【leetcode】Best Time to Buy and Sell (easy)

题目：

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.

说白了，就是找一组数字里最大差值，并且大数要在小数的后面。 因为只能做一次买卖操作。

思路：从后向前记录最大的数和该最大数前面的最小数，不断更新最大差值。

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        if(prices.empty())
        {
            return 0;
        }
        int maxNum = prices.back();
        int minNum = prices.back();
        int maxprofit = 0;
        vector<int>::iterator it;
        for(it = prices.end() - 2; it >= prices.begin(); it--)
        {
            if(*it < minNum)
            {
                minNum = *it;
            }
            else if(*it > maxNum)
            {
                maxprofit = ((maxNum - minNum) > maxprofit) ? (maxNum - minNum) : maxprofit;
                maxNum = *it;
                minNum = *it;
            }
        }
        maxprofit = ((maxNum - minNum) > maxprofit) ? (maxNum - minNum) : maxprofit;
        return maxprofit;
    }
};

int main()
{
    Solution s;
    vector<int> vec;
    int n = s.maxProfit(vec);

    return 0;
}