Delete a Linked List node at a given position

Difficulty Level : Easy
Last Updated : 21 Oct, 2020

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.

C

// A complete working C program to delete a node in a linked list
// at a given position
#include <stdio.h>
#include <stdlib.h>
  
// A linked list node
struct Node
{
    int data;
    struct Node *next;
};
  
/* Given a reference (pointer to pointer) to the head of a list
   and an int inserts a new node on the front of the list. */
void push(struct Node** head_ref, int new_data)
{
    struct Node* new_node = (struct Node*) malloc(sizeof(struct Node));
    new_node->data  = new_data;
    new_node->next = (*head_ref);
    (*head_ref)    = new_node;
}
  
/* Given a reference (pointer to pointer) to the head of a list
   and a position, deletes the node at the given position */
void deleteNode(struct Node **head_ref, int position)
{
   // If linked list is empty
   if (*head_ref == NULL)
      return;
  
   // Store head node
   struct Node* temp = *head_ref;
  
    // If head needs to be removed
    if (position == 0)
    {
        *head_ref = temp->next;   // Change head
        free(temp);               // free old head
        return;
    }
  
    // Find previous node of the node to be deleted
    for (int 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
    struct Node *next = temp->next->next;
  
    // Unlink the node from linked list
    free(temp->next);  // Free memory
  
    temp->next = next;  // Unlink the deleted node from list
}
  
// This function prints contents of linked list starting from
// the given node
void printList(struct Node *node)
{
    while (node != NULL)
    {
        printf(" %d ", node->data);
        node = node->next;
    }
}
  
/* Driver program to test above functions*/
int main()
{
    /* Start with the empty list */
    struct Node* head = NULL;
  
    push(&head, 7);
    push(&head, 1);
    push(&head, 3);
    push(&head, 2);
    push(&head, 8);
  
    puts("Created Linked List: ");
    printList(head);
    deleteNode(&head, 4);
    puts("\nLinked List after Deletion at position 4: ");
    printList(head);
    return 0;
}

Java

// A complete working Java program to delete a node in a linked list
// at a given position
class LinkedList
{
    Node head;  // head of list
  
    /* Linked list Node*/
    class Node
    {
        int data;
        Node next;
        Node(int d)
        {
            data = d;
            next = null;
        }
    }
  
    /* Inserts a new Node at front of the list. */
    public void push(int new_data)
    {
        /* 1 & 2: Allocate the Node &
                  Put in the data*/
        Node 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 */
    void deleteNode(int position)
    {
        // If linked list is empty
        if (head == null)
            return;
  
        // Store head node
        Node temp = head;
  
        // If head needs to be removed
        if (position == 0)
        {
            head = temp.next;   // Change head
            return;
        }
  
        // Find previous node of the node to be deleted
        for (int 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
        Node next = temp.next.next;
  
        temp.next = next;  // Unlink the deleted node from list
    }
  
    /* This function prints contents of linked list starting from
        the given node */
    public void printList()
    {
        Node tnode = head;
        while (tnode != null)
        {
            System.out.print(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 */
    public static void main(String[] args)
    {
        /* Start with the empty list */
        LinkedList llist = new LinkedList();
  
        llist.push(7);
        llist.push(1);
        llist.push(3);
        llist.push(2);
        llist.push(8);
  
        System.out.println("\nCreated Linked list is: ");
        llist.printList();
  
        llist.deleteNode(4);  // Delete node at position 4
  
        System.out.println("\nLinked List after Deletion at position 4: ");
        llist.printList();
    }
}

Python

# Python program to delete a node in a linked list
# at a given position
  
# Node class 
class Node:
  
    # Constructor to initialize the node object
    def __init__(self, data):
        self.data = data
        self.next = None
  
class LinkedList:
  
    # Constructor to initialize head
    def __init__(self):
        self.head = None
  
    # Function to insert a new node at the beginning
    def push(self, new_data):
        new_node = Node(new_data)
        new_node.next = self.head
        self.head = new_node
  
    # Given a reference to the head of a list 
    # and a position, delete the node at a given position
    def deleteNode(self, position):
  
        # If linked list is empty
        if self.head == None:
            return 
  
        # Store head node
        temp = self.head
  
        # If head needs to be removed
        if position == 0:
            self.head = temp.next
            temp = None
            return 
  
        # Find previous node of the node to be deleted
        for i in range(position -1 ):
            temp = temp.next
            if temp is None:
                break
  
        # If position is more than number of nodes
        if temp is None:
            return 
        if temp.next is None:
            return 
  
        # Node temp.next is the node to be deleted
        # store pointer to the next of node to be deleted
        next = temp.next.next
  
        # Unlink the node from linked list
        temp.next = None
  
        temp.next = next 
  
  
    # Utility function to print the linked LinkedList
    def printList(self):
        temp = self.head
        while(temp):
            print " %d " %(temp.data),
            temp = temp.next
  
  
# Driver program to test above function
llist = LinkedList()
llist.push(7)
llist.push(1)
llist.push(3)
llist.push(2)
llist.push(8)
  
print "Created Linked List: "
llist.printList()
llist.deleteNode(4)
print "\nLinked List after Deletion at position 4: "
llist.printList()
  
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

C#

// A complete working C# program to delete 
// a node in a linked list at a given position
using System;
  
class GFG{
      
// Head of list
Node head;  
  
// Linked list Node
public class Node
{
    public int data;
    public Node next;
      
    public Node(int d)
    {
        data = d;
        next = null;
    }
}
  
// Inserts a new Node at front of the list. 
public void Push(int new_data)
{
      
    /* 1 & 2: Allocate the Node &
              Put in the data*/
    Node 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 
void deleteNode(int position)
{
      
    // If linked list is empty
    if (head == null)
        return;
  
    // Store head node
    Node 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(int 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
    Node 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 
public void printList()
{
    Node tnode = head;
    while (tnode != null)
    {
        Console.Write(tnode.data + " ");
        tnode = tnode.next;
    }
}
  
// Driver code
public static void Main(String[] args)
{
      
    // Start with the empty list 
    GFG llist = new GFG();
  
    llist.Push(7);
    llist.Push(1);
    llist.Push(3);
    llist.Push(2);
    llist.Push(8);
  
    Console.WriteLine("\nCreated Linked list is: ");
    llist.printList();
      
    // Delete node at position 4
    llist.deleteNode(4);  
  
    Console.WriteLine("\nLinked List after " + 
                      "Deletion at position 4: ");
    llist.printList();
}
}
  
// This code is contributed by Rajput-Ji

Output:

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

Thanks to Hemanth Kumar for suggesting initial solution. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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