[leetcode] 309. Best Time to Buy and Sell Stock with Cooldown

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 = 3
transactions = [buy, sell, cooldown, buy, sell]
题意：
给定一个数组，代表了n天的股价。你可以交易任意多次，但是不允许交易之间相交，即上次的买必须卖出之后才允许再去买入。不过加了个条件，卖完之后必须是要休息一天。
思路：
状态的跳转依旧是时间之间的跳转，即第i天的收益情况依赖于第i-1天的收益情况。不过现在需要三个状态，即buy，sell，cooldown，我们记录第i-1天的这三个状态的收益情况是last_ buy,last_sell,last_cooldown。那么第i天的这三个收益情况的依赖关系是：
buy=max(last_buy, last_cooldown - price[i])
sell = max(last_sell, last_buy + price[i])
cooldown = max(cooldown, last_sell);
然后就是每天计算完之后将该天的这三个状态值赋值last。
代码如下：

class Solution {public:    int maxProfit(vector<int>& prices) {        int last_sell = 0, last_buy = INT_MIN, last_cooldown = 0, sell = 0, buy = 0, cooldown = 0;        for(auto price:prices) {            buy = max(last_buy, last_cooldown - price);            sell = max(last_sell, last_buy + price);            cooldown = max(last_cooldown, last_sell);            last_buy = buy;            last_sell = sell;            last_cooldown = cooldown;        }        return sell;    }};