1 class Solution { 2 public: 3     string multiply( string num1, string num2 ) { 4         reverse( num1.begin(), num1.end() ); 5         reverse( num2.begin(), num2.end() ); 6         string product( num1.length()+num2.length(), ‘0‘ ); 7         for( size_t i = 0; i != num1.length(); ++i ) { 8             int carry = 0; 9             for( size_t j = 0; j != num2.length(); ++j ) {10                 carry += product[i+j]-‘0‘ + (num1[i]-‘0‘) * (num2[j]-‘0‘);11                 product[i+j] = carry%10 + ‘0‘;12                 carry /= 10;13             }14             int k = i + num2.length();15             while( carry ) {16                 carry = product[k]-‘0‘ + carry;17                 product[k++] = carry%10 + ‘0‘;18                 carry /= 10;19             }20         }21         reverse( product.begin(), product.end() );22         string::size_type pos = product.find_first_not_of( ‘0‘ );23         return pos == string::npos ? "0" : product.substr(pos);24     }25 };