Given a singly linked list and an integer K, the task is to rotate the linked list clockwise to the right by K places.
Examples:
Input: 1 -> 2 -> 3 -> 4 -> 5 -> NULL, K = 2
Output: 4 -> 5 -> 1 -> 2 -> 3 -> NULL
Input: 7 -> 9 -> 11 -> 13 -> 3 -> 5 -> NULL, K = 12
Output: 7 -> 9 -> 11 -> 13 -> 3 -> 5 -> NULL
Approach: To rotate the linked list first check whether the given k is greater than the count of nodes in the linked list or not. Traverse the list and find the length of the linked list then compare it with k, if less then continue otherwise deduce it in the range of linked list size by taking modulo with the length of the list.
After that subtract the value of k from the length of the list. Now, the question has been changed to the left rotation of the linked list so follow that procedure:
- Change the next of the kth node to NULL.
- Change the next of the last node to the previous head node.
- Change the head to (k+1)th node.
In order to do that, the pointers to the kth node, (k+1)th node, and last node are required.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
/* Link list node */ class Node {
public :
int data;
Node* next;
}; /* A utility function to push a node */ 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;
} /* A utility function to print linked list */ void printList(Node* node)
{ while (node != NULL) {
cout << node->data << " -> " ;
node = node->next;
}
cout << "NULL" ;
} // Function that rotates the given linked list // clockwise by k and returns the updated // head pointer Node* rightRotate(Node* head, int k)
{ // If the linked list is empty
if (!head)
return head;
// len is used to store length of the linked list
// tmp will point to the last node after this loop
Node* tmp = head;
int len = 1;
while (tmp->next != NULL) {
tmp = tmp->next;
len++;
}
// If k is greater than the size
// of the linked list
if (k > len)
k = k % len;
// Subtract from length to convert
// it into left rotation
k = len - k;
// If no rotation needed then
// return the head node
if (k == 0 || k == len)
return head;
// current will either point to
// kth or NULL after this loop
Node* current = head;
int cnt = 1;
while (cnt < k && current != NULL) {
current = current->next;
cnt++;
}
// If current is NULL then k is equal to the
// count of nodes in the list
// Don't change the list in this case
if (current == NULL)
return head;
// current points to the kth node
Node* kthnode = current;
// Change next of last node to previous head
tmp->next = head;
// Change head to (k+1)th node
head = kthnode->next;
// Change next of kth node to NULL
kthnode->next = NULL;
// Return the updated head pointer
return head;
} // Driver code int main()
{ /* The constructed linked list is:
1->2->3->4->5 */
Node* head = NULL;
push(&head, 5);
push(&head, 4);
push(&head, 3);
push(&head, 2);
push(&head, 1);
int k = 2;
// Rotate the linked list
Node* updated_head = rightRotate(head, k);
// Print the rotated linked list
printList(updated_head);
return 0;
} |
// Java implementation of the approach class GFG
{ /* Link list node */ static class Node
{ int data;
Node next;
} /* A utility function to push a node */ static Node 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;
return head_ref;
} /* A utility function to print linked list */ static void printList(Node node)
{ while (node != null )
{
System.out.print(node.data + " -> " );
node = node.next;
}
System.out.print( "null" );
} // Function that rotates the given linked list // clockwise by k and returns the updated // head pointer static Node rightRotate(Node head, int k)
{ // If the linked list is empty
if (head == null )
return head;
// len is used to store length of the linked list
// tmp will point to the last node after this loop
Node tmp = head;
int len = 1 ;
while (tmp.next != null )
{
tmp = tmp.next;
len++;
}
// If k is greater than the size
// of the linked list
if (k > len)
k = k % len;
// Subtract from length to convert
// it into left rotation
k = len - k;
// If no rotation needed then
// return the head node
if (k == 0 || k == len)
return head;
// current will either point to
// kth or null after this loop
Node current = head;
int cnt = 1 ;
while (cnt < k && current != null )
{
current = current.next;
cnt++;
}
// If current is null then k is equal to the
// count of nodes in the list
// Don't change the list in this case
if (current == null )
return head;
// current points to the kth node
Node kthnode = current;
// Change next of last node to previous head
tmp.next = head;
// Change head to (k+1)th node
head = kthnode.next;
// Change next of kth node to null
kthnode.next = null ;
// Return the updated head pointer
return head;
} // Driver code public static void main(String args[])
{ /* The constructed linked list is:
1.2.3.4.5 */
Node head = null ;
head = push(head, 5 );
head = push(head, 4 );
head = push(head, 3 );
head = push(head, 2 );
head = push(head, 1 );
int k = 2 ;
// Rotate the linked list
Node updated_head = rightRotate(head, k);
// Print the rotated linked list
printList(updated_head);
} } // This code is contributed by Arnub Kundu |
# Python3 implementation of the approach ''' Link list node ''' class Node:
def __init__( self , data):
self .data = data
self . next = None
''' A utility function to push a node ''' def push(head_ref, new_data):
''' allocate node '''
new_node = Node(new_data)
''' 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
''' A utility function to print linked list ''' def printList(node):
while (node ! = None ):
print (node.data, end = ' -> ' )
node = node. next
print ( "NULL" )
# Function that rotates the given linked list # clockwise by k and returns the updated # head pointer def rightRotate(head, k):
# If the linked list is empty
if ( not head):
return head
# len is used to store length of the linked list
# tmp will point to the last node after this loop
tmp = head
len = 1
while (tmp. next ! = None ):
tmp = tmp. next
len + = 1
# If k is greater than the size
# of the linked list
if (k > len ):
k = k % len
# Subtract from length to convert
# it into left rotation
k = len - k
# If no rotation needed then
# return the head node
if (k = = 0 or k = = len ):
return head
# current will either point to
# kth or None after this loop
current = head
cnt = 1
while (cnt < k and current ! = None ):
current = current. next
cnt + = 1
# If current is None then k is equal to the
# count of nodes in the list
# Don't change the list in this case
if (current = = None ):
return head
# current points to the kth node
kthnode = current
# Change next of last node to previous head
tmp. next = head
# Change head to (k+1)th node
head = kthnode. next
# Change next of kth node to None
kthnode. next = None
# Return the updated head pointer
return head
# Driver code if __name__ = = '__main__' :
''' The constructed linked list is:
1.2.3.4.5 '''
head = None
head = push(head, 5 )
head = push(head, 4 )
head = push(head, 3 )
head = push(head, 2 )
head = push(head, 1 )
k = 2
# Rotate the linked list
updated_head = rightRotate(head, k)
# Print the rotated linked list
printList(updated_head)
# This code is contributed by rutvik_56
|
// C# implementation of the approach using System;
class GFG
{ /* Link list node */ public class Node
{ public int data;
public Node next;
} /* A utility function to push a node */ static Node 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;
return head_ref;
} /* A utility function to print linked list */ static void printList(Node node)
{ while (node != null )
{
Console.Write(node.data + " -> " );
node = node.next;
}
Console.Write( "null" );
} // Function that rotates the given linked list // clockwise by k and returns the updated // head pointer static Node rightRotate(Node head, int k)
{ // If the linked list is empty
if (head == null )
return head;
// len is used to store length of
// the linked list, tmp will point
// to the last node after this loop
Node tmp = head;
int len = 1;
while (tmp.next != null )
{
tmp = tmp.next;
len++;
}
// If k is greater than the size
// of the linked list
if (k > len)
k = k % len;
// Subtract from length to convert
// it into left rotation
k = len - k;
// If no rotation needed then
// return the head node
if (k == 0 || k == len)
return head;
// current will either point to
// kth or null after this loop
Node current = head;
int cnt = 1;
while (cnt < k && current != null )
{
current = current.next;
cnt++;
}
// If current is null then k is equal
// to the count of nodes in the list
// Don't change the list in this case
if (current == null )
return head;
// current points to the kth node
Node kthnode = current;
// Change next of last node
// to previous head
tmp.next = head;
// Change head to (k+1)th node
head = kthnode.next;
// Change next of kth node to null
kthnode.next = null ;
// Return the updated head pointer
return head;
} // Driver code public static void Main(String []args)
{ /* The constructed linked list is:
1.2.3.4.5 */
Node head = null ;
head = push(head, 5);
head = push(head, 4);
head = push(head, 3);
head = push(head, 2);
head = push(head, 1);
int k = 2;
// Rotate the linked list
Node updated_head = rightRotate(head, k);
// Print the rotated linked list
printList(updated_head);
} } // This code is contributed by PrinciRaj1992 |
<script> // JavaScript implementation of the approach /* Link list node */ class Node {
constructor() {
this .data = 0;
this .next = null ;
}
}
/* A utility function to push a node */
function push(head_ref , 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_ref);
/* move the head to point to the new node */
(head_ref) = new_node;
return head_ref;
}
/* A utility function to print linked list */
function printList(node) {
while (node != null ) {
document.write(node.data + " -> " );
node = node.next;
}
document.write( "null" );
}
// Function that rotates the given linked list
// clockwise by k and returns the updated
// head pointer
function rightRotate(head , k) {
// If the linked list is empty
if (head == null )
return head;
// len is used to store length
// of the linked list
// tmp will point to the last
// node after this loop
var tmp = head;
var len = 1;
while (tmp.next != null ) {
tmp = tmp.next;
len++;
}
// If k is greater than the size
// of the linked list
if (k > len)
k = k % len;
// Subtract from length to convert
// it into left rotation
k = len - k;
// If no rotation needed then
// return the head node
if (k == 0 || k == len)
return head;
// current will either point to
// kth or null after this loop
var current = head;
var cnt = 1;
while (cnt < k && current != null ) {
current = current.next;
cnt++;
}
// If current is null then k is equal to the
// count of nodes in the list
// Don't change the list in this case
if (current == null )
return head;
// current points to the kth node
var kthnode = current;
// Change next of last node to previous head
tmp.next = head;
// Change head to (k+1)th node
head = kthnode.next;
// Change next of kth node to null
kthnode.next = null ;
// Return the updated head pointer
return head;
}
// Driver code
/*
* The constructed linked list is: 1.2.3.4.5
*/
var head = null ;
head = push(head, 5);
head = push(head, 4);
head = push(head, 3);
head = push(head, 2);
head = push(head, 1);
var k = 2;
// Rotate the linked list
var updated_head = rightRotate(head, k);
// Print the rotated linked list
printList(updated_head);
// This code contributed by Rajput-Ji </script> |
4 -> 5 -> 1 -> 2 -> 3 -> NULL
Time Complexity: O(n) where n is the number of nodes in Linked List.
Auxiliary Space: O(1)
STL based approach :
This problem can also be solved using the deque data structure provided in the C++ STL
Approach :
Initialise a deque with the type Node* and push the linked list into it.Then keep popping from it’s back and adding that node to it’s front until the number of operations are not equal to k.
#include <bits/stdc++.h> using namespace std;
class Node {
public :
int val;
Node* next;
Node( int d)
{
val = d;
next = NULL;
}
}; void build(Node*& head, int val)
{ if (head == NULL) {
head = new Node(val);
}
else {
Node* temp = head;
while (temp->next != NULL) {
temp = temp->next;
}
temp->next = new Node(val);
}
} Node* rotate_clockwise(Node* head, int k)
{ if (head == NULL) {
return NULL;
}
deque<Node*> q;
Node* temp = head;
while (temp != NULL) {
q.push_back(temp);
temp = temp->next;
}
k %= q.size();
while (
k--) // popping from back and adding to it's front
{
q.back()->next = q.front();
q.push_front(q.back());
q.pop_back();
q.back()->next = NULL;
}
return q.front();
} void print(Node* head)
{ while (head != NULL) {
cout << head->val << " -> " ;
head = head->next;
}
cout << "NULL" ;
cout << endl;
} int main()
{ Node* head = NULL;
build(head, 1);
build(head, 2);
build(head, 3);
build(head, 4);
build(head, 5);
int k = 2;
Node* r = rotate_clockwise(head, k);
print(r);
return 0;
} |
# Python Program for rotating a Linked List by k from collections import deque
class Node:
def __init__( self , data):
self .data = data
self . next = None
def build(head, val):
if (head = = None ):
head = Node(val)
else :
temp = head
while (temp. next ! = None ):
temp = temp. next
temp. next = Node(val)
return head
def printList(node):
while (node ! = None ):
print (node.data, end = "->" )
node = node. next
print ( "NULL" )
def rotate_clockwise(head, k):
if (head = = None ):
return None
q = deque([])
temp = head
while (temp ! = None ):
q.append(temp)
temp = temp. next
k % = len (q)
while (k ! = 0 ):
q[ - 1 ]. next = q[ 0 ]
q.appendleft(q[ - 1 ])
q.pop()
q[ - 1 ]. next = None
k = k - 1
return q[ 0 ]
head = None
head = build(head, 1 )
head = build(head, 2 )
head = build(head, 3 )
head = build(head, 4 )
head = build(head, 5 )
k = 2
r = rotate_clockwise(head, k)
printList(r) # This code is contributed by Yash Agarwal(yashagarwal2852002) |
// C# program for the above approach using System;
using System.Collections;
using System.Collections.Generic;
using System.Runtime.InteropServices;
// class containing left and right // child of current node and key value public class Node{
public int val;
public Node next;
public Node( int d){
val = d;
next = null ;
}
} public class llist{
Node head;
public void build( int val){
if (head == null ){
head = new Node(val);
}
else {
Node temp = head;
while (temp.next != null ){
temp = temp.next;
}
temp.next = new Node(val);
}
}
public Node rotate_clockwise( int k){
if (head == null ) return null ;
LinkedList<Node> q = new LinkedList<Node>();
Node temp = head;
while (temp != null ){
q.AddLast(temp);
temp = temp.next;
}
k %= q.Count;
while (k > 0) // popping from back and adding to it's front
{
k--;
q.Last.Value.next = q.First.Value;
q.AddFirst(q.Last.Value);
q.RemoveLast();
q.Last.Value.next = null ;
}
return q.First.Value;
}
public void print(Node head){
while (head != null ){
Console.Write(head.val + " -> " );
head = head.next;
}
Console.Write( "NULL" );
Console.Write( "\n" );
}
public static void Main(String[] args){
llist list = new llist();
list.build(1);
list.build(2);
list.build(3);
list.build(4);
list.build(5);
int k = 2;
Node r = list.rotate_clockwise(k);
list.print(r);
}
} // this code is contributed by Kirti Agarwal(kirtiagarwal23121999) |
// JavaScript Program for the above approach class Node{ constructor(d){
this .val = d;
this .next = null ;
}
} function build(head, val){
if (head == null ){
head = new Node(val);
}
else {
let temp = head;
while (temp.next != null ){
temp = temp.next;
}
temp.next = new Node(val);
}
return head;
} function rotate_clockwise(head, k){
if (head == null ) return null ;
let q = [];
let temp = head;
while (temp != null ){
q.push(temp);
temp = temp.next;
}
k %= q.length;
while (k--) // popping from back and adding to it's front
{
q[q.length-1].next = q[0];
q.unshift(q[q.length-1]);
q.pop();
q[q.length-1].next = null ;
}
return q[0];
} function print(head){
while (head != null ){
console.log(head.val + " -> " );
head = head.next;
}
console.log( "NULL" );
} let head = null ;
head = build(head, 1); head = build(head, 2); head = build(head, 3); head = build(head, 4); head = build(head, 5); let k = 2; let r = rotate_clockwise(head, k); print(r); // This code contributed by Yash Agarwal(yashagawal2852002) |
import java.util.*;
class Node {
public int val;
public Node next;
public Node( int d) {
val = d;
next = null ;
}
} class GFG {
public static Node build(Node head, int data) {
Node new_node = new Node(data);
if (head == null ) {
head = new_node;
} else {
Node last = head;
while (last.next != null ) {
last = last.next;
}
last.next = new_node;
}
return head;
}
public static Node rotate_clockwise(Node head, int k) {
if (head == null ) {
return null ;
}
Deque < Node > q = new LinkedList < > ();
Node temp = head;
while (temp != null ) {
q.add(temp);
temp = temp.next;
}
k %= q.size();
while (k-- > 0 ) {
q.peekLast().next = q.peekFirst();
q.offerFirst(q.pollLast());
q.peekLast().next = null ;
}
return q.peekFirst();
}
public static void print(Node head) {
while (head != null ) {
System.out.print(head.val + " -> " );
head = head.next;
}
System.out.print( "NULL" );
System.out.println();
}
public static void main(String[] args) {
Node head = null ;
head = build(head, 1 );
head = build(head, 2 );
head = build(head, 3 );
head = build(head, 4 );
head = build(head, 5 );
int k = 2 ;
Node r = rotate_clockwise(head, k);
print(r);
}
} |
4 -> 5 -> 1 -> 2 -> 3 -> NULL
Time Complexity: O(N)
Auxiliary Space: O(N)