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

Last Updated : 15 Jun, 2022
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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.

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!
 



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