问题链接

https://leetcode.com/explore/interview/card/top-interview-questions-easy/97/dynamic-programming/572/

解题思路

思路一：两次遍历寻找前后两个数的最大差值，发现超时，时间复杂度O(n^2)
思路二：遍历一次，迭代更新最小值，并计算最小值与最新遍历点的差值，最大的即为最大交易

代码(思路二)

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