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Code Doubly linked list

Convert a doubly linked list into reversal anti clockwise spiral form

In this article, we will explore a fascinating algorithmic problem: converting a doubly linked list into a reversal anti-clockwise spiral form. We'll delve into the problem statement, walk through an illustrative example, outline the algorithmic approach, provide pseudocode, detail the algorithm's implementation, explain the resulting output, and analyze the time complexity of the solution.

Problem Statement

Given a doubly linked list, we are tasked with transforming it into a specific arrangement known as the reversal anti-clockwise spiral form. This form involves altering the links between nodes in such a way that the order of nodes in the list forms an anti-clockwise spiral pattern. Additionally, the direction of the doubly linked list should be reversed. The problem presents a unique challenge that combines linked list manipulation and spiral pattern creation.

Illustrative Example

Consider the following doubly linked list:

1 <-> 2 <-> 3 <-> 4 <-> 5 <-> 6 <-> 7 <-> 8 <-> 9

After converting the list into the reversal anti-clockwise spiral form, it will look like this:

5 <-> 4 <-> 6 <-> 3 <-> 7 <-> 2 <-> 8 <-> 1 <-> 9

This transformation involves reordering the nodes to create an anti-clockwise spiral pattern while reversing the direction of the list.

Algorithmic Approach

  1. Initialize pointers current and last to the head and tail of the linked list, respectively.
  2. Set up auxiliary pointers to assist in link swapping.
  3. Begin converting the list:
    • Swap links between current and last, considering the anti-clockwise spiral pattern.
    • Update pointers and auxiliary pointers as needed.
  4. Reverse the entire linked list to complete the transformation.

Pseudocode

reversalAntiSpiral():
    if head is null or head.next is null:
        return
    Initialize last = tail, current = head, auxiliary = null
    Set up temp and back pointers

    Set head as tail to start the new configuration

    Loop until current and last pointers meet or cross:
        Handle edge cases when current and last are adjacent
        Swap links considering the anti-clockwise spiral
        Update pointers and auxiliary pointers

    Reverse the entire linked list using reverseNode()

reverseNode():
    if head is null:
        return
    Initialize temp = head, back = null
    Update tail as head

    Traverse the list and modify links to reverse nodes

Main():
    Initialize a doubly linked list (dll)
    Insert nodes into the linked list

    Display the list before the transformation
    Convert the list to reversal anti-clockwise spiral form
    Display the transformed list

Code Solution

Time Complexity Analysis

  • The reversalAntiSpiral function iterates through the linked list once, performing constant-time operations per iteration. Therefore, its time complexity is O(n), where n is the number of nodes in the linked list.
  • The reverseNode function also traverses the entire linked list once, resulting in a time complexity of O(n).

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