Print reverse of a Linked List without extra space and modifications (original) (raw)

Last Updated : 29 Mar, 2024

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.

Recursive solution to print reverse a linked list. Requires extra space.
Reverse linked list and then print. This requires modifications to original list.
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:
Find n = count nodes in linked list.
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;

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

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.