leetcode 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]

原文地址 http://www.hrwhisper.me/leetcode-best-time-to-buy-and-sell-stock-with-cooldown/

題目地址：https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-cooldown/

題意：

給定一個數組prices，prices[i]表明第i天股票的價格。讓你進行若干次買賣，求最大利潤

• 你每次只能買一支並且在再次買入以前必須出售以前手頭上的股票（就是手頭上最多有一支股票）
• 每次出售須要休息一天才能再次買入

設sell[i] 賣出操做的最大利潤。它須要考慮的是，第i天是否賣出。（手上有stock在第i天所能得到的最大利潤）

buy[i] 買進操做的最大利潤。它須要考慮的是，第i天是否買進。（手上沒有stock在第i天所能得到的最大利潤）

因此，顯然有狀態轉移方程

• buy[i] = max(buy[i-1] , sell[i-2] – prices[i])  // 休息一天在買入，因此是sell[i-2]在狀態轉移
• sell[i] = max(sell[i-1], buy[i-1] + prices[i])

最後顯然有sell[n-1] > buy[n-1] 因此咱們返回sell[n-1]

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        if not prices or len(prices) < 2: return 0
        n = len(prices)
        buy, sell = [0] * n, [0] * n
        buy[0] = -prices[0]
        buy[1] = max(-prices[0], -prices[1])
        sell[1] = max(0, prices[1] - prices[0])
        for i in xrange(2, n):
            buy[i] = max(sell[i - 2] - prices[i], buy[i - 1])
            sell[i] = max(buy[i - 1] + prices[i], sell[i - 1])
    
        return sell[n - 1]

本文地址（就是個人新blog）： http://www.hrwhisper.me/leetcode-best-time-to-buy-and-sell-stock-with-cooldown/
本文由  hrwhisper  原創發佈，轉載請點名出處