LeetCode 2. Add Two Numbers

Problem Statement

The problem is as follows: 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 contains a single digit. Add the two numbers and return it as a linked list.

Steps to Solve/ Approach

We create a dummy node to serve as the starting point of our result linked list. This dummy node will help simplify the code.
We initialize two variables: cur (the current node) and carry. The cur variable helps us build the result linked list, and carry stores the carry-over from the previous addition (initially set to 0).
We use a while loop that continues until both linked lists have been processed and there is no remaining carry.
Within the loop, we extract the values from the current nodes of both linked lists (if available) and add them together, along with the carry.
We calculate the new value to insert in the result linked list and update the carry based on the division result of the sum by 10.
We create a new node with the calculated value and append it to the result linked list, updating the cur pointer.
We move to the next nodes in both linked lists (if available) for the next iteration.
Finally, we return the result linked list starting from the node after the dummy node.

Code

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

class Solution:
    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        dummy = ListNode()
        cur = dummy

        carry = 0
        while l1 or l2 or carry:
            v1 = l1.val if l1 else 0
            v2 = l2.val if l2 else 0

            # New digit
            val = v1 + v2 + carry
            carry = val // 10
            val = val % 10
            cur.next = ListNode(val)

            # Updating
            cur = cur.next
            l1 = l1.next if l1 else None
            l2 = l2.next if l2 else None
        return dummy.next