Best Time to Buy and Sell Stock

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 minPrice = prices[0];
        int maxProfit = 0;
        for (int i = 1; i < prices.size(); i++)
        {
            minPrice = min(prices[i], minPrice);
            maxProfit = max(prices[i] - minPrice, maxProfit);
        }

        return maxProfit;
    }
};

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        int i,j,p=0,q=0;
        if(n == 0||n == 1)
        return p;
        int min1=prices[0],max=prices[0],max1,min=prices[0];
        int res1,res2,res3;
        for(i=1;i<n;i++)
        {
            //max=(max<prices[i])?prices[i]:max;
            //min=(min>prices[i])?prices[i]:min;
            if(max<prices[i])
            {
                max=prices[i];
                q=i;
            }
            if(min>prices[i])
            {
                min=prices[i];
                p=i;
            }
        }

        if(p<=q)//最小值在最大值前面
        return (max-min);
        int p1=p,q1=q+1;
        vector<int>price1;
        while(q1<p1)
        {
            price1.push_back(prices[q1]);
            q1++;
        }

      //  return price1[1];
        res1 =  maxProfit(price1);
        return res1;
        for(i=0;i<q;i++)//最大值在前面，找最大值前面俄最小值
        {
           min1=(min1>prices[i])?prices[i]:min1;
        }
        max1=prices[p];
         for(i=n-1;i>=p;i--)
        {
           max1=(max1<prices[i])?prices[i]:max1;
        }

        res2 = ((max-min1)>(max1-min))?(max-min1):(max1-min);
        res3 = (res2>res1)?res2:res1;
        return res3;
    }
};