Leetcode Add Strings problem solution

In this Leetcode Add Strings problem solution we have given two non-negative integers, num1 and num2 represented as a string, return the sum of num1 and num2 as a string.

You must solve the problem without using any built-in library for handling large integers (such as BigInteger). You must also not convert the inputs to integers directly.

Problem solution in Python.

def addStrings(self, num1, num2):
        result = ""
        carry = 0

        def equalizeNumberOfCharacters(num1, num2):
            if len(num1) < len(num2):
                while len(num1) != len(num2):
                    num1 = "0" + num1
            else:
                while len(num2) != len(num1):
                    num2 = "0" + num2
            return [num1,num2]
        
        num1, num2 = equalizeNumberOfCharacters(num1, num2)
        
        num1Array = list(num1)
        num2Array = list(num2)
        
        while len(num1Array) != 0:
            add = int(num1Array.pop()) + int(num2Array.pop()) + int(carry)
            carry = add // 10
            result = str(add % 10) + result
        
        if carry != 0:
            result = str(carry) + result
        return result

Problem solution in Java.

class Solution {
	public String addStrings(String num1, String num2) {
		StringBuilder result = new StringBuilder();
		int r1 = num1.length();
		int r2 = num2.length();
		int carry = 0;
		while(r1>0 || r2>0) {
			int n1 = (r1 > 0) ? (num1.charAt(r1-1) - '0') : 0;
			int n2 = (r2 > 0) ?	(num2.charAt(r2-1) - '0') : 0;
			int sum = (n1 + n2 + carry) % 10;
			carry = (n1 + n2 + carry) / 10;
			result.insert(0, sum);
			r1 -= 1;
			r2 -= 1;
		}
		if(carry > 0) {
			result.insert(0, carry);
		}
		return result.toString();
	}
}

Problem solution in C++.

string addStrings(string num1, string num2) {
    if (num1.size() > num2.size())
	return addStrings(num2, num1);
    reverse(num1.begin(), num1.end());
    reverse(num2.begin(), num2.end());
    string sum;
    int carry = 0, i = 0;

    for (; i < num1.size(); i++) {
    	int curDigit = (num1[i] - '0' + num2[i] - '0' + carry) % 10;
    	carry = (num1[i] - '0' + num2[i] - '0' + carry) / 10;
    	sum += to_string(curDigit);
    }
    for (; i < num2.size(); i++) {
    	int curDigit = (num2[i] - '0' + carry) % 10;
    	carry = (num2[i] - '0' + carry) / 10;
    	sum += to_string(curDigit);
    }
    if (carry == 1)
    	sum += "1";
    reverse(sum.begin(), sum.end());
    return sum;
}

Problem solution in C.

char* addStrings(char* num1, char* num2) {
    int len1 = strlen(num1), len2 = strlen(num2);
    int i = len1 - 1, j = len2 - 1, tmp, len, k;
    char *ret = NULL;
    len = len1 > len2? len1: len2;
    ret = (char *)malloc(sizeof(char) * (len + 2));

    for (k = 0; k < len + 1; k++)
        ret[k] = '0';
    ret[k] = '\0';
    
    for (; i >= 0 || j >= 0; len--, i--, j--) {
        tmp = ret[len] - '0';
        if (i >= 0)
            tmp += num1[i] - '0';
        if (j >= 0)
            tmp += num2[j] - '0';
        if (tmp >= 10) { 
            ret[len] = tmp - 10 + '0'; 
            ret[len - 1] = '1';
        }
        else
            ret[len] = tmp + '0';
    }
    if ('0' != ret[0])
        len--;
    return ret + len + 1;
}