LeetCode解题报告（319)-- 714. Best Time to Buy and Sell Stock with Transaction Fee

Problem

You are given an array prices where prices[i] is the price of a given stock on the ith day, and an integer fee representing a transaction fee.

Find the maximum profit you can achieve. You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction.

Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

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.

Example 2:

Input: prices = [1,3,7,5,10,3], fee = 3
Output: 6

Constraints:

• 1 <= prices.length <= 5 * 104
• 1 <= prices[i] < 5 * 104
• 0 <= fee < 5 * 104

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Analysis

  股票买卖类的系列题，解题的大方向还是动态规划，因为买卖都有，所以还是使用两个状态：

• hold[i]：第i天持有股票的最大收益；
• sold[i]：第i天没有持有股票的最大收益。

   然后我们来看看状态转移方程怎么写。收益的变化只有在买卖的时候才会有，所以对于某一天来说，如果什么都不做的话，应该是和前一天的收益是一样的。然后我们要考虑买卖发生的情况，因为股票不能重复购买或重复抛售，只能一买一卖，所以交易发生只有两种情况：

1. 第i - 1天是持有状态，第i天卖出；
2. 第i - 1天是抛售状态，第i天买入。

  我们根据上面两种状态来写状态转移方程即可，买入的时候减去当天股票的价格即可，卖出的时候需要加上当前股票的价格，但是还要减去手续费。然后结果是最后一天抛售的结果。

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Solution

  无

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Code

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

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Summary

  这道题是比较简单的dp，条件和状态转移都不难，主要是理清买卖的关系和发生条件。这道题目的分享到这里，感谢你的支持！