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Print reverse of a Linked List without extra space and modifications

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  • Difficulty Level : Easy
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Given a Linked List, display the linked list in reverse without using recursion, stack or modifications to given list.
 

Examples:  

Input: 1->2->3->4->5->NULL
Output: 5 4 3 2 1

Input: 10->5->15->20->24->NULL
Output: 24 20 15 5 10

Below are different solutions that are now allowed here as we cannot use extra space and modify list.
1) Recursive solution to print reverse a linked list. Requires extra space.
2) Reverse linked list and then print. This requires modifications to original list. 
3) Stack based solution to print linked list reverse. Push all nodes one by one to a stack. Then one by one pop elements from stack and print. This also requires extra space.
Algorithms:

1) Find n = count nodes in linked list.
2) For i = n to 1, do following.
    Print i-th node using get n-th node function

C++




// C/C++ program to print reverse of linked list
// without extra space and without modifications.
#include<stdio.h>
#include<stdlib.h>
 
/* Link list node */
struct Node
{
    int data;
    struct Node* next;
};
 
/* Given a reference (pointer to pointer) to the head
  of a list and an int, push a new node on the front
  of the list. */
void push(struct Node** head_ref, int new_data)
{
    struct Node* new_node =
            (struct Node*) malloc(sizeof(struct Node));
    new_node->data  = new_data;
    new_node->next = (*head_ref);
    (*head_ref)    = new_node;
}
 
/* Counts no. of nodes in linked list */
int getCount(struct Node* head)
{
    int count = 0;  // Initialize count
    struct Node* current = head;  // Initialize current
    while (current != NULL)
    {
        count++;
        current = current->next;
    }
    return count;
}
 
/* Takes head pointer of the linked list and index
    as arguments and return data at index*/
int getNth(struct Node* head, int n)
{
    struct Node* curr = head;
    for (int i=0; i<n-1 && curr != NULL; i++)
       curr = curr->next;
    return curr->data;
}
 
void printReverse(Node *head)
{
    // Count nodes
    int n = getCount(head);
 
    for (int i=n; i>=1; i--)
        printf("%d ", getNth(head, i));
}
 
/* Driver program to test count function*/
int main()
{
    /* Start with the empty list */
    struct Node* head = NULL;
 
    /* Use push() to construct below list
     1->2->3->4->5 */
    push(&head, 5);
    push(&head, 4);
    push(&head, 3);
    push(&head, 2);
    push(&head, 1);
 
    printReverse(head);
 
    return 0;
}

Java




// Java program to print reverse of linked list
// without extra space and without modifications.
class GFG
{
     
/* Link list node */
static class Node
{
    int data;
    Node next;
};
static Node head;
 
/* Given a reference (pointer to pointer) to
the head of a list and an int, push a new node
on the front of the list. */
static void push(Node head_ref, int new_data)
{
    Node new_node = new Node();
    new_node.data = new_data;
    new_node.next = (head_ref);
    (head_ref) = new_node;
    head = head_ref;
}
 
/* Counts no. of nodes in linked list */
static int getCount(Node head)
{
    int count = 0; // Initialize count
    Node current = head; // Initialize current
    while (current != null)
    {
        count++;
        current = current.next;
    }
    return count;
}
 
/* Takes head pointer of the linked list and
index as arguments and return data at index*/
static int getNth(Node head, int n)
{
    Node curr = head;
    for (int i = 0; i < n - 1 && curr != null; i++)
    curr = curr.next;
    return curr.data;
}
 
static void printReverse()
{
    // Count nodes
    int n = getCount(head);
 
    for (int i = n; i >= 1; i--)
        System.out.printf("%d ", getNth(head, i));
}
 
// Driver Code
public static void main(String[] args)
{
    /* Start with the empty list */
    head = null;
 
    /* Use push() to construct below list
    1->2->3->4->5 */
    push(head, 5);
    push(head, 4);
    push(head, 3);
    push(head, 2);
    push(head, 1);
 
    printReverse();
}
}
 
// This code is contributed by PrinciRaj1992

Python3




# Python3 program to print reverse of linked list
# without extra space and without modifications.
  
''' Link list node '''
class Node:
     
    def __init__(self, data):
        self.data = data
        self.next = None
      
''' Given a reference (pointer to pointer) to the head
  of a list and an int, push a new node on the front
  of the list. '''
def push( head_ref, new_data):   
    new_node = Node(new_data)
    new_node.next = head_ref;
    head_ref = new_node;
    return head_ref
 
''' Counts no. of nodes in linked list '''
def getCount(head):
    count = 0# Initialize count
    current = head;  # Initialize current   
    while (current != None):   
        count += 1
        current = current.next;   
    return count;
  
''' Takes head pointer of the linked list and index
    as arguments and return data at index'''
def getNth(head, n):
    curr = head;  
    i = 0   
    while i < n - 1 and curr != None:
        curr = curr.next;
        i += 1   
    return curr.data;
  
def printReverse(head):
 
    # Count nodes
    n = getCount(head);
     
    for i in range(n, 0, -1):
        print(getNth(head, i), end = ' ');
      
# Driver code
if __name__=='__main__':
     
    ''' Start with the empty list '''
    head = None;
  
    ''' Use push() to construct below list
     1.2.3.4.5 '''
    head = push(head, 5);
    head = push(head, 4);
    head = push(head, 3);
    head = push(head, 2);
    head = push(head, 1);
  
    printReverse(head);
      
# This code is contributed by rutvik_56

C#




// C# program to print reverse of
// linked list without extra space
// and without modifications.
using System;
     
class GFG
{
     
/* Link list node */
public class Node
{
    public int data;
    public Node next;
};
static Node head;
 
/* Given a reference (pointer to pointer)
to the head of a list and an int, push a
new node on the front of the list. */
static void push(Node head_ref,
                 int new_data)
{
    Node new_node = new Node();
    new_node.data = new_data;
    new_node.next = (head_ref);
    (head_ref) = new_node;
    head = head_ref;
}
 
/* Counts no. of nodes in linked list */
static int getCount(Node head)
{
    int count = 0; // Initialize count
    Node current = head; // Initialize current
    while (current != null)
    {
        count++;
        current = current.next;
    }
    return count;
}
 
/* Takes head pointer of the linked list and
index as arguments and return data at index*/
static int getNth(Node head, int n)
{
    Node curr = head;
    for (int i = 0;
             i < n - 1 && curr != null; i++)
    curr = curr.next;
    return curr.data;
}
 
static void printReverse()
{
    // Count nodes
    int n = getCount(head);
 
    for (int i = n; i >= 1; i--)
        Console.Write("{0} ", getNth(head, i));
}
 
// Driver Code
public static void Main(String[] args)
{
    /* Start with the empty list */
    head = null;
 
    /* Use push() to construct below list
    1->2->3->4->5 */
    push(head, 5);
    push(head, 4);
    push(head, 3);
    push(head, 2);
    push(head, 1);
 
    printReverse();
}
}
 
// This code is contributed by Princi Singh

Javascript




<script>
 
// JavaScript program to implement above approach
 
 
//  Link list node
class Node{
     
    constructor(data){
        this.data = data
        this.next = null
    }
}
     
// Given a reference (pointer to pointer) to the head
// of a list and an int, push a new node on the front
// of the list.
 
function push( head_ref, new_data){
    let new_node = new Node(new_data)
    new_node.next = head_ref;
    head_ref = new_node;
    return head_ref
}
 
// Counts no. of nodes in linked list
 
function getCount(head){
    let count = 0; // Initialize count
    let current = head; // Initialize current
    while (current != null){
        count += 1
        current = current.next;
    }
    return count;
}
 
// Takes head pointer of the linked list and index
//     as arguments and return data at index
 
function getNth(head, n){
    let curr = head;
    let i = 0
    while(i < n - 1 && curr != null){
        curr = curr.next;
        i += 1
    }
    return curr.data;
}
 
function printReverse(head){
 
    // Count nodes
    let n = getCount(head);
     
    for(let i=n;i>0;i--)
        document.write(getNth(head, i),' ');
}
     
// Driver code
     
// Start with the empty list
let head = null;
 
//  Use push() to construct below list
// 1.2.3.4.5
head = push(head, 5);
head = push(head, 4);
head = push(head, 3);
head = push(head, 2);
head = push(head, 1);
 
printReverse(head);
 
 
// This code is contributed by shinjanpatra
 
</script>

Output: 
 

5 4 3 2 1

Time complexity: O(n2) where n is the number of nodes in the given linked list
Auxiliary Space: O(1), as constant extra space is required.


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