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Python Program For Reversing A Linked List In Groups Of Given Size – Set 1

  • Last Updated : 16 Dec, 2021

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

Example: 

Input: 1->2->3->4->5->6->7->8->NULL, K = 3 
Output: 3->2->1->6->5->4->8->7->NULL 
Input: 1->2->3->4->5->6->7->8->NULL, K = 5 
Output: 5->4->3->2->1->8->7->6->NULL 

Algorithm: reverse(head, k) 

  • 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 (see the diagrams of an iterative method of this post )

Below is image shows how the reverse function works: 

Below is the implementation of the above approach:

Python




# 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):      
        if head == None:
          return None
        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 List
    def printList(self):
        temp = self.head
        while(temp):
            print temp.data,
            temp = temp.next
  
# Driver code
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 "Reversed Linked list"
llist.printList()
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

Output: 

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

Complexity Analysis: 

  • Time Complexity: O(n). 
    Traversal of list is done only once and it has ‘n’ elements.
  • Auxiliary Space: O(n/k). 
    For each Linked List of size n, n/k or (n/k)+1 calls will be made during the recursion.

Please refer complete article on Reverse a Linked List in groups of given size | Set 1 for more details!


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