---
title: leetcode-121
date: 2020-03-17 18:06:45
tags:
categories: leetcode
---

### 121. 买卖股票的最佳时机

[题目](https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock/)

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原始答案：

```python
class Solution:
  def maxProfit(self, prices) -> int:
    profit = 0
    if not prices:
      return profit
    min_buyin = prices[0]
    max_sellout = prices[0]
    l = len(prices)
    i = 0
    for buyin in prices:
      if i == l:
        return profit
      if min_buyin <= buyin:
        max_sellout = max(prices[i:])
        p = max_sellout - min_buyin
        if p > profit:
          profit = p
      if buyin < min_buyin:
        min_buyin = buyin
      i += 1
    return profit
#1880 ms	14.4 MB
```

主要思路是找到波谷，如果当前价格比前一天要低，则还是在去往波谷的路上；当价格比前一天高或相同时，则到达了一个波谷，计算波谷和之后的波峰的差，就是这一段的利润。将从头至尾过一次，就能找到所有波谷和其后波峰的差，返回最大的即可。但是这个明显地在重复max，时间复杂度是O(n^2)，看起来就很傻逼。

仔细想想，其实并不需要直接找出波谷后的波峰，只要在for loop时保持波谷为最低的那个，就能算出每一个后续与波谷的差，找最大差即可。改了下代码变成这样

```python
class Solution:
  def maxProfit(self, prices) -> int:
    profit = 0
    if not prices:
      return profit
    min_buyin = prices[0]
    for i in range(1, len(prices)):
      min_buyin = min(min_buyin, prices[i-1])
      profit = max(prices[i] - min_buyin, profit)
    return profit
  #44 ms	14.4 MB
```