class Solution {public:    string multiply(string num1, string num2) {        int m = num1.size();        int n = num2.size();        //record each digit of result in reverse order        vector<int> result(m+n, 0);         //num index, the rightmost denotes as 0        int ind1 = 0;           int ind2 = 0;        for(int i = m-1; i >= 0; i --)        {            int carrier = 0;            int n1 = num1[i]-‘0‘;            //for every digit in num1, the num2 start with the rightmost digit            ind2 = 0;               for(int j = n-1; j >= 0; j --)            {                int n2 = num2[j]-‘0‘;                int product = n1*n2 + result[ind1+ind2] + carrier;                carrier = product/10;                result[ind1+ind2] = product%10;                ind2 ++;            }            //all digits in num2 finished multiplication with num1[i]            //the remaining carrier added in the right position            result[ind1+ind2] += carrier;            ind1 ++;        }                //skip 0s in the right of result        while(result.size() > 0 && result[result.size()-1] == 0)            result.pop_back();        if(result.empty())            return "0";        else        {            string s;            int i = result.size();            while(i --)                s += (‘0‘+result[i]);            return s;        }    }};