Python Program For Removing Every K-th Node Of The Linked List
Last Updated :
18 May, 2022
Given a singly linked list, Your task is to remove every K-th node of the linked list. Assume that K is always less than or equal to length of Linked List.
Examples :
Input: 1->2->3->4->5->6->7->8
k = 3
Output: 1->2->4->5->7->8
As 3 is the k-th node after its deletion list
would be 1->2->4->5->6->7->8
And now 4 is the starting node then from it, 6
would be the k-th node. So no other kth node
could be there.So, final list is:
1->2->4->5->7->8.
Input: 1->2->3->4->5->6
k = 1
Output: Empty list
All nodes need to be deleted
The idea is to traverse the list from the beginning and keep track of nodes visited after the last deletion. Whenever count becomes k, delete the current node and reset the count as 0.
Traverse list and do following
(a) Count node before deletion.
(b) If (count == k) that means current
node is to be deleted.
(i) Delete current node i.e. do
// assign address of next node of
// current node to the previous node
// of the current node.
prev->next = ptr->next i.e.
(ii) Reset count as 0, i.e., do count = 0.
(c) Update prev node if count != 0 and if
count is 0 that means that node is a
starting point.
(d) Update ptr and continue until all
k-th node gets deleted.
Below is the implementation.Â
Python3
import math
class Node:
def __init__( self , data):
self .data = data
self . next = None
def freeList(node):
while (node ! = None ):
next = node. next
node = next
return node
def deleteKthNode(head, k):
if (head = = None ):
return None
if (k = = 1 ):
freeList(head)
return None
ptr = head
prev = None
count = 0
while (ptr ! = None ):
count = count + 1
if (k = = count):
prev. next = ptr. next
count = 0
if (count ! = 0 ):
prev = ptr
ptr = prev. next
return head
def displayList(head):
temp = head
while (temp ! = None ):
print (temp.data,
end = ' ' )
temp = temp. next
def newNode( x):
temp = Node(x)
temp.data = x
temp. next = None
return temp
if __name__ = = '__main__' :
head = newNode( 1 )
head. next = newNode( 2 )
head. next . next = newNode( 3 )
head. next . next . next = newNode( 4 )
head. next . next . next . next =
newNode( 5 )
head. next . next . next . next . next =
newNode( 6 )
head. next . next . next . next . next . next =
newNode( 7 )
head. next . next . next . next . next . next . next =
newNode( 8 )
k = 3
head = deleteKthNode(head, k)
displayList(head)
|
Output:Â Â
1 2 4 5 7 8
Time Complexity: O(n)
Please refer complete article on Remove every k-th node of the linked list for more details!
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