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

# 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)