Open In App

Reverse a Linked List in groups of given size using Stack

Improve
Improve
Like Article
Like
Save
Share
Report

Given a linked list, write a function to reverse every k node (where k is an input to the function). 

Examples: 

Inputs:  1->2->3->4->5->6->7->8->NULL and k = 3 
Output:  3->2->1->6->5->4->8->7->NULL. 

Inputs:   1->2->3->4->5->6->7->8->NULL and k = 5
Output:  5->4->3->2->1->8->7->6->NULL.

We have already discussed its solution in the below post 
Reverse a Linked List in groups of given size | Set 1
In this post, we have used a stack that will store the nodes of the given linked list. Firstly, push the k elements of the linked list in the stack. Now pop elements one by one and keep track of the previously popped node. Point the next pointer of the prev node to the top element of the stack. Repeat this process, until NULL is reached.
This algorithm uses O(k) extra space. 
 

Implementation:

C++




// C++ program to reverse a linked list in groups
// of given size
#include <bits/stdc++.h>
using namespace std;
  
/* Link list node */
struct Node {
    int data;
    struct Node* next;
};
  
/* Reverses the linked list in groups of size k
   and returns the pointer to the new head node. */
struct Node* Reverse(struct Node* head, int k)
{
    // Create a stack of Node*
    stack<Node*> mystack;
    struct Node* current = head;
    struct Node* prev = NULL;
  
    while (current != NULL) {
  
        // Terminate the loop whichever comes first
        // either current == NULL or count >= k
        int count = 0;
        while (current != NULL && count < k) {
            mystack.push(current);
            current = current->next;
            count++;
        }
  
        // Now pop the elements of stack one by one
        while (mystack.size() > 0) {
  
            // If final list has not been started yet.
            if (prev == NULL) {
                prev = mystack.top();
                head = prev;
                mystack.pop();
            } else {
                prev->next = mystack.top();
                prev = prev->next;
                mystack.pop();
            }
        }
    }
  
    // Next of last element will point to NULL.
    prev->next = NULL;
  
    return head;
}
  
/* UTILITY FUNCTIONS */
/* Function to push a node */
void push(struct Node** head_ref, int new_data)
{
    /* allocate node */
    struct Node* new_node = 
          (struct Node*)malloc(sizeof(struct Node));
  
    /* put in the data  */
    new_node->data = new_data;
  
    /* link the old list of the new node */
    new_node->next = (*head_ref);
  
    /* move the head to point to the new node */
    (*head_ref) = new_node;
}
  
/* Function to print linked list */
void printList(struct Node* node)
{
    while (node != NULL) {
        printf("%d  ", node->data);
        node = node->next;
    }
}
  
/* Driver program to test above function*/
int main(void)
{
    /* Start with the empty list */
    struct Node* head = NULL;
  
    /* Created Linked list is 1->2->3->4->5->6->7->8->9 */
    push(&head, 9);
    push(&head, 8);
    push(&head, 7);
    push(&head, 6);
    push(&head, 5);
    push(&head, 4);
    push(&head, 3);
    push(&head, 2);
    push(&head, 1);
  
    printf("\nGiven linked list \n");
    printList(head);
    head = Reverse(head, 3);
  
    printf("\nReversed Linked list \n");
    printList(head);
  
    return 0;
}


Java




// Java program to reverse a linked list in groups 
// of given size 
import java.util.*;
class GfG 
{
      
/* Link list node */
static class Node 
    int data; 
    Node next; 
}
static Node head = null;
  
/* Reverses the linked list in groups of size k 
and returns the pointer to the new head node. */
static Node Reverse(Node head, int k) 
    // Create a stack of Node* 
    Stack<Node> mystack = new Stack<Node> (); 
    Node current = head; 
    Node prev = null
  
    while (current != null)
    
  
        // Terminate the loop whichever comes first 
        // either current == NULL or count >= k 
        int count = 0
        while (current != null && count < k)
        
            mystack.push(current); 
            current = current.next; 
            count++; 
        
  
        // Now pop the elements of stack one by one 
        while (mystack.size() > 0
        
  
            // If final list has not been started yet. 
            if (prev == null)
            
                prev = mystack.peek(); 
                head = prev; 
                mystack.pop(); 
            
            else
            
                prev.next = mystack.peek(); 
                prev = prev.next; 
                mystack.pop(); 
            
        
    
  
    // Next of last element will point to NULL. 
    prev.next = null
  
    return head; 
  
/* UTILITY FUNCTIONS */
/* Function to push a node */
static void push( int new_data) 
    /* allocate node */
    Node new_node = new Node(); 
  
    /* put in the data */
    new_node.data = new_data; 
  
    /* link the old list of the new node */
    new_node.next = head; 
  
    /* move the head to point to the new node */
    head = new_node; 
  
/* Function to print linked list */
static void printList(Node node) 
    while (node != null
    
        System.out.print(node.data + " "); 
        node = node.next; 
    
  
/* Driver code*/
public static void main(String[] args) 
    /* Start with the empty list */
    //Node head = null; 
  
    /* Created Linked list is 1->2->3->
    4->5->6->7->8->9 */
    push( 9); 
    push( 8); 
    push( 7); 
    push( 6); 
    push( 5); 
    push(4); 
    push(3); 
    push(2); 
    push( 1); 
  
    System.out.println("Given linked list "); 
    printList(head); 
    head = Reverse(head, 3);
    System.out.println();
  
    System.out.println("Reversed Linked list "); 
    printList(head); 
  
// This code is contributed by Prerna Saini


Python3




# Python3 program to reverse a Linked List
# in groups of given size
  
# Node class
class Node(object):
    __slots__ = 'data', 'next'
  
    # Constructor to initialize the node object
    def __init__(self, data = None, next = None):
        self.data = data
        self.next = next
  
    def __repr__(self):
        return repr(self.data)
  
class LinkedList(object):
  
    # Function to initialize head
    def __init__(self):
        self.head = None
  
    # Utility function to print nodes 
    # of LinkedList
    def __repr__(self):
        nodes = []
        curr = self.head
        while curr:
            nodes.append(repr(curr))
            curr = curr.next
        return '[' + ', '.join(nodes) + ']'
  
    # Function to insert a new node at
    # the beginning
    def prepend(self, data):
        self.head = Node(data = data,
                         next = self.head)
  
    # Reverses the linked list in groups of size k
    # and returns the pointer to the new head node.
    def reverse(self, k = 1):
        if self.head is None:
            return
  
        curr = self.head
        prev = None
        new_stack = []
        while curr is not None:
            val = 0
              
            # Terminate the loop whichever comes first
            # either current == None or value >= k
            while curr is not None and val < k:
                new_stack.append(curr.data)
                curr = curr.next
                val += 1
  
            # Now pop the elements of stack one by one
            while new_stack:
                  
                # If final list has not been started yet.
                if prev is None:
                    prev = Node(new_stack.pop())
                    self.head = prev
                else:
                    prev.next = Node(new_stack.pop())
                    prev = prev.next
                      
        # Next of last element will point to None.
        prev.next = None
        return self.head
  
# Driver Code
llist = LinkedList() 
llist.prepend(9)
llist.prepend(8)
llist.prepend(7)
llist.prepend(6)
llist.prepend(5)
llist.prepend(4)
llist.prepend(3)
llist.prepend(2)
llist.prepend(1)
  
print("Given linked list")
print(llist)
llist.head = llist.reverse(3)
  
print("Reversed Linked list")
print(llist)
  
# This code is contributed by 
# Sagar Kumar Sinha(sagarsinha7777)


C#




// C# program to reverse a linked list  
// in groups of given size 
using System;
using System.Collections;
  
class GfG 
      
/* Link list node */
public class Node 
    public int data; 
    public Node next; 
static Node head = null
  
/* Reverses the linked list in groups of size k 
and returns the pointer to the new head node. */
static Node Reverse(Node head, int k) 
    // Create a stack of Node* 
    Stack mystack = new Stack(); 
    Node current = head; 
    Node prev = null
  
    while (current != null
    
  
        // Terminate the loop whichever comes first 
        // either current == NULL or count >= k 
        int count = 0; 
        while (current != null && count < k) 
        
            mystack.Push(current); 
            current = current.next; 
            count++; 
        
  
        // Now Pop the elements of stack one by one 
        while (mystack.Count > 0) 
        
  
            // If final list has not been started yet. 
            if (prev == null
            
                prev = (Node)mystack.Peek(); 
                head = prev; 
                mystack.Pop(); 
            
            else
            
                prev.next = (Node)mystack.Peek(); 
                prev = prev.next; 
                mystack.Pop(); 
            
        
    
  
    // Next of last element will point to NULL. 
    prev.next = null
  
    return head; 
  
/* UTILITY FUNCTIONS */
/* Function to Push a node */
static void Push( int new_data) 
    /* allocate node */
    Node new_node = new Node(); 
  
    /* put in the data */
    new_node.data = new_data; 
  
    /* link the old list of the new node */
    new_node.next = head; 
  
    /* move the head to point to the new node */
    head = new_node; 
  
/* Function to print linked list */
static void printList(Node node) 
    while (node != null
    
        Console.Write(node.data + " "); 
        node = node.next; 
    
  
/* Driver code*/
public static void Main(String []args) 
    /* Start with the empty list */
    //Node head = null; 
  
    /* Created Linked list is 1->2->3-> 
    4->5->6->7->8->9 */
    Push( 9); 
    Push( 8); 
    Push( 7); 
    Push( 6); 
    Push( 5); 
    Push(4); 
    Push(3); 
    Push(2); 
    Push( 1); 
  
    Console.WriteLine("Given linked list "); 
    printList(head); 
    head = Reverse(head, 3); 
    Console.WriteLine(); 
  
    Console.WriteLine("Reversed Linked list "); 
    printList(head); 
  
// This code is contributed by Arnab Kundu


Javascript




<script>
// javascript program to reverse a linked list in groups 
// of given size class GfG {
  
    /* Link list node */
class Node {
    constructor() {
        this.data = 0;
        this.next = null;
    }
}
  
    var head = null;
  
    /*
     * Reverses the linked list in groups of size k and returns the pointer to the
     * new head node.
     */
    function Reverse(head , k) {
        // Create a stack of Node*
        var mystack = [];
var current = head;
var prev = null;
  
        while (current != null) {
  
            // Terminate the loop whichever comes first
            // either current == NULL or count >= k
            var count = 0;
            while (current != null && count < k) {
                mystack.push(current);
                current = current.next;
                count++;
            }
  
            // Now pop the elements of stack one by one
            while (mystack.length > 0) {
  
                // If final list has not been started yet.
                if (prev == null) {
                    prev = mystack.pop();
                    head = prev;
                      
                } else {
                    prev.next = mystack.pop();
                    prev = prev.next;
                      
                }
            }
        }
  
        // Next of last element will point to NULL.
        prev.next = null;
  
        return head;
    }
  
    /* UTILITY FUNCTIONS */
    /* Function to push a node */
    function push(new_data) {
        /* allocate node */
var new_node = new Node();
  
        /* put in the data */
        new_node.data = new_data;
  
        /* link the old list of the new node */
        new_node.next = head;
  
        /* move the head to point to the new node */
        head = new_node;
    }
  
    /* Function to print linked list */
    function printList(node) {
        while (node != null) {
            document.write(node.data + " ");
            node = node.next;
        }
    }
  
    /* Driver code */
      
        /* Start with the empty list */
        // Node head = null;
  
        /*
         * Created Linked list is 1->2->3-> 4->5->6->7->8->9
         */
        push(9);
        push(8);
        push(7);
        push(6);
        push(5);
        push(4);
        push(3);
        push(2);
        push(1);
  
        document.write("Given linked list <br/>");
        printList(head);
        head = Reverse(head, 3);
        document.write("<br/>");
  
        document.write("Reversed Linked list <br/>");
        printList(head);
  
// This code contributed by aashish1995 
</script>


Output

Given linked list 
1  2  3  4  5  6  7  8  9  
Reversed Linked list 
3  2  1  6  5  4  9  8  7  

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(k), as we are using extra space for the stack.

 



Last Updated : 10 Jan, 2023
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads