Clockwise rotation of Linked List

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++

// 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

// 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

# 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#

// 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

Javascript

<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>

Output:

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.

C++

#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;
}

Python3

# 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#

// 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

// 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)

Java

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);
    }
}

Output

4 -> 5 -> 1 -> 2 -> 3 -> NULL

Time Complexity: O(N)
Auxiliary Space: O(N)