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

动态规划感觉很重要，很锻炼思维。

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.

思路如下：
定义：
hold：买了石头后手头的钱
sell：卖了石头后手头的钱
状态转移：
当天卖了石头后手头的钱＝max（前一天卖了石头后手头的钱，今天做了交易之后手头的钱）
当天买了石头后手头的钱＝max（前一天买了石头后手头的钱，今天用手头的钱买了石头后手头的钱）

代码：

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

看似很短，更重要的是思路