Java Program For Removing Every K-th Node Of The Linked List
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
Java
class GFG{
static class Node
{
int data;
Node next;
}
static Node freeList(Node node)
{
while (node != null )
{
Node next = node.next;
node = next;
}
return node;
}
static Node deleteKthNode(Node head,
int k)
{
if (head == null )
return null ;
if (k == 1 )
{
head = freeList(head);
return null ;
}
Node ptr = head, prev = null ;
int count = 0 ;
while (ptr != null )
{
count++;
if (k == count)
{
prev.next = ptr.next;
count = 0 ;
}
if (count != 0 )
prev = ptr;
ptr = prev.next;
}
return head;
}
static void displayList(Node head)
{
Node temp = head;
while (temp != null )
{
System.out.print(temp.data +
" " );
temp = temp.next;
}
}
static Node newNode( int x)
{
Node temp = new Node();
temp.data = x;
temp.next = null ;
return temp;
}
public static void main(String args[])
{
Node 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 );
int 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!
Last Updated :
10 Jan, 2022
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