Leetcode 714 - Node

1. 题目要求

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.

2. Code

DP解法：

　　pay用于计算买入价格

　　sell用于计算当前利润

　　DP的算法是，假定我们当前买入第一天的股票，开销是 【第一天价格】+手续费。利润0。接下来是类似滑窗的设计，数组中，只有i-1和i中的数值是有效的，其中i-1记录着之前的开销，i位置则是当前i-1处的利润-i处的价格+手续费，如果赚到的钱在买了新股，并交了手续费后赚到了钱，就买入。i位记录新的数值，否则就不买，i位记录之前的开销。这样保证买在低点。（开销为负）

　　接下来的利润为开销+当前卖出价，i-1初始为0，i处则是i-1买入价格+i卖出价格，即当前的balance，这样计算可以保证卖在高点。

public int maxProfit(int[] prices, int fee) {
        int res = 0;
        int size = prices.length;
        int[] pay = new int[size];
        int[] sell = new int[size];

        pay[0] = 0 - prices[0] - fee;

        for(int i = 0 ; i < prices.length ; i ++) {

            pay[i] = Math.max(pay[i - 1], sell[i - 1] - prices[i] - fee);
            sell[i] = Math.max(sell[i - 1], pay[i - 1] + prices[i]);

        }

        return sell[size - 1];
    }