Skip to content
Related Articles

Related Articles

Java Program for Clockwise rotation of Linked List

View Discussion
Improve Article
Save Article
  • Last Updated : 31 May, 2022
View Discussion
Improve Article
Save Article

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: 
 

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

Output: 

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

 

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(1), as we are not using any extra space.
 

Please refer complete article on Clockwise rotation of Linked List for more details!


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!