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

You 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] = 1
Selling at prices[3] = 8
Buying at prices[4] = 4
Selling 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.

原题地址

题目的意思是，每次交易都会有手续费，可以多次交易。

buy[i]表示第i天手里有股票所能获得的最大利润，sell[i]表示第i天手里没有股票所能获得的最大利润。

buy[i] = max{buy[i-1],sell[i-1]-prices[i-1]}

sell[i] = max{sell[i-1],buy[i-1]+prices[i-1]-fee}

因为只用到前一天的状态，所以可以优化如下：

class Solution {    public int maxProfit(int[] prices, int fee) {        if (prices == null || prices.length == 0 || prices.length == 1) return 0;        int buy = Integer.MIN_VALUE, sell = 0;        for (int price : prices) {            buy = Integer.max(buy, sell - price);            sell = Integer.max(sell, buy + price - fee);        }        return sell;    }}