Python3 Program for Clockwise rotation of Linked List

Write a Python3 program for a given 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: 

Steps:

• 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.

Deque based approach :

This problem can also be solved using the deque data structure.

Approach:

Initialise a deque 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.

Below is the implementation of the above approach: 

Python3

# Python Program for rotating a Linked List by k

from collections import deque

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None


def build(head, val):
    if(head == None):
        head = Node(val)
    else:
        temp = head
        while(temp.next != None):
            temp = temp.next
        temp.next = Node(val)
    return head

def printList(node):
    while(node != None):
        print(node.data, end = "->")
        node = node.next
    print("NULL")

def rotate_clockwise(head, k):
    if(head == None):
        return None
    q = deque([])
    temp = head
    while(temp != None):
        q.append(temp)
        temp = temp.next

    k %= len(q)
    while(k != 0):
        q[-1].next = q[0]
        q.appendleft(q[-1])
        q.pop()
        q[-1].next = None
        k = k-1

    return q[0]

head = None
head = build(head, 1)
head = build(head, 2)
head = build(head, 3)
head = build(head, 4)
head = build(head, 5)
k = 2
r = rotate_clockwise(head, k)
printList(r)

# This code is contributed by Yash Agarwal(yashagarwal2852002)

Output

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

Time Complexity: O(N)
Auxiliary Space: O(N)

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