A wonderful time to buy and sell stocks

  Traverse the entire array , Deviate to No i Number of hours , Before finding out i-1 The minimum number , Subtracting , Update stock profits

step

1. Create two variables , One stores stock profits , Before a storage i-1 The minimum value of

2. Traverse the entire array for comparison

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int res = 0,m=1e4+10;
        for(int i = 0;i < prices.size(); i++){
            res=max(prices[i]-m,res);
            m=min(m,prices[i]);
        }
        return res;
    }
};