Given a linked list, write a function to reverse every alternate k nodes (where k is an input to the function) in an efficient way. Give the complexity of your algorithm.
Example:
Inputs: 1->2->3->4->5->6->7->8->9->NULL and k = 3 Output: 3->2->1->4->5->6->9->8->7->NULL.
Method 1 (Process 2k nodes and recursively call for rest of the list):
This method is basically an extension of the method discussed in this post.
kAltReverse(struct node *head, int k) 1) Reverse first k nodes. 2) In the modified list head points to the kth node. So change next of head to (k+1)th node 3) Move the current pointer to skip next k nodes. 4) Call the kAltReverse() recursively for rest of the n - 2k nodes. 5) Return new head of the list.
# Python3 program to reverse alternate # k nodes in a linked list import math
# Link list node class Node:
def __init__( self , data):
self .data = data
self . next = None
# Reverses alternate k nodes and #returns the pointer to the new # head node def kAltReverse(head, k) :
current = head
next = None
prev = None
count = 0
# 1) reverse first k nodes of the
# linked list
while (current ! = None and
count < k) :
next = current. next
current. next = prev
prev = current
current = next
count = count + 1 ;
# 2) Now head pos to the kth node.
# So change next of head to (k+1)th node
if (head ! = None ):
head. next = current
# 3) We do not want to reverse next k
# nodes. So move the current
# pointer to skip next k nodes
count = 0
while (count < k - 1 and
current ! = None ):
current = current. next
count = count + 1
# 4) Recursively call for the list
# starting from current.next. And make
# rest of the list as next of first node
if (current ! = None ):
current. next =
kAltReverse(current. next , k)
# 5) prev is the new head of the
# input list
return prev
# UTILITY FUNCTIONS # 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 of the
# new node
new_node. next = head_ref
# Move the head to point to the
# new node
head_ref = new_node
return head_ref
# Function to print linked list def printList(node):
count = 0
while (node ! = None ):
print (node.data, end = " " )
node = node. next
count = count + 1
# Driver code if __name__ = = '__main__' :
# Start with the empty list
head = None
# Create a list
# 1.2.3.4.5...... .20
for i in range ( 20 , 0 , - 1 ):
head = push(head, i)
print ( "Given linked list " )
printList(head)
head = kAltReverse(head, 3 )
print ( "Modified Linked list" )
printList(head)
# This code is contributed by Srathore |
Output:
Given linked list 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Modified Linked list 3 2 1 4 5 6 9 8 7 10 11 12 15 14 13 16 17 18 20 19
Time Complexity: O(n)
Auxiliary Space: O(n) due to recursive stack space
Method 2 (Process k nodes and recursively call for rest of the list):
The method 1 reverses the first k node and then moves the pointer to k nodes ahead. So method 1 uses two while loops and processes 2k nodes in one recursive call.
This method processes only k nodes in a recursive call. It uses a third bool parameter b which decides whether to reverse the k elements or simply move the pointer.
_kAltReverse(struct node *head, int k, bool b) 1) If b is true, then reverse first k nodes. 2) If b is false, then move the pointer k nodes ahead. 3) Call the kAltReverse() recursively for rest of the n - k nodes and link rest of the modified list with end of first k nodes. 4) Return new head of the list.
# Python code for above algorithm # Link list node class node:
def __init__( self , data):
self .data = data
self . next = next
# Function to insert a node at the # beginning of the linked list def push(head_ref, new_data):
# Allocate node
new_node = node( 0 )
# Put in the data
new_node.data = new_data
# Link the old list to the
# new node
new_node. next = (head_ref)
# Move the head to point to the
# new node
(head_ref) = new_node
return head_ref
""" Alternatively reverses the given linked list in groups of given
size k. """
def kAltReverse(head, k) :
return _kAltReverse(head, k, True )
""" Helper function for kAltReverse(). It reverses k nodes of the list only
if the third parameter b is passed as
True, otherwise moves the pointer k
nodes ahead and recursively call itself """
def _kAltReverse(Node, k, b) :
if (Node = = None ) :
return None
count = 1
prev = None
current = Node
next = None
""" The loop serves two purposes
1) If b is True,
then it reverses the k nodes
2) If b is false,
then it moves the current pointer """
while (current ! = None and count < = k) :
next = current. next
""" Reverse the nodes only if b
is True """
if (b = = True ) :
current. next = prev
prev = current
current = next
count = count + 1
""" 3) If b is True, then node is the
kth node. So attach rest of the list
after node.
4) After attaching, return the new
head """
if (b = = True ) :
Node. next =
_kAltReverse(current,
k, not b)
return prev
else :
""" If b is not True, then attach
rest of the list after prev.
So attach rest of the list
after prev """
prev. next = _kAltReverse(current,
k, not b)
return Node
""" Function to print linked list """ def printList(node) :
count = 0
while (node ! = None ) :
print ( node.data, end = " " )
node = node. next
count = count + 1
# Driver Code # Start with the empty list head = None
i = 20
# Create a list # 1->2->3->4->5...... ->20 while (i > 0 ):
head = push(head, i)
i = i - 1
print ( "Given linked list " )
printList(head) head = kAltReverse(head, 3 )
print ( "Modified Linked list " )
printList(head) # This code is contributed by Arnab Kundu |
Output:
Given linked list 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Modified Linked list 3 2 1 4 5 6 9 8 7 10 11 12 15 14 13 16 17 18 20 19
Time Complexity: O(n)
Auxiliary Space: O(n) For call stack because it is using recursion
Please refer complete article on Reverse alternate K nodes in a Singly Linked List for more details!