Javascript Program For Deleting A Linked List Node At A Given Position

0 0 2 minutes read

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.

Javascript

<script>
 
// A complete working javascript program to 
// delete a node in a linked list at a 
// given position
 
// head of list
var 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
     */
    var 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
    var 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
    var 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()
{
    var tnode = head;
    while (tnode != null) 
    {
        document.write(tnode.data + " ");
        tnode = tnode.next;
    }
}
 
/*
* 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);
 
document.write("Created Linked list is: <br/>");
printList();
 
// Delete node at position 4
deleteNode(4); 
 
document.write("<br/>Linked List after " +
               "Deletion at position 4: <br/>");
printList();
 
// This code is contributed by todaysgaurav
 
</script>

Output:

Created Linked List: 
 8  2  3  1  7 
Linked List after Deletion at position 4: 
 8  2  3  1

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!

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!