# 714. Best Time to Buy and Sell Stock with Transaction Fee 买卖股票的最佳时机含手续费

@TOC

## # 题目描述

Your are given an array of integers prices, for which the i-th element is the price of a given stock on day i; and a non-negative integer fee representing a transaction fee.

You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction. You may not buy more than 1 share of a stock at a time (ie. you must sell the stock share before you buy again.)

Return the maximum profit you can make.

``````Example 1:
Input: prices = [1, 3, 2, 8, 4, 9], fee = 2
Output: 8
Explanation: The maximum profit can be achieved by:
Buying at prices[0] = 1
Selling at prices[3] = 8
Buying at prices[4] = 4
Selling at prices[5] = 9
The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.
``````

Note:

• 0 < prices.length <= 50000.
• 0 < prices[i] < 50000.
• 0 <= fee < 50000.

## # 解题方法

### # 动态规划

1. cash 该天结束手里没有股票的情况下，已经获得的最大收益
2. hold 该天结束手里股票的情况下，已经获得的最大收益

cash 更新的策略是：既然今天结束之后手里没有股票，那么可能是今天没买（保持昨天的状态），也可能是今天把股票卖出了

hold 更新的策略是：今天今天结束之后手里有股票，那么可能是今天没卖（保持昨天的状态），也可能是今天买了股票

``````class Solution:
def maxProfit(self, prices, fee):
"""
:type prices: List[int]
:type fee: int
:rtype: int
"""
cash = 0
hold = -prices[0]
for i in range(1, len(prices)):
cash = max(cash, hold + prices[i] - fee)
hold = max(hold, cash - prices[i])
return cash
``````

``````class Solution {
public:
int maxProfit(vector<int>& prices, int fee) {
const int N = prices.size();
vector<int> cash(N, 0);
vector<int> hold(N, 0);
cash[0] = 0;
hold[0] = -prices[0];
for (int i = 1; i < N; i ++) {
cash[i] = max(cash[i - 1], prices[i] + hold[i - 1] - fee);
hold[i] = max(hold[i - 1], cash[i - 1] - prices[i]);
}
return cash[N - 1];
}
};
``````

``````class Solution {
public:
int maxProfit(vector<int>& prices, int fee) {
const int N = prices.size();
// max profit if today don't have stock
int cash = 0;
// max profit if today have stock
int hold = -prices[0];
for (int i = 1; i < N; ++i) {
cash = max(cash, prices[i] + hold - fee);
hold = max(hold, cash - prices[i]);
}
return cash;
}
};
``````

## # 日期

2018 年 4 月 10 日 —— 风很大，很舒服～～ 2018 年 12 月 5 日 —— 周三啦！ 2019 年 2 月 22 日 —— 这周结束了