Tackling the Add Two Numbers Problem in Python

In the world of programming and data structures, the way we approach a problem plays a vital role in determining the efficiency of the solution. Today, we will be navigating through a popular problem: Adding two numbers represented by linked lists in Python. Also known as the Add Two Numbers Problem (Leetcode 13). These two non-empty linked lists signify two non-negative integers, stored in reverse order with each node encapsulating a single digit. We will focus on adding these two numbers and returning the result as a linked list.

A Deep Dive into the Problem

To clarify the concept, if we have an integer 123, the linked list would present it in reverse as (3 -> 2 -> 1). The term "non-negative integers" refers to integers that are either positive or zero. While the idea of "adding" two linked lists might seem unusual, we are essentially summing up the two numbers that these lists represent, and then storing that total in a new linked list. Consider the numbers 101 and 201, the linked list would encapsulate the sum, 302, in reverse order.

The Essence of Linked Lists

A linked list is a data structure where each node holds a value and a reference (or pointer) to the next node. Navigating a linked list implies moving from one node to the following one using these pointers. The number represented by a linked list is the result of appending the values in each node from the head node to the tail node. To add a new node to a linked list, we create the node and then set the pointer of the existing tail node to our new node.

Solution Approach

Visualizing the problem with an example can provide a clearer perspective. Let's take two lists: List1 is (2 -> 4 -> 3) and List2 is (5 -> 6 -> 4).

We commence by adding the numbers in the first nodes of the two lists: 2 + 5 = 7.
Progressing to the next nodes, we add 4 + 6 = 10. Since each node can contain only a single digit, we retain the '0' in the current second node and carry forward the '1' to the next calculation.
We then add the next pair of digits: 3 + 4 = 7, not forgetting to include the carry '1' from the previous step, making the total 7 + 1 = 8.
With no more nodes left to process, the final result becomes a new linked list: (7 -> 0 -> 8).
It is essential to note that if there was a remaining carry after adding the final nodes, we would have needed to create an extra node for that carry.

Translating the Solution into Code

We can now develop a Python function that brings this solution to life:

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

def addTwoNumbers(l1, l2):
    dummy = cur = ListNode(0)
    carry = 0

    while l1 or l2 or carry:
        if l1:
            carry += l1.val
            l1 = l1.next
        if l2:
            carry += l2.val
            l2 = l2.next
        cur.next = ListNode(carry % 10)
        cur = cur.next
        carry //= 10

    return dummy.next

The code creates a dummy head node and a variable carry to keep track of the carry-over value from each addition. It then enters a loop that continues until we have processed all nodes of both lists and there's no carry left. In each

iteration, it adds the values of the current nodes and the carry, if they exist. It creates a new node with the unit digit of the result, moves on to the next nodes, and updates the carry.