題目：

Say you have an array for which the i^th 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.

解題思路：

這道題不難，最簡單的解題思路複雜度為O（n²),不過我想應該會超時，這裡採用另一種解題方法，複雜度為O（n²

採用兩個指標i和j，i用來指向所遍歷過的元素中最小值，j則用來遍歷陣列，然後用j指向的值與i指向的值相減，如果差值大於max，則替換max，當j指向的值小於i指向的元素時，將i = j,繼續之前的動作。

程式碼：

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

/**
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.

 
*/

class Solution {
public:
    int maxProfit(vector<int> &prices) {
        if(prices.empty())
            return 0;
        int i = 0;
        int j = i+1;
        int max = 0;
        while(j < prices.size())
        {
            if(prices[j] < prices[i])
            {
                i 
 
= j;

            }
            else
            {
                int t = prices[j] - prices[i];
                if(t > max)
                    max = t;
            }
            j++;
        }
        return max;
        
    }
};

int main(void)
{
    int arr[] = {2,4,5,1,7,10};
    int n = sizeof(arr) / sizeof(arr[0]);
    vector<int> stock(arr, arr+n);
    Solution solution;
    int max = solution.maxProfit(stock);
    cout<<max<<endl;
    return 0;
}