714. Best Time to Buy and Sell Stock with Transaction Fee

最新推荐文章于 2025-04-17 23:54:56 发布

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最新推荐文章于 2025-04-17 23:54:56 发布

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分类专栏： leetcode 算法 leetcode不会做的动态规划 Leetcode 文章标签： leedcode

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本文介绍了一种通过多次买卖股票并支付交易费用来最大化利润的算法。该算法使用两个数组记录每天买入和卖出的最佳利润，并通过动态规划的方法迭代计算出最终的最大利润。

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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 feerepresenting 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] = 1Selling at prices[3] = 8Buying at prices[4] = 4Selling at prices[5] = 9The total profit is ((8 - 1) - 2) + ((9 - 4) - 2) = 8.

Note:

0 < prices.length <= 50000.0 < prices[i] < 50000.

思路：

数组a表示第i天卖出时的利润，b表示第i天买入时的利润。

class Solution {
public:
    int maxProfit(vector<int>& prices, int fee) {
        int a[50001],b[50001];
        a[0]=0;
        b[0]=-prices[0];
        for(int i=1;i<prices.size();i++)
        {
            a[i]=max(a[i-1],b[i-1]+prices[i]-fee);
            b[i]=max(b[i-1],a[i-1]-prices[i]);
        }
        return a[prices.size()-1];
    }
};