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David Babalola
David Babalola

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LeetCode 2. Add Two Numbers

This year, I am solving more data structures and algorithms.
P.S.: This post was first seen on Twitter. See the link below.

The Problem is 2. Add Two Numbers from LeetCode. Check it out here.

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:

Example 1 of Add Two Numbers problem
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]

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.

Here is my solution:

    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        current_l1 = l1
        current_l2 = l2
        l1_list, l2_list = [], []    

        # Convert both LinkedLists to Python Lists
        while current_l1 is not None or current_l2 is not None:            
            if current_l1 is not None:
                current_l1 =

            if current_l2 is not None:
                current_l2 =

        l1_list.reverse() # Reverse lists e.g. [2, 4, 3] -> [3, 4, 2]
        l2_list.reverse() # Reverse lists e.g. [2, 4, 3] -> [3, 4, 2]

        # Convert the lists to integers. e.g. [3, 4, 2] -> 342
        l1_value = int(''.join(l1_list))
        l2_value = int(''.join(l2_list))      

        sum_values = l1_value + l2_value
        sum_values = str(sum_values)

        # Create the head node
        head = ListNode(sum_values[-1])
        current = head

        # Insert the elements to LinkedList in reverse order
        for digit in sum_values[-2::-1]:
   = ListNode(digit)
            current =
        return head
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Lines 2-3:
I create current_l1 and current_l2 to be nodes pointing to the heads of l1 and l2, respectively.
Line 4 creates empty lists l1_list and l2_list to store each digit of l1 and l2.

Lines 7-14:
I convert the linked lists l1 and l2 into Python lists (l1_list and l2_list) by traversing through them, extracting each digit from the linked lists and appending it to the respective lists.

Lines 16-17:
I reverse l1_list and l2_list lists. I do this because the digits are in reverse order in the linked lists

Lines 20-21:
I convert the reversed lists to integers (l1_value and l2_value) by using the join method to convert to a string and then to an integer ultimately

Lines 23-24:
I calculate the sum of both integers and then convert the sum back to a string (sum_values) to easily extract each digit.

Lines 27-34:
I create a head node (last digit of the sum), create a pointer to it, and iterate through the remaining digits of the sum in reverse order, creating new nodes for each digit in the sum string (sum_values). The head is then returned.

Here's the performance of my approach as of today:
Runtime: 54ms, beats 79.12% of users with Python3
Memory: 16.60MB, beats 67.14% of users with Python3

Do you have a better-performing approach? What problems would you want to see me solve?

Let me know in the comments.

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