Multiply Strings

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

Note:

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

　　基本思路：如上图所示，因为两个字符串相乘，长度不定，可以将结果存到一个数组中。我们可以看到num1[i]和num2[j]的乘积最后位置是在indices[i+j]和indices[i+j+1]两个位置上，同时我们知道

一个m位的数乘一个n位的数做多是m+n位。

class Solution {
public:
    string multiply(string num1, string num2) {//num1[i] * num2[j]` will be placed         
                                               //at indices `[i + j`, `i + j + 1]` m*n位数最多m+n位
        string result;
        int k = 0, sum = 0;
        int len1 = num1.size(), len2 = num2.size();
        vector<int>  res(len1 + len2);
        for (int i = len1-1; i>=0;i--)
        {
            for (int j = len2-1; j >= 0; j--)
            {
                int k = (num1[i] - '0')*(num2[j]-'0');
                sum = k + res[i + j + 1];//前一次最高位
                res[i + j + 1] = sum % 10;
                res[i + j] += sum / 10;
            }
        }
        //去除首位的0
        int j = len1 + len2 - 1;
        for (int i = 0; i <len1+len2; i++)
        {
            if (res[i] == 0) continue;
            else
            {
                j = i;
                break;
            }
        }
        for (int i = j; i < len1 + len2; i++)
        {
            result += res[i] + '0';
        }
        return result;
    }
};