Adding a Node to the Linked List is taking a longer Time when using a reference to the Tail Node I am trying to add two numbers in the form of linked List and return their result in a Linked List as given in https://leetcode.com/problems/add-two-numbers/

Question:


  You are given two non-empty linked lists representing two non-negative
  integers. The digits are stored in reverse order and each of their
  nodes contain a single digit. Add the two numbers and return it as a
  linked list.
  
  You may assume the two numbers do not contain any leading zero, except
  the number 0 itself.


Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807, my solution has solved all the test cases but after refactoring, it is taking longer than the original code

Solution 1:

//public class ListNode
//{
//    public int val;
//    public ListNode next;
//    public ListNode(int x) { val = x; }
//};

public class Solution {
    public ListNode AddTwoNumbers(ListNode l1, ListNode l2) {
        ListNode l3 = null;
        int carry = 0;
        while (l1 != null || l2 != null) {
            int first = 0,
            second = 0;
            if (l1 != null) {
                first = l1.val;
                l1 = l1.next;
            }
            if (l2 != null) {
                second = l2.val;
                l2 = l2.next;
            }
            int Digit = first + second;
            if (carry != 0) {
                Digit = Digit + carry;
                carry = 0;
            }
            if (Digit &gt; 9) {
                carry = Digit / 10;
                Digit = Digit % 10;
            }
            AddLastNode(Digit, ref l3);
        }
        if (carry != 0) {
            AddLastNode(carry, ref l3);
            carry = 0;
        }
        return l3;
    }

    /// In here I am looping through the Linked List every time,to find the tail node 
    private static void AddLastNode(int Digit, ref ListNode l3) {
        if (l3 != null) {
            AddLastNode(Digit, ref l3.next);
        }
        else {
            l3 = new ListNode(Digit);
        }
    }
}


So  to avoid looping through all the Nodes,in the below solution I am using a reference for the Tail Node

Solution 2:

public class Solution {
    public ListNode AddTwoNumbers(ListNode l1, ListNode l2) {
        ListNode l3 = null;
        ListNode tailNode = null;
        int remainder = 0;
        while (l1 != null || l2 != null) {
            int sum = 0;
            if (l1 != null) {
                sum = l1.val;
                l1 = l1.next;
            }
            if (l2 != null) {
                sum += l2.val;
                l2 = l2.next;
            }
            if (remainder != 0) {
                sum += remainder;
            }
            if (sum &gt; 9) {
                remainder = sum / 10;
                sum = sum % 10;
            }
            else {
                remainder = 0;
            }
            ///In here I am using tailNode has reference for adding new node to the end of Linked List
            if (tailNode == null) {
                l3 = new ListNode(sum);
                tailNode = l3;
            }
            else {
                tailNode.next = new ListNode(sum);
                tailNode = tailNode.next;
            }
        }
        if (remainder != 0) {
            tailNode.next = new ListNode(remainder);
        }
        return l3;
    }
}


Since I got a tail node for the end of Linked List instead of going through entire Linked List,I thought the solution 2 will have better performance.But it is still taking more time to execute than the first solution ,Any code changes would be appreciated



Solution 1 is taking 108 ms to execute while Solution 2 is taking 140 ms
In terms of the times on LeetCode, they can be dependent upon the server load as much as the code.  I tried your code and got a range of times from 100 ms to 136 ms.  Holding onto the last node is a better solution than repeatedly trying to find it but on a list of only a few nodes I don't know how much impact it would have.

In terms of the code, this strikes me as a good place to use the Null Condtional Operator (?.).  We can get around the repeated null check on the tailnode by using a dummy head for the return list.

public ListNode AddTwoNumbers(ListNode l1, ListNode l2) 
{
    ListNode l3 = new ListNode(-1);  // dummy head
    ListNode tailNode = l3;
    int remainder = 0;
    while (l1 != null || l2 != null) {

        var sum = (l1?.val ?? 0) + 
                  (l2?.val ?? 0) + 
                  remainder;
        l1 = l1?.next;
        l2 = l2?.next;

        if (sum &gt; 9) {
            remainder = sum / 10;
            sum = sum % 10;
        }
        else {
            remainder = 0;
        }
       tailNode.next = new ListNode(sum);
       tailNode = tailNode.next;

    }
    if (remainder != 0) {
        tailNode.next = new ListNode(remainder);
    }
    return l3.next; // skip the dummy head when returning
}