算法描述：

Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2, also represented as a string.

Example 1:

Input: num1 = "2", num2 = "3"Output: "6"

Example 2:

Input: num1 = "123", num2 = "456"Output: "56088"

Note:

1. The length of both num1 and num2 is < 110.
2. Both num1 and num2 contain only digits 0-9.
3. Both num1 and num2 do not contain any leading zero, except the number 0 itself.
4. You must not use any built-in BigInteger library or convert the inputs to integer directly.

解题思路：1 两个数相乘，得到结果最大长度为两个数位数之和。

　　　　    2 num1中的i 位和 num2中的j位相乘，其结果在最终结果中所属位置为 i+j和i+j+1。

                  3 注意细节，例如进位等。

string multiply(string num1, string num2) {        if(num1.size()==0 || num2.size()==0) return "0";        string sum(num1.size()+num2.size(),'0');        for(int i = num1.size()-1; i >=0; i--){            for(int j = num2.size()-1; j>=0; j--){                int temp = (num1[i] -'0') * (num2[j] - '0');                temp += sum[i+j+1] - '0';                sum[i+j+1] = temp%10 + '0';                sum[i+j] += temp/10;                            }        }        int index = sum.find_first_not_of('0');        if(string::npos != index)            return sum.substr(index);        return "0";    }