Merging Two Sorted Linked Lists

In this task, I worked on merging two sorted linked lists into a single sorted linked list. This problem helped me understand how linked lists work and how to handle pointers efficiently.

What I Did:
I created a function mergeTwoLists that takes two sorted linked lists as input and merges them into one sorted list.
Example:
List1: 1 → 3 → 5

List2: 2 → 4 → 6

Output: 1 → 2 → 3 → 4 → 5 → 6
How I Solved It:
To solve this problem, I used a dummy node technique to simplify handling the head of the new list.
Step-by-step approach:
I created a dummy node to act as the starting point of the merged list
Used a pointer called tail to keep track of the last node in the merged list
Compared nodes from both lists:
If list1.val <= list2.val, I attached list1 node
Otherwise, I attached list2 node
Moved the pointer (tail) forward each time
Continued this until one list becomes empty
Final Step:
If any nodes are left in either list, I attached them directly to the end

Code:
'''python
class ListNode:
def init(self, val=0, next=None):
self.val = val
self.next = next

class Solution:
def mergeTwoLists(self, list1, list2):
dummy = ListNode(-1)
tail = dummy

    while list1 and list2:
        if list1.val <= list2.val:
            tail.next = list1
            list1 = list1.next
        else:
            tail.next = list2
            list2 = list2.next

        tail = tail.next

    # attach remaining nodes
    if list1:
        tail.next = list1
    else:
        tail.next = list2

    return dummy.next 

'''
How It Works:
The function builds a new linked list by always selecting the smaller node from the two lists.
Since both input lists are already sorted:
The merged list automatically stays sorted
No extra sorting is required
The dummy node makes the implementation cleaner by avoiding edge cases like handling the head separately.