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:
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 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 */ 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), 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!
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