LeetCode题解：309. Best Time to Buy and Sell Stock with Cooldown

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]

原题地址

该题就是可以进行很多次的交易，但是交易和交易之间需要至少隔一天的冷却期。动态规划问题，重要的是定义状态，找到状态转移方程。该题目中，实际上有三种状态，手里有股票buy,手里没有股票sell,冷却cooldown。

状态转移方程式：

buy[i] = max{buy[i-1],cooldown[i-1]-prices[i-1]} 

表示如果第i天持有股票的话，那么有可能是之前买的了，或者是当天买的，如果是当天买的话，要求昨天必须是冷却期，取两种情况的最大值。

cooldown[i] = max{cooldown[i-1],sell[i-1]}

表示如果第i天是冷却期的话，那么有可能前一天就是冷却期，或者前一天刚把股票卖掉，而sell包括刚刚卖掉的情况。

sell[i] = max{buy[i-1]+price[i-1],sell[i-1]}

表示如果第i天手里没有股票的话，那么有可能是前一天早就卖掉了，有可能是今天刚刚卖。

然而总是用前一天的数据，没有必要用一个数组来存。优化如下：

class Solution {    public int maxProfit(int[] prices) {        if (prices == null || prices.length == 0 || prices.length == 1) return 0;        int buy = Integer.MIN_VALUE, sell = 0, cooldown = 0;        for (int price : prices) {            int buyi = buy, selli = sell, cooldowni = cooldown;            buy = Integer.max(buy, cooldowni - price);            cooldown = Integer.max(cooldown, selli);            sell = Integer.max(selli, buyi + price);        }        return Integer.max(sell, cooldown);    }}