Python Program For Reversing A Linked List In Groups Of Given Size- Set 2

Given a linked list, write a function to reverse every k nodes (where k is an input to the function). 
Examples:

Input: 1->2->3->4->5->6->7->8->NULL and k = 3 
Output: 3->2->1->6->5->4->8->7->NULL. 

Input: 1->2->3->4->5->6->7->8->NULL and k = 5
Output: 5->4->3->2->1->8->7->6->NULL.

We have already discussed its solution in below post 
Reverse a Linked List in groups of given size | Set 1
In this post, we have used a stack which will store the nodes of the given linked list. Firstly, push the k elements of the linked list in the stack. Now pop elements one by one and keep track of the previously popped node. Point the next pointer of prev node to top element of stack. Repeat this process, until NULL is reached.
This algorithm uses O(k) extra space.

Python3

# Python3 program to reverse a Linked List
# in groups of given size

# Node class
class Node(object):
    __slots__ = 'data', 'next'

    # Constructor to initialize the
    # node object
    def __init__(self, data = None,
                 next = None):
        self.data = data
        self.next = next

    def __repr__(self):
        return repr(self.data)

class LinkedList(object):

    # Function to initialize head
    def __init__(self):
        self.head = None

    # Utility function to print
    # nodes of LinkedList
    def __repr__(self):
        nodes = []
        curr = self.head
        while curr:
            nodes.append(repr(curr))
            curr = curr.next
        return '[' + ', '.join(nodes) + ']'

    # Function to insert a new node at
    # the beginning
    def prepend(self, data):
        self.head = Node(data = data,
                         next = self.head)

    # Reverses the linked list in groups
    # of size k and returns the pointer
    # to the new head node.
    def reverse(self, k = 1):
        if self.head is None:
            return

        curr = self.head
        prev = None
        new_stack = []
        while curr is not None:
            val = 0

            # Terminate the loop whichever
            # comes first either current == None
            # or value >= k
            while curr is not None and val < k:
                new_stack.append(curr.data)
                curr = curr.next
                val += 1

            # Now pop the elements of stack one
            # by one
            while new_stack:

                # If final list has not been
                # started yet.
                if prev is None:
                    prev = Node(new_stack.pop())
                    self.head = prev
                else:
                    prev.next = Node(new_stack.pop())
                    prev = prev.next

        # Next of last element will point to None.
        prev.next = None
        return self.head

# Driver Code
llist = LinkedList()
llist.prepend(9)
llist.prepend(8)
llist.prepend(7)
llist.prepend(6)
llist.prepend(5)
llist.prepend(4)
llist.prepend(3)
llist.prepend(2)
llist.prepend(1)

print("Given linked list")
print(llist)
llist.head = llist.reverse(3)

print("Reversed Linked list")
print(llist)
# This code is contributed by Sagar Kumar Sinha(sagarsinha7777)

Output: 

Given Linked List
1 2 3 4 5 6 7 8 9 
Reversed list
3 2 1 6 5 4 9 8 7

Please refer complete article on Reverse a Linked List in groups of given size | Set 2 for more details!