Python Program For Reversing Alternate K Nodes In A Singly Linked List
Last Updated :
22 Mar, 2023
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
import math
class Node:
def __init__( self , data):
self .data = data
self . next = None
def kAltReverse(head, k) :
current = head
next = None
prev = None
count = 0
while (current ! = None and
count < k) :
next = current. next
current. next = prev
prev = current
current = next
count = count + 1 ;
if (head ! = None ):
head. next = current
count = 0
while (count < k - 1 and
current ! = None ):
current = current. next
count = count + 1
if (current ! = None ):
current. next =
kAltReverse(current. next , k)
return prev
def push(head_ref, new_data):
new_node = Node(new_data)
new_node. next = head_ref
head_ref = new_node
return head_ref
def printList(node):
count = 0
while (node ! = None ):
print (node.data, end = " " )
node = node. next
count = count + 1
if __name__ = = '__main__' :
head = None
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)
|
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.
Python3
class node:
def __init__( self , data):
self .data = data
self . next = next
def push(head_ref, new_data):
new_node = node( 0 )
new_node.data = new_data
new_node. next = (head_ref)
(head_ref) = new_node
return head_ref
def kAltReverse(head, k) :
return _kAltReverse(head, k, True )
def _kAltReverse(Node, k, b) :
if (Node = = None ) :
return None
count = 1
prev = None
current = Node
next = None
while (current ! = None and count < = k) :
next = current. next
if (b = = True ) :
current. next = prev
prev = current
current = next
count = count + 1
if (b = = True ) :
Node. next =
_kAltReverse(current,
k, not b)
return prev
else :
prev. next = _kAltReverse(current,
k, not b)
return Node
def printList(node) :
count = 0
while (node ! = None ) :
print ( node.data, end = " " )
node = node. next
count = count + 1
head = None
i = 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)
|
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!
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