LeetCode Challenge (Python3). Problem 2

leetcode

Introduction of this challenge:

I recently decided to improve my programming skills, so I started using LeetCode to achieve this goal.

However, since I don't have much programming experience and this is my first time using LeetCode, there may be some difficulties in the process, but I believe I will be able to persevere.

There are currently 2230 problems on LeetCode, and I will be working on the challenge by problem number with the goal of solving all of them.

After a week from the first problem, I finally finished problem 2. I think the biggest difficulty in solving this problem was that I was not familiar with the use of recursion, so I spent a lot of time at first to understand how it works.

After a long time of research, I finally came up with a solution to this problem, although not yet very skilled, but a great sense of accomplishment.

Let's start today's problem!

Today's problem is "2. Add Two Numbers".

Link: https://leetcode.com/problems/add-two-numbers/
Difficulty: Medium
Related Topics: Linked List Math Recursion

Problem:

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.

My solution：

# 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]:
        l3=ListNode()
        
        if l1 != None and l2 != None: # Determine if l1,l2 is none
            l3.val=l1.val+l2.val # Add to l3
            if l3.val>=10: # Checking for carry
                l3.val-=10
                if l1.next != None: l1.next.val+=1
                elif l2.next != None: l2.next.val+=1
                else: l1.next=ListNode(1)
            l3.next=self.addTwoNumbers(l1.next,l2.next) # recursion
        elif l1 != None: # Determine if only l2 is none
            l3.val=l1.val
            if l3.val>=10:
                l3.val-=10
                if l1.next != None: l1.next.val+=1
                else: l1.next=ListNode(1)
            l3.next=self.addTwoNumbers(l1.next,None) # recursion
        elif l2 != None: # Determine if only l1 is none
            l3.val=l2.val
            if l3.val>=10:
                l3.val-=10
                if l2.next != None: l2.next.val+=1
                else: l2.next=ListNode(1)
            l3.next=self.addTwoNumbers(None,l2.next) # recursion
        else:
            return
        return l3