Add Two Numbers

Add Two Numbers | Medium
Here goes nothing; day 1

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.

Accepted 693,452 Submissions 2,315,625

Result:
Runtime: 96 ms, faster than 99.85% of Python3 online submissions for Add Two Numbers.

Solution:

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

class Solution:
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        carry = 0
        prev = ListNode('start')
        head = prev
        while l1.next and l2.next:
            hold = l1.val + l2.val + carry
            carry = hold // 10
            prev.next = ListNode( hold % 10 )
            l1, l2, prev = l1.next, l2.next, prev.next
        
        remain = ListNode('empty')
        if not l1.next and not l2.next:
            hold = l1.val + l2.val + carry
            carry = hold // 10
            prev.next = ListNode( hold % 10 )
            prev = prev.next                
        elif not l1.next:
            hold = l1.val + l2.val + carry
            carry = hold // 10
            prev.next = ListNode( hold % 10 )
            prev, remain = prev.next, l2.next
        elif not l2.next:
            hold = l1.val + l2.val + carry
            carry = hold // 10
            prev.next = ListNode( hold % 10 )
            prev, remain = prev.next, l1.next
        
        while remain and remain.val != 'empty':
            hold = remain.val + carry
            carry = hold // 10
            prev.next = ListNode( hold % 10 )
            prev, remain = prev.next, remain.next
        if carry:
            prev.next = ListNode( carry )
        
        
        return head.next