Javascript Program For Deleting A Linked List Node At A Given Position
Last Updated :
23 Jul, 2025
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Given a singly linked list and a position, delete a linked list node at the given position.
Example:
Input: position = 1, Linked List = 8->2->3->1->7
Output: Linked List = 8->3->1->7
Input: position = 0, Linked List = 8->2->3->1->7
Output: Linked List = 2->3->1->7
If the node to be deleted is the root, simply delete it. To delete a middle node, we must have a pointer to the node previous to the node to be deleted. So if positions are not zero, we run a loop position-1 times and get a pointer to the previous node.
Below is the implementation of the above idea.
// A complete working javascript program to
// delete a node in a linked list at a
// given position
// head of list
let head;
/* Linked list Node */
class Node {
constructor(val) {
this.data = val;
this.next = null;
}
}
/* Inserts a new Node at front of the list. */
function push(new_data) {
/*
* 1 & 2: Allocate the Node & Put in the data
*/
let new_node = new Node(new_data);
/* 3. Make next of new Node as head */
new_node.next = head;
/* 4. Move the head to point to new Node */
head = new_node;
}
/*
* Given a reference (pointer to pointer) to the
* head of a list and a position,
* deletes the node at the given position
*/
function deleteNode(position) {
// If linked list is empty
if (head == null)
return;
// Store head node
let temp = head;
// If head needs to be removed
if (position == 0) {
// Change head
head = temp.next;
return;
}
// Find previous node of the node to be deleted
for (i = 0; temp != null && i < position - 1; i++)
temp = temp.next;
// If position is more than number of nodes
if (temp == null || temp.next == null)
return;
// Node temp->next is the node to be deleted
// Store pointer to the next of node to be deleted
let next = temp.next.next;
// Unlink the deleted node from list
temp.next = next;
}
/*
* This function prints contents of linked
* list starting from the given node
*/
function printList() {
let tnode = head
let output = "";
while (tnode != null) {
output += tnode.data + "->";
tnode = tnode.next;
}
console.log(output.trim() + "NULL");
}
/*
* Driver program to test above functions.
* Ideally this function should be in a
* separate user class. It is kept here
* to keep code compact
*/
/* Start with the empty list */
push(7);
push(1);
push(3);
push(2);
push(8);
console.log("Created Linked list is: ");
printList();
// Delete node at position 4
deleteNode(4);
console.log("Linked List after Deletion at position 4:");
printList();
// This code is contributed by todaysgaurav
Output
Created Linked list is: 8->2->3->1->7->NULL Linked List after Deletion at position 4: 8->2->3->1->NULL
Complexity Analysis:
- Time Complexity: O(n), where n represents the length of the given linked list.
- Auxiliary Space: O(1), no extra space is required, so it is a constant.
Please refer complete article on Delete a Linked List node at a given position for more details!
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