Reverse a Linked List in groups of given size

version.cpp

// CPP program to reverse a linked list 
// in groups of given size 
#include <bits/stdc++.h> 
using namespace std; 

/* Link list node */
class Node 
{ 
	public: 
	int data; 
	Node* next; 
}; 

/* Reverses the linked list in groups 
of size k and returns the pointer 
to the new head node. */
Node *reverse (Node *head, int k) 
{ 
	Node* current = head; 
	Node* next = NULL; 
	Node* prev = NULL; 
	int count = 0; 
	
	/*reverse first k nodes of the linked list */
	while (current != NULL && count < k) 
	{ 
		next = current->next; 
		current->next = prev; 
		prev = current; 
		current = next; 
		count++; 
	} 
	
	/* next is now a pointer to (k+1)th node 
	Recursively call for the list starting from current. 
	And make rest of the list as next of first node */
	if (next != NULL) 
	head->next = reverse(next, k); 

	/* prev is new head of the input list */
	return prev; 
} 

/* UTILITY FUNCTIONS */
/* 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; 
} 

/* Function to print linked list */
void printList(Node *node) 
{ 
	while (node != NULL) 
	{ 
		cout<<node->data<<" "; 
		node = node->next; 
	} 
} 

/* Driver code*/
int main() 
{ 
	/* Start with the empty list */
	Node* head = NULL; 

	/* Created Linked list is 1->2->3->4->5->6->7->8->9 */
	push(&head, 9); 
	push(&head, 8); 
	push(&head, 7); 
	push(&head, 6); 
	push(&head, 5); 
	push(&head, 4); 
	push(&head, 3); 
	push(&head, 2); 
	push(&head, 1);	 

	cout<<"Given linked list \n"; 
	printList(head); 
	head = reverse(head, 3); 

	cout<<"\nReversed Linked list \n"; 
	printList(head); 

	return(0); 
} 

// This code is contributed by rathbhupendra 

version.py

# Python program to reverse a linked list in group of given size 

# Node class 
class Node: 

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

class LinkedList: 

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

	def reverse(self, head, k): 
		current = head 
		next = None
		prev = None
		count = 0
		
		# Reverse first k nodes of the linked list 
		while(current is not None and count < k): 
			next = current.next
			current.next = prev 
			prev = current 
			current = next
			count += 1

		# next is now a pointer to (k+1)th node 
		# recursively call for the list starting 
		# from current. And make rest of the list as 
		# next of first node 
		if next is not None: 
			head.next = self.reverse(next, k) 

		# prev is new head of the input list 
		return prev 

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

	# Utility function to print the linked LinkedList 
	def printList(self): 
		temp = self.head 
		while(temp): 
			print temp.data, 
			temp = temp.next


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

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

print "\nReversed Linked list"
llist.printList() 

Reverse the first sub-list of size k. While reversing keep track of the next node and previous node. Let the pointer to the next node be next and pointer to the previous node be prev. See this post for reversing a linked list.
head->next = reverse(next, k) ( Recursively call for rest of the list and link the two sub-lists )
Return prev ( prev becomes the new head of the list

O(n) where n is the number of nodes in the given list.