The Problem:

In this problem, we are given a linked list and our goal is to swap two adjacent nodes for example if we have a linked list 1->2->3->4->5 we have to return 2->1->4->3->5

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:

With an understanding of how recursion and linked lists work, let’s try to break down the problem. We don't have to worry about making our own linked list data structure as it is given. If there are at least two nodes in the list, the function swaps the first two nodes by updating their next pointers.

The first node’s next pointer is updated to point to the result of recursively calling swapPairs on the rest of the list (nextNode.next). The second node’s next pointer is updated to point to the first node (nextNode.next = head). Finally, the function returns the new head of the list, which is the second node (return nextNode).

Summary of Solution

Use recursion to swap pairs of adjacent nodes in the linked list. If there are at least two nodes in the list, swap the first two nodes by updating their next pointers. The first node’s next pointer is updated to point to the result of recursively calling swapPairs on the rest of the list. The second node’s next pointer is updated to point to the first node. Finally, return the new head of the list, which is the second node. This solution makes use of important concepts such as linked lists and recursion.

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 swapPairs(self, head: Optional[ListNode]) -> Optional[ListNode]:
        if head == None or head.next == None:
            return head
        nextNode = head.next
        head.next = self.swapPairs(nextNode.next)
        nextNode.next = head
        return nextNode

• Time Complexity O(N)
• Space Complexity O(N)

The time complexity of this solution is O(n) This is because the function needs to traverse the linked lists, and the time it takes to do so is proportional to the length of the linked list.

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

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