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 = 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.
Note:

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

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee
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思路：

1. cash表示不持有股票时的最大利润，hold表示持有股票时的最大利润

2. 在第i天，需根据dii-1天来更新cash和hold的值：

对于cash，可以保持不变，也可以卖出股票获得price[i]，但是要减去交易费用

对于hold，可以保持持有股票不变，也可以买入股票，则要减去price[i]

class Solution {
public:
    int maxProfit(vector<int>& prices, int fee) {
        int cash = 0, hold = -prices[0];
        for(int i = 1; i < prices.size(); i++){
            cash = max(cash, hold+prices[i]-fee); //要么是不变，要么是持有股票卖出
            hold = max(hold, cash-prices[i]); // 要么不变，要么买入股票
        }
        return cash;
    }
};