LeetCode-122. Best Time to Buy and Sell Stock II

最新推荐文章于 2020-11-03 23:22:31 发布

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本文介绍了一种算法，用于确定股票交易的最佳时机，以实现最大化的利润。通过分析价格波动，该算法能够智能地决定买入和卖出的时间点，即使在复杂的价格走势中也能找到盈利机会。

0.原题

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 (i.e., buy one and sell one share of the stock multiple times).

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

Example 1:

Input: [7,1,5,3,6,4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
             Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.

Example 2:

Input: [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
             Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
             engaging multiple transactions at the same time. You must sell before buying again.

Example 3:

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.

1.代码

class Solution:
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        if not prices:
            return 0
        hold_share = False
        profit = 0
        buy = prices[0]
        for i in range(len(prices)-1):
            if not hold_share:
                if prices[i] < prices[i+1]:
                    hold_share = True
                    buy = prices[i]
            else:
                if prices[i] > prices[i+1]:
                    hold_share = False
                    profit += prices[i] - buy
        if hold_share:
            profit += prices[-1] - buy
        return profit