LeetCode 121. Best Time to Buy and Sell Stock (买入和卖出股票的最佳时机)-优快云博客

本文介绍了一种简单有效的算法，用于解决股票交易中如何通过一次买卖操作获得最大利润的问题。算法仅需遍历一次价格数组，记录遍历过程中的最低价格，并计算当前价格与最低价格的差值作为潜在利润，最终得到的最大利润即为解决方案。

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原题

Say you have an array for which the ith element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Note that you cannot sell a stock before you buy one.

Example 1:

Input: [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
             Not 7-1 = 6, as selling price needs to be larger than buying price.

Example 2:

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

Reference Answer

思路分析

这道题相当简单，感觉达不到Medium的难度，只需要遍历一次数组，用一个变量记录遍历过数中的最小值，然后每次计算当前值和这个最小值之间的差值最为利润，然后每次选较大的利润来更新。当遍历完成后当前利润即为所求，代码如下：

Code

class Solution:
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        if not prices:
            return 0
        min_count = prices[0]
        max_count = prices[0]
        profit = 0
        for index,count in enumerate(prices[1:]):
            if count < min_count:
                min_count = count
            
            profit = max(profit, count - min_count)
            
        return profit