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

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

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] = 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.

解析：
此问题可以使用动态规划来解决，对于动态规划问题，我们首先要找到问题的状态变化。
从题目来看，可以分析出，在买卖股票的过程中，我只有两种状态，第一种是我持有一只股票，第二种是我没有股票。
那么，我假设

S0 = 我手头没有股票的利润S1 = 我手头有股票的利润

由此能够写出递推式

S0 = max(s0, s1 + p);S1 = max(s1, s0 - p - fee);

由于我最后一个状态一定是手头上没有股票的，因此递推的结果就是S0的值。

代码：

class Solution {public:    int maxProfit(vector<int>& prices, int fee) {        int s0 = 0, s1 = INT_MIN;         for(int p:prices) {            int tmp = s0;            s0 = max(s0, s1+p);            s1 = max(s1, tmp-p-fee);        }        return s0;    }};