LeetCode：Best Time to Buy and Sell Stock with Transaction Fee

题目：

　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.

Note:
0 < prices.length <= 50000.
0 < prices[i] < 50000.
0 <= fee < 50000

Example:

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.

思路：

　　这次的题目的不同之处是每一次卖出股票时需要交一定的交易费用，交易次数没有限制。
　　和之前一样，这次还是用动态规划的思想：维护两个变量，buy[i]（第 i+1 天为买进的最大利润）和 sell[i]（第 i+1 天为卖出股票的最大利润）。显然，第一天不可能卖出股票，所以buy[0] = -prices[0]，sell[0] = 0。它们的状态转移方程为：
　　buy[i] = max(sell[i-1] - prices[i], buy[i-1]);
　　sell[i] = max(buy[i-1] + prices[i] - fee, sell[i-1]);
　　

代码：

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

复杂度分析：

时间复杂度：O(n)
空间复杂度：O(n)