[LeetCode] 309. Best Time to Buy and Sell Stock with Cooldown 買賣股票的最佳時間有冷卻期

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]

買賣股票問題，這裏多了一個冷卻期，在賣出股票後的次日不能交易，必須隔一天才能在買。

須要維護三個一維數組buy, sell，和rest。其中：

buy[i]表示在第i天以前最後一個操做是買，此時的最大收益。

sell[i]表示在第i天以前最後一個操做是賣，此時的最大收益。

rest[i]表示在第i天以前最後一個操做是冷凍期，此時的最大收益。

咱們寫出遞推式爲：

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以前有冷凍期，買以前要賣掉以前的股票。一個小技巧是如何保證[buy, rest, buy]的狀況不會出現，這是因爲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，因爲冷凍期的存在，能夠得出rest[i] = sell[i-1]，這樣，能夠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])

作進一步優化，因爲i只依賴於i-1和i-2，因此能夠在O(1)的空間複雜度完成算法，

參考

Java：

public int maxProfit(int[] prices) {
    int sell = 0, prev_sell = 0, buy = Integer.MIN_VALUE, prev_buy;
    for (int price : prices) {
        prev_buy = buy;
        buy = Math.max(prev_sell - price, prev_buy);
        prev_sell = sell;
        sell = Math.max(prev_buy + price, prev_sell);
    }
    return sell;
}

Python：

def maxProfit(self, prices):
    if len(prices) < 2:
        return 0
    sell, buy, prev_sell, prev_buy = 0, -prices[0], 0, 0
    for price in prices:
        prev_buy = buy
        buy = max(prev_sell - price, prev_buy)
        prev_sell = sell
        sell = max(prev_buy + price, prev_sell)
    return sell

C++：

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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