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.

思路：

模拟手工加法过程，用carry表示进位。

第一个是自己写的，比较啰嗦。

第二个是参考的网上大神的，简洁。

string addStrings(string num1, string num2)     {         int n1 = num1.length()-1, n2 = num2.length()-1;         string ret="";         int digit = 0;         int carry = 0;         int sum = 0;         while (n1 >= 0 && n2 >= 0)         {             sum = carry + num1[n1--] + num2[n2--] - '0' - '0';             digit = sum %10;             carry = (sum >= 10) ? 1 : 0;             ret += (digit+'0');         }                 while (n1 >= 0)         {             sum = carry + num1[n1--] - '0';             digit = sum % 10;             carry = (sum >= 10) ? 1 : 0;             ret += (digit + '0');         }                          while (n2 >= 0)         {             sum = carry + num2[n2--] - '0';             digit = sum % 10;             carry = (sum >= 10) ? 1 : 0;             ret += (digit + '0');         }                         if (carry) ret += '1';         reverse(ret.begin(), ret.end());         return ret;              }

string addStrings(string num1, string num2) {    int i = num1.size() - 1;    int j = num2.size() - 1;    int carry = 0;    string res = "";    while(i>=0 || j>=0 || carry){        long sum = 0;        if(i >= 0){sum += (num1[i] - '0');i--;}        if(j >= 0){sum += (num2[j] - '0');j--;}        sum += carry;         carry = sum / 10;        sum = sum % 10;        res =  res + to_string(sum);    }    reverse(res.begin(), res.end());    return res;}

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