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docs: 新增No88、No412题解
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‎leetcode刷题/README.md‎

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- [No.48 旋转图像](https://github.com/Mayandev/javascript_algorithm/blob/master/leetcode%E5%88%B7%E9%A2%98/note/No48_rotate.md)
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- [No.66 加一](https://github.com/Mayandev/javascript_algorithm/blob/master/leetcode%E5%88%B7%E9%A2%98/note/No66_plus-one.md)
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- [No.70 爬楼梯](https://github.com/Mayandev/javascript_algorithm/blob/master/leetcode%E5%88%B7%E9%A2%98/note/No70_climb-stairs.md)
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- [No.88 合并两个有序数组](https://github.com/Mayandev/javascript_algorithm/blob/master/leetcode%E5%88%B7%E9%A2%98/note/No88_merge.md)
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- [No.98 验证二叉搜索树](https://github.com/Mayandev/javascript_algorithm/blob/master/leetcode%E5%88%B7%E9%A2%98/note/No98_is-valid-BST.md)
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- [No.104 二叉树的最大深度](https://github.com/Mayandev/javascript_algorithm/blob/master/leetcode%E5%88%B7%E9%A2%98/note/No104_max-depth.md)
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- [No.122 买卖股票的最佳时机 II](https://github.com/Mayandev/javascript_algorithm/blob/master/leetcode%E5%88%B7%E9%A2%98/note/No122_max-profit.md)
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- [No.326 3的幂](https://github.com/Mayandev/javascript_algorithm/blob/master/leetcode%E5%88%B7%E9%A2%98/note/No326_is-power-of-three.md)
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- [No.350 两个数组的交集 II](https://github.com/Mayandev/javascript_algorithm/blob/master/leetcode%E5%88%B7%E9%A2%98/note/No350_intersect.md)
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- [No.387 字符串中的第一个唯一字符](https://github.com/Mayandev/javascript_algorithm/blob/master/leetcode%E5%88%B7%E9%A2%98/note/No387_first-uniq-char.md)
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-[No.412 Fizz Buzz](https://github.com/Mayandev/javascript_algorithm/blob/master/leetcode%E5%88%B7%E9%A2%98/note/No412_fizz-buzz.md)
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## 分类
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# No.412 Fizz Buzz
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难度: `easy`
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写一个程序,输出从 1 到 n 数字的字符串表示。
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1. 如果 n 是3的倍数,输出"Fizz";
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2. 如果 n 是5的倍数,输出"Buzz";
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3. 如果 n 同时是3和5的倍数,输出 "FizzBuzz"。
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## 示例
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示例:
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```
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n = 15,
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返回:
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[
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"1",
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"2",
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"Fizz",
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"4",
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"Buzz",
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"Fizz",
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"7",
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"8",
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"Fizz",
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"Buzz",
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"11",
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"Fizz",
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"13",
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"14",
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"FizzBuzz"
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]
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```
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## 解题思路
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直接判断是否能够整除即可。
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代码如下:
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```javascript
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/**
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* @param {number} n
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* @return {string[]}
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*/
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var fizzBuzz = function(n) {
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let strArr = [];
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for (let i = 1; i <= n; i++) {
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if (i % 5 == 0 && i % 3 == 0) {
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strArr.push('FizzBuzz');
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continue;
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}
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if (i % 3 == 0) {
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strArr.push('Fizz');
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continue;
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}
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if (i % 5 == 0) {
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strArr.push('Buzz');
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continue;
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}
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strArr.push(`${i}`);
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}
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return strArr;
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};
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```
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‎leetcode刷题/note/No88_merge.md‎

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# No.88 合并两个有序数组
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难度:`easy`
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给定两个有序整数数组 nums1 和 nums2,将 nums2 合并到 nums1 中,使得 num1 成为一个有序数组。
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说明:
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初始化 nums1 和 nums2 的元素数量分别为 m 和 n。
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你可以假设 nums1 有足够的空间(空间大小大于或等于 m + n)来保存 nums2 中的元素。
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## 示例
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示例:
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```
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输入:
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nums1 = [1,2,3,0,0,0], m = 3
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nums2 = [2,5,6], n = 3
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输出: [1,2,2,3,5,6]
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```
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## 解题思路
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第一种很简单粗暴,直接合并两个数字,再进行排序。
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但是这样没有利用到数组原本就有序的特点。
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这里使用插入排序的思想,将num2的数组依次插入到num1中。
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代码如下:
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```javascript
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/**
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* @param {number[]} nums1
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* @param {number} m
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* @param {number[]} nums2
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* @param {number} n
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* @return {void} Do not return anything, modify nums1 in-place instead.
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*/
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var merge = function(nums1, m, nums2, n) {
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let j = 0;
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let valid = m;
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while (j < n) {
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let isInsert = false;
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let element = nums2[j];
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for (let i = 0; i < valid; i++) {
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// 插入排序
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if (nums1[i] > element) {
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// 整个数组向后移动
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for (let a = valid; a > i; a--) {
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nums1[a] = nums1[a-1];
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}
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nums1[i] = element;
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isInsert = true;
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break;
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}
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}
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if (!isInsert) {
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nums1[valid] = element;
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}
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valid++;
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j++;
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}
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};
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// merge([1,2,3,0,0,0],3,[2,5,6],3);
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```

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