Skip to content
Related Articles

Related Articles

Clockwise rotation of Linked List
  • Difficulty Level : Medium
  • Last Updated : 23 Dec, 2020

Given a singly 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 off 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 off 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 off 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 off 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
Output: 
4 -> 5 -> 1 -> 2 -> 3 -> NULL

 

Time Complexity: O(n) where n is the number of nodes in Linked List.
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

My Personal Notes arrow_drop_up
Recommended Articles
Page :