题目：

Say you have an array for which the ith element is the price of a given stock on day i.Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times) with the following restrictions:You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).After you sell your stock, you cannot buy stock on next day. (ie, cooldown 1 day)Example:prices = [1, 2, 3, 0, 2]maxProfit = 3transactions = [buy, sell, cooldown, buy, sell]

思路：维护两个数组，数组buy表示买入的最大利润，它需要考虑第i天是否买入；数组sell表达卖出的最大利润，它需要考虑第i天是否卖出。然后写出下面的递推公式：

buy[i] = max(buy[i-1], sell[i-2] - prices[i])

sell[i] = max(sell[i-1], buy[i-1] + prices[i])

其中buy[0] = -prices[0], buy[1] = max(-prices[0], -prices[1]), sell[0] = 0, sell[1] = max(0, prices[1] - prices[0])

比如：

prices：1,  2,  3, 0, 2buy:   -1, -1, -1, 1, 1 sell    0,  1,  2, 2, 3

代码：

public int maxProfit(int[] prices) {        if(prices == null || prices.length <= 1) {            return 0;        }        int len = prices.length;        int[] buy = new int[len];        int[] sell = new int[len];        buy[0] = -prices[0];        buy[1] = Math.max(-prices[0], -prices[1]);        sell[0] = 0;        sell[1] = Math.max(0, prices[1]-prices[0]);        for(int i = 2; i < len; i++) {            buy[i] = Math.max(buy[i-1], sell[i-2]-prices[i]);            sell[i] = Math.max(sell[i-1], buy[i-1]+prices[i]);        }        return sell[len-1];    }