Reverse a Doubly Linked List by swapping data
Last Updated :
16 Dec, 2022
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++
// Cpp Program to Reverse a List using Data Swapping
#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;
}
// Insert a new node at the head of the list
void insert(Node** head, int val)
{
Node* temp = newNode(val);
temp->next = *head;
(*head)->prev = temp;
(*head) = temp;
}
// Function to reverse the list
void reverseList(Node** head)
{
Node* left = *head, * right = *head;
// Traverse the list and set right pointer to
// end of list
while (right->next != nullptr)
right = right->next;
// Swap data of left and right pointer and move
// them towards each other until they meet or
// cross each other
while (left != right && left->prev != right) {
// Swap data of left and right pointer
swap(left->data, right->data);
// Advance left pointer
left = left->next;
// Advance right pointer
right = right->prev;
}
}
// Driver code
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 Program to Reverse a List using Data Swapping
#include <stdio.h>
#include <stdlib.h>
struct Node {
int data;
struct Node* prev;
struct Node* next;
};
// Insert a new node at the head of the list
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);
}
// Function to reverse the list
void reverseList(struct Node* head)
{
struct Node* left = head;
struct Node* right = head;
// Traverse the list and set right pointer to
// end of list
while (right->next != NULL) {
right = right->next;
}
// Swap data of left and right pointer and move
// them towards each other until they meet or
// cross each other
while (left != right && left->prev != right) {
// Swap data of left and right pointer
int temp = left->data;
left->data = right->data;
right->data = temp;
// Advance left pointer
left = left->next;
// Advance right pointer
right = right->prev;
}
}
int main()
{
// code
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;
// This code is contributed by lokesh (lokeshmvs21)
}
// Java Program to Reverse a List using Data Swapping
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 );
}
// Insert a new node at the head of the list
static Node insert(Node head, int val)
{
Node temp = newNode(val);
temp.next = head;
(head).prev = temp;
(head) = temp;
return head;
}
// Function to reverse the list
static Node reverseList(Node head)
{
Node left = head, right = head;
// Traverse the list and set right pointer to
// end of list
while (right.next != null)
right = right.next;
// Swap data of left and right pointer and move
// them towards each other until they meet or
// cross each other
while (left != right && left.prev != right)
{
// Swap data of left and right pointer
int t = left.data;
left.data = right.data;
right.data = t;
// Advance left pointer
left = left.next;
// Advance right pointer
right = right.prev;
}
return head;
}
// Driver code
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);
}
}
// This code is contributed by Arnab Kundu
# Python3 Program to Reverse a List
# using Data Swapping
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)
# Insert a new node at the head of the list
def insert(head, val):
temp = newNode(val)
temp.next = head
(head).prev = temp
(head) = temp
return head
# Function to reverse the list
def reverseList( head):
left = head
right = head
# Traverse the list and set right
# pointer to end of list
while (right.next != None):
right = right.next
# Swap data of left and right pointer
# and move them towards each other
# until they meet or cross each other
while (left != right and left.prev != right):
# Swap data of left and right pointer
t = left.data
left.data = right.data
right.data = t
# Advance left pointer
left = left.next
# Advance right pointer
right = right.prev
return head
# Driver code
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)
# This code is contributed by AbhiThakur
// C# Program to Reverse a List using Data Swapping
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 );
}
// Insert a new node at the head of the list
static Node insert(Node head, int val)
{
Node temp = newNode(val);
temp.next = head;
(head).prev = temp;
(head) = temp;
return head;
}
// Function to reverse the list
static Node reverseList(Node head)
{
Node left = head, right = head;
// Traverse the list and set right pointer to
// end of list
while (right.next != null)
right = right.next;
// Swap data of left and right pointer and move
// them towards each other until they meet or
// cross each other
while (left != right && left.prev != right)
{
// Swap data of left and right pointer
int t = left.data;
left.data = right.data;
right.data = t;
// Advance left pointer
left = left.next;
// Advance right pointer
right = right.prev;
}
return head;
}
// Driver code
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);
}
}
// This code has been contributed by 29AjayKumar
<script>
// javascript Program to Reverse a List using Data Swapping
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);
}
// Insert a new node at the head of the list
function insert(head , val) {
var temp = newNode(val);
temp.next = head;
(head).prev = temp;
(head) = temp;
return head;
}
// Function to reverse the list
function reverseList(head) {
var left = head, right = head;
// Traverse the list and set right pointer to
// end of list
while (right.next != null)
right = right.next;
// Swap data of left and right pointer and move
// them towards each other until they meet or
// cross each other
while (left != right && left.prev != right) {
// Swap data of left and right pointer
var t = left.data;
left.data = right.data;
right.data = t;
// Advance left pointer
left = left.next;
// Advance right pointer
right = right.prev;
}
return head;
}
// Driver code
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);
// This code contributed by umadevi9616
</script>
Output1 <--> 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.
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