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# \[714]\[中等]\[动态规划] 买卖股票的最佳时机含手续费

## 题目描述

[714. 买卖股票的最佳时机含手续费](https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/)

给定一个整数数组 prices，其中第 i 个元素代表了第 i 天的股票价格 ；非负整数 fee 代表了交易股票的手续费用。

你可以无限次地完成交易，但是你每笔交易都需要付手续费。如果你已经购买了一个股票，在卖出它之前你就不能再继续购买股票了。

返回获得利润的最大值。

注意：这里的一笔交易指买入持有并卖出股票的整个过程，每笔交易你只需要为支付一次手续费。

示例 1:

```
输入: prices = [1, 3, 2, 8, 4, 9], fee = 2
输出: 8
解释: 能够达到的最大利润:  
在此处买入 prices[0] = 1
在此处卖出 prices[3] = 8
在此处买入 prices[4] = 4
在此处卖出 prices[5] = 9
总利润: ((8 - 1) - 2) + ((9 - 4) - 2) = 8.
```

注意:

* 0 < prices.length <= 50000.
* 0 < prices\[i] < 50000.
* 0 <= fee < 50000.

## 解题思路

与[\[122\]\[简单\]\[动态规划\] 买卖股票的最佳时机 II](/garnet/suan-fa/dong-tai-gui-hua/122-mai-mai-gu-piao-de-zui-jia-shi-ji-ii.md)区别不大, 只是在状态转移的时候要考虑手续费成本.

```python
class Solution:
    def maxProfit(self, prices: List[int], fee: int) -> int:
        n = len(prices)
        if n < 2:
            return 0

        hold, bear = -prices[0] - fee, 0
        for i in range(1, n):
            old_hold = hold
            hold = max(old_hold, bear - prices[i] - fee)
            bear = max(bear, old_hold + prices[i])
        return bear
```


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