Best Time to Buy and Sell Stock with Transaction Fee

本文链接：https://blog.youkuaiyun.com/caijhBlog/article/details/78710898

本文介绍了一种在考虑交易费用的情况下，通过动态规划算法实现的最佳股票买卖策略。策略通过维护两个数组sell和hold，分别记录每天卖出和持有股票所能获得的最大利润，并最终返回最后一天的最大利润。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Best Time to Buy and Sell Stock with Transaction Fee

A.题意

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.

题目的大概意思就是你要对一只股票在n天内进行操作来获得最大利润，在n天内你可以卖出或买进股票但你手里最多只能持有这一只股票，你每次卖出股票都要交一笔手续费fee。

B.分析

很显然这是一道动态规划的问题，这n天的序列你你通过不断地卖出买进这个操作我们无法贪心，这里只能采用动态规划比较来获得最大利润的值，这里我们用两个数组sell[i]表示到第i天卖股票能得到的最大利润，hold表示第i天持有股票能获得的最大利润，于是递推式为

sell[i] = max(sell[i - 1],hold[i - 1] + prices[i] - fee);
hold[i] = max(hold[i - 1],sell[i - 1] - prices[i]);
最后的答案为sell[n]

C.代码实现

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