//方法一:暴力破解法,使用2重循环来解决,这种方法时间复杂度较高
//Time:O(n^2),Space:O(1)
//语言:经典C++
class Solution 
{
public:
    int max(int a,int b)
    {
        return a>b?a:b;
    }
    int maxProfit(vector<int>& prices) 
    {
        int maxfit=0;
        if(prices.size()<2) return 0;
        for(int i=0;i<prices.size();i++)
        {
            for(int j=i+1;j<prices.size();j++)
            {
                maxfit=max(maxfit,prices[j]-prices[i]);
            }
        }
        return maxfit;
    }
};
 
//方法二
//Time:O(n),Space:O(1)
class Solution 
{
public:
    int max(int a,int b)
    {
        return a>b?a:b;
    }
    int maxProfit(vector<int>& prices) 
    {
        if(prices.size()<2||prices.size()==0) return 0;
        int maxFit=0,buy=prices[0];
        for(int i=1;i<prices.size();i++)
        {
            if(prices[i]<buy)
            {
                buy=prices[i];
            }
            else
            {
                maxFit=max(maxFit,prices[i]-buy);
            }
        }
        return maxFit;    
    }
};