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
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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 pointerNode* 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 codeint 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 approachclass 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 pointerstatic 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 codepublic 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 pointerdef 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 codeif __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 approachusing 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 pointerstatic 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 codepublic 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 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> |
Output:
4 -> 5 -> 1 -> 2 -> 3 -> NULL
Time Complexity: O(n) where n is the number of nodes in Linked List.
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;} |
Output
4 -> 5 -> 1 -> 2 -> 3 -> NULL
