Given two numbers represented as strings, return multiplication of the numbers as a string. Note: The numbers can be arbitrarily large and are non-negative.

LeetCode-Multiply Strings

C++:
class Solution {
public:
    string multiply(string num1, string num2) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(num1.size()==0||num2.size()==0) return "";
        string ret(num1.size()+num2.size()+1,'0');
        reverse(num1.begin(),num1.end());
        reverse(num2.begin(),num2.end());
        for(int i=0;i<num2.size();i++){
            int dig2=num2[i]-'0';
            int carry=0;
            for(int j=0;j<num1.size();j++){
                int dig1=num1[j]-'0';
                int temp=ret[i+j]-'0';
                int cur=dig1*dig2+temp+carry;
                ret[i+j]=cur%10+'0';
                carry=cur/10;
            }
            ret[i+num1.size()]=carry+'0';
        }
        reverse(ret.begin(),ret.end());
        int pos=ret.find_first_not_of('0',0);
        if(pos<0||pos>=ret.size())
            pos=ret.size()-1;
        return ret.substr(pos,ret.size()-pos);
    }
};
Java:
public class Solution {
    public String multiply(String num1, String num2) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(num1.length()==0||num2.length()==0) return "";
        if(num1.equals("0")||num2.equals("0")) return "0";
        char[] c1=new StringBuilder(num1).reverse().toString().toCharArray();
        char[] c2=new StringBuilder(num2).reverse().toString().toCharArray();
        char[] c=new char[c1.length+c2.length+1];
        Arrays.fill(c,'0');
        for(int i=0;i<c2.length;i++){
            int dig2=c2[i]-'0';
            int carry=0;
            for(int j=0;j<c1.length;j++){
                int dig1=c1[j]-'0';
                int temp=c[i+j]-'0';
                int cur=dig1*dig2+temp+carry;
                c[i+j]=(char) (cur%10+'0');
                carry=cur/10;
            }
            c[i+c1.length]=(char) (carry+'0');
        }
        String ret=new StringBuilder(new String(c)).reverse().toString();
        int pos=0;
        while(ret.charAt(pos)=='0'&&pos<ret.length()) pos++;
        return ret.substring(pos);
    }
}

Thanks :) Works Perfect

what is the time complexity ?

the Time Complexity is O(m*n) where n=Len(Str1),m=Len(Str2)