Reverse a Linked List in groups of given size using Stack
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++
#include <bits/stdc++.h>
using namespace std;
struct Node {
int data;
struct Node* next;
};
struct Node* Reverse( struct Node* head, int k)
{
stack<Node*> mystack;
struct Node* current = head;
struct Node* prev = NULL;
while (current != NULL) {
int count = 0;
while (current != NULL && count < k) {
mystack.push(current);
current = current->next;
count++;
}
while (mystack.size() > 0) {
if (prev == NULL) {
prev = mystack.top();
head = prev;
mystack.pop();
} else {
prev->next = mystack.top();
prev = prev->next;
mystack.pop();
}
}
}
prev->next = NULL;
return head;
}
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;
}
void printList( struct Node* node)
{
while (node != NULL) {
printf ( "%d " , node->data);
node = node->next;
}
}
int main( void )
{
struct Node* head = NULL;
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
import java.util.*;
class GfG
{
static class Node
{
int data;
Node next;
}
static Node head = null ;
static Node Reverse(Node head, int k)
{
Stack<Node> mystack = new Stack<Node> ();
Node current = head;
Node prev = null ;
while (current != null )
{
int count = 0 ;
while (current != null && count < k)
{
mystack.push(current);
current = current.next;
count++;
}
while (mystack.size() > 0 )
{
if (prev == null )
{
prev = mystack.peek();
head = prev;
mystack.pop();
}
else
{
prev.next = mystack.peek();
prev = prev.next;
mystack.pop();
}
}
}
prev.next = null ;
return head;
}
static void push( int new_data)
{
Node new_node = new Node();
new_node.data = new_data;
new_node.next = head;
head = new_node;
}
static void printList(Node node)
{
while (node != null )
{
System.out.print(node.data + " " );
node = node.next;
}
}
public static void main(String[] args)
{
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);
}
}
|
Python3
class Node( object ):
__slots__ = 'data' , 'next'
def __init__( self , data = None , next = None ):
self .data = data
self . next = next
def __repr__( self ):
return repr ( self .data)
class LinkedList( object ):
def __init__( self ):
self .head = None
def __repr__( self ):
nodes = []
curr = self .head
while curr:
nodes.append( repr (curr))
curr = curr. next
return '[' + ', ' .join(nodes) + ']'
def prepend( self , data):
self .head = Node(data = data,
next = self .head)
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
while curr is not None and val < k:
new_stack.append(curr.data)
curr = curr. next
val + = 1
while new_stack:
if prev is None :
prev = Node(new_stack.pop())
self .head = prev
else :
prev. next = Node(new_stack.pop())
prev = prev. next
prev. next = None
return self .head
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)
|
C#
using System;
using System.Collections;
class GfG
{
public class Node
{
public int data;
public Node next;
}
static Node head = null ;
static Node Reverse(Node head, int k)
{
Stack mystack = new Stack();
Node current = head;
Node prev = null ;
while (current != null )
{
int count = 0;
while (current != null && count < k)
{
mystack.Push(current);
current = current.next;
count++;
}
while (mystack.Count > 0)
{
if (prev == null )
{
prev = (Node)mystack.Peek();
head = prev;
mystack.Pop();
}
else
{
prev.next = (Node)mystack.Peek();
prev = prev.next;
mystack.Pop();
}
}
}
prev.next = null ;
return head;
}
static void Push( int new_data)
{
Node new_node = new Node();
new_node.data = new_data;
new_node.next = head;
head = new_node;
}
static void printList(Node node)
{
while (node != null )
{
Console.Write(node.data + " " );
node = node.next;
}
}
public static void Main(String []args)
{
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);
}
}
|
Javascript
<script>
class Node {
constructor() {
this .data = 0;
this .next = null ;
}
}
var head = null ;
function Reverse(head , k) {
var mystack = [];
var current = head;
var prev = null ;
while (current != null ) {
var count = 0;
while (current != null && count < k) {
mystack.push(current);
current = current.next;
count++;
}
while (mystack.length > 0) {
if (prev == null ) {
prev = mystack.pop();
head = prev;
} else {
prev.next = mystack.pop();
prev = prev.next;
}
}
}
prev.next = null ;
return head;
}
function push(new_data) {
var new_node = new Node();
new_node.data = new_data;
new_node.next = head;
head = new_node;
}
function printList(node) {
while (node != null ) {
document.write(node.data + " " );
node = node.next;
}
}
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);
</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
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