Leetcode: Best Time to Buy and Sell Stock III

最新推荐文章于 2022-07-03 21:43:22 发布

ayst123

最新推荐文章于 2022-07-03 21:43:22 发布

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本文介绍了一个算法，用于在给定的股票价格数组中找到最多完成两次交易以获取最大利润的方法。它强调了动态规划的概念，并提供了一个解决方案，通过迭代计算每个日期的利润来优化整体策略。

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Question

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 at most two transactions.

Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

Hide Tags Array Dynamic Programming
Hide Similar Problems (M) Best Time to Buy and Sell Stock (M) Best Time to Buy and Sell Stock II (H) Best Time to Buy and Sell Stock IV

Solution

see ‘Best Time to Buy and Sell Stock IV’

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """

        if prices==[]:
            return 0

        if 2>=len(prices):
            return self.maxprofit2(prices)

        globalv = [0]*3
        localv = [0]*3

        for i in range(1,len(prices)):
            diff = prices[i] - prices[i-1]
            for j in range(2,0,-1):
                localv[j]  = max( globalv[j-1] + max(diff,0), localv[j]+diff )
                globalv[j] = max( globalv[j], localv[j] )

        return globalv[2]

    def maxprofit2(self, prices):
        maxv = 0
        for ind in range(1,len(prices)):
            if prices[ind]>prices[ind-1]:
                maxv += prices[ind] - prices[ind-1]

        return maxv