[leetcode-415-Add Strings]

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;
}

参考：

https://discuss.leetcode.com/topic/62305/c-_accepted_13ms