Convert a doubly linked list spiral form

The problem at hand involves converting a given doubly linked list into a spiral form. A doubly linked list is a data structure in which each node contains a data element and two pointers, one pointing to the next node and another pointing to the previous node. The objective is to rearrange the nodes of the doubly linked list in a spiral fashion, combining the first and last nodes, then the second and second-to-last nodes, and so on.

Problem Statement

Given a doubly linked list, the task is to rearrange its nodes such that they form a spiral pattern. This involves combining the first and last nodes, then the second and second-to-last nodes, and so on, until the middle is reached.

Example

Consider the following doubly linked list:

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

After converting the doubly linked list into a spiral form, it will look like:

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

Idea to Solve

The idea to solve this problem is to iteratively manipulate the pointers of the doubly linked list nodes to achieve the desired spiral pattern. This involves combining nodes from the beginning and end of the list, updating their pointers appropriately to form the spiral shape.

Pseudocode

function spiral():
    current = head
    while current is not null and tail is not null:
        if current equals tail:
            current.next = null
            return
        else if current.next equals tail:
            tail.next = null
            return
        
        auxiliary = current.next
        current.next = tail
        temp = tail.prev
        tail.prev = current
        tail.next = auxiliary
        
        if auxiliary is not null:
            auxiliary.prev = tail
        
        current = auxiliary
        tail = temp

Algorithm Explanation

1. Start with two pointers: current pointing to the head of the doubly linked list and tail pointing to its tail.
2. Loop while both current and tail are not null:
   • If current reaches the same node as tail, set current.next to null and return.
   • Else, if current.next reaches the same node as tail, set tail.next to null and return.
   • Create an auxiliary pointer to store current.next.
   • Update the pointers to achieve the spiral pattern: current.next points to tail, tail.prev points to current, and tail.next points to auxiliary.
   • If auxiliary is not null, set its prev pointer to tail.
   • Move current to auxiliary and update tail to its previous node.
3. Repeat this process until the entire spiral pattern is formed.

Program List

1) Convert a doubly linked list into spiral form in java 2) Convert a doubly linked list into spiral form in c++ 3) Convert a doubly linked list into spiral form in c 4) Convert a doubly linked list into spiral form in c# 5) Convert a doubly linked list into spiral form in vb.net 6) Convert a doubly linked list into spiral form in php 7) Convert a doubly linked list into spiral form in python 8) Convert a doubly linked list into spiral form in ruby 9) Convert a doubly linked list into spiral form in scala 10) Convert a doubly linked list into spiral form in swift 11) Convert a doubly linked list into spiral form in kotlin 12) Convert a doubly linked list into spiral form in golang 13) Convert a doubly linked list into spiral form in node js 14) Convert a doubly linked list into spiral form in typescript

Time Complexity

The time complexity of this algorithm is O(n), where n is the number of nodes in the doubly linked list. This is because in the worst case, we iterate through all the nodes once to perform the spiral conversion operation.