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 secondtolast 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 secondtolast 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
 Start with two pointers:
current
pointing to the head of the doubly linked list andtail
pointing to its tail.  Loop while both
current
andtail
are not null: If
current
reaches the same node astail
, setcurrent.next
to null and return.  Else, if
current.next
reaches the same node astail
, settail.next
to null and return.  Create an
auxiliary
pointer to storecurrent.next
.  Update the pointers to achieve the spiral pattern:
current.next
points totail
,tail.prev
points tocurrent
, andtail.next
points toauxiliary
.  If
auxiliary
is not null, set itsprev
pointer totail
.  Move
current
toauxiliary
and updatetail
to its previous node.
 If
 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.
Please share your knowledge to improve code and content standard. Also submit your doubts, and test case. We improve by your feedback. We will try to resolve your query as soon as possible.
New Comment