Print Reverse a linked list using Stack

Given a linked list, print reverse of it without modifying the list.

Examples:

Input : 1 2 3 4 5 6
Output : 6 5 4 3 2 1

Input : 12 23 34 45 56 67 78
Output : 78 67 56 45 34 23 12

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->NULL

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



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) A O(n2) solution to print reverse of linked list that first count nodes and then prints k-th node from end.

In this post, an efficient stack based solution is discussed.
1. First insert all the element in stack
2. Print stack till stack is not empty


Note: Instead of inserting data from each node into the stack, insert the node’s address onto the stack. This is because the size of the node’s data will be generally more than the size of the node’s address. Thus the stack would end up requiring more memory if it directly stored the data elements. Also, we cannot insert the node’s data onto the stack if each node contained more than one data member. Hence the simpler and efficient solution would be to simply insert the node’s address.

C++

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// C/C++ program to print reverse of linked list
// using stack.
#include<bits/stdc++.h>
using namespace std;
   
/* 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)
{
    // store Node addresses in stack
    stack<Node *> stk; 
    Node* ptr = head;
    while (ptr != NULL) 
    {
        stk.push(ptr);
        ptr = ptr->next;
    }
  
    // print data from stack
    while (!stk.empty())
    {
        cout << stk.top()->data << " ";
        stk.pop(); // pop after print
    }
    cout << "\n";
}
   
/* 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;
}

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Java

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// Java program to print reverse of linked list
// using stack.
import java.util.*;
  
class GFG
{
      
/* Link list node */
static class Node
{
    int data;
    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. */
static Node 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;
    return 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(Node head)
{
    // store Node addresses in stack
    Stack<Node > stk = new Stack<Node>(); 
    Node ptr = head;
    while (ptr != null
    {
        stk.push(ptr);
        ptr = ptr.next;
    }
  
    // print data from stack
    while (stk.size() > 0)
    {
        System.out.print( stk.peek().data + " ");
        stk.pop(); // pop after print
    }
    System.out.println( "\n");
}
  
/* Driver code*/
public static void main(String args[])
{
    /* Start with the empty list */
    Node head = null;
  
    /* Use push() to con 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 Arnab Kundu

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Python3

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# Python3 program to print reverse of linked list
# using stack.
  
# Node of a linked list 
class Node: 
    def __init__(self, next = None, data = None): 
        self.next = next
        self.data = data 
  
# 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_node.data = 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 = 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 = i + 1
    return curr.data
  
def printReverse(head):
  
    # store Node addresses in stack
    stk = []
    ptr = head
    while (ptr != None): 
      
        stk.append(ptr)
        ptr = ptr.next
      
    # print data from stack
    while (len(stk) > 0):
      
        print( stk[-1].data, end = " ")
        stk.pop() # pop after print
      
    print( " ")
  
# Driver code
  
# Start with the empty list 
head = None
  
# Use push() to con 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 Arnab Kundu

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C#

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// C# program to print reverse of linked list
// using stack. 
using System;
using System.Collections.Generic; 
  
class GFG
{
      
/* Link list node */
public class Node
{
    public int data;
    public 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. */
static Node 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;
    return 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(Node head)
{
    // store Node addresses in stack
    Stack<Node > stk = new Stack<Node>(); 
    Node ptr = head;
    while (ptr != null
    {
        stk.Push(ptr);
        ptr = ptr.next;
    }
  
    // print data from stack
    while (stk.Count > 0)
    {
        Console.Write( stk.Peek().data + " ");
        stk.Pop(); // pop after print
    }
    Console.WriteLine( "\n");
}
  
/* Driver code*/
public static void Main(String []args)
{
    /* Start with the empty list */
    Node head = null;
  
    /* Use push() to con 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 Rajput-Ji

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Output:

5 4 3 2 1 

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