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 (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.
解题方案：
只要计算当前值和前面最小值的差是否大于最大收益就行了，下面是本题代码：
1 class Solution { 2 public: 3 int maxProfit(vector<int> &prices) { 4 if (prices.size() == 0) { 5 return 0; 6 } 7 int MinDay = INT_MAX; 8 int MaxFit = 0; 9 for (size_t i = 0; i < prices.size(); ++i) { 10 if (prices[i] - MinDay > MaxFit) { 11 MaxFit = prices[i] - MinDay; 12 } 13 if (prices[i] < MinDay) { 14 MinDay = prices[i]; 15 } 16 } 17 return MaxFit; 18 } 19 };
Best Time to Buy and Sell Stock
原文：http://www.cnblogs.com/skycore/p/4029935.html