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Best Time to Buy and Sell Stock with Cooldown

题目戳这

题意

给定一个数组，数组中的每个元素代表某一天股票的价格，你可以买入一只股票然后卖出，当然这个在卖出前一定要买入，而且同一天之内只能买入或者卖出，当卖出以后至少得休息一天，问这样安排下能够收获的最大利益。

解法

动态规划。我们用buy[i]代表第i天买入时能够得到的最大收益，sell[i]代表第i天卖出时能够得到的最大收益。则

buy[i] = max(buy[i-1], sell[i-2]-prices[i]) # 注意限制，休息一天

sell[i] = max(sell[i-1], buy[i-1]+prices[i]) # 第一天买入，第二天可以卖出

得到状态转移方程以后就好写了。。dp就是这么简单粗暴(我指的是代码简单，，想法不简单Orz…)

代码

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class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        if not prices:
            return 0
        sell = [0]*10010
        buy = [0]*10010
        buy[0] = -prices[0]	# 第一天买入
        sell[0] = 0
        mmax = 0
        
        for i in range(1,len(prices)):
            # delta = prices[i] - prices[i-1]
            sell[i] = max(buy[i-1]+prices[i], sell[i-1])
            buy[i] = max(sell[i-2]-prices[i], buy[i-1])
            mmax = max(mmax,sell[i])
        
        return mmax
        
        # if not prices:
        #     return 0
            
        # sell,buy,prev_sell,prev_buy = 0,-prices[0],0,0
        
        # for i in range(1,len(prices)):
        #     prev_buy = buy
        #     buy = max(prev_sell-prices[i], prev_buy)
        #     prev_sell = sell
        #     sell = max(prev_buy+prices[i], prev_sell)
        
        # return sell