题目描述：
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.
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.
代码实现：
1 class Solution(object): 2 def maxProfit(self, prices): 3 """ 4 :type prices: List[int] 5 :rtype: int 6 """ 7 max_profit, min_price = 0, float('inf') 8 for price in prices: 9 min_price = min(min_price, price) 10 profit = price - min_price 11 max_profit = max(max_profit, profit) 12 return max_profit
分析：
借助一个term值，结合穷举和min操作，便可以得到序列的最小值。