【LEETCODE】121-Best Time to Buy and Sell Stock

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

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

题意：

给一个数组，第i个元素是某股票在第i天的价格，

如果只允许完成一次交易，即买一次卖一次，设计算法找到最大利润

思路：

找到最大差值

初始化low＝array[0]，ans＝0，

low当然是遇到越小的越好，因为后面的即将遇到的价格都是一样的，low越小差越大

比ans大的就可以作为新的差值，是由array[i]－low生成的

class Solution(object):    def maxProfit(self, prices):        """        :type prices: List[int]        :rtype: int        """                if len(prices)==0:            return 0                low=prices[0]        ans=0                for i in range(1,len(prices)):            if prices[i]<low:                low=prices[i]            elif prices[i]-low>ans:                ans=prices[i]-low                return ans