Given a singly linked list and an integer K, the task is to reverse every K nodes of the given linked list.
Examples:
Input: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> NULL, K = 3
Output: 3 2 1 6 5 4 8 7Input: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> NULL, K = 5
Output: 5 4 3 2 1 8 7 6
Approach: Two different approaches to solve this problem have been discussed in Set 1 and Set 2 of this article. In this article, an approach based on deque will be discussed.
- Create a deque.
- Store the address of the first k nodes in the deque.
- Pop first and the last value from the deque and swap the data values at those addresses.
- Repeat step 3 till the deque is not empty.
- Repeat step 2 for the next k nodes and till the end of the linked list is not reached.
Below is the implementation of the above approach:
C++14
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Link list node struct node {
int data;
struct node* next;
}; // Function to insert a node at // the head of the linked list void push(node** head_ref, 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_ref);
/* Move the head to point to the new node */
(*head_ref) = new_node;
} // Function to print the linked list void printList(node* head)
{ while (head != NULL) {
cout << head->data << " " ;
head = head->next;
}
} /* Function to reverse the linked list in groups of size k and return the pointer to the new head node. */ struct node* reverse( struct node* head, int k)
{ if (head == NULL)
return head;
// Create deque to store the address
// of the nodes of the linked list
deque<node*> q;
// Store head pointer in current to
// traverse the linked list
node* current = head;
int i;
// Iterate through the entire linked
// list by moving the current
while (current != NULL) {
i = 1;
// Store addresses of the k
// nodes in the deque
while (i <= k) {
if (current == NULL)
break ;
q.push_back(current);
current = current->next;
i++;
}
/* pop first and the last value from
the deque and swap the data values at
those addresses
Do this till there exist an address in
the deque or deque is not empty*/
while (!q.empty()) {
node* front = q.front();
node* last = q.back();
swap(front->data, last->data);
// pop from the front if
// the deque is not empty
if (!q.empty())
q.pop_front();
// pop from the back if
// the deque is not empty
if (!q.empty())
q.pop_back();
}
}
return head;
} // Driver code int main()
{ // Start with the empty list
node* head = NULL;
// Created Linked list is
// 1->2->3->4->5->6->7->8->9->10
push(&head, 10);
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);
int k = 2;
// Get the new head after reversing the
// linked list in groups of size k
head = reverse(head, k);
printList(head);
return 0;
} |
Java
// Java implementation of the above approach import java.util.*;
class LinkedList{
static Node head;
// Creating node class static class Node
{ int data;
Node next;
Node( int d)
{
data = d;
next = null ;
}
} // Inserts a new Node at front of the list. public void push( int new_data)
{ // Allocate the Node &
// Put in the data
Node new_node = new Node(new_data);
// Make next of new Node as head
new_node.next = head;
// Move the head to point to new Node
head = new_node;
} // Prints content of linked list void printList(Node node)
{ while (node != null )
{
System.out.print(node.data + " " );
node = node.next;
}
} // Function to reverse the linked list // in groups of size k and return the // pointer to the new head node. Node reverse(Node head, int k)
{ if (head == null )
return head;
// Create deque to store the address
// of the nodes of the linked list
Deque<Node> q = new ArrayDeque<Node>();
// Store head pointer in current to
// traverse the linked list
Node current = head;
int i;
// Iterate through the entire linked
// list by moving the current
while (current != null )
{
i = 1 ;
// Store addresses of the k
// nodes in the deque
while (i <= k)
{
if (current == null )
break ;
q.addLast(current);
current = current.next;
i++;
}
// pop first and the last value from
// the deque and swap the data values at
// those addresses
// Do this till there exist an address in
// the deque or deque is not empty
while (!q.isEmpty())
{
Node front = q.peekFirst();
Node last = q.peekLast();
// Swapping front and
// last nodes
int temp = front.data;
front.data = last.data;
last.data = temp;
// pop from the front if
// the deque is not empty
if (!q.isEmpty())
q.removeFirst();
// pop from the back if
// the deque is not empty
if (!q.isEmpty())
q.removeLast();
}
}
return head;
} // Driver code public static void main(String[] args)
{ LinkedList list = new LinkedList();
// Created Linked list is
// 1->2->3->4->5->6->7->8->9->10
list.push( 10 );
list.push( 9 );
list.push( 8 );
list.push( 7 );
list.push( 6 );
list.push( 5 );
list.push( 4 );
list.push( 3 );
list.push( 2 );
list.push( 1 );
int k = 2 ;
// Get the new head after reversing the
// linked list in groups of size k
head = list.reverse(head, k);
list.printList(head);
} } // This code is contributed by Bindu Madhav |
Python3
# Python3 implementation of the approach # Link list node class Node:
def __init__( self ):
self .data = 0
self . next = None
# Function to insert a node at # the head of the linked list def push(head_ref, new_data):
# Allocate node
new_node = 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
return head_ref
# Function to print the linked list def printList( head):
while (head ! = None ) :
print ( head.data, end = " " )
head = head. next
# Function to reverse the linked list in groups of # size k and return the pointer to the new head node. def reverse( head, k):
if (head = = None ):
return head
# Create deque to store the address
# of the nodes of the linked list
q = []
# Store head pointer in current to
# traverse the linked list
current = head
i = 0
# Iterate through the entire linked
# list by moving the current
while (current ! = None ) :
i = 1
# Store addresses of the k
# nodes in the deque
while (i < = k) :
if (current = = None ):
break
q.append(current)
current = current. next
i = i + 1
# pop first and the last value from
# the deque and swap the data values at
# those addresses
# Do this till there exist an address in
# the deque or deque is not empty
while ( len (q) > 0 ):
front = q[ - 1 ]
last = q[ 0 ]
temp = front.data
front.data = last.data
last.data = temp
# pop from the front if
# the deque is not empty
if ( len (q) > 0 ):
q.pop()
# pop from the back if
# the deque is not empty
if ( len (q)):
q.pop( 0 )
return head
# Driver code # Start with the empty list head = None
# Created Linked list is # 1.2.3.4.5.6.7.8.9.10 head = push(head, 10 )
head = push(head, 9 )
head = push(head, 8 )
head = push(head, 7 )
head = push(head, 6 )
head = push(head, 5 )
head = push(head, 4 )
head = push(head, 3 )
head = push(head, 2 )
head = push(head, 1 )
k = 2
# Get the new head after reversing the # linked list in groups of size k head = reverse(head, k)
printList(head) # This code is contributed by Arnab Kundu |
C#
// C# implementation of the above approach using System;
using System.Collections.Generic;
class List{
static Node head;
// Creating node class class Node
{ public int data;
public Node next;
public Node( int d)
{
data = d;
next = null ;
}
} // Inserts a new Node at front of the list. public void Push( int new_data)
{ // Allocate the Node &
// Put in the data
Node new_node = new Node(new_data);
// Make next of new Node as head
new_node.next = head;
// Move the head to point to new Node
head = new_node;
} // Prints content of linked list void printList(Node node)
{ while (node != null )
{
Console.Write(node.data + " " );
node = node.next;
}
} // Function to reverse the linked list // in groups of size k and return the // pointer to the new head node. Node reverse(Node head, int k)
{ if (head == null )
return head;
// Create deque to store the address
// of the nodes of the linked list
List<Node> q = new List<Node>();
// Store head pointer in current to
// traverse the linked list
Node current = head;
int i;
// Iterate through the entire linked
// list by moving the current
while (current != null )
{
i = 1;
// Store addresses of the k
// nodes in the deque
while (i <= k)
{
if (current == null )
break ;
q.Add(current);
current = current.next;
i++;
}
// pop first and the last value from
// the deque and swap the data values at
// those addresses
// Do this till there exist an address in
// the deque or deque is not empty
while (q.Count != 0)
{
Node front = q[0];
Node last = q[q.Count - 1];
// Swapping front and
// last nodes
int temp = front.data;
front.data = last.data;
last.data = temp;
// pop from the front if
// the deque is not empty
if (q.Count != 0)
q.RemoveAt(0);
// pop from the back if
// the deque is not empty
if (q.Count != 0)
q.RemoveAt(q.Count - 1);
}
}
return head;
} // Driver code public static void Main(String[] args)
{ List list = new List();
// Created Linked list is
// 1->2->3->4->5->6->7->8->9->10
list.Push(10);
list.Push(9);
list.Push(8);
list.Push(7);
list.Push(6);
list.Push(5);
list.Push(4);
list.Push(3);
list.Push(2);
list.Push(1);
int k = 2;
// Get the new head after reversing the
// linked list in groups of size k
head = list.reverse(head, k);
list.printList(head);
} } // This code is contributed by todaysgaurav |
Javascript
<script> // Javascript implementation of the above approach var head = null ;
// Creating node class class Node { constructor(d)
{
this .data = d;
this .next = null ;
}
} // Inserts a new Node at front of the list. function Push(new_data)
{ // Allocate the Node &
// Put in the data
var new_node = new Node(new_data);
// Make next of new Node as head
new_node.next = head;
// Move the head to point to new Node
head = new_node;
} // Prints content of linked list function printList(node)
{ while (node != null )
{
document.write(node.data + " " );
node = node.next;
}
} // Function to reverse the linked list // in groups of size k and return the // pointer to the new head node. function reverse(head, k)
{ if (head == null )
return head;
// Create deque to store the address
// of the nodes of the linked list
var q = [];
// Store head pointer in current to
// traverse the linked list
var current = head;
var i;
// Iterate through the entire linked
// list by moving the current
while (current != null )
{
i = 1;
// Store addresses of the k
// nodes in the deque
while (i <= k)
{
if (current == null )
break ;
q.push(current);
current = current.next;
i++;
}
// pop first and the last value from
// the deque and swap the data values at
// those addresses
// Do this till there exist an address in
// the deque or deque is not empty
while (q.length != 0)
{
var front = q[0];
var last = q[q.length - 1];
// Swapping front and
// last nodes
var temp = front.data;
front.data = last.data;
last.data = temp;
// pop from the front if
// the deque is not empty
if (q.Count != 0)
q.shift();
// pop from the back if
// the deque is not empty
if (q.Count != 0)
q.pop();
}
}
return head;
} // Driver code // Created Linked list is // 1->2->3->4->5->6->7->8->9->10 Push(10); Push(9); Push(8); Push(7); Push(6); Push(5); Push(4); Push(3); Push(2); Push(1); var k = 2;
// Get the new head after reversing the // linked list in groups of size k head = reverse(head, k); printList(head); // This code is contributed by itsok </script> |
Output:
2 1 4 3 6 5 8 7 10 9
Time Complexity: O(N)
Auxiliary Space: O(N)