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📝 update alg lc
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‎Java/alg/README.md‎

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- [3.无重复字符的最长子串](lc/3.无重复字符的最长子串.md)
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- [5.最长回文子串](lc/5.最长回文子串.md)
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- [6.Z字形变换](lc/6.Z字形变换.md)
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- [11.盛最多水的容器](lc/11.盛最多水的容器.md)
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- [15.三数之和](lc/15.三数之和.md)
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## 笔试
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‎Java/alg/lc/11.盛最多水的容器.md‎

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# 11. 盛最多水的容器
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[url](https://leetcode-cn.com/problems/container-with-most-water/)
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## 题目
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给你 n 个非负整数 a1，a2，...，an，每个数代表坐标中的一个点 (i, ai) 。在坐标内画 n 条垂直线，垂直线 i 的两个端点分别为 (i, ai) 和 (i, 0) 。找出其中的两条线，使得它们与 x 轴共同构成的容器可以容纳最多的水。
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说明：你不能倾斜容器。
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![](https://aliyun-lc-upload.oss-cn-hangzhou.aliyuncs.com/aliyun-lc-upload/uploads/2018/07/25/question_11.jpg)
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```
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输入：[1,8,6,2,5,4,8,3,7]
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输出：49
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解释：图中垂直线代表输入数组 [1,8,6,2,5,4,8,3,7]。在此情况下，容器能够容纳水（表示为蓝色部分）的最大值为 49。
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输入：height = [1,1]
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输出：1
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输入：height = [4,3,2,1,4]
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输出：16
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输入：height = [1,2,1]
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输出：2
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```
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## 方法
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## code
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### js
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```js{cmd="node"}
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let maxArea = height => {
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// 算面积
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let max = 0;
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let l = 0, r = height.length - 1;
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while (l < r) {
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let min = height[l] < height[r] ? height[l++] : height[r--];
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max = Math.max(max, (r - l + 1) * min);
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}
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return max;
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};
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```
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### go
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```go
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func maxArea(height []int) int {
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max := 0
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l, r := 0, len(height) - 1
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for l < r {
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min := 0
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if height[l] < height[r] {
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min = height[l]
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l++
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} else {
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min = height[r]
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r--
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}
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max = max9(max, (r - l + 1) * min)
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}
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return max
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}
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func max9(a int, b int) int {
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if a < b {
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return b
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} else {
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return a
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}
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}
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```
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### 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 max = 0;
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int l = 0, r = height.length - 1;
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while (l < r){
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int min = height[l] < height[r] ? height[l++] : height[r--];
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max = Math.max(max, (r - l + 1) * min);
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}
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return max;
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}
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}
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```
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‎Java/alg/lc/15.三数之和.md‎

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# 15. 三数之和
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[url](https://leetcode-cn.com/problems/3sum/)
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## 题目
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给你一个包含 n 个整数的数组 nums，判断 nums 中是否存在三个元素 a，b，c ，使得 a + b + c = 0 ？请你找出所有和为 0 且不重复的三元组。
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注意：答案中不可以包含重复的三元组。
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```
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输入：nums = [-1,0,1,2,-1,-4]
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输出：[[-1,-1,2],[-1,0,1]]
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输入：nums = []
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输出：[]
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输入：nums = [0]
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输出：[]
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```
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## 方法
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## code
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### js
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```js
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let treeSum = nums => {
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let ret = [];
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if (nums === null || nums.length === 0)
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return ret;
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nums.sort();
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for (let i = 0; i < nums.length - 2; i++) {
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if (nums[i] > 0)
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break;
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if (i !== 0 && nums[i] === nums[i - 1])
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continue;
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let l = i + 1, r = nums.length - 1;
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while (l < r) {
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let sum = nums[l] + nums[r];
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if (sum < -nums[i])
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l++;
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else if (sum > -nums[i])
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r--;
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else {
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ret.push([nums[i], nums[l], nums[r]]);
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// 防止重复
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while (l < r && nums[l] === nums[l + 1]) l++;
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while (l < r && nums[r] === nums[r - 1]) r--;
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l++;
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r--;
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}
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}
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}
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return ret;
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};
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```
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### go
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```go
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func threeSum(nums []int) [][]int {
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res := make([][]int, 0)
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if nums == nil || len(nums) == 0 {
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return res
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}
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sort.Ints(nums)
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for i := 0; i < len(nums)-2; i++ {
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if nums[i] > 0 {
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break
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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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l, r := i + 1, len(nums) - 1
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for l < r {
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sum := nums[l] + nums[r]
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if sum < -nums[i] {
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l++
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} else if sum > -nums[i] {
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r--
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} else {
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res = append(res, []int{nums[i], nums[l], nums[r]})
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for l < r && nums[l] == nums[l+1] {
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l++
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}
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for l < r && nums[r] == nums[r-1] {
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r--
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}
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l++
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r--
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}
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}
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}
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return res
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}
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```
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### 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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List<List<Integer>> ret = new ArrayList<>();
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if (nums == null || nums.length == 0)
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return ret;
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Arrays.sort(nums);
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for (int i = 0; i < nums.length - 2; i++){
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if (nums[i] > 0)
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break;
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if (i != 0 && nums[i] == nums[i - 1])
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continue;
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int l = i + 1, r = nums.length - 1;
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while (l < r){
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int sum = nums[l] + nums[r];
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if (sum < -nums[i])
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l++;
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else if (sum > -nums[i])
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r--;
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else {
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ret.add(Arrays.asList(nums[i], nums[l], nums[r]));
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// 防止重复
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while (l < r && nums[l] == nums[l + 1]) l++;
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while (l < r && nums[r] == nums[r - 1]) r--;
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l++;
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r--;
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}
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}
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}
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return ret;
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}
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}
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```
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