# [LeetCode] 714. Best Time to Buy and Sell Stock with Transaction Fee 买卖股票的最佳时间有交易费

2021年09月15日 阅读数：2

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.html

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.)java

Return the maximum profit you can make.python

Example 1:数组

```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:post

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

sell[i] = Math.max(sell[i - 1], buy[i - 1] + prices[i])

Most consistent ways of dealing with the series of stock problems

Java: pay the fee when buying the stock

```public int maxProfit(int[] prices, int fee) {
if (prices.length <= 1) return 0;
int days = prices.length, buy[] = new int[days], sell[] = new int[days];
for (int i = 1; i<days; i++) {
buy[i] = Math.max(buy[i - 1], sell[i - 1] - prices[i] - fee); // keep the same as day i-1, or buy from sell status at day i-1
sell[i] = Math.max(sell[i - 1], buy[i - 1] + prices[i]); // keep the same as day i-1, or sell from buy status at day i-1
}
return sell[days - 1];
}
```

Java: pay the fee when selling the stock

```public int maxProfit(int[] prices, int fee) {
if (prices.length <= 1) return 0;
int days = prices.length, buy[] = new int[days], sell[] = new int[days];
for (int i = 1; i<days; i++) {
buy[i] = Math.max(buy[i - 1], sell[i - 1] - prices[i]); // keep the same as day i-1, or buy from sell status at day i-1
sell[i] = Math.max(sell[i - 1], buy[i - 1] + prices[i] - fee); // keep the same as day i-1, or sell from buy status at day i-1
}
return sell[days - 1];
}　```

Python:

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

C++:

```class Solution {
public:
int maxProfit(vector<int>& prices, int fee) {
int s0 = 0, s1 = INT_MIN;
for(int p:prices) {
int tmp = s0;
s0 = max(s0, s1+p);
s1 = max(s1, tmp-p-fee);
}
return s0;
}
};
```