 合并，形成新排序块`[A | num]`，最大值仍为 $head$ ； + - 当某排序块 $num >= head$ ：原排序块保留，并新加排序块 `[num]` 。 + - 而对于整个数组的排序块，其 $head$ 大小是从左到右递增的。例如：数组 $[1,2,1,3,4,7,5,6]$ 最多可划分为 $[1|2,1|3|4|7,5,6]$ ，$head$ 为 $[1,2,3,4,7]$ 。因此，若给数组尾部加入一个随机正整数 $n$ ，尾部的排序块更容易被合并（最先满足 $num < head$ ）。当 $n$ 值较小时（ $<$ 前面多个排序块的 $head$ ），则需按尾部到首部的顺序合并多个排序块。 + - 这种先入（首部到尾部添加排序块）后出（尾部到首部判断并合并排序块）的特性，让我们联想到使用 **栈** 保存排序块最大值 $head$ 。在遍历过程中，通过维护栈的 $head$ 序列，实现排序块的动态更新。 + +- **算法流程：** + 1. 遍历数组 $arr$ 中的每个数字 $num$ ； + 2. 当栈 $stack$ 不为空且数字 $num<$栈顶值 时： *（代表此 $num$ 会改变前面排序块分布）* + - 栈顶 `pop()` 出栈，并保存栈顶值为 $head$ 。 *（此情况下，新排序块最大值还为 $head$ ，因此先暂存）* + - 当 $stack$ 不为空且数字 $num<$栈顶值 时，循环栈顶 `pop()` 出栈。 *（判断加入 $num$ 需要合并的所有排序块，每 `pop()` 一个 $head$ 代表合并一个块）* + - 将保存的栈顶值 $head$ 重新 `push()` 入栈。 *（将 $head$ 重新加入，作为新排序块的最大值）* + 3. 当栈 $stack$ 为空或数字 $num>=$栈顶值 时： *（代表此 $num$ 不影响前面排序块分布）* + - 将 $num$ 数字 `push()` 入栈。 *（加入单个元素的新排序块 `[num]`）* + 4. 遍历完成后，栈中保存 **所有排序块的对应最大值** $head$ ，因此返回栈 $stack$ 长度即可获得排序块数量。 + +- **复杂度分析：** + - 时间复杂度 $O(N)$ ：遍历一遍 $arr$ 为 $O(N)$，修正排序块最多遍历一遍 $arr$ 为 $O(N)$； + - 空间复杂度 $O(N)$ ：极端情况下排序块数量等于数组长度，此时 $stack$ 占用线性大小额外空间。 + + > 图解中每一种颜色代表一个排序块。 + + + +#### 代码： + +```python [-Python] +class Solution: + def maxChunksToSorted(self, arr: [int]) -> int: + stack = [] + for num in arr: + if stack and num < stack[-1]: + head = stack.pop() + while stack and num < stack[-1]: stack.pop() + stack.append(head) + else: stack.append(num) + return len(stack) +``` + +```java [-Java] +class Solution { + public int maxChunksToSorted(int[] arr) { + LinkedList stack = new LinkedList(); + for(int num : arr) { + if(!stack.isEmpty() && num < stack.getLast()) { + int head = stack.removeLast(); + while(!stack.isEmpty() && num < stack.getLast()) stack.removeLast(); + stack.addLast(head); + } + else stack.addLast(num); + } + return stack.size(); + } +} +``` diff --git "a/selected_coding_interview/docs/79. \345\215\225\350\257\215\346\220\234\347\264\242.md" "b/selected_coding_interview/docs/79. \345\215\225\350\257\215\346\220\234\347\264\242.md" new file mode 100644 index 0000000..d9eff42 --- /dev/null +++ "b/selected_coding_interview/docs/79. \345\215\225\350\257\215\346\220\234\347\264\242.md" @@ -0,0 +1,108 @@ +## 解题思路： + +本问题是典型的回溯问题，需要使用**深度优先搜索（DFS）+ 剪枝**解决。 + +- **深度优先搜索：** 即暴力法遍历矩阵中所有字符串可能性。DFS 通过递归，先朝一个方向搜到底，再回溯至上个节点，沿另一个方向搜索，以此类推。 +- **剪枝：** 在搜索中，遇到“这条路不可能和目标字符串匹配成功”的情况，例如当前矩阵元素和目标字符不匹配、或此元素已被访问，则应立即返回，从而避免不必要的搜索分支。 + +![Picture0.png](https://pic.leetcode-cn.com/1604944042-glmqJO-Picture0.png){:width=500} + +### 算法解析： + +- **递归参数：** 当前元素在矩阵 `board` 中的行列索引 `i` 和 `j` ，当前目标字符在 `word` 中的索引 `k` 。 +- **终止条件：** + 1. 返回 $false$ ： (1) 行或列索引越界 **或** (2) 当前矩阵元素与目标字符不同 **或** (3) 当前矩阵元素已访问过 （ (3) 可合并至 (2) ） 。 + 2. 返回 $true$ ： `k = len(word) - 1` ，即字符串 `word` 已全部匹配。 +- **递推工作：** + 1. 标记当前矩阵元素： 将 `board[i][j]` 修改为 **空字符** `''` ，代表此元素已访问过，防止之后搜索时重复访问。 + 2. 搜索下一单元格： 朝当前元素的 **上、下、左、右** 四个方向开启下层递归，使用 `或` 连接 （代表只需找到一条可行路径就直接返回，不再做后续 DFS ），并记录结果至 `res` 。 + 3. 还原当前矩阵元素： 将 `board[i][j]` 元素还原至初始值，即 `word[k]` 。 +- **返回值：** 返回布尔量 `res` ，代表是否搜索到目标字符串。 + +> 使用空字符（Python: `''` , Java/C++: `'\0'` ）做标记是为了防止标记字符与矩阵原有字符重复。当存在重复时，此算法会将矩阵原有字符认作标记字符，从而出现错误。 + + + +## 代码： + +```Python [] +class Solution: + def exist(self, board: List[List[str]], word: str) -> bool: + def dfs(i, j, k): + if not 0 <= i < len(board) or not 0 <= j < len(board[0]) or board[i][j] != word[k]: return False + if k == len(word) - 1: return True + board[i][j] = '' + res = dfs(i + 1, j, k + 1) or dfs(i - 1, j, k + 1) or dfs(i, j + 1, k + 1) or dfs(i, j - 1, k + 1) + board[i][j] = word[k] + return res + + for i in range(len(board)): + for j in range(len(board[0])): + if dfs(i, j, 0): return True + return False +``` + +```Java [] +class Solution { + public boolean exist(char[][] board, String word) { + char[] words = word.toCharArray(); + for(int i = 0; i < board.length; i++) { + for(int j = 0; j < board[0].length; j++) { + if (dfs(board, words, i, j, 0)) return true; + } + } + return false; + } + boolean dfs(char[][] board, char[] word, int i, int j, int k) { + if (i >= board.length || i < 0 || j >= board[0].length || j < 0 || board[i][j] != word[k]) return false; + if (k == word.length - 1) return true; + board[i][j] = '\0'; + boolean res = dfs(board, word, i + 1, j, k + 1) || dfs(board, word, i - 1, j, k + 1) || + dfs(board, word, i, j + 1, k + 1) || dfs(board, word, i , j - 1, k + 1); + board[i][j] = word[k]; + return res; + } +} +``` + +```C++ [] +class Solution { +public: + bool exist(vector>& board, string word) { + rows = board.size(); + cols = board[0].size(); + for(int i = 0; i < rows; i++) { + for(int j = 0; j < cols; j++) { + if (dfs(board, word, i, j, 0)) return true; + } + } + return false; + } +private: + int rows, cols; + bool dfs(vector>& board, string word, int i, int j, int k) { + if (i >= rows || i < 0 || j >= cols || j < 0 || board[i][j] != word[k]) return false; + if (k == word.size() - 1) return true; + board[i][j] = '\0'; + bool res = dfs(board, word, i + 1, j, k + 1) || dfs(board, word, i - 1, j, k + 1) || + dfs(board, word, i, j + 1, k + 1) || dfs(board, word, i , j - 1, k + 1); + board[i][j] = word[k]; + return res; + } +}; +``` + +### 复杂度分析： + +在代码中，$M, N$ 分别为矩阵行列大小， $K$ 为字符串 `word` 长度。 + +- **时间复杂度 $O(3^KMN)$ ：** 最差情况下，需要遍历矩阵中长度为 $K$ 字符串的所有方案，时间复杂度为 $O(3^K)$；矩阵中共有 $MN$ 个起点，时间复杂度为 $O(MN)$ 。 + - **方案数计算：** 设字符串长度为 $K$ ，搜索中每个字符有上、下、左、右四个方向可以选择，舍弃回头（上个字符）的方向，剩下 $3$ 种选择，因此方案数的复杂度为 $O(3^K)$ 。 +- **空间复杂度 $O(K)$ ：** 搜索过程中的递归深度不超过 $K$ ，因此系统因函数调用累计使用的栈空间占用 $O(K)$ （因为函数返回后，系统调用的[栈空间会释放](https://leetcode-cn.com/explore/orignial/card/recursion-i/259/complexity-analysis/1223/)）。最坏情况下 $K = MN$ ，递归深度为 $MN$ ，此时系统栈使用 $O(MN)$ 的额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/796. \346\227\213\350\275\254\345\255\227\347\254\246\344\270\262.md" "b/selected_coding_interview/docs/796. \346\227\213\350\275\254\345\255\227\347\254\246\344\270\262.md" new file mode 100644 index 0000000..c81cece --- /dev/null +++ "b/selected_coding_interview/docs/796. \346\227\213\350\275\254\345\255\227\347\254\246\344\270\262.md" @@ -0,0 +1,61 @@ +## 解题思路: + +输入一个字符串 $s$ ，做如下操作： + +1. 选择任意位置，将字符串切分为两个子字符串 $s = L \ R$ ； +2. 将 $R$ 移动至 $L$ 前面得到 $goal = R \ L$ ； + +此时，称 $goal$ 为 $s$ 的一个「旋转字符串」。 + +![ccw-01-09.001-004.gif](https://pic.leetcode-cn.com/1638612131-zkecdf-ccw-01-09.001-004.gif) + +如下图所示，根据旋转字符串特点，若构造一个拼接字符串 $goal \ goal$ ，则有 $goal \ goal = R \ L \ R \ L = R \ s \ L$ ，即拼接字符串 $goal \ goal$ 中**包含**原字符串 $s$ 。因此，$goal$ 为 $s$ 的旋转字符串的「充要条件」为： + +- 字符串 $s$ , $goal$ 的长度相等； +- 拼接字符串 $goal \ goal$ 中**包含**原字符串 $s$ ； + +![ccw-01-09.005.png](https://pic.leetcode-cn.com/1638612131-tGESRo-ccw-01-09.005.png) + +## 代码: + +```Python [] +class Solution: + def rotateString(self, s: str, goal: str) -> bool: + return len(s) == len(goal) and s in (goal + goal) +``` + +```Java [] +class Solution { + public boolean rotateString(String s, String goal) { + return s.length() == goal.length() && (goal + goal).contains(s); + } +} +``` + +```C++ [] +class Solution { +public: + bool rotateString(string s, string goal) { + return s.length() == goal.length() && (goal + goal).find(s) != -1; + } +}; +``` + +### 复杂度分析: + +**时间复杂度：** 设字符串 $s$ , $goal$ 的长度都为 $N$ 。 + +- 「暴力匹配」需要分别以 $goal$ 前 $N$ 个字符为起始点，遍历匹配 $s$ ，总体时间复杂度为 $O(N^2)$ 。 +- 「子串匹配 KMP 算法」的时间复杂度为 $O(N)$ 。 +- 「Boyer–Moore string-search algorithm」时间复杂度为 $O(N)$ 。 + +> 本文直接调用编程语言的库函数，时间复杂度由库函数的具体实现方法确定。 + +**空间复杂度 $O(N)$ ：** 构造拼接字符串 $goal \ goal$ 使用 $O(N)$ 大小的额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/8. \345\255\227\347\254\246\344\270\262\350\275\254\346\215\242\346\225\264\346\225\260 (atoi).md" "b/selected_coding_interview/docs/8. \345\255\227\347\254\246\344\270\262\350\275\254\346\215\242\346\225\264\346\225\260 (atoi).md" new file mode 100644 index 0000000..dc5a92f --- /dev/null +++ "b/selected_coding_interview/docs/8. \345\255\227\347\254\246\344\270\262\350\275\254\346\215\242\346\225\264\346\225\260 (atoi).md" @@ -0,0 +1,150 @@ +## 解题思路： + +根据题意，有以下四种字符需要考虑： + +1. **首部空格：** 删除之即可。 +2. **符号位：** 三种情况，即 ''$+$'' , ''$-$'' , ''无符号" ；新建一个变量保存符号位，返回前判断正负即可。 +3. **非数字字符：** 遇到首个非数字的字符时，应立即返回。 +4. **数字字符：** + 1. **字符转数字：** “此数字的 ASCII 码” 与 “ $0$ 的 ASCII 码” 相减即可。 + 2. **数字拼接：** 若从左向右遍历数字，设当前位字符为 $c$ ，当前位数字为 $x$ ，数字结果为 $res$ ，则数字拼接公式为： + +$$ +res = 10 \times res + x \\ +x = ascii(c) - ascii('0') +$$ + +![Picture1.png](https://pic.leetcode-cn.com/1600793383-jCgsGU-Picture1.png){:width=450} + +**数字越界处理：** + +> 题目要求返回的数值范围应在 $[-2^{31}, 2^{31} - 1]$ ，因此需要考虑数字越界问题。而由于题目指出 `环境只能存储 32 位大小的有符号整数` ，因此判断数字越界时，要始终保持 $res$ 在 int 类型的取值范围内。 + +在每轮数字拼接前，判断 $res$ **在此轮拼接后是否超过 $2147483647$** ，若超过则加上符号位直接返回。 +设数字拼接边界 $bndry = 2147483647 // 10 = 214748364$ ，则以下两种情况越界： + +$$ +\begin{cases} + res > bndry & 情况一：执行拼接 10 \times res \geq 2147483650 越界 \\ + res = bndry, x > 7 & 情况二：拼接后是 2147483648 或 2147483649 越界 \\ +\end{cases} +$$ + +![Picture2.png](https://pic.leetcode-cn.com/1600793383-JZRYip-Picture2.png){:width=450} + +下图展示了一个转化示例。 + + + +## 代码： + +```Python [] +class Solution: + def myAtoi(self, s: str) -> int: + s = s.strip() # 删除首尾空格 + if not s: return 0 # 字符串为空则直接返回 + res, i, sign = 0, 1, 1 + int_max, int_min, bndry = 2 ** 31 - 1, -2 ** 31, 2 ** 31 // 10 + if s[0] == '-': sign = -1 # 保存负号 + elif s[0] != '+': i = 0 # 若无符号位，则需从 i = 0 开始数字拼接 + for c in s[i:]: + if not '0' <= c <= '9' : break # 遇到非数字的字符则跳出 + if res > bndry or res == bndry and c > '7': return int_max if sign == 1 else int_min # 数字越界处理 + res = 10 * res + ord(c) - ord('0') # 数字拼接 + return sign * res +``` + +```Java [] +class Solution { + public int myAtoi(String s) { + char[] c = s.trim().toCharArray(); + if (c.length == 0) return 0; + int res = 0, bndry = Integer.MAX_VALUE / 10; + int i = 1, sign = 1; + if (c[0] == '-') sign = -1; + else if (c[0] != '+') i = 0; + for (int j = i; j < c.length; j++) { + if (c[j] < '0' || c[j] > '9') break; + if (res > bndry || res == bndry && c[j] > '7') return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE; + res = res * 10 + (c[j] - '0'); + } + return sign * res; + } +} +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为字符串长度，线性遍历字符串占用 $O(N)$ 时间。 +- **空间复杂度 $O(N)$ ：** 删除首尾空格后需建立新字符串，最差情况下占用 $O(N)$ 额外空间。 + +若不使用 `trim() / strip()` 删除首部空格，而采取遍历跳过空格的方式，则可以将空间复杂度降低至 $O(1)$ ，代码如下： + +```Python [] +class Solution: + def myAtoi(self, s: str) -> int: + res, i, sign, length = 0, 0, 1, len(s) + int_max, int_min, bndry = 2 ** 31 - 1, -2 ** 31, 2 ** 31 // 10 + if not s: return 0 # 空字符串，提前返回 + while s[i] == ' ': + i += 1 + if i == length: return 0 # 字符串全为空格，提前返回 + if s[i] == '-': sign = -1 + if s[i] in '+-': i += 1 + for j in range(i, length): + if not '0' <= s[j] <= '9' : break + if res > bndry or res == bndry and s[j] > '7': + return int_max if sign == 1 else int_min + res = 10 * res + ord(s[j]) - ord('0') + return sign * res +``` + +```Java [] +class Solution { + public int myAtoi(String s) { + int res = 0, bndry = Integer.MAX_VALUE / 10; + int i = 0, sign = 1, length = s.length(); + if(length == 0) return 0; + while(s.charAt(i) == ' ') + if(++i == length) return 0; + if(s.charAt(i) == '-') sign = -1; + if(s.charAt(i) == '-' || s.charAt(i) == '+') i++; + for(int j = i; j < length; j++) { + if(s.charAt(j) < '0' || s.charAt(j) > '9') break; + if(res > bndry || res == bndry && s.charAt(j) > '7') + return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE; + res = res * 10 + (s.charAt(j) - '0'); + } + return sign * res; + } +} +``` + +```C++ [] +class Solution { +public: + int myAtoi(string s) { + int res = 0, bndry = INT_MAX / 10; + int i = 0, sign = 1, length = s.size(); + if(length == 0) return 0; + while(s[i] == ' ') + if(++i == length) return 0; + if(s[i] == '-') sign = -1; + if(s[i] == '-' || s[i] == '+') i++; + for(int j = i; j < length; j++) { + if(s[j] < '0' || s[j] > '9') break; + if(res > bndry || res == bndry && s[j] > '7') + return sign == 1 ? INT_MAX : INT_MIN; + res = res * 10 + (s[j] - '0'); + } + return sign * res; + } +}; +``` + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/86. \345\210\206\351\232\224\351\223\276\350\241\250.md" "b/selected_coding_interview/docs/86. \345\210\206\351\232\224\351\223\276\350\241\250.md" new file mode 100644 index 0000000..644819f --- /dev/null +++ "b/selected_coding_interview/docs/86. \345\210\206\351\232\224\351\223\276\350\241\250.md" @@ -0,0 +1,174 @@ +## 解题思路： + +如下图所示，题目要求实现链表所有「值 $< x$ 节点」出现在「值 $\geq x$ 节点」前面。 + +![ccw-02-04.001.png](https://pic.leetcode-cn.com/1642327155-ZeSkst-ccw-02-04.001.png) + +根据题意，考虑通过「新建两个链表」实现原链表分割，算法流程为： + +1. 新建两个链表 `sml_dummy` , `big_dummy` ，分别用于添加所有「节点值 $< x$ 」、「节点值 $\geq x$ 」的节点。 +2. 遍历链表 `head` 并依次比较各节点值 `head.val` 和 $x$ 的大小： + 1. 若 `head.val < x` ，则将节点 `head` 添加至链表 `sml_dummy` 最后面； + 2. 若 `head.val >= x` ，则将节点 `head` 添加至链表 `big_dummy` 最后面； +3. 遍历完成后，拼接 `sml_dummy` 和 `big_dummy` 链表。 +4. 最终返回头节点 `sml_dummy.next` 即可。 + + + +## 代码： + +> 后三个 Tab 为「代码注释解析」。 + +```Python [] +class Solution: + def partition(self, head: Optional[ListNode], x: int) -> Optional[ListNode]: + sml_dummy, big_dummy = ListNode(0), ListNode(0) + sml, big = sml_dummy, big_dummy + while head: + if head.val < x: + sml.next = head + sml = sml.next + else: + big.next = head + big = big.next + head = head.next + sml.next = big_dummy.next + big.next = None + return sml_dummy.next +``` + +```Java [] +class Solution { + public ListNode partition(ListNode head, int x) { + ListNode smlDummy = new ListNode(0), bigDummy = new ListNode(0); + ListNode sml = smlDummy, big = bigDummy; + while (head != null) { + if (head.val < x) { + sml.next = head; + sml = sml.next; + } else { + big.next = head; + big = big.next; + } + head = head.next; + } + sml.next = bigDummy.next; + big.next = null; + return smlDummy.next; + } +} +``` + +```C++ [] +class Solution { +public: + ListNode* partition(ListNode* head, int x) { + ListNode *smlDummy = new ListNode(0), *bigDummy = new ListNode(0); + ListNode *sml = smlDummy, *big = bigDummy; + while (head != nullptr) { + if (head->val < x) { + sml->next = head; + sml = sml->next; + } else { + big->next = head; + big = big->next; + } + head = head->next; + } + sml->next = bigDummy->next; + big->next = nullptr; + return smlDummy->next; + } +}; +``` + +```Python [] +class Solution: + def partition(self, head: Optional[ListNode], x: int) -> Optional[ListNode]: + # 新建两个链表 + sml_dummy, big_dummy = ListNode(0), ListNode(0) + # 遍历链表 + sml, big = sml_dummy, big_dummy + while head: + # 将 < x 的节点加入 sml 节点后 + if head.val < x: + sml.next = head + sml = sml.next + # 将 >= x 的节点加入 big 节点后 + else: + big.next = head + big = big.next + head = head.next + # 拼接两链表 + sml.next = big_dummy.next + big.next = None + return sml_dummy.next +``` + +```Java [] +class Solution { + public ListNode partition(ListNode head, int x) { + // 新建两个链表 + ListNode smlDummy = new ListNode(0), bigDummy = new ListNode(0); + // 遍历链表 + ListNode sml = smlDummy, big = bigDummy; + while (head != null) { + // 将 < x 的节点加入 sml 节点后 + if (head.val < x) { + sml.next = head; + sml = sml.next; + // 将 >= x 的节点加入 big 节点后 + } else { + big.next = head; + big = big.next; + } + head = head.next; + } + // 拼接两链表 + sml.next = bigDummy.next; + big.next = null; + return smlDummy.next; + } +} +``` + +```C++ [] +class Solution { +public: + ListNode* partition(ListNode* head, int x) { + // 新建两个链表 + ListNode *smlDummy = new ListNode(0), *bigDummy = new ListNode(0); + // 遍历链表 + ListNode *sml = smlDummy, *big = bigDummy; + while (head != nullptr) { + // 将 < x 的节点加入 sml 节点后 + if (head->val < x) { + sml->next = head; + sml = sml->next; + // 将 >= x 的节点加入 big 节点后 + } else { + big->next = head; + big = big->next; + } + head = head->next; + } + // 拼接两链表 + sml->next = bigDummy->next; + big->next = nullptr; + return smlDummy->next; + } +}; +``` + +### 复杂度分析： + +**时间复杂度 $O(N)$ ：** 其中 $N$ 为链表长度；遍历链表使用线性时间。 + +**空间复杂度 $O(1)$ ：** 假头节点使用常数大小的额外空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/876. \351\223\276\350\241\250\347\232\204\344\270\255\351\227\264\347\273\223\347\202\271.md" "b/selected_coding_interview/docs/876. \351\223\276\350\241\250\347\232\204\344\270\255\351\227\264\347\273\223\347\202\271.md" new file mode 100644 index 0000000..7a43e02 --- /dev/null +++ "b/selected_coding_interview/docs/876. \351\223\276\350\241\250\347\232\204\344\270\255\351\227\264\347\273\223\347\202\271.md" @@ -0,0 +1,67 @@ +## 解题思路 + +考虑借助快慢双指针 `fast`, `slow` ，「快指针 `fast`」每轮走 2 步，「慢指针 `slow`」每轮走 1 步。`fast` 的步数恒为 `slow` 的 2 倍，因此当快指针遍历完链表时，慢指针就指向链表中间节点。而由于长度为偶数的链表**有两个中间节点**，因此需要分两种情况考虑： + +- **链表长度为奇数：** 当 `fast` 走到链表「尾节点」时，`slow` 正好走到「中间节点」。 +- **链表长度为偶数：** 当 `fast` 走到「null」时（越过「尾节点」后），`slow` 正好走到「第二个中间节点」。 + +总结以上规律，应在当 `fast` **遇到或越过尾节点** 时跳出循环，并返回 `slow` 即可。 + +![figures.gif](https://pic.leetcode-cn.com/1656953441-Kshqch-figures.gif) + +> 上为动态图，下为静态图，内容一致。 + + + +## 代码 + +**拓展思考：** 若题目要求返回「第一个中间节点」，则应在 `fast` **遇到尾节点或其前驱节点** 时跳出循环。此时，修改判断条件为 `while fast.next and fast.next.next` 即可。 + +```Python [] +class Solution: + def middleNode(self, head: ListNode) -> ListNode: + fast = slow = head + while fast and fast.next: + fast = fast.next.next + slow = slow.next + return slow +``` + +```Java [] +class Solution { + public ListNode middleNode(ListNode head) { + ListNode fast = head, slow = head; + while (fast != null && fast.next != null) { + fast = fast.next.next; + slow = slow.next; + } + return slow; + } +} +``` + +```C++ [] +class Solution { +public: + ListNode* middleNode(ListNode* head) { + ListNode *fast = head, *slow = head; + while (fast != nullptr && fast->next != nullptr) { + fast = fast->next->next; + slow = slow->next; + } + return slow; + } +}; +``` + +## 复杂度分析 + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为链表长度。 `fast` 遍历完链表需 $\frac{N}{2}$ 轮迭代，使用 $O(\frac{N}{2}) = O(N)$ 线性时间。 +- **空间复杂度 $O(1)$ ：** 节点指针 `fast` , `slow` 使用常数大小空间。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/89. \346\240\274\351\233\267\347\274\226\347\240\201.md" "b/selected_coding_interview/docs/89. \346\240\274\351\233\267\347\274\226\347\240\201.md" new file mode 100644 index 0000000..9f158f0 --- /dev/null +++ "b/selected_coding_interview/docs/89. \346\240\274\351\233\267\347\274\226\347\240\201.md" @@ -0,0 +1,40 @@ +#### 思路： +- 设 $n$ 阶格雷码集合为 $G(n)$，则 $G(n+1)$ 阶格雷码为： + - 给 $G(n)$ 阶格雷码每个元素二进制形式前面添加 $0$，得到 $G'(n)$； + - 设 $G(n)$ 集合倒序（镜像）为 $R(n)$，给 $R(n)$ 每个元素二进制形式前面添加 $1$，得到 $R'(n)$； + - $G(n+1) = G'(n) ∪ R'(n)$ 拼接两个集合即可得到下一阶格雷码。 +- 根据以上规律，可从 $0$ 阶格雷码推导致任何阶格雷码。 +- 代码解析： + - 由于最高位前默认为 $0$，因此 $G'(n) = G(n)$，只需在 `res`(即 $G(n)$ )后添加 $R'(n)$ 即可； + - 计算 $R'(n)$：执行 `head = 1 << i` 计算出对应位数，以给 $R(n)$ 前添加 $1$ 得到对应 $R'(n)$； + - 倒序遍历 `res`(即 $G(n)$ )：依次求得 $R'(n)$ 各元素添加至 `res` 尾端，遍历完成后 `res`(即 $G(n+1)$)。 + +#### 图解： + + + +#### 代码： +```Python [] +class Solution: + def grayCode(self, n: int) -> List[int]: + res, head = [0], 1 + for i in range(n): + for j in range(len(res) - 1, -1, -1): + res.append(head + res[j]) + head <<= 1 + return res +``` +```Java [] +class Solution { + public List grayCode(int n) { + List res = new ArrayList() {{ add(0); }}; + int head = 1; + for (int i = 0; i < n; i++) { + for (int j = res.size() - 1; j >= 0; j--) + res.add(head + res.get(j)); + head <<= 1; + } + return res; + } +} +``` diff --git "a/selected_coding_interview/docs/946. \351\252\214\350\257\201\346\240\210\345\272\217\345\210\227.md" "b/selected_coding_interview/docs/946. \351\252\214\350\257\201\346\240\210\345\272\217\345\210\227.md" new file mode 100644 index 0000000..4edd17e --- /dev/null +++ "b/selected_coding_interview/docs/946. \351\252\214\350\257\201\346\240\210\345\272\217\345\210\227.md" @@ -0,0 +1,89 @@ +## 解题思路： + +如下图所示，给定一个压入序列 $pushed$ 和弹出序列 $popped$ ，则压入 / 弹出操作的顺序（即排列）是 **唯一确定** 的。 + +![Picture1.png](https://pic.leetcode-cn.com/1188474dc6a377fc258004bca84e5a130b663eeb24bf22c4fa4eb998a0249f97-Picture1.png){:width=500} + +如下图所示，栈的数据操作具有 **先入后出** 的特性，因此某些弹出序列是无法实现的。 + +![Picture2.png](https://pic.leetcode-cn.com/3f43b224553bf3a37f9758dbb41655e547795e293524a148380c6f335af315e4-Picture2.png){:width=500} + +考虑借用一个辅助栈 $stack$ ，**模拟** 压入 / 弹出操作的排列。根据是否模拟成功，即可得到结果。 + +- **入栈操作：** 按照压栈序列的顺序执行。 +- **出栈操作：** 每次入栈后，循环判断 “栈顶元素 $=$ 弹出序列的当前元素” 是否成立，将符合弹出序列顺序的栈顶元素全部弹出。 + +> 由于题目规定 `栈的所有数字均不相等` ，因此在循环入栈中，每个元素出栈的位置的可能性是唯一的（若有重复数字，则具有多个可出栈的位置）。因而，在遇到 “栈顶元素 $=$ 弹出序列的当前元素” 就应立即执行出栈。 + +### 算法流程： + +1. **初始化：** 辅助栈 $stack$ ，弹出序列的索引 $i$ 。 +2. **遍历压栈序列：** 各元素记为 $num$ 。 + 1. 元素 $num$ 入栈。 + 2. 循环出栈：若 $stack$ 的栈顶元素 $=$ 弹出序列元素 $popped[i]$ ，则执行出栈与 $i++$ 。 +3. **返回值：** 若 $stack$ 为空，则此弹出序列合法。 + + + +## 代码： + +题目指出“ $pushed$ 一定是 $popped$ 的排列”。因此，无需考虑 $pushed$ 和 $popped$ **长度不同** 或 **包含元素不同** 的情况。 + +```Python [] +class Solution: + def validateStackSequences(self, pushed: List[int], popped: List[int]) -> bool: + stack, i = [], 0 + for num in pushed: + stack.append(num) # num 入栈 + while stack and stack[-1] == popped[i]: # 循环判断与出栈 + stack.pop() + i += 1 + return not stack +``` + +```Java [] +class Solution { + public boolean validateStackSequences(int[] pushed, int[] popped) { + Stack stack = new Stack<>(); + int i = 0; + for (int num : pushed) { + stack.push(num); // num 入栈 + while (!stack.isEmpty() && stack.peek() == popped[i]) { // 循环判断与出栈 + stack.pop(); + i++; + } + } + return stack.isEmpty(); + } +} +``` + +```C++ [] +class Solution { +public: + bool validateStackSequences(vector& pushed, vector& popped) { + stack stk; + int i = 0; + for (int num : pushed) { + stk.push(num); // num 入栈 + while (!stk.empty() && stk.top() == popped[i]) { // 循环判断与出栈 + stk.pop(); + i++; + } + } + return stk.empty(); + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 其中 $N$ 为列表 $pushed$ 的长度；每个元素最多入栈与出栈一次，即最多共 $2N$ 次出入栈操作。 +- **空间复杂度 $O(N)$ ：** 辅助栈 $stack$ 最多同时存储 $N$ 个元素。 + +--- + +**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** + +- 代码与测试样例，方便 VSCode 本地运行与调试代码。 +- 数据结构封装，提升刷题与 Debug 效率。 From 759b1d488da9cd56c2bb9576bd05ddc3c5e51f46 Mon Sep 17 00:00:00 2001 From: krahets Date: Tue, 10 Oct 2023 21:45:16 +0800 Subject: [PATCH 13/23] Update README and gitignore --- .gitignore | 3 +- README.md | 50 +++++--- ...67\351\242\230\350\256\241\345\210\222.md" | 119 ++++++++++++++++++ ...30\347\233\256\345\210\206\347\261\273.md" | 81 ++++++++++++ 4 files changed, 233 insertions(+), 20 deletions(-) create mode 100755 "sword_for_offer/\345\211\221\346\214\207 Offer \345\210\267\351\242\230\350\256\241\345\210\222.md" create mode 100755 "sword_for_offer/\345\211\221\346\214\207 Offer \351\242\230\347\233\256\345\210\206\347\261\273.md" diff --git a/.gitignore b/.gitignore index 4b64967..4069360 100644 --- a/.gitignore +++ b/.gitignore @@ -1,3 +1,4 @@ __pycache__ -.vscode .DS_Store +.vscode +*.xlsx diff --git a/README.md b/README.md index de7f97e..5258c3d 100755 --- a/README.md +++ b/README.md @@ -1,10 +1,10 @@ -

+

-

+

@@ -13,39 +13,51 @@ [《图解算法数据结构》](https://leetcode-cn.com/leetbook/detail/illustration-of-algorithm/)是一本面向算法初学者和互联网求职者编写的 LeetBook 手册。 -- 图文详解 75 道题目，覆盖主要算法知识点，非常适合作为算法学习的第一份题库。 +- 图文详解 75 道题目，覆盖主要算法知识点。 - 题目活跃于各大互联网公司招聘中，可使笔面试准备事半功倍。 - 致力于行文深入浅出、图文搭配，提供简洁的 Python, Java, C++ 解题代码。 > 若本 LeetBook 对您有所帮助，请在页面右上角点个 Star :star: 支持一下，谢谢！ +本仓库包含“图解算法数据结构”、“Krahets 笔面试精选 88 题”和“剑指 Offer”的题解内容，所有代码皆可一键运行： + +```shell +LeetCode-Book +├── leetbook_ioa # 《图解算法数据结构》题解和专栏文档、代码 +├── selected_coding_interview # 「Krahets 笔面试精选 88 题」题解文档、代码 +└── sword_for_offer # 《剑指 Offer》题解文档、代码、刷题计划 +``` + ## 如何入门算法？ ### 第一步：看入门书 -

- +

+

-

- 《Hello 算法》：动画图解、一键运行的数据结构与算法教程 -

+[《Hello 算法》](https://github.com/krahets/hello-algo)：动画图解、一键运行的数据结构与算法教程。 -

- - 前往仓库 > - -     - - 在线阅读 > - -

+这本书旨在引导初学者探索数据结构与算法的知识地图，掌握完整的刷题工具库。 ### 第二步：刷算法题 -1. [Krahets 笔面试精选 88 题](https://leetcode.cn/studyplan/selected-coding-interview/)：由“剑指 Offer”和“Hot 100”组成的高频面试题单。 -2. [图解算法数据结构](https://leetcode-cn.com/leetbook/detail/illustration-of-algorithm/)或[Interview-75](https://leetcode.cn/studyplan/coding-interviews/)：题目更贴近实际应用，相对“剑指 Offer”难度有所增加。 +推荐以下 LeetCode 题单（前两项 [@Krahets](https://leetcode.cn/u/jyd/) 题解全覆盖）： + +1. [Krahets 笔面试精选 88 题](https://leetcode.cn/studyplan/selected-coding-interview/)：从“剑指 Offer”和“热题 100”精选出的 88 道高频算法笔试题。 +2. [图解算法数据结构](https://leetcode-cn.com/leetbook/detail/illustration-of-algorithm/)（[Interview-75](https://leetcode.cn/studyplan/coding-interviews/)）：题目更贴近实际应用，相对“剑指 Offer”难度有所增加。 +3. [LeetCode 热题 100](https://leetcode.cn/studyplan/top-100-liked/)：力扣用户最喜爱的 100 道题。 +4. [面试经典 150 题](https://leetcode.cn/studyplan/top-interview-150/)：150 道经典面试力扣题。 +5. [LeetCode-75](https://leetcode.cn/studyplan/leetcode-75/)：精选面试必备的 75 道核心题目。 + +对初学者的刷题建议： + +1. 建议每日刷 2~3 题。若能轻松完成，可以尝试增加至 5～8 题。请注意，刷题的质量远重要于数量。务必确保你真正理解了每个题目的解法及背后的算法原理。 +2. 建议你按照目录顺序逐题解答。如果碰到某些难以解决的题目，可以先跳过，稍后回顾时再挑战。 +3. 很多题目都有不止一种解法，请你注意比较和探讨各种方法的特点和适用情况。 +4. 如果你发现自己忘记了某个题目的解法，不必灰心。艾宾浩斯遗忘曲线指出，通常需要复习至少三次，才能真正掌握一个知识点。 +5. 行百里者半九十。坚持至关重要，加油，相信你可以做到！ ## License diff --git "a/sword_for_offer/\345\211\221\346\214\207 Offer \345\210\267\351\242\230\350\256\241\345\210\222.md" "b/sword_for_offer/\345\211\221\346\214\207 Offer \345\210\267\351\242\230\350\256\241\345\210\222.md" new file mode 100755 index 0000000..5f112ff --- /dev/null +++ "b/sword_for_offer/\345\211\221\346\214\207 Offer \345\210\267\351\242\230\350\256\241\345\210\222.md" @@ -0,0 +1,119 @@ +# 剑指 Offer 刷题计划 + +剑指 Offer 系列题目非常适合作为算法入门题目集，尤其适合有互联网求职需求的同学们，能够让我们全面练习，快速搭建起算法与数据结构的知识框架。 + +- 算法： 动态规划、回溯算法、查找算法、搜索算法、贪心算法、分治算法、位运算、双指针、排序、模拟、数学、…… +- 数据结构： 数组、栈、队列、字符串、链表、树、图、堆、哈希表、…… + +笔者整理了本刷题计划，核心理念为从易到难、从基础类题目到综合类题目，供希望按照知识点类型顺序刷题的小伙伴们参考。 + +> 行百里者半九十。坚持一个月刷完，一起加油！ + +| 日程 | 题目 | +| :--------: | :-------------------------------------------- | +| **Day 1** | **栈与队列（简单）** | +| | 剑指 Offer 09. 用两个栈实现队列 | +| | 剑指 Offer 30. 包含 min 函数的栈 | +| **Day 2** | **链表（简单）** | +| | 剑指 Offer 06. 从尾到头打印链表 | +| | 剑指 Offer 24. 反转链表 | +| | 剑指 Offer 35. 复杂链表的复制 | +| **Day 3** | **字符串（简单）** | +| | 剑指 Offer 05. 替换空格 | +| | 剑指 Offer 58 - II. 左旋转字符串 | +| **Day 4** | **查找算法（简单）** | +| | 剑指 Offer 03. 数组中重复的数字 | +| | 剑指 Offer 53 - I. 在排序数组中查找数字 I | +| | 剑指 Offer 53 - II. 0～n-1 中缺失的数字 | +| **Day 5** | **查找算法（中等）** | +| | 剑指 Offer 04. 二维数组中的查找 | +| | 剑指 Offer 11. 旋转数组的最小数字 | +| | 剑指 Offer 50. 第一个只出现一次的字符 | +| **Day 6** | **搜索与回溯算法（简单）** | +| | 剑指 Offer 32 - I. 从上到下打印二叉树 | +| | 剑指 Offer 32 - II. 从上到下打印二叉树 II | +| | 剑指 Offer 32 - III. 从上到下打印二叉树 III | +| **Day 7** | **搜索与回溯算法（简单）** | +| | 剑指 Offer 26. 树的子结构 | +| | 剑指 Offer 27. 二叉树的镜像 | +| | 剑指 Offer 28. 对称的二叉树 | +| **Day 8** | **动态规划（简单）** | +| | 剑指 Offer 10- I. 斐波那契数列 | +| | 剑指 Offer 10- II. 青蛙跳台阶问题 | +| | 剑指 Offer 63. 股票的最大利润 | +| **Day 9** | **动态规划（中等）** | +| | 剑指 Offer 42. 连续子数组的最大和 | +| | 剑指 Offer 47. 礼物的最大价值 | +| **Day 10** | **动态规划（中等）** | +| | 剑指 Offer 46. 把数字翻译成字符串 | +| | 剑指 Offer 48. 最长不含重复字符的子字符串 | +| **Day 11** | **双指针（简单）** | +| | 剑指 Offer 18. 删除链表的节点 | +| | 剑指 Offer 22. 链表中倒数第 k 个节点 | +| **Day 12** | **双指针（简单）** | +| | 剑指 Offer 25. 合并两个排序的链表 | +| | 剑指 Offer 52. 两个链表的第一个公共节点 | +| **Day 13** | **双指针（简单）** | +| | 剑指 Offer 21. 调整数组顺序使奇数位于偶数前面 | +| | 剑指 Offer 57. 和为 s 的两个数字 | +| | 剑指 Offer 58 - I. 翻转单词顺序 | +| **Day 14** | **搜索与回溯算法（中等）** | +| | 剑指 Offer 12. 矩阵中的路径 | +| | 剑指 Offer 13. 机器人的运动范围 | +| **Day 15** | **搜索与回溯算法（中等）** | +| | 剑指 Offer 34. 二叉树中和为某一值的路径 | +| | 剑指 Offer 36. 二叉搜索树与双向链表 | +| | 剑指 Offer 54. 二叉搜索树的第 k 大节点 | +| **Day 16** | **排序（简单）** | +| | 剑指 Offer 45. 把数组排成最小的数 | +| | 剑指 Offer 61. 扑克牌中的顺子 | +| **Day 17** | **排序（中等）** | +| | 剑指 Offer 40. 最小的 k 个数 | +| | 剑指 Offer 41. 数据流中的中位数 | +| **Day 18** | **搜索与回溯算法（中等）** | +| | 剑指 Offer 55 - I. 二叉树的深度 | +| | 剑指 Offer 55 - II. 平衡二叉树 | +| **Day 19** | **搜索与回溯算法（中等）** | +| | 剑指 Offer 64. 求 1 + 2 + … + n | +| | 剑指 Offer 68 - I. 二叉搜索树的最近公共祖先 | +| | 剑指 Offer 68 - II. 二叉树的最近公共祖先 | +| **Day 20** | **分治算法（中等）** | +| | 剑指 Offer 07. 重建二叉树 | +| | 剑指 Offer 16. 数值的整数次方 | +| | 剑指 Offer 33. 二叉搜索树的后序遍历序列 | +| **Day 21** | **位运算（简单）** | +| | 剑指 Offer 15. 二进制中 1 的个数 | +| | 剑指 Offer 65. 不用加减乘除做加法 | +| **Day 22** | **位运算（中等）** | +| | 剑指 Offer 56 - I. 数组中数字出现的次数 | +| | 剑指 Offer 56 - II. 数组中数字出现的次数 II | +| **Day 23** | **数学（简单）** | +| | 剑指 Offer 39. 数组中出现次数超过一半的数字 | +| | 剑指 Offer 66. 构建乘积数组 | +| **Day 24** | **数学（中等）** | +| | 剑指 Offer 14- I. 剪绳子 | +| | 剑指 Offer 57 - II. 和为 s 的连续正数序列 | +| | 剑指 Offer 62. 圆圈中最后剩下的数字 | +| **Day 25** | **模拟（中等）** | +| | 剑指 Offer 29. 顺时针打印矩阵 | +| | 剑指 Offer 31. 栈的压入、弹出序列 | +| **Day 26** | **字符串（中等）** | +| | 剑指 Offer 20. 表示数值的字符串 | +| | 剑指 Offer 67. 把字符串转换成整数 | +| **Day 27** | **栈与队列（困难）** | +| | 剑指 Offer 59 - I. 滑动窗口的最大值 | +| | 剑指 Offer 59 - II. 队列的最大值 | +| **Day 28** | **搜索与回溯算法（困难）** | +| | 剑指 Offer 37. 序列化二叉树 | +| | 剑指 Offer 38. 字符串的排列 | +| **Day 29** | **动态规划（困难）** | +| | 剑指 Offer 19. 正则表达式匹配 | +| | 剑指 Offer 49. 丑数 | +| | 剑指 Offer 60. n 个骰子的点数 | +| **Day 30** | **分治算法（困难）** | +| | 剑指 Offer 17. 打印从 1 到最大的 n 位数 | +| | 剑指 Offer 51. 数组中的逆序对 | +| **Day 31** | **数学（困难）** | +| | 剑指 Offer 14- II. 剪绳子 II | +| | 剑指 Offer 43. 1～n 整数中 1 出现的次数 | +| | 剑指 Offer 44. 数字序列中某一位的数字 | diff --git "a/sword_for_offer/\345\211\221\346\214\207 Offer \351\242\230\347\233\256\345\210\206\347\261\273.md" "b/sword_for_offer/\345\211\221\346\214\207 Offer \351\242\230\347\233\256\345\210\206\347\261\273.md" new file mode 100755 index 0000000..31c0b21 --- /dev/null +++ "b/sword_for_offer/\345\211\221\346\214\207 Offer \351\242\230\347\233\256\345\210\206\347\261\273.md" @@ -0,0 +1,81 @@ +# 剑指 Offer 题目分类 + +下表划分了各题使用的算法与数据结构。由于题目可能存在多种解法，本表格只列举最优解法（时间与空间复杂度最低）的对应算法。 + +| 题目 | 算法分类 | 数据结构分类 | +| ---------------------------------- | ------------------------ | --------------- | +| 03. 数组中重复的数字 | 查找算法 | 数组 | +| 04. 二维数组中的查找 | 查找算法 | 数组 | +| 05. 替换空格 | | 字符串 | +| 06. 从尾到头打印链表 | | 栈 / 队列，链表 | +| 07. 重建二叉树 | 分治算法 | 树，哈希表 | +| 09. 用两个栈实现队列 | | 栈 / 队列 | +| 10-I. 斐波那契数列 | 动态规划 | | +| 10-II. 青蛙跳台阶问题 | 动态规划 | | +| 11. 旋转数组的最小数字 | 查找算法 | 数组 | +| 12. 矩阵中的路径 | 回溯算法，搜索算法 | 数组，图 | +| 13. 机器人的运动范围 | 回溯算法，搜索算法 | 数组，图 | +| 14-I. 剪绳子 | 贪心算法，数学 | | +| 14- II. 剪绳子 II | 贪心算法，分治算法，数学 | | +| 15. 二进制中 1 的个数 | 位运算 | | +| 16. 数值的整数次方 | 分治算法，位运算 | | +| 17. 打印从 1 到最大的 n 位数 | | 数组 | +| 18. 删除链表的节点 | 双指针 | 链表 | +| 19. 正则表达式匹配 | 动态规划 | 字符串 | +| 20. 表示数值的字符串 | | 字符串 | +| 21. 调整数组顺序使奇数位于偶数前面 | 双指针 | 数组 | +| 22. 链表中倒数第 k 个节点 | 双指针 | 链表 | +| 24. 反转链表 | 双指针 | 链表 | +| 25. 合并两个排序的链表 | 双指针 | 链表 | +| 26. 树的子结构 | 搜索算法 | 树 | +| 27. 二叉树的镜像 | 搜索算法 | 栈 / 队列，树 | +| 28. 对称的二叉树 | 搜索算法 | 树 | +| 29. 顺时针打印矩阵 | 模拟 | 数组 | +| 30. 包含 min 函数的栈 | 排序 | 栈 / 队列 | +| 31. 栈的压入、弹出序列 | 模拟 | 栈 / 队列 | +| 32-I. 从上到下打印二叉树 | 搜索算法 | 栈 / 队列，树 | +| 32-II. 从上到下打印二叉树 II | 搜索算法 | 栈 / 队列，树 | +| 32-III. 从上到下打印二叉树 III | 搜索算法 | 栈 / 队列，树 | +| 33. 二叉搜索树的后序遍历序列 | 分治算法 | 栈 / 队列，树 | +| 34. 二叉树中和为某一值的路径 | 回溯算法，搜索算法 | 树 | +| 35. 复杂链表的复制 | | 链表 | +| 36. 二叉搜索树与双向链表 | 搜索算法，双指针 | 树 | +| 37. 序列化二叉树 | 搜索算法 | 树 | +| 38. 字符串的排列 | 回溯算法 | 字符串，哈希表 | +| 39. 数组中出现次数超过一半的数字 | | 数组 | +| 40. 最小的 k 个数 | 排序 | 数组，堆 | +| 41. 数据流中的中位数 | 排序 | 堆 | +| 42. 连续子数组的最大和 | 动态规划 | 数组 | +| 43. 1 ～ n整数中 1 出现的次数 | 数学 | | +| 44. 数字序列中某一位的数字 | 数学 | | +| 45. 把数组排成最小的数 | 排序 | 字符串 | +| 46. 把数字翻译成字符串 | 动态规划 | 字符串 | +| 47. 礼物的最大价值 | 动态规划 | 数组 | +| 48. 最长不含重复字符的子字符串 | 动态规划，双指针 | 哈希表 | +| 49. 丑数 | 动态规划 | | +| 50. 第一个只出现一次的字符 | | 哈希表 | +| 51. 数组中的逆序对 | 分治算法 | 数组 | +| 52. 两个链表的第一个公共节点 | 双指针 | 链表 | +| 53-I. 在排序数组中查找数字 I | 查找算法 | 数组 | +| 53-II. 0 ～ n - 1 中缺失的数字 | 查找算法 | 数组 | +| 54. 二叉搜索树的第 k 大节点 | 搜索算法 | 树 | +| 55-I. 二叉树的深度 | 搜索算法 | 树 | +| 55-II. 平衡二叉树 | 搜索算法 | 树 | +| 56-I. 数组中数字出现的次数 | 位运算 | 数组 | +| 56-II. 数组中数字出现的次数 II | 位运算 | 数组 | +| 57. 和为 s 的两个数字 | 双指针 | 数组 | +| 57-II. 和为 s 的连续正数序列 | 双指针 | 数组 | +| 58-I. 翻转单词顺序 | 双指针 | 字符串 | +| 58-II. 左旋转字符串 | | 字符串 | +| 59-I. 滑动窗口的最大值 | 排序 | 数组，栈 / 队列 | +| 59-II. 队列的最大值 | 排序 | 数组，栈 / 队列 | +| 60. n 个骰子的点数 | 动态规划 | | +| 61. 扑克牌中的顺子 | 排序 | 数组，哈希表 | +| 62. 圆圈中最后剩下的数字 | 数学 | | +| 63. 股票的最大利润 | 动态规划 | 数组 | +| 64. 求 1 + 2 + … + n | | | +| 65. 不用加减乘除做加法 | 位运算 | | +| 66. 构建乘积数组 | 数学 | 数组 | +| 67. 把字符串转换成整数 | | 字符串 | +| 68-I. 二叉搜索树的最近公共祖先 | 搜索算法 | 树 | +| 68-II. 二叉树的最近公共祖先 | 搜索算法 | 树 | From 038a3fd068a6dd693a1d6e63289346857f48799b Mon Sep 17 00:00:00 2001 From: Yudong Jin Date: Mon, 16 Oct 2023 09:37:39 -0500 Subject: [PATCH 14/23] Update README.md --- README.md | 32 +++++++++++++++++--------------- 1 file changed, 17 insertions(+), 15 deletions(-) diff --git a/README.md b/README.md index 5258c3d..726675c 100755 --- a/README.md +++ b/README.md @@ -1,10 +1,10 @@ -

+

-

+

@@ -17,24 +17,26 @@ - 题目活跃于各大互联网公司招聘中，可使笔面试准备事半功倍。 - 致力于行文深入浅出、图文搭配，提供简洁的 Python, Java, C++ 解题代码。 -> 若本 LeetBook 对您有所帮助，请在页面右上角点个 Star :star: 支持一下，谢谢！ - -本仓库包含“图解算法数据结构”、“Krahets 笔面试精选 88 题”和“剑指 Offer”的题解内容，所有代码皆可一键运行： +> **Note** +> +> 本仓库包含“图解算法数据结构”、“Krahets 笔面试精选 88 题”和“剑指 Offer”的题解内容，所有代码皆可一键运行： +> +> ```python +> LeetCode-Book +> ├── leetbook_ioa # 《图解算法数据结构》题解和专栏文档、代码 +> ├── selected_coding_interview # 《Krahets 笔面试精选 88 题》题解文档、代码 +> └── sword_for_offer # 《剑指 Offer》题解文档、代码、刷题计划 +> ``` -```shell -LeetCode-Book -├── leetbook_ioa # 《图解算法数据结构》题解和专栏文档、代码 -├── selected_coding_interview # 「Krahets 笔面试精选 88 题」题解文档、代码 -└── sword_for_offer # 《剑指 Offer》题解文档、代码、刷题计划 -``` +> 若本 LeetBook 对您有所帮助，请在页面右上角点个 Star :star: 支持一下，谢谢！ -## 如何入门算法？ +## 如何学习算法 ### 第一步：看入门书

- - + +

[《Hello 算法》](https://github.com/krahets/hello-algo)：动画图解、一键运行的数据结构与算法教程。 @@ -61,4 +63,4 @@ LeetCode-Book ## License -The texts, codes and images in this repository are licensed under [CC BY-NC-SA-4.0](https://creativecommons.org/licenses/by-nc-sa/4.0/). +The texts, code and images in this repository are licensed under [CC BY-NC-SA-4.0](https://creativecommons.org/licenses/by-nc-sa/4.0/). From f6aa76b30e5f871361e044957908e9b5d07edc53 Mon Sep 17 00:00:00 2001 From: Yudong Jin Date: Tue, 24 Oct 2023 23:00:23 +0800 Subject: [PATCH 15/23] Update README.md --- README.md | 3 +-- 1 file changed, 1 insertion(+), 2 deletions(-) diff --git a/README.md b/README.md index 726675c..7246339 100755 --- a/README.md +++ b/README.md @@ -35,8 +35,7 @@ ### 第一步：看入门书

- - +

[《Hello 算法》](https://github.com/krahets/hello-algo)：动画图解、一键运行的数据结构与算法教程。 From cff02766bfaff296867717b23fea9f0209eb5940 Mon Sep 17 00:00:00 2001 From: Yudong Jin Date: Wed, 25 Oct 2023 18:55:22 +0800 Subject: [PATCH 16/23] Update README.md --- README.md | 19 ++++++++++--------- 1 file changed, 10 insertions(+), 9 deletions(-) diff --git a/README.md b/README.md index 7246339..dddae9e 100755 --- a/README.md +++ b/README.md @@ -19,7 +19,7 @@ > **Note** > -> 本仓库包含“图解算法数据结构”、“Krahets 笔面试精选 88 题”和“剑指 Offer”的题解内容，所有代码皆可一键运行： +> 本仓库包含“图解算法数据结构”、“Krahets 笔面试精选 88 题”和“剑指 Offer”的题解内容： > > ```python > LeetCode-Book @@ -34,13 +34,13 @@ ### 第一步：看入门书 -

- +

+

[《Hello 算法》](https://github.com/krahets/hello-algo)：动画图解、一键运行的数据结构与算法教程。 -这本书旨在引导初学者探索数据结构与算法的知识地图，掌握完整的刷题工具库。 +这本书旨在引导初学者探索数据结构与算法的知识地图，掌握刷题需要的前置知识与工具库。 ### 第二步：刷算法题 @@ -54,11 +54,12 @@ 对初学者的刷题建议： -1. 建议每日刷 2~3 题。若能轻松完成，可以尝试增加至 5～8 题。请注意，刷题的质量远重要于数量。务必确保你真正理解了每个题目的解法及背后的算法原理。 -2. 建议你按照目录顺序逐题解答。如果碰到某些难以解决的题目，可以先跳过，稍后回顾时再挑战。 -3. 很多题目都有不止一种解法，请你注意比较和探讨各种方法的特点和适用情况。 -4. 如果你发现自己忘记了某个题目的解法，不必灰心。艾宾浩斯遗忘曲线指出，通常需要复习至少三次，才能真正掌握一个知识点。 -5. 行百里者半九十。坚持至关重要，加油，相信你可以做到！ +1. 建议每日刷 2~3 题。若能轻松完成，可以尝试增加至 5～8 题。 +2. 刷题的质量远重要于数量。务必确保你真正理解了每个题目的解法及背后的算法原理。 +3. 建议你按照目录顺序做题。如果感觉某题很难，可以先跳过，稍后再回顾挑战。 +4. 很多题目都有不止一种解法，请你注意比较和探讨各种方法的特点和适用情况。 +5. 如果你发现自己遗忘了题目解法，不必灰心。艾宾浩斯遗忘曲线指出，通常需要复习至少三次，才能真正掌握一个知识点。 +6. 行百里者半九十。坚持至关重要，加油，相信你可以做到！ ## License From 589bc1a15d582f1f7c2016b2ad4e1971ef3ee295 Mon Sep 17 00:00:00 2001 From: krahets Date: Sat, 4 Nov 2023 20:42:39 +0800 Subject: [PATCH 17/23] Add a solution to problem 287. Remove the footers. --- ...76\345\274\217\345\214\271\351\205\215.md" | 7 -- ...60\344\272\214\345\217\211\346\240\221.md" | 7 -- ...02\345\272\217\351\201\215\345\216\206.md" | 7 -- ...02\345\272\217\351\201\215\345\216\206.md" | 7 -- ...00\345\244\247\346\267\261\345\272\246.md" | 7 -- ...40\344\272\214\345\217\211\346\240\221.md" | 7 -- ...345\276\204\346\200\273\345\222\214 II.md" | 7 -- ...00\344\275\263\346\227\266\346\234\272.md" | 7 -- ...41\347\232\204\346\225\260\345\255\227.md" | 7 -- ...10\347\232\204\351\223\276\350\241\250.md" | 7 -- ...04\345\212\250\346\200\201\345\222\214.md" | 7 -- ...55\347\232\204\345\215\225\350\257\215.md" | 7 -- ...346\234\200\345\260\217\345\200\274 II.md" | 7 -- ...346\225\260\344\271\213\345\222\214 II.md" | 7 -- ...32\346\225\260\345\205\203\347\264\240.md" | 7 -- .... \346\234\200\345\244\247\346\225\260.md" | 7 -- ...04\350\216\267\350\203\234\350\200\205.md" | 7 -- ...51\347\232\204\344\270\252\346\225\260.md" | 7 -- ...04\345\255\227\347\254\246\344\270\262.md" | 7 -- ...15\350\275\254\351\223\276\350\241\250.md" | 7 -- ...11\345\272\217\351\223\276\350\241\250.md" | 7 -- ...00\345\244\247\345\205\203\347\264\240.md" | 7 -- ...54\344\272\214\345\217\211\346\240\221.md" | 7 -- ...17\347\232\204\345\205\203\347\264\240.md" | 7 -- ...36\347\216\260\351\230\237\345\210\227.md" | 7 -- ...1 \347\232\204\344\270\252\346\225\260.md" | 7 -- ...54\345\205\261\347\245\226\345\205\210.md" | 7 -- ...55\347\232\204\350\212\202\347\202\271.md" | 7 -- ...04\347\232\204\344\271\230\347\247\257.md" | 7 -- ...43\346\234\200\345\244\247\345\200\274.md" | 7 -- ...347\273\264\347\237\251\351\230\265 II.md" | 7 -- ...15\345\274\202\344\275\215\350\257\215.md" | 7 -- ...4. \344\270\221\346\225\260 II\302\240.md" | 7 -- ...36\346\226\207\346\216\222\345\210\227.md" | 7 -- ...57\347\232\204\347\211\210\346\234\254.md" | 7 -- ...76\351\207\215\345\244\215\346\225\260.md" | 100 +++++++++++++++++- ...04\344\270\255\344\275\215\346\225\260.md" | 7 -- ...15\345\272\217\345\210\227\345\214\226.md" | 7 -- ...00\351\225\277\345\255\220\344\270\262.md" | 7 -- ...64\346\225\260\344\271\213\345\222\214.md" | 7 -- ...57\344\270\200\345\255\227\347\254\246.md" | 7 -- ...04\345\220\210\346\200\273\345\222\214.md" | 7 -- ...55\345\255\220\345\272\217\345\210\227.md" | 7 -- ...345\220\210\346\200\273\345\222\214 II.md" | 7 -- ...N \344\275\215\346\225\260\345\255\227.md" | 7 -- ...77\345\233\236\346\226\207\344\270\262.md" | 7 -- ...14\345\220\221\351\223\276\350\241\250.md" | 7 -- .... \345\205\250\346\216\222\345\210\227.md" | 7 -- ...345\205\250\346\216\222\345\210\227 II.md" | 7 -- ...13\350\275\254\345\233\276\345\203\217.md" | 7 -- ...42\351\202\243\345\245\221\346\225\260.md" | 7 -- ...20\346\225\260\347\273\204\345\222\214.md" | 7 -- ...72\346\227\213\347\237\251\351\230\265.md" | 7 -- ...11\346\225\210\346\225\260\345\255\227.md" | 7 -- .... \347\210\254\346\245\274\346\242\257.md" | 7 -- ...14\345\210\206\346\237\245\346\211\276.md" | 7 -- ...55\345\277\203\344\270\213\346\240\207.md" | 7 -- ...25\350\257\215\346\220\234\347\264\242.md" | 7 -- ...54\345\255\227\347\254\246\344\270\262.md" | 7 -- ...215\242\346\225\264\346\225\260 (atoi).md" | 7 -- ...06\351\232\224\351\223\276\350\241\250.md" | 7 -- ...55\351\227\264\347\273\223\347\202\271.md" | 7 -- ...01\346\240\210\345\272\217\345\210\227.md" | 7 -- 63 files changed, 96 insertions(+), 438 deletions(-) diff --git "a/selected_coding_interview/docs/10. \346\255\243\345\210\231\350\241\250\350\276\276\345\274\217\345\214\271\351\205\215.md" "b/selected_coding_interview/docs/10. \346\255\243\345\210\231\350\241\250\350\276\276\345\274\217\345\214\271\351\205\215.md" index 0811314..ed1ca02 100644 --- "a/selected_coding_interview/docs/10. \346\255\243\345\210\231\350\241\250\350\276\276\345\274\217\345\214\271\351\205\215.md" +++ "b/selected_coding_interview/docs/10. \346\255\243\345\210\231\350\241\250\350\276\276\345\274\217\345\214\271\351\205\215.md" @@ -179,10 +179,3 @@ public: - **时间复杂度 $O(MN)$ ：** 其中 $M, N$ 分别为 `s` 和 `p` 的长度，状态转移需遍历整个 `dp` 矩阵。 - **空间复杂度 $O(MN)$ ：** 状态矩阵 `dp` 使用 $O(MN)$ 的额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/101. \345\257\271\347\247\260\344\272\214\345\217\211\346\240\221.md" "b/selected_coding_interview/docs/101. \345\257\271\347\247\260\344\272\214\345\217\211\346\240\221.md" index 047cf26..0324eb7 100644 --- "a/selected_coding_interview/docs/101. \345\257\271\347\247\260\344\272\214\345\217\211\346\240\221.md" +++ "b/selected_coding_interview/docs/101. \345\257\271\347\247\260\344\272\214\345\217\211\346\240\221.md" @@ -77,10 +77,3 @@ private: - **空间复杂度 $O(N)$ ：** 如下图所示，最差情况下（二叉树退化为链表），系统使用 $O(N)$ 大小的空间。 ![Picture2.png](https://pic.leetcode-cn.com/1599398062-LmPbix-Picture2.png){:width=450} - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/102. \344\272\214\345\217\211\346\240\221\347\232\204\345\261\202\345\272\217\351\201\215\345\216\206.md" "b/selected_coding_interview/docs/102. \344\272\214\345\217\211\346\240\221\347\232\204\345\261\202\345\272\217\351\201\215\345\216\206.md" index afab2d0..7ea36cd 100644 --- "a/selected_coding_interview/docs/102. \344\272\214\345\217\211\346\240\221\347\232\204\345\261\202\345\272\217\351\201\215\345\216\206.md" +++ "b/selected_coding_interview/docs/102. \344\272\214\345\217\211\346\240\221\347\232\204\345\261\202\345\272\217\351\201\215\345\216\206.md" @@ -90,10 +90,3 @@ public: - **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数量，即 BFS 需循环 $N$ 次。 - **空间复杂度 $O(N)$ ：** 最差情况下，即当树为平衡二叉树时，最多有 $N/2$ 个树节点**同时**在 `queue` 中，使用 $O(N)$ 大小的额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/103. \344\272\214\345\217\211\346\240\221\347\232\204\351\224\257\351\275\277\345\275\242\345\261\202\345\272\217\351\201\215\345\216\206.md" "b/selected_coding_interview/docs/103. \344\272\214\345\217\211\346\240\221\347\232\204\351\224\257\351\275\277\345\275\242\345\261\202\345\272\217\351\201\215\345\216\206.md" index 7a9070e..b9bf399 100644 --- "a/selected_coding_interview/docs/103. \344\272\214\345\217\211\346\240\221\347\232\204\351\224\257\351\275\277\345\275\242\345\261\202\345\272\217\351\201\215\345\216\206.md" +++ "b/selected_coding_interview/docs/103. \344\272\214\345\217\211\346\240\221\347\232\204\351\224\257\351\275\277\345\275\242\345\261\202\345\272\217\351\201\215\345\216\206.md" @@ -278,10 +278,3 @@ public: - **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数量，即 BFS 需循环 $N$ 次，占用 $O(N)$ 。**共完成** 少于 $N$ 个节点的倒序操作，占用 $O(N)$ 。 - **空间复杂度 $O(N)$ ：** 最差情况下，即当树为满二叉树时，最多有 $N/2$ 个树节点**同时**在 `queue` 中，使用 $O(N)$ 大小的额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/104. \344\272\214\345\217\211\346\240\221\347\232\204\346\234\200\345\244\247\346\267\261\345\272\246.md" "b/selected_coding_interview/docs/104. \344\272\214\345\217\211\346\240\221\347\232\204\346\234\200\345\244\247\346\267\261\345\272\246.md" index 7664055..adc5ba2 100644 --- "a/selected_coding_interview/docs/104. \344\272\214\345\217\211\346\240\221\347\232\204\346\234\200\345\244\247\346\267\261\345\272\246.md" +++ "b/selected_coding_interview/docs/104. \344\272\214\345\217\211\346\240\221\347\232\204\346\234\200\345\244\247\346\267\261\345\272\246.md" @@ -140,10 +140,3 @@ public: - **时间复杂度 $O(N)$ ：** $N$ 为树的节点数量，计算树的深度需要遍历所有节点。 - **空间复杂度 $O(N)$ ：** 最差情况下（当树平衡时），队列 `queue` 同时存储 $N/2$ 个节点。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/105. \344\273\216\345\211\215\345\272\217\344\270\216\344\270\255\345\272\217\351\201\215\345\216\206\345\272\217\345\210\227\346\236\204\351\200\240\344\272\214\345\217\211\346\240\221.md" "b/selected_coding_interview/docs/105. \344\273\216\345\211\215\345\272\217\344\270\216\344\270\255\345\272\217\351\201\215\345\216\206\345\272\217\345\210\227\346\236\204\351\200\240\344\272\214\345\217\211\346\240\221.md" index 16fce3f..02615eb 100644 --- "a/selected_coding_interview/docs/105. \344\273\216\345\211\215\345\272\217\344\270\216\344\270\255\345\272\217\351\201\215\345\216\206\345\272\217\345\210\227\346\236\204\351\200\240\344\272\214\345\217\211\346\240\221.md" +++ "b/selected_coding_interview/docs/105. \344\273\216\345\211\215\345\272\217\344\270\216\344\270\255\345\272\217\351\201\215\345\216\206\345\272\217\345\210\227\346\236\204\351\200\240\344\272\214\345\217\211\346\240\221.md" @@ -114,10 +114,3 @@ private: - **时间复杂度 $O(N)$ ：** 其中 $N$ 为树的节点数量。初始化 HashMap 需遍历 `inorder` ，占用 $O(N)$ 。递归共建立 $N$ 个节点，每层递归中的节点建立、搜索操作占用 $O(1)$ ，因此使用 $O(N)$ 时间。 - **空间复杂度 $O(N)$ ：** HashMap 使用 $O(N)$ 额外空间；最差情况下（输入二叉树为链表时），递归深度达到 $N$ ，占用 $O(N)$ 的栈帧空间；因此总共使用 $O(N)$ 空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/113. \350\267\257\345\276\204\346\200\273\345\222\214 II.md" "b/selected_coding_interview/docs/113. \350\267\257\345\276\204\346\200\273\345\222\214 II.md" index 6af756a..4e60b8d 100644 --- "a/selected_coding_interview/docs/113. \350\267\257\345\276\204\346\200\273\345\222\214 II.md" +++ "b/selected_coding_interview/docs/113. \350\267\257\345\276\204\346\200\273\345\222\214 II.md" @@ -103,10 +103,3 @@ private: - **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数，先序遍历需要遍历所有节点。 - **空间复杂度 $O(N)$ ：** 最差情况下，即树退化为链表时，`path` 存储所有树节点，使用 $O(N)$ 额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/121. \344\271\260\345\215\226\350\202\241\347\245\250\347\232\204\346\234\200\344\275\263\346\227\266\346\234\272.md" "b/selected_coding_interview/docs/121. \344\271\260\345\215\226\350\202\241\347\245\250\347\232\204\346\234\200\344\275\263\346\227\266\346\234\272.md" index fc49258..6e95115 100644 --- "a/selected_coding_interview/docs/121. \344\271\260\345\215\226\350\202\241\347\245\250\347\232\204\346\234\200\344\275\263\346\227\266\346\234\272.md" +++ "b/selected_coding_interview/docs/121. \344\271\260\345\215\226\350\202\241\347\245\250\347\232\204\346\234\200\344\275\263\346\227\266\346\234\272.md" @@ -58,10 +58,3 @@ public: - **时间复杂度 $O(N)$ ：** 其中 $N$ 为数组 `prices` 长度。遍历 `prices` 使用线性时间。 - **空间复杂度 $O(1)$ ：** 变量 `cost` , `profit` 使用 $O(1)$ 空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/136. \345\217\252\345\207\272\347\216\260\344\270\200\346\254\241\347\232\204\346\225\260\345\255\227.md" "b/selected_coding_interview/docs/136. \345\217\252\345\207\272\347\216\260\344\270\200\346\254\241\347\232\204\346\225\260\345\255\227.md" index cf14b76..ac296e3 100644 --- "a/selected_coding_interview/docs/136. \345\217\252\345\207\272\347\216\260\344\270\200\346\254\241\347\232\204\346\225\260\345\255\227.md" +++ "b/selected_coding_interview/docs/136. \345\217\252\345\207\272\347\216\260\344\270\200\346\254\241\347\232\204\346\225\260\345\255\227.md" @@ -60,10 +60,3 @@ public: - **时间复杂度 $O(N)$ ：** 线性遍历 $nums$ 使用 $O(N)$ 时间，异或操作使用 $O(1)$ 时间。 - **空间复杂度 $O(1)$ ：** 辅助变量 $a$ , $b$ , $x$ , $y$ 使用常数大小额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/138. \345\244\215\345\210\266\345\270\246\351\232\217\346\234\272\346\214\207\351\222\210\347\232\204\351\223\276\350\241\250.md" "b/selected_coding_interview/docs/138. \345\244\215\345\210\266\345\270\246\351\232\217\346\234\272\346\214\207\351\222\210\347\232\204\351\223\276\350\241\250.md" index 7def256..b59501f 100644 --- "a/selected_coding_interview/docs/138. \345\244\215\345\210\266\345\270\246\351\232\217\346\234\272\346\214\207\351\222\210\347\232\204\351\223\276\350\241\250.md" +++ "b/selected_coding_interview/docs/138. \345\244\215\345\210\266\345\270\246\351\232\217\346\234\272\346\214\207\351\222\210\347\232\204\351\223\276\350\241\250.md" @@ -351,10 +351,3 @@ public: - **时间复杂度 $O(N)$ ：** 三轮遍历链表，使用 $O(N)$ 时间。 - **空间复杂度 $O(1)$ ：** 节点引用变量使用常数大小的额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/1480. \344\270\200\347\273\264\346\225\260\347\273\204\347\232\204\345\212\250\346\200\201\345\222\214.md" "b/selected_coding_interview/docs/1480. \344\270\200\347\273\264\346\225\260\347\273\204\347\232\204\345\212\250\346\200\201\345\222\214.md" index e6d66b3..cc7ed3a 100644 --- "a/selected_coding_interview/docs/1480. \344\270\200\347\273\264\346\225\260\347\273\204\347\232\204\345\212\250\346\200\201\345\222\214.md" +++ "b/selected_coding_interview/docs/1480. \344\270\200\347\273\264\346\225\260\347\273\204\347\232\204\345\212\250\346\200\201\345\222\214.md" @@ -58,10 +58,3 @@ public: - **时间复杂度 $O(N)$ ：** 遍历 `nums` 使用线性时间。 - **空间复杂度 $O(1)$ ：** 用于保存结果的 `dp` 是必须使用的空间，此处不计入。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/151. \345\217\215\350\275\254\345\255\227\347\254\246\344\270\262\344\270\255\347\232\204\345\215\225\350\257\215.md" "b/selected_coding_interview/docs/151. \345\217\215\350\275\254\345\255\227\347\254\246\344\270\262\344\270\255\347\232\204\345\215\225\350\257\215.md" index de14f07..1ffc668 100644 --- "a/selected_coding_interview/docs/151. \345\217\215\350\275\254\345\255\227\347\254\246\344\270\262\344\270\255\347\232\204\345\215\225\350\257\215.md" +++ "b/selected_coding_interview/docs/151. \345\217\215\350\275\254\345\255\227\347\254\246\344\270\262\344\270\255\347\232\204\345\215\225\350\257\215.md" @@ -101,10 +101,3 @@ class Solution: - [`join()` 方法：](https://stackoverflow.com/questions/37133547/time-complexity-of-string-concatenation-in-python) 为 $O(N)$ 。 - [`reverse()` 方法：](https://stackoverflow.com/questions/37606159/what-is-the-time-complexity-of-python-list-reverse) 为 $O(N)$ 。 - **空间复杂度 $O(N)$ ：** 单词列表 $strs$ 占用线性大小的额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/154. \345\257\273\346\211\276\346\227\213\350\275\254\346\216\222\345\272\217\346\225\260\347\273\204\344\270\255\347\232\204\346\234\200\345\260\217\345\200\274 II.md" "b/selected_coding_interview/docs/154. \345\257\273\346\211\276\346\227\213\350\275\254\346\216\222\345\272\217\346\225\260\347\273\204\344\270\255\347\232\204\346\234\200\345\260\217\345\200\274 II.md" index ccd5e78..66f68e8 100644 --- "a/selected_coding_interview/docs/154. \345\257\273\346\211\276\346\227\213\350\275\254\346\216\222\345\272\217\346\225\260\347\273\204\344\270\255\347\232\204\346\234\200\345\260\217\345\200\274 II.md" +++ "b/selected_coding_interview/docs/154. \345\257\273\346\211\276\346\227\213\350\275\254\346\216\222\345\272\217\346\225\260\347\273\204\344\270\255\347\232\204\346\234\200\345\260\217\345\200\274 II.md" @@ -157,10 +157,3 @@ public: - **时间复杂度 $O(\log_2 N)$ ：** 在特例情况下（例如 $[1, 1, 1, 1]$），会退化到 $O(N)$。 - **空间复杂度 $O(1)$ ：** $i$ , $j$ , $m$ 变量使用常数大小的额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/167. \344\270\244\346\225\260\344\271\213\345\222\214 II.md" "b/selected_coding_interview/docs/167. \344\270\244\346\225\260\344\271\213\345\222\214 II.md" index 764a781..0811bd0 100644 --- "a/selected_coding_interview/docs/167. \344\270\244\346\225\260\344\271\213\345\222\214 II.md" +++ "b/selected_coding_interview/docs/167. \344\270\244\346\225\260\344\271\213\345\222\214 II.md" @@ -76,10 +76,3 @@ public: - **时间复杂度 $O(N)$ ：** $N$ 为数组 $numbers$ 的长度；双指针共同线性遍历整个数组。 - **空间复杂度 $O(1)$ ：** 变量 $i$, $j$ 使用常数大小的额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/169. \345\244\232\346\225\260\345\205\203\347\264\240.md" "b/selected_coding_interview/docs/169. \345\244\232\346\225\260\345\205\203\347\264\240.md" index c639d38..a84c19c 100644 --- "a/selected_coding_interview/docs/169. \345\244\232\346\225\260\345\205\203\347\264\240.md" +++ "b/selected_coding_interview/docs/169. \345\244\232\346\225\260\345\205\203\347\264\240.md" @@ -135,10 +135,3 @@ public: ``` 时间复杂度和空间复杂度都不变，仍为 $O(N)$ 和 $O(1)$ 。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/179. \346\234\200\345\244\247\346\225\260.md" "b/selected_coding_interview/docs/179. \346\234\200\345\244\247\346\225\260.md" index e8be5d8..06738bd 100644 --- "a/selected_coding_interview/docs/179. \346\234\200\345\244\247\346\225\260.md" +++ "b/selected_coding_interview/docs/179. \346\234\200\345\244\247\346\225\260.md" @@ -212,10 +212,3 @@ private: - **时间复杂度 $O(N \log N)$ ：** $N$ 为最终返回值的字符数量（ $strs$ 列表的长度 $\leq N$ ）；使用快排或内置函数的平均时间复杂度为 $O(N \log N)$ ，最差为 $O(N^2)$ 。 - **空间复杂度 $O(N)$ ：** 字符串列表 $strs$ 占用线性大小的额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/1823. \346\211\276\345\207\272\346\270\270\346\210\217\347\232\204\350\216\267\350\203\234\350\200\205.md" "b/selected_coding_interview/docs/1823. \346\211\276\345\207\272\346\270\270\346\210\217\347\232\204\350\216\267\350\203\234\350\200\205.md" index fc129fb..594457f 100644 --- "a/selected_coding_interview/docs/1823. \346\211\276\345\207\272\346\270\270\346\210\217\347\232\204\350\216\267\350\203\234\350\200\205.md" +++ "b/selected_coding_interview/docs/1823. \346\211\276\345\207\272\346\270\270\346\210\217\347\232\204\350\216\267\350\203\234\350\200\205.md" @@ -119,10 +119,3 @@ public: - **时间复杂度 $O(n)$ ：** 状态转移循环 $n - 1$ 次使用 $O(n)$ 时间，状态转移方程计算使用 $O(1)$ 时间。 - **空间复杂度 $O(1)$ ：** 使用常数大小的额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/191. \344\275\2151\347\232\204\344\270\252\346\225\260.md" "b/selected_coding_interview/docs/191. \344\275\2151\347\232\204\344\270\252\346\225\260.md" index 3f55d9d..b4fe629 100644 --- "a/selected_coding_interview/docs/191. \344\275\2151\347\232\204\344\270\252\346\225\260.md" +++ "b/selected_coding_interview/docs/191. \344\275\2151\347\232\204\344\270\252\346\225\260.md" @@ -125,10 +125,3 @@ public: - **时间复杂度 $O(M)$ ：** $n \& (n - 1)$ 操作仅有减法和与运算，占用 $O(1)$ ；设 $M$ 为二进制数字 $n$ 中 $1$ 的个数，则需循环 $M$ 次（每轮消去一个 $1$ ），占用 $O(M)$ 。 - **空间复杂度 $O(1)$ ：** 变量 $res$ 使用常数大小额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/205. \345\220\214\346\236\204\345\255\227\347\254\246\344\270\262.md" "b/selected_coding_interview/docs/205. \345\220\214\346\236\204\345\255\227\347\254\246\344\270\262.md" index 1f3f1d4..63927de 100644 --- "a/selected_coding_interview/docs/205. \345\220\214\346\236\204\345\255\227\347\254\246\344\270\262.md" +++ "b/selected_coding_interview/docs/205. \345\220\214\346\236\204\345\255\227\347\254\246\344\270\262.md" @@ -81,10 +81,3 @@ public: - **时间复杂度 $O(N)$ ：** 其中 $N$ 为字符串 `s` , `t` 的长度。遍历字符串 `s` , `t` 使用线性时间，hashmap 查询操作使用 $O(1)$ 时间。 - **空间复杂度 $O(1)$ ：** 题目说明 `s` 和 `t` 由任意有效的 ASCII 字符组成。由于 ASCII 字符共 128 个，因此 hashmap `s2t` , `t2s` 使用 $O(128) = O(1)$ 空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/206. \345\217\215\350\275\254\351\223\276\350\241\250.md" "b/selected_coding_interview/docs/206. \345\217\215\350\275\254\351\223\276\350\241\250.md" index 57ea5f9..7645abd 100644 --- "a/selected_coding_interview/docs/206. \345\217\215\350\275\254\351\223\276\350\241\250.md" +++ "b/selected_coding_interview/docs/206. \345\217\215\350\275\254\351\223\276\350\241\250.md" @@ -136,10 +136,3 @@ private: - **时间复杂度 $O(N)$ ：** 遍历链表使用线性大小时间。 - **空间复杂度 $O(N)$ ：** 遍历链表的递归深度达到 $N$ ，系统使用 $O(N)$ 大小额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/21. \345\220\210\345\271\266\344\270\244\344\270\252\346\234\211\345\272\217\351\223\276\350\241\250.md" "b/selected_coding_interview/docs/21. \345\220\210\345\271\266\344\270\244\344\270\252\346\234\211\345\272\217\351\223\276\350\241\250.md" index 658e418..a495ff2 100644 --- "a/selected_coding_interview/docs/21. \345\220\210\345\271\266\344\270\244\344\270\252\346\234\211\345\272\217\351\223\276\350\241\250.md" +++ "b/selected_coding_interview/docs/21. \345\220\210\345\271\266\344\270\244\344\270\252\346\234\211\345\272\217\351\223\276\350\241\250.md" @@ -86,10 +86,3 @@ public: - **时间复杂度 $O(M+N)$ ：** $M, N$ 分别为链表 $l_1$, $l_2$ 的长度，合并操作需遍历两链表。 - **空间复杂度 $O(1)$ ：** 节点引用 $dum$ , $cur$ 使用常数大小的额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/215. \346\225\260\347\273\204\344\270\255\347\232\204\347\254\254K\344\270\252\346\234\200\345\244\247\345\205\203\347\264\240.md" "b/selected_coding_interview/docs/215. \346\225\260\347\273\204\344\270\255\347\232\204\347\254\254K\344\270\252\346\234\200\345\244\247\345\205\203\347\264\240.md" index 2a70c69..469531a 100644 --- "a/selected_coding_interview/docs/215. \346\225\260\347\273\204\344\270\255\347\232\204\347\254\254K\344\270\252\346\234\200\345\244\247\345\205\203\347\264\240.md" +++ "b/selected_coding_interview/docs/215. \346\225\260\347\273\204\344\270\255\347\232\204\347\254\254K\344\270\252\346\234\200\345\244\247\345\205\203\347\264\240.md" @@ -181,10 +181,3 @@ private: - **时间复杂度 $O(N)$ ：** 其中 $N$ 为数组元素数量；对于长度为 $N$ 的数组执行哨兵划分操作的时间复杂度为 $O(N)$ ；每轮哨兵划分后根据 $k$ 和 $i$ 的大小关系选择递归，由于 $i$ 分布的随机性，则向下递归子数组的平均长度为 $\frac{N}{2}$ ；因此平均情况下，哨兵划分操作一共有 $N + \frac{N}{2} + \frac{N}{4} + ... + \frac{N}{N} = \frac{N - \frac{1}{2}}{1 - \frac{1}{2}} = 2N - 1$ （等比数列求和），即总体时间复杂度为 $O(N)$ 。 - **空间复杂度 $O(\log N)$ ：** 划分函数的平均递归深度为 $O(\log N)$ 。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/226. \347\277\273\350\275\254\344\272\214\345\217\211\346\240\221.md" "b/selected_coding_interview/docs/226. \347\277\273\350\275\254\344\272\214\345\217\211\346\240\221.md" index 98eba7c..5a12937 100644 --- "a/selected_coding_interview/docs/226. \347\277\273\350\275\254\344\272\214\345\217\211\346\240\221.md" +++ "b/selected_coding_interview/docs/226. \347\277\273\350\275\254\344\272\214\345\217\211\346\240\221.md" @@ -151,10 +151,3 @@ public: - **空间复杂度 $O(N)$ ：** 如下图所示，最差情况下，栈 $stack$ 最多同时存储 $\frac{N + 1}{2}$ 个节点，占用 $O(N)$ 额外空间。 ![Picture0.png](https://pic.leetcode-cn.com/1614450330-bTAcyj-Picture0.png){:width=450} - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/230. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\344\270\255\347\254\254K\345\260\217\347\232\204\345\205\203\347\264\240.md" "b/selected_coding_interview/docs/230. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\344\270\255\347\254\254K\345\260\217\347\232\204\345\205\203\347\264\240.md" index 2cc5043..cb7d02b 100644 --- "a/selected_coding_interview/docs/230. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\344\270\255\347\254\254K\345\260\217\347\232\204\345\205\203\347\264\240.md" +++ "b/selected_coding_interview/docs/230. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\344\270\255\347\254\254K\345\260\217\347\232\204\345\205\203\347\264\240.md" @@ -75,10 +75,3 @@ private: - **时间复杂度 $O(N)$ ：** 当树退化为链表，即全部为左子节点时，无论 $k$ 的值大小，递归深度都为 $N$ ，使用 $O(N)$ 时间。 - **空间复杂度 $O(N)$ ：** 当树退化为链表时，系统使用 $O(N)$ 大小的栈空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/232. \347\224\250\346\240\210\345\256\236\347\216\260\351\230\237\345\210\227.md" "b/selected_coding_interview/docs/232. \347\224\250\346\240\210\345\256\236\347\216\260\351\230\237\345\210\227.md" index a574868..ef4d6ad 100644 --- "a/selected_coding_interview/docs/232. \347\224\250\346\240\210\345\256\236\347\216\260\351\230\237\345\210\227.md" +++ "b/selected_coding_interview/docs/232. \347\224\250\346\240\210\345\256\236\347\216\260\351\230\237\345\210\227.md" @@ -126,10 +126,3 @@ public: - **时间复杂度：** `push()`, `empty()` 函数的时间复杂度为 $O(1)$ ；`peek()` , `pop()` 函数在 $N$ 次队首元素删除操作中总共需完成 $N$ 个元素的倒序，均摊时间复杂度为 $O(1)$ 。 - **空间复杂度 $O(N)$ ：** 最差情况下，栈 `A` 和 `B` 共保存 $N$ 个元素。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/233. \346\225\260\345\255\227 1 \347\232\204\344\270\252\346\225\260.md" "b/selected_coding_interview/docs/233. \346\225\260\345\255\227 1 \347\232\204\344\270\252\346\225\260.md" index 6ea21df..0d19277 100644 --- "a/selected_coding_interview/docs/233. \346\225\260\345\255\227 1 \347\232\204\344\270\252\346\225\260.md" +++ "b/selected_coding_interview/docs/233. \346\225\260\345\255\227 1 \347\232\204\344\270\252\346\225\260.md" @@ -159,10 +159,3 @@ public: - **时间复杂度 $O(\log n)$ ：** 循环内的计算操作使用 $O(1)$ 时间；循环次数为数字 $n$ 的位数，即 $\log_{10}{n}$ ，因此循环使用 $O(\log n)$ 时间。 - **空间复杂度 $O(1)$ ：** 几个变量使用常数大小的额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/235. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\347\232\204\346\234\200\350\277\221\345\205\254\345\205\261\347\245\226\345\205\210.md" "b/selected_coding_interview/docs/235. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\347\232\204\346\234\200\350\277\221\345\205\254\345\205\261\347\245\226\345\205\210.md" index de41558..60f97e1 100644 --- "a/selected_coding_interview/docs/235. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\347\232\204\346\234\200\350\277\221\345\205\254\345\205\261\347\245\226\345\205\210.md" +++ "b/selected_coding_interview/docs/235. \344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\347\232\204\346\234\200\350\277\221\345\205\254\345\205\261\347\245\226\345\205\210.md" @@ -181,10 +181,3 @@ public: - **时间复杂度 $O(N)$ ：** 其中 $N$ 为二叉树节点数；每循环一轮排除一层，二叉搜索树的层数最小为 $\log N$ （满二叉树），最大为 $N$ （退化为链表）。 - **空间复杂度 $O(N)$ ：** 最差情况下，即树退化为链表时，递归深度达到树的层数 $N$ 。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/237. \345\210\240\351\231\244\351\223\276\350\241\250\344\270\255\347\232\204\350\212\202\347\202\271.md" "b/selected_coding_interview/docs/237. \345\210\240\351\231\244\351\223\276\350\241\250\344\270\255\347\232\204\350\212\202\347\202\271.md" index 999a3ce..200affa 100644 --- "a/selected_coding_interview/docs/237. \345\210\240\351\231\244\351\223\276\350\241\250\344\270\255\347\232\204\350\212\202\347\202\271.md" +++ "b/selected_coding_interview/docs/237. \345\210\240\351\231\244\351\223\276\350\241\250\344\270\255\347\232\204\350\212\202\347\202\271.md" @@ -82,10 +82,3 @@ public: **时间复杂度 $O(1)$ ：** 使用常数时间。 **空间复杂度 $O(1)$ ：** 使用常数大小额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/238. \351\231\244\350\207\252\350\272\253\344\273\245\345\244\226\346\225\260\347\273\204\347\232\204\344\271\230\347\247\257.md" "b/selected_coding_interview/docs/238. \351\231\244\350\207\252\350\272\253\344\273\245\345\244\226\346\225\260\347\273\204\347\232\204\344\271\230\347\247\257.md" index 3f58f1b..8e46474 100644 --- "a/selected_coding_interview/docs/238. \351\231\244\350\207\252\350\272\253\344\273\245\345\244\226\346\225\260\347\273\204\347\232\204\344\271\230\347\247\257.md" +++ "b/selected_coding_interview/docs/238. \351\231\244\350\207\252\350\272\253\344\273\245\345\244\226\346\225\260\347\273\204\347\232\204\344\271\230\347\247\257.md" @@ -76,10 +76,3 @@ public: - **时间复杂度 $O(N)$ ：** 其中 $N$ 为数组长度，两轮遍历数组 $nums$ ，使用 $O(N)$ 时间。 - **空间复杂度 $O(1)$ ：** 变量 $tmp$ 使用常数大小额外空间（数组 $ans$ 作为返回值，不计入复杂度考虑）。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/239. \346\273\221\345\212\250\347\252\227\345\217\243\346\234\200\345\244\247\345\200\274.md" "b/selected_coding_interview/docs/239. \346\273\221\345\212\250\347\252\227\345\217\243\346\234\200\345\244\247\345\200\274.md" index fa28595..1eaf142 100644 --- "a/selected_coding_interview/docs/239. \346\273\221\345\212\250\347\252\227\345\217\243\346\234\200\345\244\247\345\200\274.md" +++ "b/selected_coding_interview/docs/239. \346\273\221\345\212\250\347\252\227\345\217\243\346\234\200\345\244\247\345\200\274.md" @@ -224,10 +224,3 @@ public: - **时间复杂度 $O(N^2)$ ：** 其中 $N$ 为输入矩阵的行（列）数。需要将矩阵中每个元素旋转到新的位置，即对矩阵所有元素操作一次，使用 $O(N^2)$ 时间。 - **空间复杂度 $O(1)$ ：** 临时变量 $tmp$ 使用常数大小的额外空间。值得注意，当循环中进入下轮迭代，上轮迭代初始化的 $tmp$ 占用的内存就会被自动释放，因此无累计使用空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/240. \346\220\234\347\264\242\344\272\214\347\273\264\347\237\251\351\230\265 II.md" "b/selected_coding_interview/docs/240. \346\220\234\347\264\242\344\272\214\347\273\264\347\237\251\351\230\265 II.md" index aa81196..62301c4 100644 --- "a/selected_coding_interview/docs/240. \346\220\234\347\264\242\344\272\214\347\273\264\347\237\251\351\230\265 II.md" +++ "b/selected_coding_interview/docs/240. \346\220\234\347\264\242\344\272\214\347\273\264\347\237\251\351\230\265 II.md" @@ -71,10 +71,3 @@ public: - 时间复杂度 $O(M+N)$ ：其中，$N$ 和 $M$ 分别为矩阵行数和列数，此算法最多循环 $M+N$ 次。 - 空间复杂度 $O(1)$ : `i`, `j` 指针使用常数大小额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/242. \346\234\211\346\225\210\347\232\204\345\255\227\346\257\215\345\274\202\344\275\215\350\257\215.md" "b/selected_coding_interview/docs/242. \346\234\211\346\225\210\347\232\204\345\255\227\346\257\215\345\274\202\344\275\215\350\257\215.md" index d480c82..d70ad80 100644 --- "a/selected_coding_interview/docs/242. \346\234\211\346\225\210\347\232\204\345\255\227\346\257\215\345\274\202\344\275\215\350\257\215.md" +++ "b/selected_coding_interview/docs/242. \346\234\211\346\225\210\347\232\204\345\255\227\346\257\215\345\274\202\344\275\215\350\257\215.md" @@ -169,10 +169,3 @@ public: **时间复杂度 $O(M + N)$ ：** 其 $M$ , $N$ 分别为字符串 $s_1$ , $s_2$ 长度。当 $s_1$ , $s_2$ 无相同字符时，三轮循环的总迭代次数最多为 $2M + 2N$ ，使用 $O(M + N)$ 线性时间。 **空间复杂度 $O(1)$ ：** 由于字符种类是有限的（常量），一般 ASCII 码共包含 128 个字符，因此可假设使用 $O(1)$ 大小的额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/264. \344\270\221\346\225\260 II\302\240.md" "b/selected_coding_interview/docs/264. \344\270\221\346\225\260 II\302\240.md" index d6f1d72..f095e80 100644 --- "a/selected_coding_interview/docs/264. \344\270\221\346\225\260 II\302\240.md" +++ "b/selected_coding_interview/docs/264. \344\270\221\346\225\260 II\302\240.md" @@ -80,10 +80,3 @@ public: - **时间复杂度 $O(n)$ ：** 计算 $res$ 列表需遍历 $n-1$ 轮。 - **空间复杂度 $O(n)$ ：** 长度为 $n$ 的 $res$ 列表使用 $O(n)$ 的额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/266. \345\233\236\346\226\207\346\216\222\345\210\227.md" "b/selected_coding_interview/docs/266. \345\233\236\346\226\207\346\216\222\345\210\227.md" index 9116451..e5aa648 100644 --- "a/selected_coding_interview/docs/266. \345\233\236\346\226\207\346\216\222\345\210\227.md" +++ "b/selected_coding_interview/docs/266. \345\233\236\346\226\207\346\216\222\345\210\227.md" @@ -149,10 +149,3 @@ public: **时间复杂度 $O(N)$ ：** 其中 $N$ 为字符串长度；哈希表统计字符数量迭代 $N$ 次，判断是否为回文串最多迭代 $N$ 次，总体使用 $O(N + N) = O(N)$ 时间。 **空间复杂度 $O(N)$ ：** 哈希表 `dic` 使用 $O(N)$ 大小的额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/278. \347\254\254\344\270\200\344\270\252\351\224\231\350\257\257\347\232\204\347\211\210\346\234\254.md" "b/selected_coding_interview/docs/278. \347\254\254\344\270\200\344\270\252\351\224\231\350\257\257\347\232\204\347\211\210\346\234\254.md" index ce9efc9..7a4283b 100644 --- "a/selected_coding_interview/docs/278. \347\254\254\344\270\200\344\270\252\351\224\231\350\257\257\347\232\204\347\211\210\346\234\254.md" +++ "b/selected_coding_interview/docs/278. \347\254\254\344\270\200\344\270\252\351\224\231\350\257\257\347\232\204\347\211\210\346\234\254.md" @@ -68,10 +68,3 @@ public: - **时间复杂度 $O(\log n)$ ：** 其中 $n$ 为版本数；二分查找使用对数级别时间。 - **空间复杂度 $O(1)$ ：** 变量 $i$ , $j$ 使用常数大小空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/287. \345\257\273\346\211\276\351\207\215\345\244\215\346\225\260.md" "b/selected_coding_interview/docs/287. \345\257\273\346\211\276\351\207\215\345\244\215\346\225\260.md" index 5d0010d..a3f76d2 100644 --- "a/selected_coding_interview/docs/287. \345\257\273\346\211\276\351\207\215\345\244\215\346\225\260.md" +++ "b/selected_coding_interview/docs/287. \345\257\273\346\211\276\351\207\215\345\244\215\346\225\260.md" @@ -1,3 +1,10 @@ +## 解题思路： + +本文介绍三种时间复杂度为 $O(N)$ 的解法，其中只有方法三完全满足题目要求，其中： + +- 方法一使用了哈希表额外空间； +- 方法二需要修改原数组 `nums` ； + ## 方法一：哈希表 利用数据结构特点，容易想到使用哈希表（Set）记录数组的各个数字，当查找到重复数字则直接返回。 @@ -137,9 +144,94 @@ public: - **时间复杂度 $O(N)$ ：** 遍历数组使用 $O(N)$ ，每轮遍历的判断和交换操作使用 $O(1)$ 。 - **空间复杂度 $O(1)$ ：** 使用常数复杂度的额外空间。 ---- +## 方法三：环形链表 + +根据题意，数组索引和元素的取值范围 $\in [1, n]$ 。我们考虑建立一个 $n$ 个节点的链表： + +- $n$ 个节点的值：$1$ , $2$ , $\cdots$ , $n$ ； +- 对于每个节点 $i$ ，其 `next` 引用指向节点 $nums[i]$ 。 + +![image.png](https://pic.leetcode.cn/1699101015-eXfiQU-image.png) + +假设重复元素为 $x$ ，那么在这个链表中，一定同时有两条边指向节点 $x$ 。例如在上图中，有两条边都指向节点 $2$ 。因此可得到推论：**此链表中一定存在环，且节点 $x$ 是环的入口**。 + +换句话说，**找出重复元素 $x$ 等价于找出链表中环的入口**。这个问题实际上就是[环形链表 II](https://leetcode.cn/problems/linked-list-cycle-ii/description/)，唯一的不同点在于我们需要在数组中进行链表操作。 + +### 代码： -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** +```Python [] +class Solution: + def findDuplicate(self, nums: List[int]) -> int: + def next(i): + return nums[i] + slow = fast = 0 + # 第一次相遇 + while True: + slow = next(slow) + fast = next(next(fast)) + if slow == fast: + break + slow = 0 + # 第二次相遇 + while slow != fast: + slow = next(slow) + fast = next(fast) + return slow +``` -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 +```Java [] +public class Solution { + private int next(int[] nums, int index) { + return nums[index]; + } + public int findDuplicate(int[] nums) { + int slow = 0; + int fast = 0; + // 第一次相遇 + do { + slow = next(nums, slow); + fast = next(nums, next(nums, fast)); + } while (slow != fast); + slow = 0; + // 第二次相遇 + while (slow != fast) { + slow = next(nums, slow); + fast = next(nums, fast); + } + return slow; + } +} +``` + +```C++ [] +class Solution { +public: + vector nums; + int next(int index) { + // 直接返回当前索引处的值作为下一个索引 + return nums[index]; + } + int findDuplicate(vector& nums) { + this->nums = nums; + int slow = 0; + int fast = 0; + // 第一次相遇 + do { + slow = next(slow); + fast = next(next(fast)); + } while (slow != fast); + slow = 0; + // 第二次相遇 + while (slow != fast) { + slow = next(slow); + fast = next(fast); + } + return slow; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$ ：** 遍历链表使用 $O(N)$ 。 +- **空间复杂度 $O(1)$ ：** 使用常数复杂度的额外空间。 diff --git "a/selected_coding_interview/docs/295. \346\225\260\346\215\256\346\265\201\347\232\204\344\270\255\344\275\215\346\225\260.md" "b/selected_coding_interview/docs/295. \346\225\260\346\215\256\346\265\201\347\232\204\344\270\255\344\275\215\346\225\260.md" index 5b1ed19..4bcc08c 100644 --- "a/selected_coding_interview/docs/295. \346\225\260\346\215\256\346\265\201\347\232\204\344\270\255\344\275\215\346\225\260.md" +++ "b/selected_coding_interview/docs/295. \346\225\260\346\215\256\346\265\201\347\232\204\344\270\255\344\275\215\346\225\260.md" @@ -132,10 +132,3 @@ class MedianFinder: - **查找中位数 $O(1)$ ：** 获取堆顶元素使用 $O(1)$ 时间。 - **添加数字 $O(\log N)$ ：** 堆的插入和弹出操作使用 $O(\log N)$ 时间。 - **空间复杂度 $O(N)$ ：** 其中 $N$ 为数据流中的元素数量，小顶堆 $A$ 和大顶堆 $B$ 最多同时保存 $N$ 个元素。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/297. \344\272\214\345\217\211\346\240\221\347\232\204\345\272\217\345\210\227\345\214\226\344\270\216\345\217\215\345\272\217\345\210\227\345\214\226.md" "b/selected_coding_interview/docs/297. \344\272\214\345\217\211\346\240\221\347\232\204\345\272\217\345\210\227\345\214\226\344\270\216\345\217\215\345\272\217\345\210\227\345\214\226.md" index 92bee38..c17abca 100644 --- "a/selected_coding_interview/docs/297. \344\272\214\345\217\211\346\240\221\347\232\204\345\272\217\345\210\227\345\214\226\344\270\216\345\217\215\345\272\217\345\210\227\345\214\226.md" +++ "b/selected_coding_interview/docs/297. \344\272\214\345\217\211\346\240\221\347\232\204\345\272\217\345\210\227\345\214\226\344\270\216\345\217\215\345\272\217\345\210\227\345\214\226.md" @@ -155,10 +155,3 @@ public class Codec { - **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数，按层构建二叉树需要遍历整个 $vals$ ，其长度最大为 $2N+1$ 。 - **空间复杂度 $O(N)$ ：** 最差情况下，队列 `queue` 同时存储 $\frac{N + 1}{2}$ 个节点，因此使用 $O(N)$ 额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/3. \346\227\240\351\207\215\345\244\215\345\255\227\347\254\246\347\232\204\346\234\200\351\225\277\345\255\220\344\270\262.md" "b/selected_coding_interview/docs/3. \346\227\240\351\207\215\345\244\215\345\255\227\347\254\246\347\232\204\346\234\200\351\225\277\345\255\220\344\270\262.md" index b843447..5a69021 100644 --- "a/selected_coding_interview/docs/3. \346\227\240\351\207\215\345\244\215\345\255\227\347\254\246\347\232\204\346\234\200\351\225\277\345\255\220\344\270\262.md" +++ "b/selected_coding_interview/docs/3. \346\227\240\351\207\215\345\244\215\345\255\227\347\254\246\347\232\204\346\234\200\351\225\277\345\255\220\344\270\262.md" @@ -165,10 +165,3 @@ public: - **时间复杂度 $O(N)$ ：** 其中 $N$ 为字符串长度，动态规划需遍历计算 $dp$ 列表。 - **空间复杂度 $O(1)$ ：** 字符的 ASCII 码范围为 $0$ ~ $127$ ，哈希表 $dic$ 最多使用 $O(128) = O(1)$ 大小的额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/371. \344\270\244\346\225\264\346\225\260\344\271\213\345\222\214.md" "b/selected_coding_interview/docs/371. \344\270\244\346\225\264\346\225\260\344\271\213\345\222\214.md" index 8d215ee..63e121d 100644 --- "a/selected_coding_interview/docs/371. \344\270\244\346\225\264\346\225\260\344\271\213\345\222\214.md" +++ "b/selected_coding_interview/docs/371. \344\270\244\346\225\264\346\225\260\344\271\213\345\222\214.md" @@ -101,10 +101,3 @@ print(hex(-1 & 0xffffffff)) # = 0xffffffff 负数补码 print(-1 & 0xffffffff) # = 4294967295 （ Python 将其认为正数） ``` - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/387. \345\255\227\347\254\246\344\270\262\344\270\255\347\232\204\347\254\254\344\270\200\344\270\252\345\224\257\344\270\200\345\255\227\347\254\246.md" "b/selected_coding_interview/docs/387. \345\255\227\347\254\246\344\270\262\344\270\255\347\232\204\347\254\254\344\270\200\344\270\252\345\224\257\344\270\200\345\255\227\347\254\246.md" index bb24a9b..366d18d 100644 --- "a/selected_coding_interview/docs/387. \345\255\227\347\254\246\344\270\262\344\270\255\347\232\204\347\254\254\344\270\200\344\270\252\345\224\257\344\270\200\345\255\227\347\254\246.md" +++ "b/selected_coding_interview/docs/387. \345\255\227\347\254\246\344\270\262\344\270\255\347\232\204\347\254\254\344\270\200\344\270\252\345\224\257\344\270\200\345\255\227\347\254\246.md" @@ -64,10 +64,3 @@ public: - **时间复杂度 $O(N)$ ：** $N$ 为字符串 `s` 的长度；需遍历 `s` 两轮，使用 $O(N)$ ；HashMap 查找操作的复杂度为 $O(1)$ 。 - **空间复杂度 $O(1)$ ：** 由于题目指出 `s` 只包含小写字母，因此最多有 26 个不同字符，HashMap 存储需占用 $O(26) = O(1)$ 的额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/39. \347\273\204\345\220\210\346\200\273\345\222\214.md" "b/selected_coding_interview/docs/39. \347\273\204\345\220\210\346\200\273\345\222\214.md" index ee4af2d..d0a2726 100644 --- "a/selected_coding_interview/docs/39. \347\273\204\345\220\210\346\200\273\345\222\214.md" +++ "b/selected_coding_interview/docs/39. \347\273\204\345\220\210\346\200\273\345\222\214.md" @@ -152,10 +152,3 @@ class Solution: 如下图所示，为将数组 $[3, 4, 5]$ 和目标元素 $9$ 输入到以上代码后的整体回溯过程。 ![subset_sum_i.png](https://pic.leetcode.cn/1690624990-TxtFOY-subset_sum_i.png) - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/392. \345\210\244\346\226\255\345\255\220\345\272\217\345\210\227.md" "b/selected_coding_interview/docs/392. \345\210\244\346\226\255\345\255\220\345\272\217\345\210\227.md" index cdb4556..c44527c 100644 --- "a/selected_coding_interview/docs/392. \345\210\244\346\226\255\345\255\220\345\272\217\345\210\227.md" +++ "b/selected_coding_interview/docs/392. \345\210\244\346\226\255\345\255\220\345\272\217\345\210\227.md" @@ -67,10 +67,3 @@ public: - **时间复杂度 $O(N)$ ：** 其中 $N$ 为字符串 `t` 的长度。最差情况下需完整遍历 `t` 。 - **空间复杂度 $O(1)$ ：** `i` , `j` 变量使用常数大小空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/40. \347\273\204\345\220\210\346\200\273\345\222\214 II.md" "b/selected_coding_interview/docs/40. \347\273\204\345\220\210\346\200\273\345\222\214 II.md" index 36592a7..fa6d24a 100644 --- "a/selected_coding_interview/docs/40. \347\273\204\345\220\210\346\200\273\345\222\214 II.md" +++ "b/selected_coding_interview/docs/40. \347\273\204\345\220\210\346\200\273\345\222\214 II.md" @@ -139,10 +139,3 @@ private: 下图展示了数组 $[4, 4, 5]$ 和目标元素 $9$ 的回溯过程，共包含四种剪枝操作。请你将图示与代码注释相结合，理解整个搜索过程，以及每种剪枝操作是如何工作的。 ![subset_sum_ii.png](https://pic.leetcode.cn/1690625346-mkEVHR-subset_sum_ii.png) - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/400. \347\254\254 N \344\275\215\346\225\260\345\255\227.md" "b/selected_coding_interview/docs/400. \347\254\254 N \344\275\215\346\225\260\345\255\227.md" index a0c511a..fe8c59a 100644 --- "a/selected_coding_interview/docs/400. \347\254\254 N \344\275\215\346\225\260\345\255\227.md" +++ "b/selected_coding_interview/docs/400. \347\254\254 N \344\275\215\346\225\260\345\255\227.md" @@ -167,10 +167,3 @@ public: - **时间复杂度 $O(\log n)$ ：** 所求数位 $n$ 对应数字 $num$ 的位数 $digit$ 最大为 $O(\log n)$ ；第一步最多循环 $O(\log n)$ 次；第三步中将 $num$ 转化为字符串使用 $O(\log n)$ 时间；因此总体为 $O(\log n)$ 。 - **空间复杂度 $O(\log n)$ ：** 将数字 $num$ 转化为字符串 `str(num)` ，占用 $O(\log n)$ 的额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/409. \346\234\200\351\225\277\345\233\236\346\226\207\344\270\262.md" "b/selected_coding_interview/docs/409. \346\234\200\351\225\277\345\233\236\346\226\207\344\270\262.md" index 15ed9a2..83fea6d 100644 --- "a/selected_coding_interview/docs/409. \346\234\200\351\225\277\345\233\236\346\226\207\344\270\262.md" +++ "b/selected_coding_interview/docs/409. \346\234\200\351\225\277\345\233\236\346\226\207\344\270\262.md" @@ -84,10 +84,3 @@ public: - **时间复杂度 $O(N)$ ：** 其中 $N$ 为字符串 `s` 长度。遍历字符串 `s` 和哈希表 `counter` 皆使用线性时间。 - **空间复杂度 $O(1)$ ：** 由于 ASCII 字符数量为 128 ，哈希表 `counter` 最多使用 $O(128) = O(1)$ 空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/426. \345\260\206\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\350\275\254\345\214\226\344\270\272\346\216\222\345\272\217\347\232\204\345\217\214\345\220\221\351\223\276\350\241\250.md" "b/selected_coding_interview/docs/426. \345\260\206\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\350\275\254\345\214\226\344\270\272\346\216\222\345\272\217\347\232\204\345\217\214\345\220\221\351\223\276\350\241\250.md" index 32486f8..d7c8301 100644 --- "a/selected_coding_interview/docs/426. \345\260\206\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\350\275\254\345\214\226\344\270\272\346\216\222\345\272\217\347\232\204\345\217\214\345\220\221\351\223\276\350\241\250.md" +++ "b/selected_coding_interview/docs/426. \345\260\206\344\272\214\345\217\211\346\220\234\347\264\242\346\240\221\350\275\254\345\214\226\344\270\272\346\216\222\345\272\217\347\232\204\345\217\214\345\220\221\351\223\276\350\241\250.md" @@ -140,10 +140,3 @@ private: - **时间复杂度 $O(N)$ ：** $N$ 为二叉树的节点数，中序遍历需要访问所有节点。 - **空间复杂度 $O(N)$ ：** 最差情况下，即树退化为链表时，递归深度达到 $N$，系统使用 $O(N)$ 栈空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/46. \345\205\250\346\216\222\345\210\227.md" "b/selected_coding_interview/docs/46. \345\205\250\346\216\222\345\210\227.md" index a7fa6e3..a5baa2d 100644 --- "a/selected_coding_interview/docs/46. \345\205\250\346\216\222\345\210\227.md" +++ "b/selected_coding_interview/docs/46. \345\205\250\346\216\222\345\210\227.md" @@ -106,10 +106,3 @@ private: - **时间复杂度 $O(N!N)$ ：** $N$ 为数组 `nums` 的长度；时间复杂度和数组排列的方案数成线性关系，方案数为 $N \times (N-1) \times (N-2) … \times 2 \times 1$ ，即复杂度为 $O(N!)$ ；数组拼接操作 `join()` 使用 $O(N)$ ；因此总体时间复杂度为 $O(N!N)$ 。 - **空间复杂度 $O(N)$ ：** 全排列的递归深度为 $N$ ，系统累计使用栈空间大小为 $O(N)$ 。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/47. \345\205\250\346\216\222\345\210\227 II.md" "b/selected_coding_interview/docs/47. \345\205\250\346\216\222\345\210\227 II.md" index cba1780..9d0ff1a 100644 --- "a/selected_coding_interview/docs/47. \345\205\250\346\216\222\345\210\227 II.md" +++ "b/selected_coding_interview/docs/47. \345\205\250\346\216\222\345\210\227 II.md" @@ -115,10 +115,3 @@ private: - **时间复杂度 $O(N!N)$ ：** 与上一题相同 。 - **空间复杂度 $O(N^2)$ ：** 递归中辅助 Set 累计存储的元素数量最多为 $N + (N-1) + ... + 2 + 1 = (N+1)N/2$ ，因此占用 $O(N^2)$ 的额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/48. \346\227\213\350\275\254\345\233\276\345\203\217.md" "b/selected_coding_interview/docs/48. \346\227\213\350\275\254\345\233\276\345\203\217.md" index ae20ed3..dd746ec 100644 --- "a/selected_coding_interview/docs/48. \346\227\213\350\275\254\345\233\276\345\203\217.md" +++ "b/selected_coding_interview/docs/48. \346\227\213\350\275\254\345\233\276\345\203\217.md" @@ -224,10 +224,3 @@ public: - **时间复杂度 $O(N^2)$ ：** 其中 $N$ 为输入矩阵的行（列）数。需要将矩阵中每个元素旋转到新的位置，即对矩阵所有元素操作一次，使用 $O(N^2)$ 时间。 - **空间复杂度 $O(1)$ ：** 临时变量 $tmp$ 使用常数大小的额外空间。值得注意，当循环中进入下轮迭代，上轮迭代初始化的 $tmp$ 占用的内存就会被自动释放，因此无累计使用空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/509. \346\226\220\346\263\242\351\202\243\345\245\221\346\225\260.md" "b/selected_coding_interview/docs/509. \346\226\220\346\263\242\351\202\243\345\245\221\346\225\260.md" index b6a88e8..9839507 100644 --- "a/selected_coding_interview/docs/509. \346\226\220\346\263\242\351\202\243\345\245\221\346\225\260.md" +++ "b/selected_coding_interview/docs/509. \346\226\220\346\263\242\351\202\243\345\245\221\346\225\260.md" @@ -74,10 +74,3 @@ public: - **时间复杂度 $O(n)$ ：** 计算 $f(n)$ 需循环 $n$ 次，每轮循环内计算操作使用 $O(1)$ 。 - **空间复杂度 $O(1)$ ：** 几个标志变量使用常数大小的额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/53. \346\234\200\345\244\247\345\255\220\346\225\260\347\273\204\345\222\214.md" "b/selected_coding_interview/docs/53. \346\234\200\345\244\247\345\255\220\346\225\260\347\273\204\345\222\214.md" index fe42a24..3e1fb62 100644 --- "a/selected_coding_interview/docs/53. \346\234\200\345\244\247\345\255\220\346\225\260\347\273\204\345\222\214.md" +++ "b/selected_coding_interview/docs/53. \346\234\200\345\244\247\345\255\220\346\225\260\347\273\204\345\222\214.md" @@ -77,10 +77,3 @@ public: - **时间复杂度 $O(N)$ ：** 线性遍历数组 $nums$ 即可获得结果，使用 $O(N)$ 时间。 - **空间复杂度 $O(1)$ ：** 使用常数大小的额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/54. \350\236\272\346\227\213\347\237\251\351\230\265.md" "b/selected_coding_interview/docs/54. \350\236\272\346\227\213\347\237\251\351\230\265.md" index ef91bda..159691e 100644 --- "a/selected_coding_interview/docs/54. \350\236\272\346\227\213\347\237\251\351\230\265.md" +++ "b/selected_coding_interview/docs/54. \350\236\272\346\227\213\347\237\251\351\230\265.md" @@ -103,10 +103,3 @@ public: - **时间复杂度 $O(MN)$ ：** $M, N$ 分别为矩阵行数和列数。 - **空间复杂度 $O(1)$ ：** 四个边界 `l` , `r` , `t` , `b` 使用常数大小的额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/65. \346\234\211\346\225\210\346\225\260\345\255\227.md" "b/selected_coding_interview/docs/65. \346\234\211\346\225\210\346\225\260\345\255\227.md" index 7cfeaa2..b71993d 100644 --- "a/selected_coding_interview/docs/65. \346\234\211\346\225\210\346\225\260\345\255\227.md" +++ "b/selected_coding_interview/docs/65. \346\234\211\346\225\210\346\225\260\345\255\227.md" @@ -137,10 +137,3 @@ public: - **时间复杂度 $O(N)$ ：** 其中 $N$ 为字符串 `s` 的长度，判断需遍历字符串，每轮状态转移的使用 $O(1)$ 时间。 - **空间复杂度 $O(1)$ ：** `states` 和 `p` 使用常数大小的额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/70. \347\210\254\346\245\274\346\242\257.md" "b/selected_coding_interview/docs/70. \347\210\254\346\245\274\346\242\257.md" index 8d2c3e5..5195a3c 100644 --- "a/selected_coding_interview/docs/70. \347\210\254\346\245\274\346\242\257.md" +++ "b/selected_coding_interview/docs/70. \347\210\254\346\245\274\346\242\257.md" @@ -69,10 +69,3 @@ public: - **时间复杂度 $O(n)$ ：** 计算 $f(n)$ 需循环 $n$ 次，每轮循环内计算操作使用 $O(1)$ 。 - **空间复杂度 $O(1)$ ：** 几个标志变量使用常数大小的额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/704. \344\272\214\345\210\206\346\237\245\346\211\276.md" "b/selected_coding_interview/docs/704. \344\272\214\345\210\206\346\237\245\346\211\276.md" index 8fcc61b..4d01d55 100644 --- "a/selected_coding_interview/docs/704. \344\272\214\345\210\206\346\237\245\346\211\276.md" +++ "b/selected_coding_interview/docs/704. \344\272\214\345\210\206\346\237\245\346\211\276.md" @@ -77,10 +77,3 @@ public: - **时间复杂度 $O(\log N)$ ：** 其中 $N$ 为数组 $nums$ 长度。二分查找使用对数级别时间。 - **空间复杂度 $O(1)$ ：** 变量 $i$ , $j$ 使用常数大小空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/724. \345\257\273\346\211\276\346\225\260\347\273\204\347\232\204\344\270\255\345\277\203\344\270\213\346\240\207.md" "b/selected_coding_interview/docs/724. \345\257\273\346\211\276\346\225\260\347\273\204\347\232\204\344\270\255\345\277\203\344\270\213\346\240\207.md" index 1ff9265..8bf9ad9 100644 --- "a/selected_coding_interview/docs/724. \345\257\273\346\211\276\346\225\260\347\273\204\347\232\204\344\270\255\345\277\203\344\270\213\346\240\207.md" +++ "b/selected_coding_interview/docs/724. \345\257\273\346\211\276\346\225\260\347\273\204\347\232\204\344\270\255\345\277\203\344\270\213\346\240\207.md" @@ -74,10 +74,3 @@ public: - **时间复杂度 $O(N)$ ：** 其中 $N$ 为数组 `nums` 长度。求和操作使用 $O(N)$ 线性时间，遍历 `nums` 最差使用 $O(N)$ 线性时间。 - **空间复杂度 $O(1)$ ：** 变量 `sum_left` , `sum_right` 使用常数大小空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/79. \345\215\225\350\257\215\346\220\234\347\264\242.md" "b/selected_coding_interview/docs/79. \345\215\225\350\257\215\346\220\234\347\264\242.md" index d9eff42..f070a1e 100644 --- "a/selected_coding_interview/docs/79. \345\215\225\350\257\215\346\220\234\347\264\242.md" +++ "b/selected_coding_interview/docs/79. \345\215\225\350\257\215\346\220\234\347\264\242.md" @@ -99,10 +99,3 @@ private: - **时间复杂度 $O(3^KMN)$ ：** 最差情况下，需要遍历矩阵中长度为 $K$ 字符串的所有方案，时间复杂度为 $O(3^K)$；矩阵中共有 $MN$ 个起点，时间复杂度为 $O(MN)$ 。 - **方案数计算：** 设字符串长度为 $K$ ，搜索中每个字符有上、下、左、右四个方向可以选择，舍弃回头（上个字符）的方向，剩下 $3$ 种选择，因此方案数的复杂度为 $O(3^K)$ 。 - **空间复杂度 $O(K)$ ：** 搜索过程中的递归深度不超过 $K$ ，因此系统因函数调用累计使用的栈空间占用 $O(K)$ （因为函数返回后，系统调用的[栈空间会释放](https://leetcode-cn.com/explore/orignial/card/recursion-i/259/complexity-analysis/1223/)）。最坏情况下 $K = MN$ ，递归深度为 $MN$ ，此时系统栈使用 $O(MN)$ 的额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/796. \346\227\213\350\275\254\345\255\227\347\254\246\344\270\262.md" "b/selected_coding_interview/docs/796. \346\227\213\350\275\254\345\255\227\347\254\246\344\270\262.md" index c81cece..0ad5a69 100644 --- "a/selected_coding_interview/docs/796. \346\227\213\350\275\254\345\255\227\347\254\246\344\270\262.md" +++ "b/selected_coding_interview/docs/796. \346\227\213\350\275\254\345\255\227\347\254\246\344\270\262.md" @@ -52,10 +52,3 @@ public: > 本文直接调用编程语言的库函数，时间复杂度由库函数的具体实现方法确定。 **空间复杂度 $O(N)$ ：** 构造拼接字符串 $goal \ goal$ 使用 $O(N)$ 大小的额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/8. \345\255\227\347\254\246\344\270\262\350\275\254\346\215\242\346\225\264\346\225\260 (atoi).md" "b/selected_coding_interview/docs/8. \345\255\227\347\254\246\344\270\262\350\275\254\346\215\242\346\225\264\346\225\260 (atoi).md" index dc5a92f..266010b 100644 --- "a/selected_coding_interview/docs/8. \345\255\227\347\254\246\344\270\262\350\275\254\346\215\242\346\225\264\346\225\260 (atoi).md" +++ "b/selected_coding_interview/docs/8. \345\255\227\347\254\246\344\270\262\350\275\254\346\215\242\346\225\264\346\225\260 (atoi).md" @@ -141,10 +141,3 @@ public: } }; ``` - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/86. \345\210\206\351\232\224\351\223\276\350\241\250.md" "b/selected_coding_interview/docs/86. \345\210\206\351\232\224\351\223\276\350\241\250.md" index 644819f..8f7d9dc 100644 --- "a/selected_coding_interview/docs/86. \345\210\206\351\232\224\351\223\276\350\241\250.md" +++ "b/selected_coding_interview/docs/86. \345\210\206\351\232\224\351\223\276\350\241\250.md" @@ -165,10 +165,3 @@ public: **时间复杂度 $O(N)$ ：** 其中 $N$ 为链表长度；遍历链表使用线性时间。 **空间复杂度 $O(1)$ ：** 假头节点使用常数大小的额外空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/876. \351\223\276\350\241\250\347\232\204\344\270\255\351\227\264\347\273\223\347\202\271.md" "b/selected_coding_interview/docs/876. \351\223\276\350\241\250\347\232\204\344\270\255\351\227\264\347\273\223\347\202\271.md" index 7a43e02..d1bf064 100644 --- "a/selected_coding_interview/docs/876. \351\223\276\350\241\250\347\232\204\344\270\255\351\227\264\347\273\223\347\202\271.md" +++ "b/selected_coding_interview/docs/876. \351\223\276\350\241\250\347\232\204\344\270\255\351\227\264\347\273\223\347\202\271.md" @@ -58,10 +58,3 @@ public: - **时间复杂度 $O(N)$ ：** 其中 $N$ 为链表长度。 `fast` 遍历完链表需 $\frac{N}{2}$ 轮迭代，使用 $O(\frac{N}{2}) = O(N)$ 线性时间。 - **空间复杂度 $O(1)$ ：** 节点指针 `fast` , `slow` 使用常数大小空间。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 diff --git "a/selected_coding_interview/docs/946. \351\252\214\350\257\201\346\240\210\345\272\217\345\210\227.md" "b/selected_coding_interview/docs/946. \351\252\214\350\257\201\346\240\210\345\272\217\345\210\227.md" index 4edd17e..94a5e17 100644 --- "a/selected_coding_interview/docs/946. \351\252\214\350\257\201\346\240\210\345\272\217\345\210\227.md" +++ "b/selected_coding_interview/docs/946. \351\252\214\350\257\201\346\240\210\345\272\217\345\210\227.md" @@ -80,10 +80,3 @@ public: - **时间复杂度 $O(N)$ ：** 其中 $N$ 为列表 $pushed$ 的长度；每个元素最多入栈与出栈一次，即最多共 $2N$ 次出入栈操作。 - **空间复杂度 $O(N)$ ：** 辅助栈 $stack$ 最多同时存储 $N$ 个元素。 - ---- - -**本学习计划的配套代码请见「[代码仓](https://github.com/krahets/selected-coding-interview)」，包括：** - -- 代码与测试样例，方便 VSCode 本地运行与调试代码。 -- 数据结构封装，提升刷题与 Debug 效率。 From c12faa39f70b5ddbcd2830db88ac74fea599442b Mon Sep 17 00:00:00 2001 From: Yudong Jin Date: Tue, 21 Nov 2023 21:54:33 +0800 Subject: [PATCH 18/23] Update README.md --- README.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/README.md b/README.md index dddae9e..cd258f3 100755 --- a/README.md +++ b/README.md @@ -17,7 +17,7 @@ - 题目活跃于各大互联网公司招聘中，可使笔面试准备事半功倍。 - 致力于行文深入浅出、图文搭配，提供简洁的 Python, Java, C++ 解题代码。 -> **Note** +> [!NOTE] > > 本仓库包含“图解算法数据结构”、“Krahets 笔面试精选 88 题”和“剑指 Offer”的题解内容： > From cda99e77dbf2fc5bcf9f51628af905b096ef2411 Mon Sep 17 00:00:00 2001 From: Yudong Jin Date: Thu, 21 Dec 2023 00:44:47 +0800 Subject: [PATCH 19/23] Update README.md --- README.md | 16 ++++++++-------- 1 file changed, 8 insertions(+), 8 deletions(-) diff --git a/README.md b/README.md index cd258f3..514cee8 100755 --- a/README.md +++ b/README.md @@ -23,11 +23,11 @@ > > ```python > LeetCode-Book -> ├── leetbook_ioa # 《图解算法数据结构》题解和专栏文档、代码 -> ├── selected_coding_interview # 《Krahets 笔面试精选 88 题》题解文档、代码 +> ├── leetbook_ioa # 《图解算法数据结构》题解和专栏文档 +> ├── selected_coding_interview # 《Krahets 笔面试精选 88 题》题解文档 > └── sword_for_offer # 《剑指 Offer》题解文档、代码、刷题计划 > ``` - +> > 若本 LeetBook 对您有所帮助，请在页面右上角点个 Star :star: 支持一下，谢谢！ ## 如何学习算法 @@ -46,8 +46,8 @@ 推荐以下 LeetCode 题单（前两项 [@Krahets](https://leetcode.cn/u/jyd/) 题解全覆盖）： -1. [Krahets 笔面试精选 88 题](https://leetcode.cn/studyplan/selected-coding-interview/)：从“剑指 Offer”和“热题 100”精选出的 88 道高频算法笔试题。 -2. [图解算法数据结构](https://leetcode-cn.com/leetbook/detail/illustration-of-algorithm/)（[Interview-75](https://leetcode.cn/studyplan/coding-interviews/)）：题目更贴近实际应用，相对“剑指 Offer”难度有所增加。 +1. [Krahets 笔面试精选 88 题](https://leetcode.cn/studyplan/selected-coding-interview/)：从“剑指 Offer”和“热题 100”精选出的 88 道高频算法笔试题，适合初学者入门。 +2. [图解算法数据结构](https://leetcode-cn.com/leetbook/detail/illustration-of-algorithm/)（[Interview-75](https://leetcode.cn/studyplan/coding-interviews/)）：题目更贴近实际应用，相较于“剑指 Offer”难度有所增加。 3. [LeetCode 热题 100](https://leetcode.cn/studyplan/top-100-liked/)：力扣用户最喜爱的 100 道题。 4. [面试经典 150 题](https://leetcode.cn/studyplan/top-interview-150/)：150 道经典面试力扣题。 5. [LeetCode-75](https://leetcode.cn/studyplan/leetcode-75/)：精选面试必备的 75 道核心题目。 @@ -56,11 +56,11 @@ 1. 建议每日刷 2~3 题。若能轻松完成，可以尝试增加至 5～8 题。 2. 刷题的质量远重要于数量。务必确保你真正理解了每个题目的解法及背后的算法原理。 -3. 建议你按照目录顺序做题。如果感觉某题很难，可以先跳过，稍后再回顾挑战。 -4. 很多题目都有不止一种解法，请你注意比较和探讨各种方法的特点和适用情况。 +3. 建议你按照目录顺序做题。如果感觉某题很难，可以先跳过，后续再攻克。 +4. 题目通常有不止一种解法，请你注意比较和探讨各种方法的特点和适用情况。 5. 如果你发现自己遗忘了题目解法，不必灰心。艾宾浩斯遗忘曲线指出，通常需要复习至少三次，才能真正掌握一个知识点。 6. 行百里者半九十。坚持至关重要，加油，相信你可以做到！ ## License -The texts, code and images in this repository are licensed under [CC BY-NC-SA-4.0](https://creativecommons.org/licenses/by-nc-sa/4.0/). +The texts, code, and images in this repository are licensed under [CC BY-NC-SA-4.0](https://creativecommons.org/licenses/by-nc-sa/4.0/). From 897833b40f9609d79afeb18256ff80b984ef6edf Mon Sep 17 00:00:00 2001 From: Yudong Jin Date: Thu, 4 Jan 2024 13:13:08 +0800 Subject: [PATCH 20/23] Update README.md --- README.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/README.md b/README.md index 514cee8..bea7826 100755 --- a/README.md +++ b/README.md @@ -50,11 +50,11 @@ 2. [图解算法数据结构](https://leetcode-cn.com/leetbook/detail/illustration-of-algorithm/)（[Interview-75](https://leetcode.cn/studyplan/coding-interviews/)）：题目更贴近实际应用，相较于“剑指 Offer”难度有所增加。 3. [LeetCode 热题 100](https://leetcode.cn/studyplan/top-100-liked/)：力扣用户最喜爱的 100 道题。 4. [面试经典 150 题](https://leetcode.cn/studyplan/top-interview-150/)：150 道经典面试力扣题。 -5. [LeetCode-75](https://leetcode.cn/studyplan/leetcode-75/)：精选面试必备的 75 道核心题目。 +5. [LeetCode-75](https://leetcode.cn/studyplan/leetcode-75/)：精选 75 道面试核心题目。 对初学者的刷题建议： -1. 建议每日刷 2~3 题。若能轻松完成，可以尝试增加至 5～8 题。 +1. 建议每日刷 2～3 题。若能轻松完成，可以尝试增加至 5～8 题。 2. 刷题的质量远重要于数量。务必确保你真正理解了每个题目的解法及背后的算法原理。 3. 建议你按照目录顺序做题。如果感觉某题很难，可以先跳过，后续再攻克。 4. 题目通常有不止一种解法，请你注意比较和探讨各种方法的特点和适用情况。 From e0636c8424c762989b3e3baad70514b85c2fd77f Mon Sep 17 00:00:00 2001 From: Yudong Jin Date: Fri, 23 Feb 2024 13:46:38 +0800 Subject: [PATCH 21/23] Update README.md --- README.md | 24 ++++++++++++++++-------- 1 file changed, 16 insertions(+), 8 deletions(-) diff --git a/README.md b/README.md index bea7826..0a62dce 100755 --- a/README.md +++ b/README.md @@ -38,13 +38,18 @@

-[《Hello 算法》](https://github.com/krahets/hello-algo)：动画图解、一键运行的数据结构与算法教程。 +

+ 《Hello 算法》 —— 动画图解、一键运行的数据结构与算法教程 +

-这本书旨在引导初学者探索数据结构与算法的知识地图，掌握刷题需要的前置知识与工具库。 +此书旨在引导初学者探索数据结构与算法的知识地图，掌握刷题需要的前置知识与工具库。 + +> [!TIP] +> 建议先读完这本书（概括且全面地了解数据结构与算法），再开始刷题（深入探索各类算法和数据结构）。 ### 第二步：刷算法题 -推荐以下 LeetCode 题单（前两项 [@Krahets](https://leetcode.cn/u/jyd/) 题解全覆盖）： +推荐以下 LeetCode 题单： 1. [Krahets 笔面试精选 88 题](https://leetcode.cn/studyplan/selected-coding-interview/)：从“剑指 Offer”和“热题 100”精选出的 88 道高频算法笔试题，适合初学者入门。 2. [图解算法数据结构](https://leetcode-cn.com/leetbook/detail/illustration-of-algorithm/)（[Interview-75](https://leetcode.cn/studyplan/coding-interviews/)）：题目更贴近实际应用，相较于“剑指 Offer”难度有所增加。 @@ -52,14 +57,17 @@ 4. [面试经典 150 题](https://leetcode.cn/studyplan/top-interview-150/)：150 道经典面试力扣题。 5. [LeetCode-75](https://leetcode.cn/studyplan/leetcode-75/)：精选 75 道面试核心题目。 -对初学者的刷题建议： +> [!TIP] +> 第 `1.`、`2.` 项的题目分类与《Hello 算法》的章节内容对应，且包含 [@krahets](https://leetcode.cn/u/jyd/) 写的题目解析，适合作为配套习题。 + +对初学者刷题的一些建议： 1. 建议每日刷 2～3 题。若能轻松完成，可以尝试增加至 5～8 题。 -2. 刷题的质量远重要于数量。务必确保你真正理解了每个题目的解法及背后的算法原理。 -3. 建议你按照目录顺序做题。如果感觉某题很难，可以先跳过，后续再攻克。 +2. 刷题的质量比数量更加重要。请确保你真正理解了每个题目的解法及背后的算法原理。 +3. 建议你按照题单目录的顺序做题。如果感觉一道题很难，可以先跳过，后续再攻克。 4. 题目通常有不止一种解法，请你注意比较和探讨各种方法的特点和适用情况。 -5. 如果你发现自己遗忘了题目解法，不必灰心。艾宾浩斯遗忘曲线指出，通常需要复习至少三次，才能真正掌握一个知识点。 -6. 行百里者半九十。坚持至关重要，加油，相信你可以做到！ +5. 如果你发现自己遗忘了题目解法，不必灰心；我们通常需要复习三次以上，才能真正掌握一个知识点。 +6. 行百里者半九十。坚持至关重要，加油！ ## License From af01eae09e4ca78ebca324071c2e707d33c3d0b2 Mon Sep 17 00:00:00 2001 From: Yudong Jin Date: Fri, 23 Feb 2024 13:53:13 +0800 Subject: [PATCH 22/23] Update README.md --- README.md | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) diff --git a/README.md b/README.md index 0a62dce..1163fa5 100755 --- a/README.md +++ b/README.md @@ -27,8 +27,8 @@ > ├── selected_coding_interview # 《Krahets 笔面试精选 88 题》题解文档 > └── sword_for_offer # 《剑指 Offer》题解文档、代码、刷题计划 > ``` -> -> 若本 LeetBook 对您有所帮助，请在页面右上角点个 Star :star: 支持一下，谢谢！ + +> 若本仓库对您有所帮助，请在页面右上角点个 **Star** :star: 支持一下，谢谢！ ## 如何学习算法 @@ -58,7 +58,7 @@ 5. [LeetCode-75](https://leetcode.cn/studyplan/leetcode-75/)：精选 75 道面试核心题目。 > [!TIP] -> 第 `1.`、`2.` 项的题目分类与《Hello 算法》的章节内容对应，且包含 [@krahets](https://leetcode.cn/u/jyd/) 写的题目解析，适合作为配套习题。 +> 第 `1.` `2.` 项的题目分类与《Hello 算法》的章节内容对应，且提供 [@krahets](https://leetcode.cn/u/jyd/) 写的题解，适合作为配套习题。 对初学者刷题的一些建议： From c27d12f2d23d43fd283cba11d6150b51b05ad6dc Mon Sep 17 00:00:00 2001 From: krahets Date: Sat, 11 May 2024 15:33:42 +0800 Subject: [PATCH 23/23] Fix 6. and 239. in selected_coding_interview --- ...43\346\234\200\345\244\247\345\200\274.md" | 302 ++++++++---------- ...27\345\275\242\345\217\230\346\215\242.md" | 57 +++- 2 files changed, 180 insertions(+), 179 deletions(-) diff --git "a/selected_coding_interview/docs/239. \346\273\221\345\212\250\347\252\227\345\217\243\346\234\200\345\244\247\345\200\274.md" "b/selected_coding_interview/docs/239. \346\273\221\345\212\250\347\252\227\345\217\243\346\234\200\345\244\247\345\200\274.md" index 1eaf142..4c24875 100644 --- "a/selected_coding_interview/docs/239. \346\273\221\345\212\250\347\252\227\345\217\243\346\234\200\345\244\247\345\200\274.md" +++ "b/selected_coding_interview/docs/239. \346\273\221\345\212\250\347\252\227\345\217\243\346\234\200\345\244\247\345\200\274.md" @@ -1,140 +1,89 @@ -# 方法一：辅助矩阵 +## 解题思路： -如下图所示，矩阵顺时针旋转 90º 后，可找到以下规律： +设窗口区间为 $[i, j]$ ，最大值为 $x_j$ 。当窗口向前移动一格，则区间变为 $[i+1,j+1]$ ，即添加了 $nums[j + 1]$ ，删除了 $nums[i]$ 。 -- 「第 $i$ 行」元素旋转到「第 $n - 1 - i$ 列」元素； -- 「第 $j$ 列」元素旋转到「第 $j$ 行」元素； - -因此，对于矩阵任意第 $i$ 行、第 $j$ 列元素 $matrix[i][j]$ ，矩阵旋转 90º 后「元素位置旋转公式」为： +若只向窗口 $[i, j]$ 右边添加数字 $nums[j + 1]$ ，则新窗口最大值可以 **通过一次对比** 使用 $O(1)$ 时间得到，即： $$ -\begin{aligned} -matrix[i][j] & \rightarrow matrix[j][n - 1 - i] \\ -原索引位置 & \rightarrow 旋转后索引位置 -\end{aligned} +x_{j+1} = \max(x_{j}, nums[j + 1]) $$ -![ccw-01-07.001.png](https://pic.leetcode-cn.com/1638557961-AVzCQb-ccw-01-07.001.png) - -根据以上「元素旋转公式」，考虑遍历矩阵，将各元素依次写入到旋转后的索引位置。但仍**存在问题**：在写入一个元素 $matrix[i][j] \rightarrow matrix[j][n - 1 - i]$ 后，原矩阵元素 $matrix[j][n - 1 - i]$ 就会**被覆盖（即丢失）**，而此丢失的元素就无法被写入到旋转后的索引位置了。 - -为解决此问题，考虑借助一个「辅助矩阵」暂存原矩阵，通过遍历辅助矩阵所有元素，将各元素填入「原矩阵」旋转后的新索引位置即可。 - -```Python [] -class Solution: - def rotate(self, matrix: List[List[int]]) -> None: - n = len(matrix) - # 深拷贝 matrix -> tmp - tmp = copy.deepcopy(matrix) - # 根据元素旋转公式，遍历修改原矩阵 matrix 的各元素 - for i in range(n): - for j in range(n): - matrix[j][n - 1 - i] = tmp[i][j] -``` - -```Java [] -class Solution { - public void rotate(int[][] matrix) { - int n = matrix.length; - // 深拷贝 matrix -> tmp - int[][] tmp = new int[n][]; - for (int i = 0; i < n; i++) - tmp[i] = matrix[i].clone(); - // 根据元素旋转公式，遍历修改原矩阵 matrix 的各元素 - for (int i = 0; i < n; i++) { - for (int j = 0; j < n; j++) { - matrix[j][n - 1 - i] = tmp[i][j]; - } - } - } -} -``` - -```C++ [] -class Solution { -public: - void rotate(vector>& matrix) { - int n = matrix.size(); - // 深拷贝 matrix -> tmp - vector> tmp = matrix; - // 根据元素旋转公式，遍历修改原矩阵 matrix 的各元素 - for (int i = 0; i < n; i++) { - for (int j = 0; j < n; j++) { - matrix[j][n - 1 - i] = tmp[i][j]; - } - } - } -}; -``` - -如以上代码所示，遍历矩阵所有元素的时间复杂度为 $O(N^2)$ ；由于借助了一个辅助矩阵，**空间复杂度**为 $O(N^2)$ 。 - -# 方法二：原地修改 - -考虑不借助辅助矩阵，通过在原矩阵中直接「原地修改」，实现空间复杂度 $O(1)$ 的解法。 - -以位于矩阵四个角点的元素为例，设矩阵左上角元素 $A$ 、右上角元素 $B$ 、右下角元素 $C$ 、左下角元素 $D$ 。矩阵旋转 90º 后，相当于依次先后执行 $D \rightarrow A$ , $C \rightarrow D$ , $B \rightarrow C$ , $A \rightarrow B$ 修改元素，即如下「首尾相接」的元素旋转操作： +而由于删除的 $nums[i]$ 可能恰好是窗口内唯一的最大值 $x_j$ ，因此不能通过以上方法计算 $x_{j+1}$ ，而必须使用 $O(j-i)$ 时间， **遍历整个窗口区间** 获取最大值，即： $$ -A \leftarrow D \leftarrow C \leftarrow B \leftarrow A +x_{j+1} = \max(nums(i+1), \cdots , num(j+1)) $$ -![ccw-01-07.002.png](https://pic.leetcode-cn.com/1638557961-BSxFQQ-ccw-01-07.002.png) +根据以上分析，可得 **暴力法** 的时间复杂度为 $O((n-k+1)k) \approx O(nk)$ 。 -如上图所示，由于第 $1$ 步 $D \rightarrow A$ 已经将 $A$ 覆盖（导致 $A$ 丢失），此丢失导致最后第 $4$ 步 $A \rightarrow B$ 无法赋值。为解决此问题，考虑借助一个「辅助变量 $tmp$ 」预先存储 $A$ ，此时的旋转操作变为： +- 设数组 $nums$ 的长度为 $n$ ，则共有 $(n-k+1)$ 个窗口； +- 获取每个窗口最大值需线性遍历，时间复杂度为 $O(k)$ 。 -$$ -暂存 \ tmp = A \\ -A \leftarrow D \leftarrow C \leftarrow B \leftarrow tmp -$$ +![Picture1.png](https://pic.leetcode-cn.com/1600878237-pBiBdf-Picture1.png){:width=650} -![ccw-01-07.003.png](https://pic.leetcode-cn.com/1638557961-hYpOoH-ccw-01-07.003.png) +> **本题难点：** 如何在每次窗口滑动后，将 “获取窗口内最大值” 的时间复杂度从 $O(k)$ 降低至 $O(1)$ 。 -如上图所示，一轮可以完成矩阵 4 个元素的旋转。因而，只要分别以矩阵左上角 $1/4$ 的各元素为起始点执行以上旋转操作，即可完整实现矩阵旋转。 +回忆 [最小栈](https://leetcode.cn/problems/min-stack/) ，其使用 **单调栈** 实现了随意入栈、出栈情况下的 $O(1)$ 时间获取 “栈内最小值” 。本题同理，不同点在于 “出栈操作” 删除的是 “列表尾部元素” ，而 “窗口滑动” 删除的是 “列表首部元素” 。 -具体来看，当矩阵大小 $n$ 为偶数时，取前 $\frac{n}{2}$ 行、前 $\frac{n}{2}$ 列的元素为起始点；当矩阵大小 $n$ 为奇数时，取前 $\frac{n}{2}$ 行、前 $\frac{n + 1}{2}$ 列的元素为起始点。 +窗口对应的数据结构为 **双端队列** ，本题使用 **单调队列** 即可解决以上问题。遍历数组时，每轮保证单调队列 $deque$ ： -令 $matrix[i][j] = A$ ，根据文章开头的元素旋转公式，可推导得适用于任意起始点的元素旋转操作： +1. $deque$ 内 **仅包含窗口内的元素** $\Rightarrow$ 每轮窗口滑动移除了元素 $nums[i - 1]$ ，需将 $deque$ 内的对应元素一起删除。 +2. $deque$ 内的元素 **非严格递减** $\Rightarrow$ 每轮窗口滑动添加了元素 $nums[j + 1]$ ，需将 $deque$ 内所有 $< nums[j + 1]$ 的元素删除。 -$$ -暂存 tmp = matrix[i][j] \\ -matrix[i][j] \leftarrow matrix[n - 1 - j][i] \leftarrow matrix[n - 1 - i][n - 1 - j] \leftarrow matrix[j][n - 1 - i] \leftarrow tmp -$$ +### 算法流程： -> 如下图所示，为示例矩阵的算法执行流程。 +1. **初始化：** 双端队列 $deque$ ，结果列表 $res$ ，数组长度 $n$ ； +2. **滑动窗口：** 左边界范围 $i \in [1 - k, n - k]$ ，右边界范围 $j \in [0, n - 1]$ ； + 1. 若 $i > 0$ 且 队首元素 $deque[0]$ $=$ 被删除元素 $nums[i - 1]$ ：则队首元素出队； + 2. 删除 $deque$ 内所有 $< nums[j]$ 的元素，以保持 $deque$ 递减； + 3. 将 $nums[j]$ 添加至 $deque$ 尾部； + 4. 若已形成窗口（即 $i \geq 0$ ）：将窗口最大值（即队首元素 $deque[0]$ ）添加至列表 $res$ ； +3. **返回值：** 返回结果列表 $res$ ； - + -## 代码 +## 代码： -> 后三个 Tab 为「代码注释解析」。 +Python 通过 `zip(range(), range())` 可实现滑动窗口的左右边界 `i, j` 同时遍历。 ```Python [] class Solution: - def rotate(self, matrix: List[List[int]]) -> None: - n = len(matrix) - for i in range(n // 2): - for j in range((n + 1) // 2): - tmp = matrix[i][j] - matrix[i][j] = matrix[n - 1 - j][i] - matrix[n - 1 - j][i] = matrix[n - 1 - i][n - 1 - j] - matrix[n - 1 - i][n - 1 - j] = matrix[j][n - 1 - i] - matrix[j][n - 1 - i] = tmp + def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]: + deque = collections.deque() + res, n = [], len(nums) + for i, j in zip(range(1 - k, n + 1 - k), range(n)): + # 删除 deque 中对应的 nums[i-1] + if i > 0 and deque[0] == nums[i - 1]: + deque.popleft() + # 保持 deque 递减 + while deque and deque[-1] < nums[j]: + deque.pop() + deque.append(nums[j]) + # 记录窗口最大值 + if i >= 0: + res.append(deque[0]) + return res ``` ```Java [] class Solution { - public void rotate(int[][] matrix) { - int n = matrix.length; - for (int i = 0; i < n / 2; i++) { - for (int j = 0; j < (n + 1) / 2; j++) { - int tmp = matrix[i][j]; - matrix[i][j] = matrix[n - 1 - j][i]; - matrix[n - 1 - j][i] = matrix[n - 1 - i][n - 1 - j]; - matrix[n - 1 - i][n - 1 - j] = matrix[j][n - 1 - i]; - matrix[j][n - 1 - i] = tmp; - } + public int[] maxSlidingWindow(int[] nums, int k) { + if(nums.length == 0 || k == 0) return new int[0]; + Deque deque = new LinkedList<>(); + int[] res = new int[nums.length - k + 1]; + for(int j = 0, i = 1 - k; j < nums.length; i++, j++) { + // 删除 deque 中对应的 nums[i-1] + if(i > 0 && deque.peekFirst() == nums[i - 1]) + deque.removeFirst(); + // 保持 deque 递减 + while(!deque.isEmpty() && deque.peekLast() < nums[j]) + deque.removeLast(); + deque.addLast(nums[j]); + // 记录窗口最大值 + if(i >= 0) + res[i] = deque.peekFirst(); } + return res; } } ``` @@ -142,57 +91,75 @@ class Solution { ```C++ [] class Solution { public: - void rotate(vector>& matrix) { - int n = matrix.size(); - for (int i = 0; i < n / 2; i++) { - for (int j = 0; j < (n + 1) / 2; j++) { - int tmp = matrix[i][j]; - matrix[i][j] = matrix[n - 1 - j][i]; - matrix[n - 1 - j][i] = matrix[n - 1 - i][n - 1 - j]; - matrix[n - 1 - i][n - 1 - j] = matrix[j][n - 1 - i]; - matrix[j][n - 1 - i] = tmp; - } + vector maxSlidingWindow(vector& nums, int k) { + if(nums.size() == 0 || k == 0) return {}; + deque deque; + vector res(nums.size() - k + 1); + for(int j = 0, i = 1 - k; j < nums.size(); i++, j++) { + // 删除 deque 中对应的 nums[i-1] + if(i > 0 && deque.front() == nums[i - 1]) + deque.pop_front(); + // 保持 deque 递减 + while(!deque.empty() && deque.back() < nums[j]) + deque.pop_back(); + deque.push_back(nums[j]); + // 记录窗口最大值 + if(i >= 0) + res[i] = deque.front(); } + return res; + } } }; ``` +可以将 “未形成窗口” 和 “形成窗口后” 两个阶段拆分到两个循环里实现。代码虽变长，但减少了冗余的判断操作。 + ```Python [] class Solution: - def rotate(self, matrix: List[List[int]]) -> None: - # 设矩阵行列数为 n - n = len(matrix) - # 起始点范围为 0 <= i < n // 2 , 0 <= j < (n + 1) // 2 - # 其中 '//' 为整数除法 - for i in range(n // 2): - for j in range((n + 1) // 2): - # 暂存 A 至 tmp - tmp = matrix[i][j] - # 元素旋转操作 A <- D <- C <- B <- tmp - matrix[i][j] = matrix[n - 1 - j][i] - matrix[n - 1 - j][i] = matrix[n - 1 - i][n - 1 - j] - matrix[n - 1 - i][n - 1 - j] = matrix[j][n - 1 - i] - matrix[j][n - 1 - i] = tmp + def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]: + if not nums or k == 0: return [] + deque = collections.deque() + # 未形成窗口 + for i in range(k): + while deque and deque[-1] < nums[i]: + deque.pop() + deque.append(nums[i]) + res = [deque[0]] + # 形成窗口后 + for i in range(k, len(nums)): + if deque[0] == nums[i - k]: + deque.popleft() + while deque and deque[-1] < nums[i]: + deque.pop() + deque.append(nums[i]) + res.append(deque[0]) + return res ``` ```Java [] class Solution { - public void rotate(int[][] matrix) { - // 设矩阵行列数为 n - int n = matrix.length; - // 起始点范围为 0 <= i < n / 2 , 0 <= j < (n + 1) / 2 - // 其中 '/' 为整数除法 - for (int i = 0; i < n / 2; i++) { - for (int j = 0; j < (n + 1) / 2; j++) { - // 暂存 A 至 tmp - int tmp = matrix[i][j]; - // 元素旋转操作 A <- D <- C <- B <- tmp - matrix[i][j] = matrix[n - 1 - j][i]; - matrix[n - 1 - j][i] = matrix[n - 1 - i][n - 1 - j]; - matrix[n - 1 - i][n - 1 - j] = matrix[j][n - 1 - i]; - matrix[j][n - 1 - i] = tmp; - } + public int[] maxSlidingWindow(int[] nums, int k) { + if(nums.length == 0 || k == 0) return new int[0]; + Deque deque = new LinkedList<>(); + int[] res = new int[nums.length - k + 1]; + // 未形成窗口 + for(int i = 0; i < k; i++) { + while(!deque.isEmpty() && deque.peekLast() < nums[i]) + deque.removeLast(); + deque.addLast(nums[i]); } + res[0] = deque.peekFirst(); + // 形成窗口后 + for(int i = k; i < nums.length; i++) { + if(deque.peekFirst() == nums[i - k]) + deque.removeFirst(); + while(!deque.isEmpty() && deque.peekLast() < nums[i]) + deque.removeLast(); + deque.addLast(nums[i]); + res[i - k + 1] = deque.peekFirst(); + } + return res; } } ``` @@ -200,27 +167,32 @@ class Solution { ```C++ [] class Solution { public: - void rotate(vector>& matrix) { - // 设矩阵行列数为 n - int n = matrix.size(); - // 起始点范围为 0 <= i < n / 2 , 0 <= j < (n + 1) / 2 - // 其中 '/' 为整数除法 - for (int i = 0; i < n / 2; i++) { - for (int j = 0; j < (n + 1) / 2; j++) { - // 暂存 A 至 tmp - int tmp = matrix[i][j]; - // 元素旋转操作 A <- D <- C <- B <- tmp - matrix[i][j] = matrix[n - 1 - j][i]; - matrix[n - 1 - j][i] = matrix[n - 1 - i][n - 1 - j]; - matrix[n - 1 - i][n - 1 - j] = matrix[j][n - 1 - i]; - matrix[j][n - 1 - i] = tmp; - } + vector maxSlidingWindow(vector& nums, int k) { + if(nums.size() == 0 || k == 0) return {}; + deque deque; + vector res(nums.size() - k + 1); + // 未形成窗口 + for(int i = 0; i < k; i++) { + while(!deque.empty() && deque.back() < nums[i]) + deque.pop_back(); + deque.push_back(nums[i]); + } + res[0] = deque.front(); + // 形成窗口后 + for(int i = k; i < nums.size(); i++) { + if(deque.front() == nums[i - k]) + deque.pop_front(); + while(!deque.empty() && deque.back() < nums[i]) + deque.pop_back(); + deque.push_back(nums[i]); + res[i - k + 1] = deque.front(); } + return res; } }; ``` -## 复杂度分析 +### 复杂度分析： -- **时间复杂度 $O(N^2)$ ：** 其中 $N$ 为输入矩阵的行（列）数。需要将矩阵中每个元素旋转到新的位置，即对矩阵所有元素操作一次，使用 $O(N^2)$ 时间。 -- **空间复杂度 $O(1)$ ：** 临时变量 $tmp$ 使用常数大小的额外空间。值得注意，当循环中进入下轮迭代，上轮迭代初始化的 $tmp$ 占用的内存就会被自动释放，因此无累计使用空间。 +- **时间复杂度 $O(n)$ ：** 其中 $n$ 为数组 $nums$ 长度；线性遍历 $nums$ 占用 $O(n)$ ；每个元素最多仅入队和出队一次，因此单调队列 $deque$ 占用 $O(2n)$ 。 +- **空间复杂度 $O(k)$ ：** 双端队列 $deque$ 中最多同时存储 $k$ 个元素（即窗口大小）。 diff --git "a/selected_coding_interview/docs/6. N \345\255\227\345\275\242\345\217\230\346\215\242.md" "b/selected_coding_interview/docs/6. N \345\255\227\345\275\242\345\217\230\346\215\242.md" index b586706..5c88b7e 100644 --- "a/selected_coding_interview/docs/6. N \345\255\227\345\275\242\345\217\230\346\215\242.md" +++ "b/selected_coding_interview/docs/6. N \345\255\227\345\275\242\345\217\230\346\215\242.md" @@ -1,20 +1,22 @@ -#### 解题思路： - -- **题目理解：** - - 字符串 `s` 是以 $Z$ 字形为顺序存储的字符串，目标是按行打印。 - - 设 `numRows` 行字符串分别为 $s_1$ , $s_2$ ,..., $s_n$，则容易发现：按顺序遍历字符串 `s` 时，每个字符 `c` 在 $Z$ 字形中对应的 **行索引** 先从 $s_1$ 增大至 $s_n$，再从 $s_n$ 减小至 $s_1$ …… 如此反复。 - - 因此，解决方案为：模拟这个行索引的变化，在遍历 `s` 中把每个字符填到正确的行 `res[i]` 。 -- **算法流程：** 按顺序遍历字符串 `s`； - 1. **`res[i] += c`：** 把每个字符 `c` 填入对应行 $s_i$； - 2. **`i += flag`：** 更新当前字符 `c` 对应的行索引； - 3. **`flag = - flag`：** 在达到 $Z$ 字形转折点时，执行反向。 -- **复杂度分析：** - - 时间复杂度 $O(N)$ ：遍历一遍字符串 `s`； - - 空间复杂度 $O(N)$ ：各行字符串共占用 $O(N)$ 额外空间。 +## 解题思路： + +字符串 `s` 是以 $Z$ 字形为顺序存储的字符串，目标是按行打印。 + +设 `numRows` 行字符串分别为 $s_1$ , $s_2$ , $\dots$ , $s_n$，则容易发现：按顺序遍历字符串 `s` 时，每个字符 `c` 在 N 字形中对应的 **行索引** 先从 $s_1$ 增大至 $s_n$，再从 $s_n$ 减小至 $s_1$ …… 如此反复。 + +因此解决方案为：模拟这个行索引的变化，在遍历 `s` 中把每个字符填到正确的行 `res[i]` 。 + +### 算法流程： + +按顺序遍历字符串 `s` ： + +1. **`res[i] += c`：** 把每个字符 `c` 填入对应行 $s_i$； +2. **`i += flag`：** 更新当前字符 `c` 对应的行索引； +3. **`flag = - flag`：** 在达到 $Z$ 字形转折点时，执行反向。 -#### 代码： +## 代码： ```Python [] class Solution: @@ -47,3 +49,30 @@ class Solution { } } ``` + +```C++ [] +class Solution { +public: + string convert(string s, int numRows) { + if (numRows < 2) + return s; + vector rows(numRows); + int i = 0, flag = -1; + for (char c : s) { + rows[i].push_back(c); + if (i == 0 || i == numRows -1) + flag = - flag; + i += flag; + } + string res; + for (const string &row : rows) + res += row; + return res; + } +}; +``` + +### 复杂度分析： + +- **时间复杂度 $O(N)$** ：遍历一遍字符串 `s`； +- **空间复杂度 $O(N)$** ：各行字符串共占用 $O(N)$ 额外空间。