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

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

Description

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 = 2Output: 8Explanation: 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] < 50000.0 <= fee < 50000.

分析

题目的意思是： 求股票交易的最大收益，其实这道题也是股票买卖的变形题目，这个题目加了一个交易费用，这样的话，跟以前的稍稍不一样。然后在你进行下一次买入前，必须把当前的股票交易。

sell[i]表示第i天卖掉股票此时的最大利润，buy[i]表示第i天保留手里的股票此时的最大利润

在第i天，我们要卖掉手中的股票，那么此时的利润是前一天手里有股票的利润+此时卖掉股票的价格+交易费用；跟前一天卖出的利润相比，取较大值。如果前一天的利润较大，那么我们就在前一天卖了，不留在今天了。如果第i天不卖的利润，就是昨天股票卖了的利润然后今天再买入股票，得减去今天的价格，得到的值和昨天股票保留时的利润相比，取其中的较大值，如果昨天保留股票的利润大，那么我们就继续保留到今天，所以递推时可以得到：

sell[i] = max(sell[i - 1], buy[i - 1] + prices[i] - fee);

buy[i] = max(buy[i - 1], sell[i - 1] - prices[i]);

代码

class Solution {public: int maxProfit(vector& prices, int fee) { int n=prices.size(); vector buy(n,0); vector sell(n,0); buy[0]=-prices[0]; for(int i=1;i

参考文献

​​[LeetCode] Best Time to Buy and Sell Stock with Transaction Fee 买股票的最佳时间含交易费​​

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