Adding a Node to the Linked List is taking a longer Time when using a reference to the Tail Node

Question

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 > 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 > 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

Answer 1

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 > 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 }