# Python3 Program for 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 = 2Output:4 -> 5 -> 1 -> 2 -> 3 -> NULLInput:7 -> 9 -> 11 -> 13 -> 3 -> 5 -> NULL, K = 12Output: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:

## 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` |

**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!