The Problem:

In this problem, we are given 2 sorted singly linked list and we have to merge and return them into 1 singly linked list while maintaining the sorted order.

Before we begin trying to figure out a solution for this problem, let’s first revisit an important concepts Linked List and Recursion.

Understanding Recursion:

Recursion is essentially a function calling itself. It is frequently used to solve problems that can be divided into smaller, similar sub-problems and can be very useful in dynamic programming. To understand recursion, it can be helpful to break it down into 3 parts:

1. Base Condition: This condition prevents the recursive function from calling itself infinitely.
2. Recursive Call: This is responsible for calling the recursive function itself.
3. Small Calculation: This step, which can come before or after the Recursive Call, is responsible for performing some calculation that we need for the recursive function or for returning some result after a recursive call has been completed.

A recursive tree is a visual representation of the recursive calls made by a function. It can be very helpful in understanding how recursion works and in analyzing the time and space complexity of recursive algorithms.

Understanding Linked List:

A linked list is a linear data structure that consists of a sequence of nodes, each containing data and a reference to the next node in the list. It is often used as an alternative to arrays, as it can be more efficient in certain operations such as insertion and deletion.

To understand linked lists, it can be helpful to break it down into 3 parts:

1. Nodes: Each node in a linked list contains some data and a reference to the next node in the list.
2. Head: The head of a linked list is the first node in the list. It is used as the starting point for many operations on the list.
3. Traversal: Traversing a linked list involves starting at the head and following the references from one node to the next until the desired node is reached or the end of the list is reached.

Linked lists can be very useful in solving problems that involve dynamic data structures, as they allow for efficient insertion and deletion of elements. They can also be used in combination with other data structures, such as stacks and queues, to solve more complex problems.

Intuition & Algorithm:

With an understanding of how recursion and linked lists work, let’s try to break down the problem. We don’t need to create our own linked list data structure because it is already provided. This is a fairly straightforward problem, so we will begin by figuring out the base condition, which in this case would be when the recursion reaches the end of either of the linked lists (head == None) this would mean that there are remaining nodes in the other linked list, and they would be returned.

At each recursive step, we can compare the values of the first nodes of each list and selects the node with the smaller value to be part of the merged list. We could then update the next pointer of this node to point to the result of a recursive call with the next node of this list and the other list as input.

Finally, we return the head of the merged list, which is the result of merging the two input lists. This new head is either the head of list1 or the head of list2, depending on which node had the smaller value. Let's visualize the algorithm below.

Summary of Solution

Use recursion to merge two sorted linked lists. At each recursive step, compare the values of the first nodes of each list and select the node with the smaller value to be part of the merged list. Update its next pointer to point to the result of a recursive call with its next node and other list as input. Return one of lists if other is empty.

Code

This can be translated to code in the following way:

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next  
class Solution:
    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
        # here list1 and list2 are the heads of list 1 and list 2

        if list1 == None: return list2 # base

        elif list2 == None: return list1 # base

        elif (list1.val <= list2.val): # small calculation
            list1.next = self.mergeTwoLists(list1.next, list2) # recursive call
            return list1
            
        else:
            list2.next = self.mergeTwoLists(list1, list2.next) # recursive call
            return list2

• Time Complexity O(n+m)
• Space Complexity O(n+m)

The time complexity of this solution is O(n+m) This is because the function needs to traverse both the linked lists to merge them.

This approach is simple and efficient, making it a good solution to this problem.

I hope you enjoyed this post. Stay tuned for daily LeetCode blog posts!