Reverse a Doubly Linked List by swapping data
Given a Doubly Linked List, we are asked to reverse the list in place without using any extra space.
Examples:
Input : 1 <--> 2 <--> 5 <--> 6 <--> 7
Output : 7 <--> 6 <--> 5 <--> 2 <--> 1
Input : 11 <--> 22 <--> 33 <--> 22 <--> 1
Output : 1 <--> 22 <--> 33 <--> 22 <--> 11
We have discussed three methods to reverse a doubly-linked list: Reverse a doubly-linked list, Reverse a Doubly Linked List (Set 2) and Reverse a Doubly linked list using recursion.
The first two methods work in O(n) time and require no extra space. The first method works by swapping the next and previous pointers of each node. The second method takes each node from the list and adds it to the beginning of the list.
There is another approach that is a bit more intuitive, but also a bit more costly.
This method is similar to the reverse array. To reverse an array, we put two pointers-one at the beginning and another at the end of the list. We then swap the data of the two pointers and advance both pointers toward each other. We stop either when the two pointers meet or when they cross each other. We perform exactly n/2 swaps, and the time complexity is also O(N).
A doubly linked list has both a previous and a next pointer, which means we can traverse in both forward and backward directions in the list. So if we put a pointer( say left pointer) at the beginning of the list and another right pointer at the end of the list, we can move these pointers toward each other by advancing the left pointer and receding the right pointer.
Algorithm:
Step 1: Set LEFT to head of list
Step 2: Traverse the list and set RIGHT to end of the list
Step 3: Repeat following steps while LEFT != RIGHT and
LEFT->PREV != RIGHT
Step 4: Swap LEFT->DATA and RIGHT->DATA
Step 5: Advance LEFT pointer by one, LEFT = LEFT->NEXT
Step 6: Recede RIGHT pointer by one, i.e RIGHT = RIGHT->PREV
[END OF LOOP]
Step 7: End
A Note on the comparative efficiency of the three methods
A few things must be mentioned. This method is simple to implement, but it is also more costly when compared to the pointer-exchange method. This is because we swap data and not pointers. Swapping data can be more costly if the nodes are large complex data types with multiple data members. In contrast, the pointer to the node will always be a simpler data type and either 4 or 8 bytes.
Below is the implementation of the algorithm.
C++
#include <bits/stdc++.h>
using namespace std;
struct Node {
int data;
struct Node *prev, *next;
};
Node* newNode( int val)
{
Node* temp = new Node;
temp->data = val;
temp->prev = temp->next = nullptr;
return temp;
}
void printList(Node* head)
{
while (head->next != nullptr) {
cout << head->data << " <--> " ;
head = head->next;
}
cout << head->data << endl;
}
void insert(Node** head, int val)
{
Node* temp = newNode(val);
temp->next = *head;
(*head)->prev = temp;
(*head) = temp;
}
void reverseList(Node** head)
{
Node* left = *head, * right = *head;
while (right->next != nullptr)
right = right->next;
while (left != right && left->prev != right) {
swap(left->data, right->data);
left = left->next;
right = right->prev;
}
}
int main()
{
Node* head = newNode(5);
insert(&head, 4);
insert(&head, 3);
insert(&head, 2);
insert(&head, 1);
printList(head);
cout << "List After Reversing" << endl;
reverseList(&head);
printList(head);
return 0;
}
|
C
#include <stdio.h>
#include <stdlib.h>
struct Node {
int data;
struct Node* prev;
struct Node* next;
};
void insert( struct Node** head, int val)
{
struct Node* temp;
temp = ( struct Node*) malloc ( sizeof ( struct Node));
temp->data = val;
temp->next = *head;
temp->prev = NULL;
if ((*head) != NULL) {
(*head)->prev = temp;
}
(*head) = temp;
}
void printList( struct Node* head)
{
struct Node* temp = head;
while (temp->next != NULL) {
printf ( "%d <--> " , temp->data);
temp = temp->next;
}
printf ( "%d\n" , temp->data);
}
void reverseList( struct Node* head)
{
struct Node* left = head;
struct Node* right = head;
while (right->next != NULL) {
right = right->next;
}
while (left != right && left->prev != right) {
int temp = left->data;
left->data = right->data;
right->data = temp;
left = left->next;
right = right->prev;
}
}
int main()
{
struct Node* head = NULL;
insert(&head, 5);
insert(&head, 4);
insert(&head, 3);
insert(&head, 2);
insert(&head, 1);
printList(head);
printf ( "List After Reversing\n" );
reverseList(head);
printList(head);
return 0;
}
|
Java
import java.util.*;
import java.io.*;
class GFG
{
static class Node
{
int data;
Node prev, next;
};
static Node newNode( int val)
{
Node temp = new Node();
temp.data = val;
temp.prev = temp.next = null ;
return temp;
}
static void printList(Node head)
{
while (head.next != null )
{
System.out.print(head.data+ " <--> " );
head = head.next;
}
System.out.println( head.data );
}
static Node insert(Node head, int val)
{
Node temp = newNode(val);
temp.next = head;
(head).prev = temp;
(head) = temp;
return head;
}
static Node reverseList(Node head)
{
Node left = head, right = head;
while (right.next != null )
right = right.next;
while (left != right && left.prev != right)
{
int t = left.data;
left.data = right.data;
right.data = t;
left = left.next;
right = right.prev;
}
return head;
}
public static void main(String args[])
{
Node head = newNode( 5 );
head = insert(head, 4 );
head = insert(head, 3 );
head = insert(head, 2 );
head = insert(head, 1 );
printList(head);
System.out.println( "List After Reversing" );
head=reverseList(head);
printList(head);
}
}
|
Python3
import math
class Node:
def __init__( self , data):
self .data = data
self . next = None
def newNode(val):
temp = Node(val)
temp.data = val
temp.prev = None
temp. next = None
return temp
def printList( head):
while (head. next ! = None ):
print (head.data, end = "<-->" )
head = head. next
print (head.data)
def insert(head, val):
temp = newNode(val)
temp. next = head
(head).prev = temp
(head) = temp
return head
def reverseList( head):
left = head
right = head
while (right. next ! = None ):
right = right. next
while (left ! = right and left.prev ! = right):
t = left.data
left.data = right.data
right.data = t
left = left. next
right = right.prev
return head
if __name__ = = '__main__' :
head = newNode( 5 )
head = insert(head, 4 )
head = insert(head, 3 )
head = insert(head, 2 )
head = insert(head, 1 )
printList(head)
print ( "List After Reversing" )
head = reverseList(head)
printList(head)
|
C#
using System;
class GFG
{
public class Node
{
public int data;
public Node prev, next;
};
static Node newNode( int val)
{
Node temp = new Node();
temp.data = val;
temp.prev = temp.next = null ;
return temp;
}
static void printList(Node head)
{
while (head.next != null )
{
Console.Write(head.data+ " <--> " );
head = head.next;
}
Console.WriteLine( head.data );
}
static Node insert(Node head, int val)
{
Node temp = newNode(val);
temp.next = head;
(head).prev = temp;
(head) = temp;
return head;
}
static Node reverseList(Node head)
{
Node left = head, right = head;
while (right.next != null )
right = right.next;
while (left != right && left.prev != right)
{
int t = left.data;
left.data = right.data;
right.data = t;
left = left.next;
right = right.prev;
}
return head;
}
public static void Main(String []args)
{
Node head = newNode(5);
head = insert(head, 4);
head = insert(head, 3);
head = insert(head, 2);
head = insert(head, 1);
printList(head);
Console.WriteLine( "List After Reversing" );
head=reverseList(head);
printList(head);
}
}
|
Javascript
<script>
class Node {
constructor(val) {
this .data = val;
this .prev = null ;
this .next = null ;
}
}
function newNode(val) {
var temp = new Node();
temp.data = val;
temp.prev = temp.next = null ;
return temp;
}
function printList(head) {
while (head.next != null ) {
document.write(head.data + " <-> " );
head = head.next;
}
document.write(head.data);
}
function insert(head , val) {
var temp = newNode(val);
temp.next = head;
(head).prev = temp;
(head) = temp;
return head;
}
function reverseList(head) {
var left = head, right = head;
while (right.next != null )
right = right.next;
while (left != right && left.prev != right) {
var t = left.data;
left.data = right.data;
right.data = t;
left = left.next;
right = right.prev;
}
return head;
}
var head = newNode(5);
head = insert(head, 4);
head = insert(head, 3);
head = insert(head, 2);
head = insert(head, 1);
printList(head);
document.write( "<br/>List After Reversing<br/>" );
head = reverseList(head);
printList(head);
</script>
|
Output
1 <--> 2 <--> 3 <--> 4 <--> 5
List After Reversing
5 <--> 4 <--> 3 <--> 2 <--> 1
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.
Last Updated :
16 Dec, 2022
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