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🐱(dynamic): 123. 买卖股票的最佳时机 III

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`@@ -485,6 +485,57 @@ class Solution(object):`
`485`	`485`	`return dp[n]`
`486`	`486`	```
`487`	`487`
	`488`	`+## 123. 买卖股票的最佳时机 III`
	`489`	`+`
	`490`	`+[原题链接](https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock-iii/)`
	`491`	`+`
	`492`	`+### 思路`
	`493`	`+`
	`494`	`+这个系列的题真是折磨人呐。首先还是要先确定需要记录几个状态值。`
	`495`	`+`
	`496`	`+对于某一天来说,可能会发生的情况:`
	`497`	`+`
	`498`	`+1. 第一次被买入`
	`499`	`+2. 第一次被卖出`
	`500`	`+3. 第二次被买入`
	`501`	`+4. 第二次被卖出`
	`502`	`+`
	`503`	`+按题目要求,我们最重要求出的是第二次被卖出所能获得的最大收益。`
	`504`	`+`
	`505`	`+我们分别用 4 个变量来表示这 4 种情况下所能获得的最大收益:`
	`506`	`+`
	`507`	+1. 在某天完成第一次买入的最大收益 `first_buy`
	`508`	+2. 在某天完成第一次被卖出的最大收益 `first_sell`
	`509`	+3. 在某天完成第二次被买入的最大收益 `second_buy`
	`510`	+4. 在某天完成第二次被卖出最大收益 `second_sell`
	`511`	`+`
	`512`	`+可以得出状态转移公式:`
	`513`	`+`
	`514`	+```
	`515`	`+first_buy = max(first_buy, -price)`
	`516`	`+first_sell = max(first_sell, first_buy + price)`
	`517`	`+second_buy = max(second_buy, first_sell - price)`
	`518`	`+second_sell = max(second_sell, second_buy + price)`
	`519`	+```
	`520`	`+`
	`521`	+其中 `price` 为某天股票的价格。
	`522`	`+`
	`523`	+```python
	`524`	`+class Solution:`
	`525`	`+ def maxProfit(self, prices: List[int]) -> int:`
	`526`	`+ first_buy = float('-inf')`
	`527`	`+ first_sell = 0`
	`528`	`+ second_buy = float('-inf')`
	`529`	`+ second_sell = 0`
	`530`	`+`
	`531`	`+ for price in prices:`
	`532`	`+ first_buy = max(first_buy, -price)`
	`533`	`+ first_sell = max(first_sell, first_buy + price)`
	`534`	`+ second_buy = max(second_buy, first_sell - price)`
	`535`	`+ second_sell = max(second_sell, second_buy + price)`
	`536`	`+`
	`537`	`+ return second_sell`
	`538`	+```
`488`	`539`
`489`	`540`	`## 152. 乘积最大子序列`
`490`	`541`

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