LeetCode121:Best Time to Buy and Sell Stock

题目：

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.

解题思路：

这道题不难，最简单的解题思路复杂度为O（n²),不过我想应该会超时，这里采用另一种解题方法，复杂度为O（n²）。

采用两个指针i和j，i用来指向所遍历过的元素中最小值，j则用来遍历数组，然后用j指向的值与i指向的值相减，如果差值大于max，则替换max，当j指向的值小于i指向的元素时，将i = j,继续之前的动作。

代码：

#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;

}