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‎README.md

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# 和小浩学算法
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# Hi, Boy
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> 本项目于 2020.7.22 位于 github 日总榜第一！截图留念！(还没有助力的，右上角快来个 star )
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> On July 22, 2020, the project ranked first in GitHub daily list! Take a screenshot as a souvenir! (star in the upper right corner! Thank you)
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> [![图片展示失败，点击查看](https://www.geekxh.com/mark.png)](https://www.geekxh.com/mark.png)
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## 简介
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Xiaohao algorithm is a graphic algorithm problem dictionary completed by me during the epidemic period! At present, it has completed 140 + high-frequency interview algorithm questions, with a total of 30W words. All of them are in the way of cartoon illustration, which is easy to understand and suitable for the primary and intermediate readers.
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> [![图片展示失败，点击查看](https://www.geekxh.com/book.png)](https://www.geekxh.com/book.png)
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小浩算法是我在疫情期间完成的一部图解算法题典！<b> 目前共完成 140+ 道高频面试算法题目，总计 30w 字，全部采用漫画图解的方式，简单易懂，适合初中级读者。</b>
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> If you are willing to help me translate, welcome to contact ~
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## 如何使用
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# Related Topics
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1、因为本教程完全免费，但现在被一些不良商家拿去卖钱，<b> 所以我需要你先帮我点一个 star </b>，助力原创，防止更多人上当受骗，也顺便支持我一下。
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1、my_book（No translation yet.....）
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2、如果你想直接下载我的这本电子书，[点击这里就行](https://www.geekxh.com/0.0.%E5%AD%A6%E4%B9%A0%E9%A1%BB%E7%9F%A5/03.html?3pVd)
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2、code_100
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3、如果你还想加入我们的<b>万人刷题群</b>，和我们一起交流，遇见更优秀的人。请下方扫码回复【进群】
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3、interview_100
2020

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> [![图片展示失败，点击查看](https://www.geekxh.com/code.png)](https://www.geekxh.com/code.png)
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4、当然，也许你就想加我的私人微信: [个人名片](https://www.geekxh.com/contact.jpeg)
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## 本项目除此之外，还包括（持续更新）：
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1、100 张超清思维导图
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2、100 篇大厂面试面经
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3、100 本清晰版电子书
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## 内容展示
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我这里截取了思维导图以及算法题解的内容图片，可以看一下效果：
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导图目录：[点击查看](https://www.geekxh.com/daotu-1.png)
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导图效果展示：[点击查看](https://www.geekxh.com/daotu-2.png)
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电子书效果展示: [点击查看](https://www.geekxh.com/book.png)
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网站效果：[点击查看](https://www.geekxh.com/demo.png)
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## 获取全部内容方式（任选其一）
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1、通过网站下载（推荐，网速快）：[下载地址](https://www.geekxh.com/0.0.%E5%AD%A6%E4%B9%A0%E9%A1%BB%E7%9F%A5/03.html?3pVd)
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2、直接通过 github clone 项目。（国内不推荐）
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3、公众号扫码回复 999。（直接通过百度云下载）
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## 题解目录
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* [开始学习](http://www.geekxh.com)
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* [阅读指南](http://www.geekxh.com/c99/hello.html)
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* [1.数组]()
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* [01.两个数组的交集(350)](http://www.geekxh.com/1.0.数组系列/001.html)
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* [02.最长公共前缀(14)](http://www.geekxh.com/1.0.数组系列/002.html)
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* [03.买卖股票的最佳时机(122)](http://www.geekxh.com/1.0.数组系列/003.html)
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* [04.旋转数组(189)](http://www.geekxh.com/1.0.数组系列/004.html)
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* [05.原地删除(27)](http://www.geekxh.com/1.0.数组系列/005.html)
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* [06.加一(66)](http://www.geekxh.com/1.0.数组系列/006.html)
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* [07.两数之和(1)](http://www.geekxh.com/1.0.数组系列/007.html)
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* [08.三数之和(15)](http://www.geekxh.com/1.0.数组系列/008.html)
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* [09.Z字形变换(6)](http://www.geekxh.com/1.0.数组系列/009.html)
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* [2.链表]()
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* [01.删除链表倒数第N个节点(19)](http://www.geekxh.com/1.1.链表系列/101.html)
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* [02.合并两个有序链表(21)](http://www.geekxh.com/1.1.链表系列/102.html)
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* [03.环形链表(21)](http://www.geekxh.com/1.1.链表系列/103.html)
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* [04.两数相加(2)](http://www.geekxh.com/1.1.链表系列/104.html)
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* [3.动态规划]()
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* [01.爬楼梯(70)](http://www.geekxh.com/1.2.动态规划系列/201.html)
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* [02.最大子序和(53)](http://www.geekxh.com/1.2.动态规划系列/202.html)
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* [03.最长上升子序列(300)](http://www.geekxh.com/1.2.动态规划系列/203.html)
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* [04.三角形最小路径和(120)](http://www.geekxh.com/1.2.动态规划系列/204.html)
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* [05.最小路径和(64)](http://www.geekxh.com/1.2.动态规划系列/205.html)
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* [06.打家劫舍(198)](http://www.geekxh.com/1.2.动态规划系列/206.html)
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* [4.字符串]()
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* [01.反转字符串(344)](http://www.geekxh.com/1.3.字符串系列/301.html)
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* [02.字符串中的第一个唯一字符(387)](http://www.geekxh.com/1.3.字符串系列/302.html)
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* [03.实现 Sunday 匹配](http://www.geekxh.com/1.3.字符串系列/303.html)
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* [04.大数打印](http://www.geekxh.com/1.3.字符串系列/304.html)
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* [05.验证回文串(125)](http://www.geekxh.com/1.3.字符串系列/305.html)
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* [06.KMP 精讲](http://www.geekxh.com/1.3.字符串系列/306.html)
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* [07.旋转字符串(796)](http://www.geekxh.com/1.3.字符串系列/307.html)
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* [08.最后一个单词的长度(58)](http://www.geekxh.com/1.3.字符串系列/308.html)
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* [5.二叉树]()
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* [01.最大深度与DFS(104)](http://www.geekxh.com/1.4.二叉树系列/401.html)
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* [02.层次遍历与BFS(102)](http://www.geekxh.com/1.4.二叉树系列/402.html)
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* [03.BST与其验证(98)](http://www.geekxh.com/1.4.二叉树系列/403.html)
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* [04.BST 的查找(700)](http://www.geekxh.com/1.4.二叉树系列/404.html)
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* [05.BST 的删除(450)](http://www.geekxh.com/1.4.二叉树系列/405.html)
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* [06.平衡二叉树(110)](http://www.geekxh.com/1.4.二叉树系列/406.html)
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* [07.完全二叉树(222)](http://www.geekxh.com/1.4.二叉树系列/407.html)
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* [08.二叉树的剪枝(814)](http://www.geekxh.com/1.4.二叉树系列/408.html)
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* [6.滑动窗口]()
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* [01.滑动窗口最大值（239）](http://www.geekxh.com/1.5.滑动窗口系列/501.html)
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* [02.无重复字符的最长子串（3）](http://www.geekxh.com/1.5.滑动窗口系列/502.html)
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* [03.找到字符串中所有字母异位词（438）](http://www.geekxh.com/1.5.滑动窗口系列/503.html)
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* [04.和为s的连续正数序列](http://www.geekxh.com/1.5.滑动窗口系列/504.html)
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* [7.博弈论系列]()
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* [01.囚徒困境](http://www.geekxh.com/1.6.博弈论系列/601.html)
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* [02.辛普森悖论](http://www.geekxh.com/1.6.博弈论系列/602.html)
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* [03.红眼睛和蓝眼睛](http://www.geekxh.com/1.6.博弈论系列/603.html)
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* [04.海盗分金币](http://www.geekxh.com/1.6.博弈论系列/604.html)
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* [05.智猪博弈](http://www.geekxh.com/1.6.博弈论系列/605.html)
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* [06.生男生女的问题](http://www.geekxh.com/1.6.博弈论系列/606.html)
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* [07.硬币问题](http://www.geekxh.com/1.6.博弈论系列/607.html)
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* [08.画圈圈的问题](http://www.geekxh.com/1.6.博弈论系列/608.html)
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* [09.巧克力问题](http://www.geekxh.com/1.6.博弈论系列/609.html)
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* [10.大鱼和小鱼的问题](http://www.geekxh.com/1.6.博弈论系列/610.html)
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* [8.位运算系列]()
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* [01.使用位运算求和](http://www.geekxh.com/1.8.位运算系列/801.html)
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* [02.2的幂(231)](http://www.geekxh.com/1.8.位运算系列/802.html)
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* [03.返回一个数二进制中1的个数](http://www.geekxh.com/1.8.位运算系列/803.html)
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* [04.只出现一次的数字](http://www.geekxh.com/1.8.位运算系列/804.html)
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* [05.只出现一次的数字Ⅱ](http://www.geekxh.com/1.8.位运算系列/805.html)
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* [06.缺失数字(268)](http://www.geekxh.com/1.8.位运算系列/806.html)
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* [9.二分法系列]()
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* [01.爱吃香蕉的珂珂（875）](http://www.geekxh.com/1.9.二分法系列/901.html)
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* [02.x的平方根（69）](http://www.geekxh.com/1.9.二分法系列/902.html)
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* [03.x的平方根（69）](http://www.geekxh.com/1.9.二分法系列/903.html)
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* [04.旋转排序数组中的最小值Ⅰ（153）](http://www.geekxh.com/1.9.二分法系列/904.html)
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* [05.旋转排序数组中的最小值Ⅱ(154)](http://www.geekxh.com/1.9.二分法系列/905.html)
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* [06.供暖器（475）](http://www.geekxh.com/1.9.二分法系列/906.html)
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* [其他补充题目]()
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* [01.螺旋矩阵(54)](http://www.geekxh.com/1.99.其他补充题目/01.html)
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* [02.只有两个键的键盘(650)](http://www.geekxh.com/1.99.其他补充题目/02.html)
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* [03.24点游戏(679)](http://www.geekxh.com/1.99.其他补充题目/03.html)
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* [04.飞机座位分配概率(1227)](http://www.geekxh.com/1.99.其他补充题目/04.html)
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* [05.水分子的产生](http://www.geekxh.com/1.99.其他补充题目/05.html)
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* [06.救生艇(881)](http://www.geekxh.com/1.99.其他补充题目/06.html)
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* [07.救生艇(881)](http://www.geekxh.com/1.99.其他补充题目/07.html)
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* [08.灯泡开关(319)](http://www.geekxh.com/1.99.其他补充题目/08.html)
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* [09.三门问题](http://www.geekxh.com/1.99.其他补充题目/09.html)
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* [10.猜数字游戏(299)](http://www.geekxh.com/1.99.其他补充题目/10.html)
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* [11.LRU缓存机制(146)](http://www.geekxh.com/1.99.其他补充题目/11.html)
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* [12.最小的k个数](http://www.geekxh.com/1.99.其他补充题目/12.html)
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* [13.不同路径](http://www.geekxh.com/1.99.其他补充题目/13.html)
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* [14.不同路径-障碍物](http://www.geekxh.com/1.99.其他补充题目/14.html)
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* [15.连续n个数的和](http://www.geekxh.com/1.99.其他补充题目/15.html)
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* [16.盛水最多的容器](http://www.geekxh.com/1.99.其他补充题目/16.html)
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* [17.扑克牌中的顺子容器](http://www.geekxh.com/1.99.其他补充题目/17.html)
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* [18.整数拆分(343)](http://www.geekxh.com/1.99.其他补充题目/18.html)
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* [19.移动石子到连续(1033)](http://www.geekxh.com/1.99.其他补充题目/19.html)
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* [20.Nim游戏(292)](http://www.geekxh.com/1.99.其他补充题目/20.html)
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* [21.寻找两个正序数组的中位数(4)](http://www.geekxh.com/1.99.其他补充题目/21.html)
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* [22.第k个最大元素（215）](http://www.geekxh.com/1.99.其他补充题目/22.html)
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* [23.镜面反射（858）](http://www.geekxh.com/1.99.其他补充题目/23.html)
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* [25.整数转罗马数字（12）](http://www.geekxh.com/1.99.其他补充题目/25.html)
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* [26.荷兰国旗问题](http://www.geekxh.com/1.99.其他补充题目/26.html)
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* [27.六九问题](http://www.geekxh.com/1.99.其他补充题目/27.html)
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* [28.有效的数独](http://www.geekxh.com/1.99.其他补充题目/28.html)
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* [29.费米估算](http://www.geekxh.com/1.99.其他补充题目/29.html)
167-
* [30.分发饼干](http://www.geekxh.com/1.99.其他补充题目/30.html)
168-
* [31.生命游戏（289）](http://www.geekxh.com/1.99.其他补充题目/31.html)
169-
* [32.搜索二维矩阵（74）](http://www.geekxh.com/1.99.其他补充题目/32.html)
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* [33.子集（78）](http://www.geekxh.com/1.99.其他补充题目/33.html)
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* [34.面试中的智力题](http://www.geekxh.com/1.99.其他补充题目/34.html)
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* [35.旋转图像（48）](http://www.geekxh.com/1.99.其他补充题目/51.html)
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* [其他：排序专栏]()
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* [排序算法](http://www.geekxh.com/2.0.排序系列/0.hello.html)
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* [冒泡排序](http://www.geekxh.com/2.0.排序系列/1.bubbleSort.html)
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* [选择排序](http://www.geekxh.com/2.0.排序系列/2.selectionSort.html)
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* [插入排序](http://www.geekxh.com/2.0.排序系列/3.insertionSort.html)
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* [希尔排序](http://www.geekxh.com/2.0.排序系列/4.shellSort.html)
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* [归并排序](http://www.geekxh.com/2.0.排序系列/5.mergeSort.html)
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* [快速排序](http://www.geekxh.com/2.0.排序系列/6.quickSort.html)
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* [堆排序](http://www.geekxh.com/2.0.排序系列/7.heapSort.html)
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* [计数排序](http://www.geekxh.com/2.0.排序系列/8.countingSort.html)
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* [桶排序](http://www.geekxh.com/2.0.排序系列/9.bucketSort.html)
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* [基数排序](http://www.geekxh.com/2.0.排序系列/10.radixSort.html)
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4、mindMap_100（42 HD mind maps，No translation yet.....）

‎code_100/C++/Arrays/Trapping the rain water.cpp

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#include <iostream>
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using namespace std;
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int trapped_water(int array[],int size){
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int amount = 0;
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int left[size],right[size];
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left[0] = array[0]; right[size-1] = array[size-1];
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for(int i = 1; i < size; i++){
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left[i] = max(left[i-1],array[i]);
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}
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for(int i = size-2; i >=0; i--){
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right[i] = max(right[i+1],array[i]);
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}
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for(int i = 0 ; i < size;i++){
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amount += min(left[i],right[i]) - array[i];
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}
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return amount;
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}
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int main(){
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int array[] = {1,0,3,4,5,0,5,7,7,8,9,0};
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int size = sizeof(array) / sizeof(int);
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cout << trapped_water(array,size);
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return 0;
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}

‎code_100/C++/Bit Manipulation/Checking Whether K-th Bit is Set or Not.cpp

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#include<iostream>
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using namespace std;
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int main(){
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int n,k;
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cout << "Enter the number and the value of K : ";
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cin >> n >> k;
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int mask = 1 << (k-1);
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if(n & mask){
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cout << "Yes K-th bit is set" << endl;
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}
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else{
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cout << "No K-th bit is not set" << endl;
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}
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return 0;
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}

‎code_100/C++/Bit Manipulation/Clearing the K-th bit of a number.cpp

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#include<iostream>
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using namespace std;
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int main(){
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int n,k,mask;
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cout << "Enter the number and the value of K : ";
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cin >> n >> k;
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mask = ~(1 << (k-1));
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n = n&mask;
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cout << "The number after clearing the K-th bit is : " << n << endl;
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return 0;
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}

‎code_100/C++/Bit Manipulation/Setting the K-th bit of a number.cpp

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#include<iostream>
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using namespace std;
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int main(){
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int n,k;
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cout << "Enter the number and the value of K :";
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cin >> n >> k;
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int mask = 1 << (k - 1);
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n = n | mask;
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cout << "The number after setting the K-th bit is:" << n;
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return 0;
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}

‎code_100/C++/Bit Manipulation/Toggling Rightmost Set Bit of a number.cpp

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#include<iostream>
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using namespace std;
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int main(){
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int n;
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cout << "Enter the number : ";
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cin >> n ;
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n = n & (n-1);
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cout << "The number after toggling right most set bit : " << n << endl;
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return 0;
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}

‎code_100/C++/Bit Manipulation/Toggling the K-th bit of a number.cpp

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#include<iostream>
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using namespace std;
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int main(){
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int n,k,mask;
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cout << "Enter the number and the value of K : ";
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cin >> n >> k;
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mask = 1 << (k-1);
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n = n ^ mask;
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cout << "The number after toggling the K-th bit is : " << n << endl;
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return 0;
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}

‎code_100/C++/Dynamic Programming/Edit Distance.cpp

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int editDistance(string s1, string s2){
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int m = s1.length();
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int n = s2.length();
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int dp[m+1][n+1];
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for (int i = 0; i <= m; i++) {
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dp[i][0] = i;
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}
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for (int j = 0; j <= n; j++) {
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dp[0][j] = j;
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}
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for (int i = 1; i <= m; i++) {
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for (int j = 1; j <= n; j++) {
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if (s1[i-1] == s2[j-1]) dp[i][j] = dp[i-1][j-1];
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else dp[i][j] = 1 + min(min(dp[i][j-1],dp[i-1][j]),dp[i-1][j-1]);
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}
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}
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return dp[m][n];
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}

‎code_100/C++/Dynamic Programming/Longest Common Subsequence.cpp

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int lcs(string x,string y){
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int m = x.size(),n = y.size();
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int dp[m+1][n+1];
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for(int i=0;i<=m;i++){
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dp[i][0] = 0;
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}
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for(int j=0;j<=m;j++){
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dp[0][j] = 0;
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}
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for(int i=1;i<=m;i++){
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for(int j=1;j<=n;j++){
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if(x[i-1] == y[j-1]){
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dp[i][j] = dp[i-1][j-1]+1;
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}
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else{
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dp[i][j] = max(dp[i][j-1],dp[i-1][j]);
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}
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}
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}
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return dp[m][n];
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}

‎code_100/C++/Dynamic Programming/Longest Common Substring.cpp

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#include <iostream>
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using namespace std;
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int longest_common_substring(string x,string y){
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int m = x.size();
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int n = y.size();
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int lcs[m+1][n+1];
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for(int i = 0 ; i < m; i++){
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lcs[i][0] = 0;
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}
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for(int j = 0; j < n; j++){
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lcs[0][j] = 0;
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}
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for(int i = 1; i <= m; i++){
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for(int j = 1; j <=n; j++){
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if(x[i-1] == y[j-1]){
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lcs[i][j] = 1 + lcs[i-1][j-1];
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}
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else{
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lcs[i][j] = 0;
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}
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}
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}
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return lcs[m][n];
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}
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int main(){
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string x,y;
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cin >> x >> y;
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cout << longest_common_substring(x,y);
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return 0;
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}

‎code_100/C++/Dynamic Programming/Longest Increasing Subsequence.cpp

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int lis(int array[],int n){
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int dp[n],lis_value = -1;
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for(int i=0;i<n;i++){
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dp[i] = 1;
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}
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for(int i=1;i<n;i++){
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for(int j=0;j<i;j++){
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if(array[i] > array[j] and dp[i] < dp[j]+1){
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dp[i] = dp[j] + 1;
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}
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}
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}
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for(int i=0;i<n;i++){
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if(lis_value < dp[i]){
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lis_value = dp[i];
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}
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}
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return lis_value;
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}

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