leetcode oj java 309. Best Time to Buy and Sell Stock with Cooldown

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一、问题描述：

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]

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

二、解决方案：

动态规划，每一天有三种选择：买入、卖出、冷却（被动）。

维护两个数组buys[] 和 sells[] ： buys[i] 表示在第i天的时候的买入的最大利润，sells[i] 表示在第i天的时候的卖出的最大利润。

buys[0] = -prices[0];
buys[1] = Math.max(-prices[0], -prices[1]);
sells[1] = Math.max(0, prices[1] - prices[0]);

递归公式：

buys[i] = Math.max( buys[i-1] , sells[i-2]-prices[i] ); 第i天是否买入如果buys[i-1]>sells[i-2]-prices[i] 说明第i天不应该买入，否则买入。

sells[i] = Math.max( sells[i-1] , buys[i-1] + prices[i]);第i天是否卖出如果sells[i-1]>buys[i-1] + prices[i] 说明第i天不应该卖出，否则卖出。

最终返回sells的最大值即可。

三、代码：

package T01;/** * @author 作者 : xcy * @version 创建时间：2017年1月13日 下午10:12:15 *          类说明 */public class t309 {    public static void main(String[] args) {        // TODO Auto-generated method stub        int[] prices = { 1, 2, 3, 0, 2 };        System.out.println(maxProfit(prices));    }    public static int maxProfit(int[] prices) {        int day = prices.length;        int[] sells = new int[day];        int[] buys = new int[day];        buys[0] = -prices[0];        buys[1] = Math.max(-prices[0], -prices[1]);        sells[1] = Math.max(0, prices[1] - prices[0]);        int max = 0;        for (int i = 2; i < day; i++) {            buys[i] = Math.max(buys[i - 1], sells[i - 2] - prices[i]);            sells[i] = Math.max(sells[i - 1], buys[i - 1] + prices[i]);            max = max > sells[i] ? max : sells[i];        }        return max;    }}

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