leetcode 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 = 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.

动态规划：dp[i][1]表示第天持有能得到的最大值，dp[i][0]代表第i天不持有能得到的最大值。
那么

dp[i][0] = max(dp[i-1][0], (prices[i] + dp[i-1][1] - fee));
dp[i][1] = max(dp[i-1][0] - prices[i], dp[i-1][1]);

代码如下：

class Solution {
public:
    int maxProfit(vector<int>& prices, int fee) {
        int n = prices.size();
        vector<vector<int> > dp(n, vector<int>(2));
        for (int i = 0; i < n; ++i) {
            if (i == 0) {
                dp[i][1] = -prices[i];
                dp[i][0] = 0;
            } else {
                dp[i][0] = max(dp[i-1][0], (prices[i] + dp[i-1][1] - fee));
                dp[i][1] = max(dp[i-1][0] - prices[i], dp[i-1][1]);
            }
        }
        return dp[n-1][0];
    }
};