415. Add Strings 计算两个数字字符串的和

Given two non-negative integers num1 and num2 represented as string, return the sum of num1 and num2.
Note:
1. The length of both num1 and num2 is < 5100.
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.

class Solution {public:    string addStrings(string num1, string num2)     {         int len1 = num1.size();        int len2 = num2.size();        if (len1 <= 0 && num2.size() <= 0)            return "";        else if (len1 <= 0)            return num2;        else if (len2 <= 0)            return num1;        else        {            string result = ""; //记录结果，将从个位开始计算的结果依次添加到后边，因而最后需要将前后翻转            int c = 0;//每个位上的进位            int sum;            int minlen = min(len1, len2);            int i = len1 - 1;            int j = len2 - 1;            while (minlen > 0)            {                sum = c + (num1[i] - '0') + (num2[j] - '0');//本位上的和等于两个家数在本位上的数字与上一位的进位之和                c = sum / 10;//计算本位上的进位                result += sum % 10 + '0';//计算本位上除了进位之后的个位数，直接添加到结果的后面                i--;                j--;                minlen--;            }                 // 当num1较长            while (i != -1)            {                sum = c + (num1[i] - '0');                c = sum / 10;                result += sum % 10 + '0';                i--;            }            // 当num2较长            while (j != -1)            {                sum = c + (num2[j] - '0');                c = sum / 10;                result += sum % 10 + '0';                j--;            }             // 处理最后的进位            if (i == -1 && j == -1 && c != 0)                result += c + '0';            reverse(result.begin(), result.end());            return result;        }    }};