Introduction

In this blog post, we will explore how to solve the LeetCode problem "24" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given a linked list, swap every two adjacent nodes and return its head. You must solve the problem without modifying the values in the list's nodes (i.e., only nodes themselves may be changed.) Example 1: Input: head = [1,2,3,4] Output: [2,1,4,3] Explanation: Example 2: Input: head = [] Output: [] Example 3: Input: head = [1] Output: [1] Example 4: Input: head = [1,2,3] Output: [2,1,3] Constraints: The number of nodes in the list is in the range [0, 100]. 0 <= Node.val <= 100

Explanation

Here's the approach, complexity, and Python code for swapping adjacent nodes in a linked list:

• Iterative Swapping: Traverse the linked list, swapping pairs of adjacent nodes at each step.
• Pointer Manipulation: Use temporary pointers to re-arrange the next pointers of the nodes to achieve the swapping.

• Handle Edge Cases: Correctly handle empty lists, lists with only one node, and lists with an odd number of nodes.

• Time Complexity: O(n), where n is the number of nodes in the linked list. Space Complexity: O(1) - constant extra space.

Code

    class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def swapPairs(head: ListNode) -> ListNode:
    """
    Swaps every two adjacent nodes in a linked list.

    Args:
        head: The head of the linked list.

    Returns:
        The head of the modified linked list.
    """
    if not head or not head.next:
        return head

    dummy = ListNode(0)  # Dummy node to simplify head manipulation
    dummy.next = head
    prev = dummy
    curr = head

    while curr and curr.next:
        # Nodes to be swapped: curr and curr.next
        next_node = curr.next

        # Swapping
        curr.next = next_node.next
        next_node.next = curr
        prev.next = next_node

        # Move pointers for the next pair
        prev = curr
        curr = curr.next

    return dummy.next