Leetcode 2. Add Two Numbers

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 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 = [2,4,3], l2 = [5,6,4] Output: [7,0,8] Explanation: 342 + 465 = 807.

Example 2:Input: l1 = [0], l2 = [0] Output: [0]

Example 3:Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9] Output: [8,9,9,9,0,0,0,1]

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.

Since it’s a sorted linked list, then the solution is very straight forward. We just need to iterate through each node and carry over.

# 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]:
        next_digit = 0

        head = None
        current_node = None

        while l1 != None or l2 != None:
            l1_val = l1.val if l1 else 0
            l2_val = l2.val if l2 else 0

            total = l1_val + l2_val + next_digit

            next_digit = int(total / 10)

            if not head:
                head = ListNode(total%10, None)
                current_node = head
            else:
                current_node.next = ListNode(total%10, None)
                current_node = current_node.next

            l1 = l1.next if l1 else None
            l2 = l2.next if l2 else None

        if next_digit > 0:
            current_node.next = ListNode(next_digit, None)

        return head