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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] = 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.

solution

dp, sell[i]当前i sell状态的最大Profit, buy[i]当前i buy状态的最大收益

2 solutions, 2 states DP solutions, clear explanation!

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