Add Two Numbers II

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example 1:

Input: l1 = [7,2,4,3], l2 = [5,6,4]
Output: [7,8,0,7]

Example 2:

Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [8,0,7]

Example 3:

Input: l1 = [0], l2 = [0]
Output: [0]

Constraints:

• The number of nodes in each linked list is in the range [1, 100].
• 0 <= Node.val <= 9
• It is guaranteed that the list represents a number that does not have leading zeros.

Follow up: Could you solve it without reversing the input lists?

SOLUTION:

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def add(self, l1, l2, carry = 0):
        if l1 or l2:
            l1val = l1.val if l1 else 0
            l2val = l2.val if l2 else 0
            currval = l1val + l2val + carry
            curr = ListNode(val = currval % 10)
            curr.next = self.add(l1.next if l1 else None, l2.next if l2 else None, carry = currval // 10)
            return curr
        else:
            if carry > 0:
                return ListNode(val = carry)
            return None

    def reverse(self, head):
        curr = head
        prev = None
        while curr:
            nxt = curr.next
            curr.next = prev
            prev = curr
            curr = nxt
        return prev

    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        l1 = self.reverse(l1)
        l2 = self.reverse(l2)
        return self.reverse(self.add(l1, l2, carry = 0))