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Commit 71a9081

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docs: 新增No121题解
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‎leetcode刷题/README.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.121 买卖股票的最佳时机](https://github.com/Mayandev/javascript_algorithm/blob/master/leetcode%E5%88%B7%E9%A2%98/note/No121_max-profit.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.125 验证回文串](https://github.com/Mayandev/javascript_algorithm/blob/master/leetcode%E5%88%B7%E9%A2%98/note/No125_is-palindrome.md)
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- [No.136 只出现一次的数字](https://github.com/Mayandev/javascript_algorithm/blob/master/leetcode%E5%88%B7%E9%A2%98/note/No136_single-number.md)

‎leetcode刷题/note/No121_max-profit.md

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# No.121 买卖股票的最佳时机
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难度:`easy`
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给定一个数组,它的第 i 个元素是一支给定股票第 i 天的价格。
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如果你最多只允许完成一笔交易(即买入和卖出一支股票),设计一个算法来计算你所能获取的最大利润。
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注意你不能在买入股票前卖出股票。
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## 示例
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示例 1:
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```
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输入: [7,1,5,3,6,4]
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输出: 5
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解释: 在第 2 天(股票价格 = 1)的时候买入,在第 5 天(股票价格 = 6)的时候卖出,最大利润 = 6-1 = 5 。
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注意利润不能是 7-1 = 6, 因为卖出价格需要大于买入价格。
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```
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示例 2:
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```
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输入: [7,6,4,3,1]
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输出: 0
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解释: 在这种情况下, 没有交易完成, 所以最大利润为 0。
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```
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## 解题思路
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**解题思路1:**
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暴力法,两重循环,找到利润最大。
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代码如下:
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```javascript
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/**
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* @param {number[]} prices
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* @return {number}
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*/
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var maxProfit = function(prices) {
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let maxProfit = 0;
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// 暴力法
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for (let i = 0, n = prices.length; i < n; i++) {
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for (let j = i+1; j < n; j++) {
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let profit = prices[j] - prices[i];
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if (profit > maxProfit)
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maxProfit = profit;
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}
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}
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return maxProfit;
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};
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```
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**解题思路二:**
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一遍循环,循环的时候就计算出当前最大的利润,最后取利润的最大值。
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代码如下:
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```javascript
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/**
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* @param {number[]} prices
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* @return {number}
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*/
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var maxProfit = function(prices) {
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let maxProfit = 0;
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let lastPrice = prices[0];
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let profitArr = [];
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for (let i = 1, n = prices.length; i < n; i++) {
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if (prices[i] > lastPrice) {
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profitArr.push(prices[i]-lastPrice);
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} else {
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lastPrice = prices[i];
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}
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}
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console.log(profitArr);
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return profitArr.length == 0 ? 0 : Math.max(...profitArr);
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};
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```

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