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1. Add leetcode solutions.
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‎leetcode/157. Read N Characters Given Read4.md

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## 157. Read N Characters Given Read4
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### Question
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Given a file and assume that you can only read the file using a given method read4, implement a method to read n characters.
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Method read4:
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The API read4 reads 4 consecutive characters from the file, then writes those characters into the buffer array buf.
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The return value is the number of actual characters read.
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Note that read4() has its own file pointer, much like FILE *fp in C.
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Definition of read4:
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Parameter: char[] buf
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Returns: int
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Note: buf[] is destination not source, the results from read4 will be copied to buf[]
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Below is a high level example of how read4 works:
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File file("abcdefghijk"); // File is "abcdefghijk", initially file pointer (fp) points to 'a'
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char[] buf = new char[4]; // Create buffer with enough space to store characters
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read4(buf); // read4 returns 4. Now buf = "abcd", fp points to 'e'
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read4(buf); // read4 returns 4. Now buf = "efgh", fp points to 'i'
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read4(buf); // read4 returns 3. Now buf = "ijk", fp points to end of file
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Method read:
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By using the read4 method, implement the method read that reads n characters from the file and store it in the buffer array buf. Consider that you cannot manipulate the file directly.
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The return value is the number of actual characters read.
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Definition of read:
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Parameters: char[] buf, int n
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Returns: int
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Note: buf[] is destination not source, you will need to write the results to buf[]
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```
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Example 1:
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Input: file = "abc", n = 4
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Output: 3
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Explanation: After calling your read method, buf should contain "abc". We read a total of 3 characters from the file, so return 3. Note that "abc" is the file's content, not buf. buf is the destination buffer that you will have to write the results to.
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Example 2:
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Input: file = "abcde", n = 5
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Output: 5
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Explanation: After calling your read method, buf should contain "abcde". We read a total of 5 characters from the file, so return 5.
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Example 3:
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Input: file = "abcdABCD1234", n = 12
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Output: 12
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Explanation: After calling your read method, buf should contain "abcdABCD1234". We read a total of 12 characters from the file, so return 12.
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Example 4:
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Input: file = "leetcode", n = 5
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Output: 5
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Explanation: After calling your read method, buf should contain "leetc". We read a total of 5 characters from the file, so return 5.
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```
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Note:
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1. Consider that you cannot manipulate the file directly, the file is only accesible for read4 but not for read.
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2. The read function will only be called once for each test case.
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3. You may assume the destination buffer array, buf, is guaranteed to have enough space for storing n characters.
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### Solutions:
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* Method 1: 14m39s
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```
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/**
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* The read4 API is defined in the parent class Reader4.
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* int read4(char[] buf);
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*/
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public class Solution extends Reader4 {
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/**
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* @param buf Destination buffer
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* @param n Number of characters to read
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* @return The number of actual characters read
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*/
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public int read(char[] buf, int n) {
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int read = 0;
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int index = 0;
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while(read < n){
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char[] buffer = new char[4];
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int cnt = read4(buffer);
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if(cnt < 4 && read + cnt <= n){
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for(int i = 0; i < cnt; i++){
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buf[index++] = buffer[i];
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}
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return read + cnt;
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}else if(cnt == 4 && read + cnt < n){
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for(int i = 0; i < 4; i++){
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buf[index++] = buffer[i];
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}
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read += 4;
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}else if(read + cnt >= n){
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while(index < n){
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buf[index] = buffer[index - read];
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index++;
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}
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return n;
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}
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}
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return n;
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}
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}
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```

‎leetcode/158. Read N Characters Given Read4 II - Call multiple times.md

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## 158. Read N Characters Given Read4 II - Call multiple times
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### Question
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Given a file and assume that you can only read the file using a given method read4, implement a method read to read n characters. Your method read may be called multiple times.
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Method read4:
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The API read4 reads 4 consecutive characters from the file, then writes those characters into the buffer array buf.
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The return value is the number of actual characters read.
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Note that read4() has its own file pointer, much like FILE *fp in C.
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Definition of read4:
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Parameter: char[] buf
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Returns: int
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Note: buf[] is destination not source, the results from read4 will be copied to buf[]
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Below is a high level example of how read4 works:
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File file("abcdefghijk"); // File is "abcdefghijk", initially file pointer (fp) points to 'a'
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char[] buf = new char[4]; // Create buffer with enough space to store characters
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read4(buf); // read4 returns 4. Now buf = "abcd", fp points to 'e'
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read4(buf); // read4 returns 4. Now buf = "efgh", fp points to 'i'
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read4(buf); // read4 returns 3. Now buf = "ijk", fp points to end of file
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Method read:
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By using the read4 method, implement the method read that reads n characters from the file and store it in the buffer array buf. Consider that you cannot manipulate the file directly.
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The return value is the number of actual characters read.
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Definition of read:
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Parameters: char[] buf, int n
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Returns: int
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Note: buf[] is destination not source, you will need to write the results to buf[]
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```
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Example 1:
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File file("abc");
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Solution sol;
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// Assume buf is allocated and guaranteed to have enough space for storing all characters from the file.
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sol.read(buf, 1); // After calling your read method, buf should contain "a". We read a total of 1 character from the file, so return 1.
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sol.read(buf, 2); // Now buf should contain "bc". We read a total of 2 characters from the file, so return 2.
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sol.read(buf, 1); // We have reached the end of file, no more characters can be read. So return 0.
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Example 2:
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File file("abc");
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Solution sol;
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sol.read(buf, 4); // After calling your read method, buf should contain "abc". We read a total of 3 characters from the file, so return 3.
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sol.read(buf, 1); // We have reached the end of file, no more characters can be read. So return 0.
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```
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Note:
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1. Consider that you cannot manipulate the file directly, the file is only accesible for read4 but not for read.
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2. The read function may be called multiple times.
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3. Please remember to RESET your class variables declared in Solution, as static/class variables are persisted across multiple test cases. Please see here for more details.
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4. You may assume the destination buffer array, buf, is guaranteed to have enough space for storing n characters.
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5. It is guaranteed that in a given test case the same buffer buf is called by read.
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### Solutions:
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* Method 1:
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```
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/**
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* The read4 API is defined in the parent class Reader4.
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* int read4(char[] buf);
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*/
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public class Solution extends Reader4 {
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private String s;
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private int index = 0;
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/**
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* @param buf Destination buffer
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* @param n Number of characters to read
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* @return The number of actual characters read
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*/
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public int read(char[] buf, int n) {
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if(s == null){
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s = "";
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char[] buffer = new char[4];
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int cnt = 0;
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while((cnt = read4(buffer)) != 0){
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for(int i = 0; i < cnt; i++)
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s += buffer[i];
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}
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}
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if(index >= s.length()) return 0;
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if(index + n <= s.length()){
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String sub = s.substring(index, index + n);
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for(int i = 0; i < n; i++){
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buf[i] = sub.charAt(i);
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}
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index += n;
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return n;
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}else{
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for(int i = index; i < s.length(); i++){
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buf[i - index] = s.charAt(i);
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}
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int res = s.length() - index;
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index = s.length();
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return res;
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}
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}
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}
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```

‎leetcode/170. Two Sum III - Data structure design.md

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## 170. Two Sum III - Data structure design
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### Question
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Design and implement a TwoSum class. It should support the following operations: add and find.
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add - Add the number to an internal data structure.
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find - Find if there exists any pair of numbers which sum is equal to the value.
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```
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Example 1:
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add(1); add(3); add(5);
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find(4) -> true
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find(7) -> false
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Example 2:
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add(3); add(1); add(2);
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find(3) -> true
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find(6) -> false
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```
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### Solutions:
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* Method 1: HashMap + List
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```Java
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class TwoSum {
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private List<Integer> list;
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private Map<Integer, Integer> map;
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/** Initialize your data structure here. */
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public TwoSum() {
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this.list = new ArrayList<>();
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this.map = new HashMap<>();
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}
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/** Add the number to an internal data structure.. */
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public void add(int number) {
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this.list.add(number);
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this.map.put(number, map.getOrDefault(number, 0) + 1);
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}
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/** Find if there exists any pair of numbers which sum is equal to the value. */
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public boolean find(int value) {
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for(int i = 0; i < list.size(); i++){
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int num = list.get(i);
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int another = value - num;
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if(num == another && map.get(num) >= 2) return true;
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else if(num != another && map.containsKey(another)) return true;
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}
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return false;
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}
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}
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/**
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* Your TwoSum object will be instantiated and called as such:
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* TwoSum obj = new TwoSum();
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* obj.add(number);
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* boolean param_2 = obj.find(value);
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*/
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```

‎leetcode/3. Longest Substring Without Repeating Characters.md

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return max;
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}
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}
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```
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```
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### Third Time
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* Method 1: Two pointers + Set
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```Java
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class Solution {
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public int lengthOfLongestSubstring(String s) {
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if(s == null || s.length() == 0) return 0;
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int slow = 0, fast = 0;
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Set<Character> set = new HashSet<>();
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char[] arr = s.toCharArray();
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int res = 0;
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while(fast < arr.length){
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if(!set.contains(arr[fast])){
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set.add(arr[fast]);
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res = Math.max(res, fast - slow + 1);
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}else{ // set already contains key of arr[fast]
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while(arr[slow] != arr[fast] && slow < fast){
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set.remove(arr[slow++]);
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}
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slow++;
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}
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fast++;
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}
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return res;
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}
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}
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```

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