Description：

• Difficulty:Medium
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• Contributor: Jiachen_

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

Thought：

给定的任何一天, 它的最大利润状态归结为以下两种状态之一:
(1) buy状态:
buy[i] 表示在第 i 天利润最大，而且手中持有一只股票。之后可以在 i+1 天的时候卖出, 或者不卖出。
(2) sell状况:
sell[i] 表示在第 i 天利润最大，手中没有股票。在第 i+1 的时候可以买一只, 或者不买。

基本情况:
在第 0 天时有
buy[0] = -prices[0] – 手续费;//如果之后买入所花费用更低，状态转换后就会变成第i天买入
sell[0] = 0;

状态转换:
buy[i] = max(buy[i-1], sell[i-1] – prices[i] – 手续费);
sell[i] = max(sell[i-1], buy[i-1] + prices[i]);

最后:
return sell[last_day];

Code：

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