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‎markdown/11. Container With Most Water.md

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### [11\. Container With Most Water](https://leetcode.com/problems/container-with-most-water/)
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Difficulty: **Medium**
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Given _n_ non-negative integers _a<sub style="display: inline;">1</sub>_, _a<sub style="display: inline;">2</sub>_, ..., _a<sub style="display: inline;">n </sub>_, where each represents a point at coordinate (_i_, _a<sub style="display: inline;">i</sub>_). _n_ vertical lines are drawn such that the two endpoints of line _i_ is at (_i_, _a<sub style="display: inline;">i</sub>_) and (_i_, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.
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**Note: **You may not slant the container and _n_ is at least 2.
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![](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/07/17/question_11.jpg)
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<small style="display: inline;">The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49\.</small>
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**Example:**
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```
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Input: [1,8,6,2,5,4,8,3,7]
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Output: 49```
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#### Solution
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Language: **Java**
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```java
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class Solution {
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   public int maxArea(int[] height) {
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       int i = 0;
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       int j = height.length - 1;
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       int water = 0;
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       while (i < j) {
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           water = Math.max((j - i) * Math.min(height[i], height[j]), water);
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           if (height[i] < height[j]) {
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               i++;
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          } else {
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               j--;
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          }
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      }
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       return water;
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  }
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}
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```

‎markdown/12. Integer to Roman.md

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### [12\. Integer to Roman](https://leetcode.com/problems/integer-to-roman/)
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Difficulty: **Medium**
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Roman numerals are represented by seven different symbols: `I`, `V`, `X`, `L`, `C`, `D` and `M`.
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```
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Symbol Value
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I 1
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V 5
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X 10
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L 50
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C 100
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D 500
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M 1000```
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For example, two is written as `II` in Roman numeral, just two one's added together. Twelve is written as, `XII`, which is simply `X` + `II`. The number twenty seven is written as `XXVII`, which is `XX` + `V` + `II`.
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Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not `IIII`. Instead, the number four is written as `IV`. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as `IX`. There are six instances where subtraction is used:
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* `I` can be placed before `V` (5) and `X` (10) to make 4 and 9. 
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* `X` can be placed before `L` (50) and `C` (100) to make 40 and 90. 
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* `C` can be placed before `D` (500) and `M` (1000) to make 400 and 900.
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Given an integer, convert it to a roman numeral. Input is guaranteed to be within the range from 1 to 3999.
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**Example 1:**
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```
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Input: 3
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Output: "III"```
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**Example 2:**
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```
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Input: 4
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Output: "IV"```
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**Example 3:**
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```
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Input: 9
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Output: "IX"```
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**Example 4:**
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```
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Input: 58
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Output: "LVIII"
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Explanation: L = 50, V = 5, III = 3.
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```
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**Example 5:**
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```
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Input: 1994
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Output: "MCMXCIV"
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Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.```
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#### Solution
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Language: **Java**
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```java
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class Solution {
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   public String intToRoman(int num) {
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       String M[] = {"", "M", "MM", "MMM"};
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       String C[] = {"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"};
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       String X[] = {"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"};
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       String I[] = {"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"};
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       return M[num/1000] + C[(num%1000)/100] + X[(num%100)/10] + I[num%10];
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  }
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}
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```

‎markdown/13. Roman to Integer.md

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### [13\. Roman to Integer](https://leetcode.com/problems/roman-to-integer/)
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Difficulty: **Easy**
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Roman numerals are represented by seven different symbols: `I`, `V`, `X`, `L`, `C`, `D` and `M`.
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```
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Symbol Value
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I 1
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V 5
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X 10
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L 50
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C 100
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D 500
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M 1000```
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For example, two is written as `II` in Roman numeral, just two one's added together. Twelve is written as, `XII`, which is simply `X` + `II`. The number twenty seven is written as `XXVII`, which is `XX` + `V` + `II`.
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Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not `IIII`. Instead, the number four is written as `IV`. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as `IX`. There are six instances where subtraction is used:
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* `I` can be placed before `V` (5) and `X` (10) to make 4 and 9. 
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* `X` can be placed before `L` (50) and `C` (100) to make 40 and 90. 
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* `C` can be placed before `D` (500) and `M` (1000) to make 400 and 900.
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Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999.
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**Example 1:**
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```
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Input: "III"
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Output: 3```
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**Example 2:**
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```
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Input: "IV"
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Output: 4```
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**Example 3:**
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```
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Input: "IX"
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Output: 9```
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**Example 4:**
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```
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Input: "LVIII"
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Output: 58
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Explanation: L = 50, V= 5, III = 3.
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```
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**Example 5:**
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```
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Input: "MCMXCIV"
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Output: 1994
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Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.```
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#### Solution
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Language: **Java**
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```java
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class Solution {
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   public int romanToInt(String s) {
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       int result = 0;
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       for (int i = 0; i < s.length(); i++) {
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           int cur = getInt(s.charAt(i));
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           if (i + 1 < s.length() && getInt(s.charAt(i + 1)) > getInt(s.charAt(i))) {
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               result -= cur;
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          } else {
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               result += cur;
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          }
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      }
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       return result;
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  }
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​
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   private int getInt(char ch) {
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       switch (ch) {
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           case 'I':
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               return 1;
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           case 'V':
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               return 5;
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           case 'X':
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               return 10;
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           case 'L':
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               return 50;
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           case 'C':
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               return 100;
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           case 'D':
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               return 500;
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           case 'M':
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               return 1000;
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           default:
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               return 0;
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      }
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  }
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}
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```

‎markdown/14. Longest Common Prefix.md

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### [14\. Longest Common Prefix](https://leetcode.com/problems/longest-common-prefix/)
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Difficulty: **Easy**
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Write a function to find the longest common prefix string amongst an array of strings.
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If there is no common prefix, return an empty string `""`.
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**Example 1:**
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```
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Input: ["flower","flow","flight"]
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Output: "fl"
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```
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**Example 2:**
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```
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Input: ["dog","racecar","car"]
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Output: ""
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Explanation: There is no common prefix among the input strings.
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```
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**Note:**
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All given inputs are in lowercase letters `a-z`.
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#### Solution
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Language: **Java**
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```java
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class Solution {
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   public String longestCommonPrefix(String[] strs) {
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       if (strs.length == 0) {
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           return "";
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      }
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​
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       if (strs.length == 1) {
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           return strs[0];
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      }
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       int minLenth = strs[0].length();
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​
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       for (String str : strs) {
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           if (minLenth > str.length()) {
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               minLenth = str.length();
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          }
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      }
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​
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       for (int longPrefixIndex = 0; longPrefixIndex < minLenth; longPrefixIndex++) {
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           for (int j = 0; j < strs.length - 1; j++) {
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               if (strs[j].charAt(longPrefixIndex) != strs[j + 1].charAt(longPrefixIndex)) {
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                   return strs[0].substring(0, longPrefixIndex);
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              }
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          }
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      }
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       return "";
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  }
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}
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```

‎markdown/15. 3Sum.md

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### [15\. 3Sum](https://leetcode.com/problems/3sum/)
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Difficulty: **Medium**
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Given an array `nums` of _n_ integers, are there elements _a_, _b_, _c_ in `nums` such that _a_ + _b_ + _c_ = 0? Find all unique triplets in the array which gives the sum of zero.
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**Note:**
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The solution set must not contain duplicate triplets.
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**Example:**
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```
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Given array nums = [-1, 0, 1, 2, -1, -4],
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A solution set is:
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[
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[-1, 0, 1],
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[-1, -1, 2]
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]
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```
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#### Solution
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Language: **Java**
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```java
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class Solution {
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   public List<List<Integer>> threeSum(int[] nums) {
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       if (nums.length < 3) {
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           return Collections.emptyList();
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      }
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​
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       Arrays.sort(nums); // 排序用于剔除重复项
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       Set<List<Integer>> resultSet = new HashSet<>();
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       Map<Integer, List<Integer>> index = new HashMap<>();
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       // List<Integer> numList = new ArrayList<>();
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       for (int i = 0; i < nums.length; i++) {
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           index.computeIfAbsent(nums[i], k -> new ArrayList<>());
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           index.get(nums[i]).add(i);
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      }
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​
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       for (int i = 0; i < nums.length; i++) {
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           if (nums[i] > 0) {
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               continue;
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          }
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           if (i > 0 && nums[i] == nums[i - 1]) {
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               continue;
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          }
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​
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           for (int j = i + 1; j < nums.length; j++) {
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               int sum = nums[i] + nums[j];
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               if (-sum < nums[j] || -sum > nums[nums.length - 1]) {
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                   continue;
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              }
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               List<Integer> integers = Arrays.asList(nums[i], nums[j], -sum);
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               if (resultSet.contains(integers)) {
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                   continue;
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              }
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               if (index.get(-sum) != null) {
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                   for (Integer integer : index.get(-sum)) {
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                       if (integer > j) {
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                           resultSet.add(integers);
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                      }
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                  }
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              }
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          }
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      }
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       return new ArrayList<>(resultSet);
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  }
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}
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​
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```

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