Reverse a Doubly Linked List | Set-2

Write a program to reverse the given Doubly Linked List.

See below diagrams for example.

          (a) Original Doubly Linked List  



          (b) Reversed Doubly Linked List  

Approach: In the previous post, doubly linked list is being reversed by swapping prev and next pointers for all nodes, changing prev of the head (or start) and then changing the head pointer in the end. In this post, we create a push function that adds the given node at the beginning of the given list. We traverse the original list and one by one pass the current node pointer to the push function. This process will reverse the list. Finally return the new head of this reversed list.

C++

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// C++ implementation to reverse
// a doubly linked list
#include <bits/stdc++.h>
  
using namespace std;
  
// a node of the doubly linked list
struct Node {
    int data;
    Node *next, *prev;
};
  
// function to get a new node
Node* getNode(int data)
{
    // allocate space
    Node* new_node = (Node*)malloc(sizeof(Node));
  
    // put in the data
    new_node->data = data;
    new_node->next = new_node->prev = NULL;
    return new_node;
}
  
// function to insert a node at the beginging
// of the Doubly Linked List
void push(Node** head_ref, Node* new_node)
{
    // since we are adding at the beginning,
    // prev is always NULL
    new_node->prev = NULL;
  
    // link the old list off the new node
    new_node->next = (*head_ref);
  
    // change prev of head node to new node
    if ((*head_ref) != NULL)
        (*head_ref)->prev = new_node;
  
    // move the head to point to the new node
    (*head_ref) = new_node;
}
  
// function to reverse a doubly linked list
void reverseList(Node** head_ref)
{
    // if list is empty or it contains
    // a single node only
    if (!(*head_ref) || !((*head_ref)->next))
        return;
  
    Node* new_head = NULL;
    Node *curr = *head_ref, *next;
  
    while (curr != NULL) {
  
        // get pointer to next node
        next = curr->next;
  
        // push 'curr' node at the beginning of the
        // list with starting with 'new_head'
        push(&new_head, curr);
  
        // update 'curr'
        curr = next;
    }
  
    // update 'head_ref'
    *head_ref = new_head;
}
  
// Function to print nodes in a
// given doubly linked list
void printList(Node* head)
{
    while (head != NULL) {
        cout << head->data << " ";
        head = head->next;
    }
}
  
// Driver program to test above
int main()
{
    // Start with the empty list
    Node* head = NULL;
  
    // Create doubly linked: 10<->8<->4<->2 */
    push(&head, getNode(2));
    push(&head, getNode(4));
    push(&head, getNode(8));
    push(&head, getNode(10));
  
    cout << "Original list: ";
    printList(head);
  
    // Reverse doubly linked list
    reverseList(&head);
  
    cout << "\nReversed list: ";
    printList(head);
  
    return 0;
}

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Java

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// Java implementation to reverse 
// a doubly linked list 
class GFG
{
  
// a node of the doubly linked list 
static class Node 
    int data; 
    Node next, prev; 
}; 
  
// function to get a new node 
static Node getNode(int data) 
    // allocate space 
    Node new_node = new Node(); 
  
    // put in the data 
    new_node.data = data; 
    new_node.next = new_node.prev = null
    return new_node; 
  
// function to insert a node at the beginging 
// of the Doubly Linked List 
static Node push(Node head_ref, Node new_node) 
    // since we are adding at the beginning, 
    // prev is always null 
    new_node.prev = null
  
    // link the old list off the new node 
    new_node.next = (head_ref); 
  
    // change prev of head node to new node 
    if ((head_ref) != null
        (head_ref).prev = new_node; 
  
    // move the head to point to the new node 
    (head_ref) = new_node;
      
    return head_ref;
  
// function to reverse a doubly linked list 
static Node reverseList(Node head_ref) 
    // if list is empty or it contains 
    // a single node only 
    if ((head_ref) == null || ((head_ref).next) == null
        return null
  
    Node new_head = null
    Node curr = head_ref, next; 
  
    while (curr != null)
    
  
        // get pointer to next node 
        next = curr.next; 
  
        // push 'curr' node at the beginning of the 
        // list with starting with 'new_head' 
        new_head = push(new_head, curr); 
  
        // update 'curr' 
        curr = next; 
    
  
    // update 'head_ref' 
    head_ref = new_head;
      
    return head_ref;
  
// Function to print nodes in a 
// given doubly linked list 
static void printList(Node head) 
    while (head != null
    
        System.out.print(head.data + " "); 
        head = head.next; 
    
  
// Driver program to test above 
public static void main(String args[])
    // Start with the empty list 
    Node head = null
  
    // Create doubly linked: 10< - >8< - >4< - >2 / 
    head = push(head, getNode(2)); 
    head = push(head, getNode(4)); 
    head = push(head, getNode(8)); 
    head = push(head, getNode(10)); 
  
    System.out.print("Original list: "); 
    printList(head); 
  
    // Reverse doubly linked list 
    head = reverseList(head); 
  
    System.out.print("\nReversed list: "); 
    printList(head); 
}
  
// This code is contributed by Arnab Kundu

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C#

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// C# implementation to reverse 
// a doubly linked list 
using System;
  
class GFG 
  
// a node of the doubly linked list 
public class Node 
    public int data; 
    public Node next, prev; 
}; 
  
// function to get a new node 
static Node getNode(int data) 
    // allocate space 
    Node new_node = new Node(); 
  
    // put in the data 
    new_node.data = data; 
    new_node.next = new_node.prev = null
    return new_node; 
  
// function to insert a node at the beginging 
// of the Doubly Linked List 
static Node push(Node head_ref, Node new_node) 
    // since we are adding at the beginning, 
    // prev is always null 
    new_node.prev = null
  
    // link the old list off the new node 
    new_node.next = (head_ref); 
  
    // change prev of head node to new node 
    if ((head_ref) != null
        (head_ref).prev = new_node; 
  
    // move the head to point to the new node 
    (head_ref) = new_node; 
      
    return head_ref; 
  
// function to reverse a doubly linked list 
static Node reverseList(Node head_ref) 
    // if list is empty or it contains 
    // a single node only 
    if ((head_ref) == null || ((head_ref).next) == null
        return null
  
    Node new_head = null
    Node curr = head_ref, next; 
  
    while (curr != null
    
  
        // get pointer to next node 
        next = curr.next; 
  
        // push 'curr' node at the beginning of the 
        // list with starting with 'new_head' 
        new_head = push(new_head, curr); 
  
        // update 'curr' 
        curr = next; 
    
  
    // update 'head_ref' 
    head_ref = new_head; 
      
    return head_ref; 
  
// Function to print nodes in a 
// given doubly linked list 
static void printList(Node head) 
    while (head != null
    
        Console.Write(head.data + " "); 
        head = head.next; 
    
  
// Driver code
public static void Main(String []args) 
    // Start with the empty list 
    Node head = null
  
    // Create doubly linked: 10< - >8< - >4< - >2 / 
    head = push(head, getNode(2)); 
    head = push(head, getNode(4)); 
    head = push(head, getNode(8)); 
    head = push(head, getNode(10)); 
  
    Console.Write("Original list: "); 
    printList(head); 
  
    // Reverse doubly linked list 
    head = reverseList(head); 
  
    Console.Write("\nReversed list: "); 
    printList(head); 
  
// This code has been contributed by 29AjayKumar

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Output:

Original list: 10 8 4 2
Reversed list: 2 4 8 10

Time Complexity: O(n).



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