Best Time to Buy and Sell Stock with Transaction Fee问题及解法

问题描述：

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.

示例:

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.

问题分析：

有题可知，每一天都会存在股票交易的两种状态-----buy or sell，我们定义如下的状态转换：

buy[i] --- 第i天之前已buy为结束状态的序列所获得的最大利益

sell[i] --- 第i天之前以sell为结束状态的序列所获得的最大利益

那么

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

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

过程详见代码：

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