Given a pointer to a node to be deleted, delete the node. Note that we don’t have a pointer to the head node.
A simple solution is to traverse the linked list until you find the node you want to delete. But this solution requires a pointer to the head node, which contradicts the problem statement.
The fast solution is to copy the data from the next node to the node to be deleted and delete the next node. Something like the following.
// Find next node using next pointer
struct Node *temp = node_ptr->next;
// Copy data of next node to this node
node_ptr->data = temp->data;
// Unlink next node
node_ptr->next = temp->next;
// Delete next node
free(temp);
Program:
#include <bits/stdc++.h> using namespace std;
/* Link list node */ class Node {
public :
int data;
Node* next;
}; /* Given a reference (pointer to pointer) to the head of a list and an int, push a new node on the front
of the list. */
void push(Node** head_ref, int new_data)
{ /* allocate node */
Node* new_node = new Node();
/* put in the data */
new_node->data = new_data;
/* link the old list of the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
} void printList(Node* head)
{ Node* temp = head;
while (temp != NULL) {
cout << temp->data << " " ;
temp = temp->next;
}
} void deleteNode(Node* node)
{ Node* prev;
if (node == NULL)
return ;
else {
Node* temp = node->next;
node->data = temp->data;
node->next = temp->next;
temp = NULL;
}
} /* Driver code*/ int main()
{ /* Start with the empty list */
Node* head = NULL;
/* Use push() to construct below list
1->12->1->4->1 */
push(&head, 1);
push(&head, 4);
push(&head, 1);
push(&head, 12);
push(&head, 1);
cout << "Before deleting \n" ;
printList(head);
/* I m deleting the head itself.
You can check for more cases */
deleteNode(head);
cout << "\nAfter deleting \n" ;
printList(head);
return 0;
} // This is code is contributed by rathbhupendra |
#include <assert.h> #include <stdio.h> #include <stdlib.h> /* Link list node */ struct Node {
int data;
struct Node* next;
}; /* Given a reference (pointer to pointer) to the head of a list and an int, push a new node on the front
of the list. */
void push( struct Node** head_ref, int new_data)
{ /* allocate node */
struct Node* new_node = ( struct Node*) malloc ( sizeof ( struct Node));
/* put in the data */
new_node->data = new_data;
/* link the old list of the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
} void printList( struct Node* head)
{ struct Node* temp = head;
while (temp != NULL) {
printf ( "%d " , temp->data);
temp = temp->next;
}
} void deleteNode( struct Node* node)
{ struct Node* prev;
if (node == NULL)
return ;
else {
Node* temp = node->next;
node->data = temp->data;
node->next = temp->next;
temp = NULL;
}
} /* Driver program to test above function*/ int main()
{ /* Start with the empty list */
struct Node* head = NULL;
/* Use push() to construct below list
1->12->1->4->1 */
push(&head, 1);
push(&head, 4);
push(&head, 1);
push(&head, 12);
push(&head, 1);
printf ( "Before deleting \n" );
printList(head);
/* I m deleting the head itself.
You can check for more cases */
deleteNode(head);
printf ( "\nAfter deleting \n" );
printList(head);
getchar ();
return 0;
} |
class LinkedList {
Node head; // head of the list
class Node {
int data;
Node next;
Node( int d)
{
data = d;
next = null ;
}
}
/* Given a reference to the head of a list and an int,
inserts a new Node on the front of the list. */
public void push( int new_data)
{
/* 1. alloc the Node and put the data */
Node new_Node = new Node(new_data);
/* 2. Make next of new Node as head */
new_Node.next = head;
/* 3. Move the head to point to new Node */
head = new_Node;
}
/* 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;
}
}
public void deleteNode(Node Node_ptr)
{
Node temp = Node_ptr.next;
Node_ptr.data = temp.data;
Node_ptr.next = temp.next;
temp = null ;
}
public static void main(String[] args)
{
LinkedList llist = new LinkedList();
/* Use push() to construct below list
1->12->1->4->1 */
llist.push( 1 );
llist.push( 4 );
llist.push( 1 );
llist.push( 12 );
llist.push( 1 );
System.out.println( "Before deleting" );
llist.printList();
/* I m deleting the head itself.
You can check for more cases */
llist.deleteNode(llist.head);
System.out.println( "\nAfter Deleting" );
llist.printList();
}
} // This code is contributed by Rajat Mishra |
# a class to define a node with # data and next pointer class Node():
# constructor to initialize a new node
def __init__( self , val = None ):
self .data = val
self . next = None
# push a node to the front of the list def push(head, val):
# allocate new node
newnode = Node(val)
# link the first node of the old list to the new node
newnode. next = head. next
# make the new node as head of the linked list
head. next = newnode
# function to print the list def print_list(head):
temp = head. next
while (temp ! = None ):
print (temp.data, end = ' ' )
temp = temp. next
print ()
# function to delete the node # the main logic is in this def delete_node(node):
prev = Node()
if (node = = None ):
return
else :
temp = Node()
temp = node. next ;
node.data = temp.data;
node. next = temp. next ;
temp = None ;
if __name__ = = '__main__' :
# allocate an empty header node
# this is a node that simply points to the
# first node in the list
head = Node()
# construct the below linked list
# 1->12->1->4->1
push(head, 1 )
push(head, 4 )
push(head, 1 )
push(head, 12 )
push(head, 1 )
print ( 'list before deleting:' )
print_list(head)
# deleting the first node in the list
delete_node(head. next )
print ( 'list after deleting: ' )
print_list(head)
# This code is contributed by Adith Bharadwaj |
using System;
public class LinkedList
{ Node head; // head of the list
public class Node
{
public int data;
public Node next;
public Node( int d)
{
data = d;
next = null ;
}
}
/* Given a reference to the head of a list and an int,
inserts a new Node on the front of the list. */
public void push( int new_data)
{
/* 1. alloc the Node and put the data */
Node new_Node = new Node(new_data);
/* 2. Make next of new Node as head */
new_Node.next = head;
/* 3. Move the head to point to new Node */
head = new_Node;
}
/* 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;
}
}
public void deleteNode(Node Node_ptr)
{
Node temp = Node_ptr.next;
Node_ptr.data = temp.data;
Node_ptr.next = temp.next;
temp = null ;
}
// Driver code
public static void Main(String[] args)
{
LinkedList llist = new LinkedList();
/* Use push() to construct below list
1->12->1->4->1 */
llist.push(1);
llist.push(4);
llist.push(1);
llist.push(12);
llist.push(1);
Console.WriteLine( "Before deleting" );
llist.printList();
/* I m deleting the head itself.
You can check for more cases */
llist.deleteNode(llist.head);
Console.WriteLine( "\nAfter Deleting" );
llist.printList();
}
} // This code is contributed by 29AjayKumar |
<script> var head; // head of the list
class Node {
constructor(val) {
this .data = val;
this .next = null ;
}
}
/*
* Given a reference to the head of a list and an int, inserts a new Node on the
* front of the
*/
function push(new_data) {
/* 1. alloc the Node and put the data */
var new_Node = new Node(new_data);
/* 2. Make next of new Node as head */
new_Node.next = head;
/* 3. Move the head to point to new Node */
head = new_Node;
}
/*
* 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;
}
}
function deleteNode(Node_ptr) {
var temp = Node_ptr.next;
Node_ptr.data = temp.data;
Node_ptr.next = temp.next;
temp = null ;
}
/*
* Use push() to construct below list 1->12->1->4->1
*/
push(1);
push(4);
push(1);
push(12);
push(1);
document.write( "Before deleting<br/>" );
printList();
/*
* I m deleting the head itself. You can check for more cases
*/
deleteNode(head);
document.write( "<br/>After Deleting <br/>" );
printList();
// This code is contributed by todaysgaurav </script> |
Output:
Before deleting
1 12 1 4 1
After deleting
12 1 4 1
Time Complexity:
- For printing linked list: O(N)
- For inserting node: O(1)
- For deleting node: O(N)
- Auxiliary Space: O(1)
This solution doesn’t work if the node to be deleted is the last node of the list. To make this solution work, we can mark the end node as a dummy node. But the programs/functions that are using this function should also be modified.
Exercise: Try this problem with the doubly linked list.
One line in the function deletenode():
void deleteNode(Node *node)
{ *node = *(node->next);
} |
static void deleteNode(Node node)
{ node = (node.next);
} // This code is contributed by umadevi9616 |
def deleteNode(Node Node):
Node = (Node. next );
# This code is contributed by gauravrajput1 |
void deleteNode(Node *node)
{ *node = *(node->next);
} // This code is contributed by shubhamsingh10 |
function deleteNode(node)
{ node = (node.next);
} // This code is contributed by gauravrajput1 </script> |