[leetcode] 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 = 3
transactions = [buy, sell, cooldown, buy, sell]

解法一：

此题需要维护三个一维数组buy, sell，和rest。其中：

buy[i]表示在第i天之前最后一个操作是买，此时的最大收益。

sell[i]表示在第i天之前最后一个操作是卖，此时的最大收益。

rest[i]表示在第i天之前最后一个操作是冷冻期，此时的最大收益。

我们写出递推式为：

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])

另外，由于冷冻期的存在，我们可以得出rest[i] = sell[i-1]，这样，我们可以将上面三个递推式精简到两个：

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

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