問(wèn)題描述

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.

思路

1. 記錄兩個(gè)值，min_price【代碼中用p表示】和max_profit【代碼中用pf表示】
2. 用for loop過(guò)一遍list，對(duì)于每個(gè)新的元素：

• 如果小于min_price，則不可能創(chuàng)造新的max_profit，于是只更新min_price
• 如果不是新的min_price，則有可能創(chuàng)造新的max_profit，于是對(duì)比當(dāng)前數(shù)字cur_price - min_price 與max_profit的值，若大于max_profit，更新max_profit

3. 返回max_profit

    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        pf = 0
        p = 999999999
        
        for i in range (0, len(prices)):
            if prices[i] < p:
                p = prices[i]
            elif prices[i]-p > pf:
                pf = prices[i]-p
            
        return pf