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# Reverse a doubly circular linked list

The problem involves reversing a circular doubly linked list by swapping the values of the nodes. A circular doubly linked list is a data structure in which each node has pointers to both its previous and next nodes, and the last node's next pointer points back to the first node, creating a loop.

## Problem Statement

Given a circular doubly linked list, the task is to reverse the order of its elements using a swap of node values.

## Example

Consider the circular doubly linked list: 1 <-> 2 <-> 3 <-> 4 <-> 5 <-> 6. After reversing the elements, the list will become: 6 <-> 5 <-> 4 <-> 3 <-> 2 <-> 1.

## Idea to Solve the Problem

To reverse the order of elements in a circular doubly linked list using a swap of node values, we need to follow these steps:

1. Get references to the first node (`front`) and the last node (`tail`).
2. Initialize a temporary variable (`temp`) to store the value during the swap operation.
3. Traverse the list using two pointers (`first` and `last`), moving towards each other from the front and the tail respectively.
4. Swap the values of the nodes pointed to by `first` and `last`.
5. Continue swapping and moving the pointers until `first` and `last` meet or cross each other.
6. The reversal is complete when `first` and `last` cross each other.

## Pseudocode

Here's the pseudocode to reverse a circular doubly linked list by swapping node values:

``````function reverseElement():
first = front
last = tail
while first is not NULL and last is not NULL:
if first is not last:
// Swap values of first and last nodes
temp = first.data
first.data = last.data
last.data = temp
if first.next is last:
// Stop reversal
first = NULL
last = NULL
else:
first = first.next
last = last.back
else:
// Stop reversal
first = NULL
last = NULL``````

## Algorithm Explanation

1. Get references to the first node (`front`) and the last node (`tail`).
2. Initialize a temporary variable (`temp`) to store the value during the swap operation.
3. Traverse the list using two pointers (`first` and `last`), moving towards each other from the front and the tail respectively.
4. Swap the values of the nodes pointed to by `first` and `last`.
5. Continue swapping and moving the pointers until `first` and `last` meet or cross each other.
6. The reversal is complete when `first` and `last` cross each other.

## Time Complexity

The time complexity of reversing a circular doubly linked list using a swap of node values is O(n), where n is the number of nodes in the list. This is because we iterate through the list once to perform the swap operation.

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