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:

Input: [1,2,3,0,2]
Output: 3 
Explanation: transactions = [buy, sell, cooldown, buy, sell]

方法1: dynamic programming

思路：

花花酱： https://www.youtube.com/watch?v=oL6mRyTn56M
维持三个变量，标示第 i 天的状态：buy，sell，cooldown。这三个状态的转移关系如下：

也就是说sold只能从hold来，sold只能变成rest，上面的关系可以表达为如下的转移方程：
buy[i] = max(rest[i-1] - price, buy[i-1])
sell[i] = max(buy[i-1] + price, sell[i-1])
rest[i] = max(sell[i-1], buy[i-1], rest[i-1])

观察上面的方程发现，因为rest[i] = sold[i - 1]，进一步化简
buy[i] = max(sell[i-2] - price, buy[i-1])
sell[i] = max(buy[i-1] + price, sell[i-1])

最终可以降维成如下代码：

class Solution {
public:
    int maxProfit(vector<int>& prices) {
         int hold = INT_MIN;
         int sold = 0;
         int rest = 0;
        
        for (int p: prices){
            int prev_sold = sold;
            sold = hold + p;
            hold = max(hold, rest - p);
            rest = max(rest, prev_sold);
        }
        return max(rest, sold);
    }
};