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 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.

Example 1:

Input: [7, 1, 5, 3, 6, 4]Output: 5max. difference = 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: 0In this case, no transaction is done, i.e. max profit = 0.

求出数列中后面的数与前面的数最大的差值；刚开始就想到了暴力解法，结果超时；随后进一步思考，在一次遍历中完成整个计算。
代码入下：

 1 class Solution { 2 public: 3     int maxProfit(vector<int> &prices) { 4         int m = prices.size(); 5         if(prices.empty()) 6         { 7             return 0; 8         }    9         int curMin = prices[0];10         int profit = 0;11         for(int i = 1; i < m; i ++)12         {13             curMin = min(curMin, prices[i]);14             profit = max(profit, prices[i]-curMin);15         }16         return profit;17     }18 };

leetcode 121