# Reverse a doubly circular linked list

Last Updated : 01 Sep, 2022

The problem is to reverse the given doubly circular linked list.

Examples:

Input:

Output:

Algorithm:

```insertEnd(head, new_node)
Declare last

new_node->next = new_node->prev = new_node
return

new_node->prev = last
last->next = new_node

Declare last

Initialize curr = last, prev

while curr->prev != last
prev = curr->prev
curr = prev

Explanation: The variable head in the parameter list of insertEnd() is a pointer to a pointer variable. reverse() traverses the doubly circular linked list starting with the head pointer in the backward direction and one by one gets the node in the traversal. It inserts those nodes at the end of the list that starts with the new_head pointer with the help of the function insertEnd() and finally returns the new_head

Implementation:

## C++

 `// C++ implementation to reverse a` `// doubly circular linked list` `#include `   `using` `namespace` `std;`   `// structure of a node of linked list` `struct` `Node {` `    ``int` `data;` `    ``Node *next, *prev;` `};`   `// function to create and return a new node` `Node* getNode(``int` `data)` `{` `    ``Node* newNode = (Node*)``malloc``(``sizeof``(Node));` `    ``newNode->data = data;` `    ``return` `newNode;` `}`   `// Function to insert at the end` `void` `insertEnd(Node** head, Node* new_node)` `{` `    ``// If the list is empty, create a single node` `    ``// circular and doubly list` `    ``if` `(*head == NULL) {` `        ``new_node->next = new_node->prev = new_node;` `        ``*head = new_node;` `        ``return``;` `    ``}`   `    ``// If list is not empty`   `    ``/* Find last node */` `    ``Node* last = (*head)->prev;`   `    ``// Start is going to be next of new_node` `    ``new_node->next = *head;`   `    ``// Make new node previous of start` `    ``(*head)->prev = new_node;`   `    ``// Make last previous of new node` `    ``new_node->prev = last;`   `    ``// Make new node next of old last` `    ``last->next = new_node;` `}`   `// Utility function to reverse a` `// doubly circular linked list` `Node* reverse(Node* head)` `{` `    ``if` `(!head)` `        ``return` `NULL;`   `    ``// Initialize a new head pointer` `    ``Node* new_head = NULL;`   `    ``// get pointer to the last node` `    ``Node* last = head->prev;`   `    ``// set 'curr' to last node` `    ``Node *curr = last, *prev;`   `    ``// traverse list in backward direction` `    ``while` `(curr->prev != last) {` `        ``prev = curr->prev;`   `        ``// insert 'curr' at the end of the list` `        ``// starting with the 'new_head' pointer` `        ``insertEnd(&new_head, curr);` `        ``curr = prev;` `    ``}` `    ``insertEnd(&new_head, curr);`   `    ``// head pointer of the reversed list` `    ``return` `new_head;` `}`   `// function to display a doubly circular list in` `// forward and backward direction` `void` `display(Node* head)` `{` `    ``if` `(!head)` `        ``return``;`   `    ``Node* temp = head;`   `    ``cout << ``"Forward direction: "``;` `    ``while` `(temp->next != head) {` `        ``cout << temp->data << ``" "``;` `        ``temp = temp->next;` `    ``}` `    ``cout << temp->data;`   `    ``Node* last = head->prev;` `    ``temp = last;`   `    ``cout << ``"\nBackward direction: "``;` `    ``while` `(temp->prev != last) {` `        ``cout << temp->data << ``" "``;` `        ``temp = temp->prev;` `    ``}` `    ``cout << temp->data;` `}`   `// Driver program to test above` `int` `main()` `{` `    ``Node* head = NULL;`   `    ``insertEnd(&head, getNode(1));` `    ``insertEnd(&head, getNode(2));` `    ``insertEnd(&head, getNode(3));` `    ``insertEnd(&head, getNode(4));` `    ``insertEnd(&head, getNode(5));`   `    ``cout << ``"Current list:\n"``;` `    ``display(head);`   `    ``head = reverse(head);`   `    ``cout << ``"\n\nReversed list:\n"``;` `    ``display(head);`   `    ``return` `0;` `}`

## Java

 `// Java implementation to reverse a ` `// doubly circular linked list ` `class` `GFG` `{`   `// structure of a node of linked list ` `static` `class` `Node` `{ ` `    ``int` `data; ` `    ``Node next, prev; ` `}; `   `// function to create and return a new node ` `static` `Node getNode(``int` `data) ` `{ ` `    ``Node newNode = ``new` `Node(); ` `    ``newNode.data = data; ` `    ``return` `newNode; ` `} `   `// Function to insert at the end ` `static` `Node insertEnd(Node head, Node new_node) ` `{ ` `    ``// If the list is empty, create a single node ` `    ``// circular and doubly list ` `    ``if` `(head == ``null``) ` `    ``{ ` `        ``new_node.next = new_node.prev = new_node; ` `        ``head = new_node; ` `        ``return` `head; ` `    ``} `   `    ``// If list is not empty `   `    ``// Find last node /` `    ``Node last = (head).prev; `   `    ``// Start is going to be next of new_node ` `    ``new_node.next = head; `   `    ``// Make new node previous of start ` `    ``(head).prev = new_node; `   `    ``// Make last previous of new node ` `    ``new_node.prev = last; `   `    ``// Make new node next of old last ` `    ``last.next = new_node; ` `    ``return` `head;` `} `   `// Utility function to reverse a ` `// doubly circular linked list ` `static` `Node reverse(Node head) ` `{ ` `    ``if` `(head==``null``) ` `        ``return` `null``; `   `    ``// Initialize a new head pointer ` `    ``Node new_head = ``null``; `   `    ``// get pointer to the last node ` `    ``Node last = head.prev; `   `    ``// set 'curr' to last node ` `    ``Node curr = last, prev; `   `    ``// traverse list in backward direction ` `    ``while` `(curr.prev != last) ` `    ``{ ` `        ``prev = curr.prev; `   `        ``// insert 'curr' at the end of the list ` `        ``// starting with the 'new_head' pointer ` `        ``new_head=insertEnd(new_head, curr); ` `        ``curr = prev; ` `    ``} ` `    ``new_head=insertEnd(new_head, curr); `   `    ``// head pointer of the reversed list ` `    ``return` `new_head; ` `} `   `// function to display a doubly circular list in ` `// forward and backward direction ` `static` `void` `display(Node head) ` `{ ` `    ``if` `(head==``null``) ` `        ``return``; `   `    ``Node temp = head; `   `    ``System.out.print( ``"Forward direction: "``); ` `    ``while` `(temp.next != head) ` `    ``{ ` `        ``System.out.print( temp.data + ``" "``); ` `        ``temp = temp.next; ` `    ``} ` `        ``System.out.print( temp.data + ``" "``); `   `    ``Node last = head.prev; ` `    ``temp = last; `   `    ``System.out.print( ``"\nBackward direction: "``); ` `    ``while` `(temp.prev != last) ` `    ``{ ` `        ``System.out.print( temp.data + ``" "``); ` `        ``temp = temp.prev; ` `    ``} ` `        ``System.out.print( temp.data + ``" "``); ` `} `   `// Driver code ` `public` `static` `void` `main(String args[])` `{ ` `    ``Node head = ``null``; `   `    ``head =insertEnd(head, getNode(``1``)); ` `    ``head =insertEnd(head, getNode(``2``)); ` `    ``head =insertEnd(head, getNode(``3``)); ` `    ``head =insertEnd(head, getNode(``4``)); ` `    ``head =insertEnd(head, getNode(``5``)); `   `    ``System.out.print( ``"Current list:\n"``); ` `    ``display(head); `   `    ``head = reverse(head); `   `    ``System.out.print( ``"\n\nReversed list:\n"``); ` `    ``display(head); ` `}` `} `   `// This code is contributed by Arnab Kundu`

## Python3

 `# Python3 implementation to reverse a` `# doubly circular linked list` `import` `math`   `# structure of a node of linked list` `class` `Node: ` `    ``def` `__init__(``self``, data): ` `        ``self``.data ``=` `data ` `        ``self``.``next` `=` `None`   `# function to create and return a new node` `def` `getNode(data):` `    ``newNode ``=` `Node(data)` `    ``newNode.data ``=` `data` `    ``return` `newNode`   `# Function to insert at the end` `def` `insertEnd(head, new_node):` `    `  `    ``# If the list is empty, create a single node` `    ``# circular and doubly list` `    ``if` `(head ``=``=` `None``) :` `        ``new_node.``next` `=` `new_node` `        ``new_node.prev ``=` `new_node` `        ``head ``=` `new_node` `        ``return` `head` `    `  `    ``# If list is not empty`   `    ``# Find last node ` `    ``last ``=` `head.prev`   `    ``# Start is going to be next of new_node` `    ``new_node.``next` `=` `head`   `    ``# Make new node previous of start` `    ``head.prev ``=` `new_node`   `    ``# Make last previous of new node` `    ``new_node.prev ``=` `last`   `    ``# Make new node next of old last` `    ``last.``next` `=` `new_node` `    ``return` `head`   `# Utility function to reverse a` `# doubly circular linked list` `def` `reverse(head):` `    ``if` `(head ``=``=` `None``):` `        ``return` `None`   `    ``# Initialize a new head pointer` `    ``new_head ``=` `None`   `    ``# get pointer to the last node` `    ``last ``=` `head.prev`   `    ``# set 'curr' to last node` `    ``curr ``=` `last` `    ``#*prev`   `    ``# traverse list in backward direction` `    ``while` `(curr.prev !``=` `last):` `        ``prev ``=` `curr.prev`   `        ``# insert 'curr' at the end of the list` `        ``# starting with the 'new_head' pointer` `        ``new_head ``=` `insertEnd(new_head, curr)` `        ``curr ``=` `prev` `    `  `    ``new_head ``=` `insertEnd(new_head, curr)`   `    ``# head pointer of the reversed list` `    ``return` `new_head`   `# function to display a doubly circular list in` `# forward and backward direction` `def` `display(head):` `    ``if` `(head ``=``=` `None``):` `        ``return` `    `  `    ``temp ``=` `head`   `    ``print``(``"Forward direction: "``, end ``=` `"")` `    ``while` `(temp.``next` `!``=` `head):` `        ``print``(temp.data, end ``=` `" "``)` `        ``temp ``=` `temp.``next` `    `  `    ``print``(temp.data)`   `    ``last ``=` `head.prev` `    ``temp ``=` `last`   `    ``print``(``"Backward direction: "``, end ``=` `"")` `    ``while` `(temp.prev !``=` `last):` `        ``print``(temp.data, end ``=` `" "``)` `        ``temp ``=` `temp.prev` `    `  `    ``print``(temp.data)`   `# Driver Code` `if` `__name__``=``=``'__main__'``: `   `    ``head ``=` `None`   `    ``head ``=` `insertEnd(head, getNode(``1``))` `    ``head ``=` `insertEnd(head, getNode(``2``))` `    ``head ``=` `insertEnd(head, getNode(``3``))` `    ``head ``=` `insertEnd(head, getNode(``4``))` `    ``head ``=` `insertEnd(head, getNode(``5``))`   `    ``print``(``"Current list:"``)` `    ``display(head)`   `    ``head ``=` `reverse(head)`   `    ``print``(``"\nReversed list:"``)` `    ``display(head)`   `# This code is contributed by Srathore`

## C#

 `// C# implementation to reverse a ` `// doubly circular linked list` `using` `System;`   `class` `GFG ` `{ `   `// structure of a node of linked list ` `public` `class` `Node ` `{ ` `    ``public` `int` `data; ` `    ``public` `Node next, prev; ` `}; `   `// function to create and return a new node ` `static` `Node getNode(``int` `data) ` `{ ` `    ``Node newNode = ``new` `Node(); ` `    ``newNode.data = data; ` `    ``return` `newNode; ` `} `   `// Function to insert at the end ` `static` `Node insertEnd(Node head, Node new_node) ` `{ ` `    ``// If the list is empty, create a single node ` `    ``// circular and doubly list ` `    ``if` `(head == ``null``) ` `    ``{ ` `        ``new_node.next = new_node.prev = new_node; ` `        ``head = new_node; ` `        ``return` `head; ` `    ``} `   `    ``// If list is not empty `   `    ``// Find last node / ` `    ``Node last = (head).prev; `   `    ``// Start is going to be next of new_node ` `    ``new_node.next = head; `   `    ``// Make new node previous of start ` `    ``(head).prev = new_node; `   `    ``// Make last previous of new node ` `    ``new_node.prev = last; `   `    ``// Make new node next of old last ` `    ``last.next = new_node; ` `    ``return` `head; ` `} `   `// Utility function to reverse a ` `// doubly circular linked list ` `static` `Node reverse(Node head) ` `{ ` `    ``if` `(head == ``null``) ` `        ``return` `null``; `   `    ``// Initialize a new head pointer ` `    ``Node new_head = ``null``; `   `    ``// get pointer to the last node ` `    ``Node last = head.prev; `   `    ``// set 'curr' to last node ` `    ``Node curr = last, prev; `   `    ``// traverse list in backward direction ` `    ``while` `(curr.prev != last) ` `    ``{ ` `        ``prev = curr.prev; `   `        ``// insert 'curr' at the end of the list ` `        ``// starting with the 'new_head' pointer ` `        ``new_head=insertEnd(new_head, curr); ` `        ``curr = prev; ` `    ``} ` `    ``new_head=insertEnd(new_head, curr); `   `    ``// head pointer of the reversed list ` `    ``return` `new_head; ` `} `   `// function to display a doubly circular list in ` `// forward and backward direction ` `static` `void` `display(Node head) ` `{ ` `    ``if` `(head == ``null``) ` `        ``return``; `   `    ``Node temp = head; `   `    ``Console.Write( ``"Forward direction: "``); ` `    ``while` `(temp.next != head) ` `    ``{ ` `        ``Console.Write( temp.data + ``" "``); ` `        ``temp = temp.next; ` `    ``} ` `        ``Console.Write( temp.data + ``" "``); `   `    ``Node last = head.prev; ` `    ``temp = last; `   `    ``Console.Write( ``"\nBackward direction: "``); ` `    ``while` `(temp.prev != last) ` `    ``{ ` `        ``Console.Write( temp.data + ``" "``); ` `        ``temp = temp.prev; ` `    ``} ` `        ``Console.Write( temp.data + ``" "``); ` `} `   `// Driver code ` `public` `static` `void` `Main(String []args) ` `{ ` `    ``Node head = ``null``; `   `    ``head = insertEnd(head, getNode(1)); ` `    ``head = insertEnd(head, getNode(2)); ` `    ``head = insertEnd(head, getNode(3)); ` `    ``head = insertEnd(head, getNode(4)); ` `    ``head = insertEnd(head, getNode(5)); `   `    ``Console.Write( ``"Current list:\n"``); ` `    ``display(head); `   `    ``head = reverse(head); `   `    ``Console.Write( ``"\n\nReversed list:\n"``); ` `    ``display(head); ` `} ` `} `   `// This code contributed by Rajput-Ji`

## Javascript

 ``

Output

```Current list:
Forward direction: 1 2 3 4 5
Backward direction: 5 4 3 2 1

Reversed list:
Forward direction: 5 4 3 2 1
Backward direction: 1 2 3 4 5```

Complexity Analysis:

• Time Complexity: O(n), as we are using a loop to traverse n times. Where n is the number of nodes in the linked list.
• Auxiliary Space: O(1), as we are not using any extra space.

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