Multiply Strings

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.
Converting the input string to integer is NOT allowed.
You should NOT use internal library such as BigInteger.

说明：模拟手工计算两个整数的乘积，注意每轮乘积最高位的进位和乘积为0的情况。

代码：

char* multiply(char* num1, char* num2) {
    char *zero = "0";
    if(strcmp(num1, zero) == 0 || strcmp(num2, zero) == 0)  return zero;

    int len1 = strlen(num1), len2 = strlen(num2);
   //逆序存放乘积
    char *ret = (char *)malloc(sizeof(char) * (len1+len2+1));    
    memset(ret, '0', sizeof(char) * (len1+len2+1));

    int i = 0, j = 0;

    for(i = 0; i <= len1-1; i++)
    {
    //乘数为0则跳过这次循环，开始下一次循环
        if(num1[len1-1-i] == '0')    continue;
        int carry = 0;
        int n1 = num1[len1-1-i] - '0';
        for(j = 0; j <= len2-1; j++)
        {
            int n2 = num2[len2-1-j] - '0';
            int sum = n1 * n2 + ret[i+j] - '0' + carry;
            carry = sum / 10;
            ret[i+j] = sum % 10 + '0';
        }
        if(carry)   ret[i+len2] = carry + '0';
    }

    ret[len1+len2] = '\0';
  //  printf("%s\n", ret);

    char *p = ret, *q = &ret[len1+len2-1];
    char tmp;
    //将逆序的乘积转换成正序
    while(p < q)
    {
        tmp = *p;
        *p = *q;
        *q = tmp;
        p++;
        q--;
    }
    //如100*100，再ret所指字符数组里原来存放"000010",翻转后变成"010000"，需要去掉开头的'0'
    if(*ret == '0')
        return ret+1;
    else
        return ret;
}