Javascript Program To Delete Middle Of Linked List
Given a singly linked list, delete the middle of the linked list. For example, if the given linked list is 1->2->3->4->5 then the linked list should be modified to 1->2->4->5
If there are even nodes, then there would be two middle nodes, we need to delete the second middle element. For example, if given linked list is 1->2->3->4->5->6 then it should be modified to 1->2->3->5->6.
If the input linked list is NULL, then it should remain NULL.
If the input linked list has 1 node, then this node should be deleted and a new head should be returned.Â
Efficient solution:Â
Approach: The above solution requires two traversals of the linked list. The middle node can be deleted using one traversal. The idea is to use two pointers, slow_ptr, and fast_ptr. Both pointers start from the head of list. When fast_ptr reaches the end, slow_ptr reaches middle. This idea is same as the one used in method 2 of this post. The additional thing in this post is to keep track of the previous middle so the middle node can be deleted.
Below is the implementation. Â
Javascript
<script>
class Node
{
constructor()
{
this .data = 0;
this .next = null ;
}
}
function deleteMid( head)
{
if (head == null )
return null ;
if (head.next == null )
{
return null ;
}
var slow_ptr = head;
var fast_ptr = head;
var prev = null ;
while (fast_ptr != null &&
fast_ptr.next != null )
{
fast_ptr = fast_ptr.next.next;
prev = slow_ptr;
slow_ptr = slow_ptr.next;
}
prev.next = slow_ptr.next;
return head;
}
function printList(ptr)
{
while (ptr != null )
{
document.write(ptr.data + "->" );
ptr = ptr.next;
}
document.write( "NULL<br/>" );
}
function newNode(data)
{
temp = new Node();
temp.data = data;
temp.next = null ;
return temp;
}
head = newNode(1);
head.next = newNode(2);
head.next.next = newNode(3);
head.next.next.next = newNode(4);
document.write( "Given Linked List<br/>" );
printList(head);
head = deleteMid(head);
document.write(
"Linked List after deletion of middle<br/>" );
printList(head);
</script>
|
Output:
Given Linked List
1->2->3->4->NULL
Linked List after deletion of middle
1->2->4->NULL
Complexity Analysis:Â
- Time Complexity: O(n).Â
Only one traversal of the linked list is needed
- Auxiliary Space: O(1).Â
As no extra space is needed.
Please refer complete article on Delete middle of linked list for more details!
Last Updated :
30 Dec, 2021
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