62 lines
1.7 KiB
Markdown
62 lines
1.7 KiB
Markdown
---
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title: leetcode-121
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date: 2020-03-17 18:06:45
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tags:
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- leetcode
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categories:
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- leetcode
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---
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### 121. 买卖股票的最佳时机
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[题目](https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock/)
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<!--more-->
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原始答案：
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```python
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class Solution:
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def maxProfit(self, prices) -> int:
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profit = 0
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if not prices:
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return profit
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min_buyin = prices[0]
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max_sellout = prices[0]
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l = len(prices)
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i = 0
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for buyin in prices:
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if i == l:
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return profit
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if min_buyin <= buyin:
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max_sellout = max(prices[i:])
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p = max_sellout - min_buyin
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if p > profit:
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profit = p
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if buyin < min_buyin:
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min_buyin = buyin
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i += 1
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return profit
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#1880 ms 14.4 MB
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```
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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。
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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
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```python
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class Solution:
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def maxProfit(self, prices) -> int:
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profit = 0
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if not prices:
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return profit
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min_buyin = prices[0]
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for i in range(1, len(prices)):
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min_buyin = min(min_buyin, prices[i-1])
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profit = max(prices[i] - min_buyin, profit)
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return profit
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#44 ms 14.4 MB
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```
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