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📝 update alg lc
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‎Java/alg/README.md‎

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- [543.二叉树的直径](lc/543.二叉树的直径.md)
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- [566.重塑矩阵](lc/566.重塑矩阵.md)
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- [572.另一个树的子树](lc/572.另一个树的子树.md)
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- [594.最长和谐子序列](lc/594.最长和谐子序列.md)
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- [594.最长和谐子序列](lc/594.最长和谐子序列.md)
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- [605.种花问题](lc/605.种花问题.md)
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- [617.合并二叉树](lc/617.合并二叉树.md)
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- [628.三个数的最大乘积](lc/628.三个数的最大乘积.md)
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- [637.二叉树的层平均值](lc/637.二叉树的层平均值.md)
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- [645.错误的集合](lc/645.错误的集合.md)

‎Java/alg/lc/594.最长和谐子序列.md‎

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# 594. 最长和谐子序列
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[url](https://leetcode-cn.com/problems/longest-harmonious-subsequence/)
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[url](https://leetcode-cn.com/problems/can-place-flowers/)
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## 题目
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输入：nums = [1,1,1,1]
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输出：0
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```
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返回 true，因为 t 与 s 的一个子树拥有相同的结构和节点值。
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## 方法
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‎Java/alg/lc/605.种花问题.md‎

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# 605.种花问题
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[url](https://leetcode-cn.com/problems/longest-harmonious-subsequence/)
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## 题目
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假设有一个很长的花坛，一部分地块种植了花，另一部分却没有。可是，花不能种植在相邻的地块上，它们会争夺水源，两者都会死去。
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给你一个整数数组  flowerbed 表示花坛，由若干 0 和 1 组成，其中 0 表示没种植花，1 表示种植了花。另有一个数 n ，能否在不打破种植规则的情况下种入 n 朵花？能则返回 true ，不能则返回 false。
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```
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输入：flowerbed = [1,0,0,0,1], n = 1
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输出：true
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输入：flowerbed = [1,0,0,0,1], n = 2
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输出：false
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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 canPlaceFlowers = (flowerbed, n) => {
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if (n < 0)
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return false;
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let cnt = 0;
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let len = flowerbed.length;
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for (let i = 0; i < len; i++) {
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if (flowerbed[i] === 1)
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continue;
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let pre = i === 0 ? 0 : flowerbed[i - 1];
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let next = i === len - 1 ? 0 : flowerbed[i + 1];
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if (pre === 0 && next === 0) {
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cnt++;
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flowerbed[i] = 1;
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}
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}
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return cnt >= n;
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};
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console.log(canPlaceFlowers([1,0,0,0,1], 1));
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console.log(canPlaceFlowers([1,0,0,0,1], 2));
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```
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### go
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```go
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func canPlaceFlower(flowerbed []int, n int) bool {
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if n < 0 {
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return false
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}
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cnt := 0
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l := len(flowerbed)
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for i := 0; i < l; i++ {
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if flowerbed[i] == 1{
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continue
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}
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pre, next := 0, 0
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if i == 0 {
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pre = 0
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} else {
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pre = flowerbed[i - 1]
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}
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if i == l - 1 {
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next = 0
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} else {
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next = flowerbed[i + 1]
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}
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if pre == 0 && next == 0 {
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cnt++
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flowerbed[i] = 1
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}
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}
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return cnt >= n
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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 boolean canPlaceFlowers(int[] flowerbed, int n) {
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// 边界？
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if (n < 0) return false;
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int cnt = 0;
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int len = flowerbed.length;
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for (int i = 0; i < len; i++) {
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// 判断是1的
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if (flowerbed[i] == 1) continue;
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int pre = i == 0 ? 0 : flowerbed[i - 1];
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int next = i == len - 1 ? 0 : flowerbed[i + 1];
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if (pre == 0 && next == 0) {
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cnt++;
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flowerbed[i] = 1;
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}
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}
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return cnt >= n;
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}
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}
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```
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‎Java/alg/lc/617.合并二叉树.md‎

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# 617. 合并二叉树
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[url](https://leetcode-cn.com/problems/merge-two-binary-trees/)
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## 题目
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给定两个二叉树，想象当你将它们中的一个覆盖到另一个上时，两个二叉树的一些节点便会重叠。
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你需要将他们合并为一个新的二叉树。合并的规则是如果两个节点重叠，那么将他们的值相加作为节点合并后的新值，否则不为 NULL 的节点将直接作为新二叉树的节点。
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```
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输入:
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Tree 1 Tree 2
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1 2
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/ \ / \
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3 2 1 3
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/ \ \
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5 4 7
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输出:
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合并后的树:
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3
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/ \
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4 5
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/ \ \
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5 4 7
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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 mergeTrees = (l1, l2) => {
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if (l1 === null && l2 === null)
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return null;
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if (l1 === null)
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return l2;
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if (l2 === null)
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return l1;
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let root = TreeNode(l1.val + l2.val);
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root.left = mergeTrees(l1.left, l2.left);
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root.right = mergeTrees(l1.right, l2.right);
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return root;
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};
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```
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### go
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```go
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func mergeTrees(t1 *TreeNode, t2 *TreeNode) *TreeNode {
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if t1 == nil && t2 == nil {
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return nil
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}
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if t1 == nil {
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return t2
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}
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if t2 == nil {
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return t1
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}
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root := &TreeNode{Val: t1.Val + t2.Val}
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root.Left = mergeTrees(t1.Left, t2.Left)
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root.Right = mergeTrees(t1.Right, t2.Right)
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return root
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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 TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
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if (t1 == null && t2 == null) return null;
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if (t1 == null) return t2;
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if (t2 == null) return t1;
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TreeNode root = new TreeNode(t1.val + t2.val);
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root.left = mergeTrees(t1.left, t2.left);
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root.right = mergeTrees(t1.right, t2.right);
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return root;
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}
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}
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```
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‎Java/alg/lc/628.三个数的最大乘积.md‎

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# 628. 三个数的最大乘积
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[url](https://leetcode-cn.com/problems/maximum-product-of-three-numbers/)
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## 题目
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给你一个整型数组 nums ，在数组中找出由三个数组成的最大乘积，并输出这个乘积。
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```
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输入：nums = [1,2,3]
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输出：6
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输入：nums = [1,2,3,4]
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输出：24
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输入：nums = [-1,-2,-3]
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输出：-6
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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 maximumProduct = nums => {
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let minValue = -Infinity;
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let max1 = minValue, max2 = minValue, max3 = minValue;
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let min1 = -minValue, min2 = -minValue;
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nums.forEach(n => {
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if (n > max1) {
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max3 = max2;
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max2 = max1;
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max1 = n;
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} else if (n > max2) {
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max3 = max2;
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max2 = n;
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} else if (n > max3) {
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max3 = n;
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}
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if (n < min1) {
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min2 = min1;
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min1 = n;
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} else if (n < min2) {
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min2 = n;
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}
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});
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return Math.max(max1 * max2 * max3, max1 * min1 * min2);
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};
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console.log(maximumProduct([1, 2, 3]));
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console.log(maximumProduct([1, 2, 3, 4]));
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```
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### go
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```go
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func maximumProduct(nums []int) int {
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INT_MAX := int(^uint((0)) >> 1)
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INT_MIN := ^INT_MAX
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max1, max2, max3 := INT_MIN, INT_MIN, INT_MIN
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min1, min2 := INT_MAX, INT_MAX
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for _, num := range nums {
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if num > max1 {
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max3 = max2
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max2 = max1
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max1 = num
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} else if num > max2 {
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max3 = max2
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max2 = num
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} else if num > max3 {
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max3 = num
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}
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if num < min1 {
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min2 = min1
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min1 = num
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} else if num < min2 {
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min2 = num
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}
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}
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return max4(max1 * max2 * max3, max1 * min1 * min2)
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}
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func max4(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 maximumProduct(int[] nums) {
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int max1 = Integer.MIN_VALUE, max2 = Integer.MIN_VALUE, max3 = Integer.MIN_VALUE,min1 = Integer.MAX_VALUE, min2 = Integer.MAX_VALUE;
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for (int n : nums) {
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if (n > max1) {
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max3 = max2;
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max2 = max1;
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max1 = n;
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} else if (n > max2) {
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max3 = max2;
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max2 = n;
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} else if (n > max3) {
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max3 = n;
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}
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if (n < min1) {
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min2 = min1;
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min1 = n;
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} else if (n < min2) {
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min2 = n;
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}
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}
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return Math.max(max1*max2*max3, max1*min1*min2);
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}
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}
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```
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