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.

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这个题是贪心题，可以一次遍历得到最小值和它的右边的最大值（比这个最小值大），然后求出两者之差。

官方题解：https://leetcode.com/articles/best-time-to-buy-and-sell-stock/

C++代码：

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int maxnum = 0;
        int minnum = 0x3f3f3f3f;
        for(int i = 0; i < prices.size(); i++){
            if(minnum > prices[i]){
                minnum = prices[i];  //为了得到最小值
            }
            if(prices[i] - minnum > maxnum){
                maxnum = prices[i] - minnum;  //得到两者之差的最大值。
            }
        }
        return maxnum;
    }
};