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.
这题的题意其实看标题基本就明白了，要我们买股和售股，返回最大利润。
Example 1:
Input: [7, 1, 5, 3, 6, 4] Output: 5 max. 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: 0 In this case, no transaction is done, i.e. max profit = 0.
解题思路：
因为本题是要我们找最大的差值，而且不用考虑前面的数，所以我们可以在每次的比较中找较小值，直接从较小值开始往下比。
代码：
1 class Solution { 2 public: 3 int maxProfit(vector<int>& prices) { 4 if(prices.size()<=1) 5 return 0; 6 int result=0; 7 int min=prices[0]; 8 for(int i=1;i<prices.size();i++){ 9 int temp=prices[i]-min; 10 if(temp<=0){ 11 min=prices[i]; 12 continue; 13 } 14 result=(result>temp)?result:temp; 15 } 16 return result; 17 } 18 };
