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.

分析：要买入和卖出的差价最大，仅仅要算出今天与历史最低价的差价，再与前面记录的最大差价进行比較，从第一天遍历到第N天就可以

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        int n = prices.size();
        if(n < 2)return 0;

        int min = prices[0];
        int ret = 0;
        
        for(int i = 0; i < n; i++)
        {
            if(ret < prices[i]-min)ret = prices[i]-min;
            if(min > prices[i])min = prices[i];
        }
        
        return ret;
    }
};