Say you have an array for which the i^th 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]

buy[i]表示在第i天之前最後一個操作是買，此時的最大收益。

sell[i]表示在第i天之前最後一個操作是賣，此時的最大收益。

rest[i]表示在第i天之前最後一個操作是冷凍期，此時的最大收益。

我們寫出遞推式為：

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])

上述遞推式很好的表示了在買之前有冷凍期，買之前要賣掉之前的股票。一個小技巧是如何保證[buy, rest, buy]的情況不會出現，這是由於buy[i] <= rest[i]， 即rest[i] = max(sell[i-1], rest[i-1])，這保證了[buy, rest, buy]不會出現。

另外，由於冷凍期的存在，我們可以得出rest[i] = sell[i-1]，這樣，我們可以將上面三個遞推式精簡到兩個：

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

我們還可以做進一步優化，由於i只依賴於i-1和i-2，所以我們可以在O(1)的空間複雜度完成演算法，參見程式碼如下：

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;
    }
};

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