Given a linked list and an integer N, the task is to delete the Nth node from the end of the given linked list.
Examples:
Input: 2 -> 3 -> 1 -> 7 -> NULL, N = 1
Output:
The created linked list is:
2 3 1 7
The linked list after deletion is:
2 3 1
Input: 1 -> 2 -> 3 -> 4 -> NULL, N = 4
Output:
The created linked list is:
1 2 3 4
The linked list after deletion is:
2 3 4
Intuition:
Lets K be the total nodes in the linked list.
Observation : The Nth node from the end is (K-N+1)th node from the beginning.
So the problem simplifies down to that we have to find (K-N+1)th node from the beginning.
- One way of doing it is to find the length (K) of the linked list in one pass and then in the second pass move (K-N+1) step from the beginning to reach the Nth node from the end.
- To do it in one pass. Let’s take the first pointer and move N step from the beginning. Now the first pointer is (K-N+1) steps away from the last node, which is the same number of steps the second pointer require to move from the beginning to reach the Nth node from the end.
// C++ code for the deleting a node from end
// in two traversal
#include <bits/stdc++.h>
using namespace std;
struct Node {
int data;
struct Node* next;
Node(int value)
{
this->data = value;
this->next = NULL;
}
};
int length(Node* head)
{
Node* temp = head;
int count = 0;
while (temp != NULL) {
count++;
temp = temp->next;
}
return count;
}
void printList(Node* head)
{
Node* ptr = head;
while (ptr != NULL) {
cout << ptr->data << " ";
ptr = ptr->next;
}
cout << endl;
}
Node* deleteNthNodeFromEnd(Node* head, int n)
{
int Length = length(head);
int nodeFromBeginning = Length - n + 1;
Node* prev = NULL;
Node* temp = head;
for (int i = 1; i < nodeFromBeginning; i++) {
prev = temp;
temp = temp->next;
}
if (prev == NULL) {
head = head->next;
return head;
}
else {
prev->next = prev->next->next;
return head;
}
}
int main()
{
Node* head = new Node(1);
head->next = new Node(2);
head->next->next = new Node(3);
head->next->next->next = new Node(4);
head->next->next->next->next = new Node(5);
cout<<"Linked List before Deletion:"<<endl;
printList(head);
head = deleteNthNodeFromEnd(head, 4);
cout<<"Linked List after Deletion: "<<endl;
printList(head);
return 0;
}
// This code is contributed by Yash Agarwal(yashagarwal2852002)
class Node {
int data;
Node next;
Node(int value)
{
this.data = value;
this.next = null;
}
}
class Main {
static int length(Node head)
{
Node temp = head;
int count = 0;
while (temp != null) {
count++;
temp = temp.next;
}
return count;
}
static void printList(Node head)
{
Node ptr = head;
while (ptr != null) {
System.out.print(ptr.data + " ");
ptr = ptr.next;
}
System.out.println();
}
static Node deleteNthNodeFromEnd(Node head, int n)
{
int Length = length(head);
int nodeFromBeginning = Length - n + 1;
Node prev = null;
Node temp = head;
for (int i = 1; i < nodeFromBeginning; i++) {
prev = temp;
temp = temp.next;
}
if (prev == null) {
head = head.next;
return head;
}
else {
prev.next = prev.next.next;
return head;
}
}
public static void main(String[] args)
{
Node head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.next.next.next = new Node(4);
head.next.next.next.next = new Node(5);
System.out.println("Linked List before Deletion:");
printList(head);
head = deleteNthNodeFromEnd(head, 4);
System.out.println("Linked List after Deletion:");
printList(head);
}
}
// This code is contributed by user_dtewbxkn77n
# Python code for the deleting a node from end
# in two traversal
class Node:
def __init__(self, value):
self.data = value
self.next = None
def length(head):
temp = head
count = 0
while(temp != None):
count += 1
temp = temp.next
return count
def printList(head):
ptr = head
while(ptr != None):
print (ptr.data, end =" ")
ptr = ptr.next
print()
def deleteNthNodeFromEnd(head, n):
Length = length(head)
nodeFromBeginning = Length - n + 1
prev = None
temp = head
for i in range(1, nodeFromBeginning):
prev = temp
temp = temp.next
if(prev == None):
head = head.next
return head
else:
prev.next = prev.next.next
return head
if __name__ == '__main__':
head = Node(1)
head.next = Node(2)
head.next.next = Node(3)
head.next.next.next = Node(4)
head.next.next.next.next = Node(5)
print("Linked List before Deletion:")
printList(head)
head = deleteNthNodeFromEnd(head, 4)
print("Linked List after Deletion:")
printList(head)
// This C# code defines a class Node which represents a node
// in a linked list. Each node has an integer value 'data'
// and a reference to the next node in the list 'next'. The
// class also defines a static method 'Length' that takes
// the head of a linked list as input and returns the number
// of nodes in the list. Another static method 'PrintList'
// is defined to print the elements of a linked list.
using System;
public class Node {
public int data; // data stored in the node
public Node next; // reference to the next node
// constructor to create a new node with the given data
public Node(int value)
{
this.data = value;
this.next = null;
}
}
public class Program {
// static method to find the length of a linked list
// given its head node
public static int Length(Node head)
{
Node temp = head;
int count = 0;
while (temp != null) // loop until the end of the
// list is reached
{
count++; // increment count for each node
// visited
temp = temp.next; // move to the next node
}
return count;
}
// static method to print the elements of a linked list
public static void PrintList(Node head)
{
Node ptr = head;
while (ptr != null) // loop until the end of the
// list is reached
{
Console.Write(ptr.data
+ " "); // print the data of the
// current node
ptr = ptr.next; // move to the next node
}
Console.WriteLine(); // move to the next line after
// printing the list
}
// static method to delete the nth node from the end of
// a linked list
public static Node DeleteNthNodeFromEnd(Node head,
int n)
{
int Length = Program.Length(
head); // find the length of the list
int nodeFromBeginning
= Length - n
+ 1; // find the index of the node to be
// deleted from the beginning
Node prev = null;
Node temp = head;
for (int i = 1; i < nodeFromBeginning;
i++) // loop until the node before the one to
// be deleted is reached
{
prev = temp;
temp = temp.next;
}
if (prev
== null) // if the first node is to be deleted
{
head = head.next; // update the head node to the
// next node
return head; // return the updated head node
}
else // if any other node is to be deleted
{
prev.next
= prev.next
.next; // skip the node to be deleted
// by updating the reference of
// the previous node
return head; // return the head node
}
}
public static void Main()
{
// create a linked list with 5 nodes
Node head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.next.next.next = new Node(4);
head.next.next.next.next = new Node(5);
// print the linked list before deletion
Console.WriteLine("Linked List before Deletion:");
Program.PrintList(head);
// delete the 4th node from the end of the linked
// list
head = Program.DeleteNthNodeFromEnd(head, 4);
// print the linked list after deletion
Console.WriteLine("Linked List after Deletion:");
Program.PrintList(head);
}
}
class Node {
constructor(value) {
this.data = value;
this.next = null;
}
}
function length(head) {
let temp = head;
let count = 0;
while (temp != null) {
count++;
temp = temp.next;
}
return count;
}
function printList(head) {
let ptr = head;
while (ptr != null) {
process.stdout.write(ptr.data + " ");
ptr = ptr.next;
}
console.log();
}
function deleteNthNodeFromEnd(head, n) {
let Length = length(head);
let nodeFromBeginning = Length - n + 1;
let prev = null;
let temp = head;
for (let i = 1; i < nodeFromBeginning; i++) {
prev = temp;
temp = temp.next;
}
if (prev == null) {
head = head.next;
return head;
} else {
prev.next = prev.next.next;
return head;
}
}
let head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.next.next.next = new Node(4);
head.next.next.next.next = new Node(5);
console.log("Linked List before Deletion:");
printList(head);
head = deleteNthNodeFromEnd(head, 4);
console.log("Linked List after Deletion:");
printList(head);
Output
Linked List before Deletion: 1 2 3 4 5 Linked List after Deletion: 1 3 4 5
Approach:
- Take two pointers; the first will point to the head of the linked list and the second will point to the Nth node from the beginning.
- Now keep incrementing both the pointers by one at the same time until the second is pointing to the last node of the linked list.
- After the operations from the previous step, the first pointer should point to the Nth node from the end now. So, delete the node the first pointer is pointing to.
Below is the implementation of the above approach:
/* C program to merge two sorted linked lists */
#include <assert.h>
#include <stdio.h>
#include <stdlib.h>
/* Link list node */
typedef struct Node {
int data;
struct Node* next;
} Node;
Node* deleteNode(Node* head, int key)
{
// We will be using this pointer for holding address
// temporarily while we delete the node
Node* temp;
// First pointer will point to the head of the linked
// list
Node* first = head;
// Second pointer will point to the Nth node from the
// beginning
Node* second = head;
for (int i = 0; i < key; i++) {
// If count of nodes in the given linked list is <=N
if (second->next == NULL) {
// If count = N i.e. delete the head node
if (i == key - 1) {
temp = head;
head = head->next;
free(temp);
}
return head;
}
second = second->next;
}
// Increment both the pointers by one until
// second pointer reaches the end
while (second->next != NULL) {
first = first->next;
second = second->next;
}
// First must be pointing to the Nth node from the end
// by now So, delete the node first is pointing to
temp = first->next;
first->next = first->next->next;
free(temp);
return head;
}
/* Function to insert a node at the beginning of the
linked list */
void push(Node** head_ref, int new_data)
{
/* allocate node */
Node* new_node = (Node*)malloc(sizeof(Node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* Function to print nodes in a given linked list */
void printList(struct Node* node)
{
while (node != NULL) {
printf("%d ", node->data);
node = node->next;
}
}
// Driver program
int main()
{
struct Node* head = NULL;
push(&head, 7);
push(&head, 1);
push(&head, 3);
push(&head, 2);
printf("Created Linked list is:\n");
printList(head);
int n = 1;
deleteNode(head, n);
printf("\nLinked List after Deletion is:\n");
printList(head);
return 0;
}
// This code is contributed by Sania Kumari Gupta
// Java implementation of the approach
class LinkedList {
// Head of list
Node head;
// Linked list Node
class Node {
int data;
Node next;
Node(int d)
{
data = d;
next = null;
}
}
// Function to delete the nth node from
// the end of the given linked list
void deleteNode(int key)
{
// First pointer will point to
// the head of the linked list
Node first = head;
// Second pointer will point to the
// Nth node from the beginning
Node second = head;
for (int i = 0; i < key; i++) {
// If count of nodes in the given
// linked list is <= N
if (second.next == null) {
// If count = N i.e. delete the head node
if (i == key - 1)
head = head.next;
return;
}
second = second.next;
}
// Increment both the pointers by one until
// second pointer reaches the end
while (second.next != null) {
first = first.next;
second = second.next;
}
// First must be pointing to the
// Nth node from the end by now
// So, delete the node first is pointing to
first.next = first.next.next;
}
// Function to insert a new Node at front of the list
public void push(int new_data)
{
Node new_node = new Node(new_data);
new_node.next = head;
head = new_node;
}
// Function to print the linked list
public void printList()
{
Node tnode = head;
while (tnode != null) {
System.out.print(tnode.data + " ");
tnode = tnode.next;
}
}
// Driver code
public static void main(String[] args)
{
LinkedList llist = new LinkedList();
llist.push(7);
llist.push(1);
llist.push(3);
llist.push(2);
System.out.println("\nCreated Linked list is:");
llist.printList();
int N = 1;
llist.deleteNode(N);
System.out.println("\nLinked List after Deletion is:");
llist.printList();
}
}
# Python3 implementation of the approach
class Node:
def __init__(self, new_data):
self.data = new_data
self.next = None
class LinkedList:
def __init__(self):
self.head = None
# createNode and make linked list
def push(self, new_data):
new_node = Node(new_data)
new_node.next = self.head
self.head = new_node
def deleteNode(self, n):
first = self.head
second = self.head
for i in range(n):
# If count of nodes in the
# given list is less than 'n'
if(second.next == None):
# If index = n then
# delete the head node
if(i == n - 1):
self.head = self.head.next
return self.head
second = second.next
while(second.next != None):
second = second.next
first = first.next
first.next = first.next.next
def printList(self):
tmp_head = self.head
while(tmp_head != None):
print(tmp_head.data, end = ' ')
tmp_head = tmp_head.next
# Driver Code
llist = LinkedList()
llist.push(7)
llist.push(1)
llist.push(3)
llist.push(2)
print("Created Linked list is:")
llist.printList()
llist.deleteNode(1)
print("\nLinked List after Deletion is:")
llist.printList()
# This code is contributed by RaviParkash
// C# implementation of the approach
using System;
public class LinkedList
{
// Head of list
public Node head;
// Linked list Node
public class Node
{
public int data;
public Node next;
public Node(int d)
{
data = d;
next = null;
}
}
// Function to delete the nth node from
// the end of the given linked list
void deleteNode(int key)
{
// First pointer will point to
// the head of the linked list
Node first = head;
// Second pointer will point to the
// Nth node from the beginning
Node second = head;
for (int i = 0; i < key; i++)
{
// If count of nodes in the given
// linked list is <= N
if (second.next == null)
{
// If count = N i.e. delete the head node
if (i == key - 1)
head = head.next;
return;
}
second = second.next;
}
// Increment both the pointers by one until
// second pointer reaches the end
while (second.next != null)
{
first = first.next;
second = second.next;
}
// First must be pointing to the
// Nth node from the end by now
// So, delete the node first is pointing to
first.next = first.next.next;
}
// Function to insert a new Node at front of the list
public void push(int new_data)
{
Node new_node = new Node(new_data);
new_node.next = head;
head = new_node;
}
// Function to print the linked list
public void printList()
{
Node tnode = head;
while (tnode != null)
{
Console.Write(tnode.data + " ");
tnode = tnode.next;
}
}
// Driver code
public static void Main(String[] args)
{
LinkedList llist = new LinkedList();
llist.push(7);
llist.push(1);
llist.push(3);
llist.push(2);
Console.WriteLine("\nCreated Linked list is:");
llist.printList();
int N = 1;
llist.deleteNode(N);
Console.WriteLine("\nLinked List after Deletion is:");
llist.printList();
}
}
// This code is contributed by 29AjayKumar
<script>
// javascript implementation of the approach
// Head of list
var head;
// Linked list Node
class Node {
constructor(val) {
this.data = val;
this.next = null;
}
}
// Function to delete the nth node from
// the end of the given linked list
function deleteNode(key) {
// First pointer will point to
// the head of the linked list
var first = head;
// Second pointer will point to the
// Nth node from the beginning
var second = head;
for (i = 0; i < key; i++) {
// If count of nodes in the given
// linked list is <= N
if (second.next == null) {
// If count = N i.e. delete the head node
if (i == key - 1)
head = head.next;
return;
}
second = second.next;
}
// Increment both the pointers by one until
// second pointer reaches the end
while (second.next != null) {
first = first.next;
second = second.next;
}
// First must be pointing to the
// Nth node from the end by now
// So, delete the node first is pointing to
first.next = first.next.next;
}
// Function to insert a new Node at front of the list
function push(new_data) {
var new_node = new Node(new_data);
new_node.next = head;
head = new_node;
}
// Function to print the linked list
function printList() {
var tnode = head;
while (tnode != null) {
document.write(tnode.data + " ");
tnode = tnode.next;
}
}
// Driver code
push(7);
push(1);
push(3);
push(2);
document.write("\nCreated Linked list is:<br/>");
printList();
var N = 1;
deleteNode(N);
document.write("<br/>Linked List after Deletion is:<br/>");
printList();
// This code is contributed by todaysgaurav
</script>
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
class LinkedList {
public:
// Linked list Node
class Node {
public:
int data;
Node* next;
Node(int d)
{
data = d;
next = NULL;
}
};
// Head of list
Node* head;
// Function to delete the nth node from the end of the
// given linked list
Node* deleteNode(int key)
{
// We will be using this pointer for holding address
// temporarily while we delete the node
Node* temp;
// First pointer will point to the head of the
// linked list
Node* first = head;
// Second pointer will point to the Nth node from
// the beginning
Node* second = head;
for (int i = 0; i < key; i++) {
// If count of nodes in the given linked list is <= N
if (second->next == NULL) {
// If count = N i.e. delete the head node
if (i == key - 1) {
temp = head;
head = head->next;
free(temp);
}
return head;
}
second = second->next;
}
// Increment both the pointers by one until second
// pointer reaches the end
while (second->next != NULL) {
first = first->next;
second = second->next;
}
// First must be pointing to the Nth node from the
// end by now So, delete the node first is pointing to
temp = first->next;
first->next = first->next->next;
free(temp);
return head;
}
// Function to insert a new Node at front of the list
Node* push(int new_data)
{
Node* new_node = new Node(new_data);
new_node->next = head;
head = new_node;
return head;
}
// Function to print the linked list
void printList()
{
Node* tnode = head;
while (tnode != NULL) {
cout << (tnode->data) << (" ");
tnode = tnode->next;
}
}
};
// Driver code
int main()
{
LinkedList* llist = new LinkedList();
llist->head = llist->push(7);
llist->head = llist->push(1);
llist->head = llist->push(3);
llist->head = llist->push(2);
cout << ("Created Linked list is:\n");
llist->printList();
int N = 1;
llist->head = llist->deleteNode(N);
cout << ("\nLinked List after Deletion is:\n");
llist->printList();
}
// This code is contributed by Sania Kumari Gupta
Output
Created Linked list is: 2 3 1 7 Linked List after Deletion is: 2 3 1
Time Complexity: O(N) where N is the number of nodes in the given Linked List.
Auxiliary Space: O(1)
we already covered the iterative version above,
Now let us see its recursive approach as well,
Recursive Approach :
1) Create a dummy node and create a link from dummy node to head node. i.e, dummy->next = head
2) Then we will use the recursion stack to keep track of elements that are being pushed in recursion calls.
3) While popping the elements from recursion stack, we will decrement the N(position of target node from the end of linked list) i.e, N = N-1.
4) When we reach (N==0) that means we have reached at the target node,
5) But here is the catch, to delete the target node we require its previous node,
6) So we will now stop when (N==-1) i.e, we reached the previous node.
7) Now it is very simple to delete the node by using previousNode->next = previousNode->next->next.
// C++ implementation of the approach
// Code is contributed by Paras Saini
#include <bits/stdc++.h>
using namespace std;
class LinkedList {
public:
int val;
LinkedList* next;
LinkedList()
{
this->next = NULL;
this->val = 0;
}
LinkedList(int val)
{
this->next = NULL;
this->val = val;
}
LinkedList* addNode(int val)
{
if (this == NULL) {
return new LinkedList(val);
}
else {
LinkedList* ptr = this;
while (ptr->next) {
ptr = ptr->next;
}
ptr->next = new LinkedList(val);
return this;
}
}
void removeNthNodeFromEndHelper(LinkedList* head,
int& n)
{
if (!head)
return;
// Adding the elements in the recursion
// stack
removeNthNodeFromEndHelper(head->next, n);
// Popping the elements from recursion stack
n -= 1;
// If we reach the previous of target node
if (n == -1){
LinkedList* temp = head->next;
head->next = head->next->next;
free (temp);
}
}
LinkedList* removeNthNodeFromEnd(int n)
{
// return NULL if we have NULL head or only
// one node.
if (!this or !this->next)
return NULL;
// Create a dummy node and point its next to
// head.
LinkedList* dummy = new LinkedList();
dummy->next = this;
// Call function to remove Nth node from end
removeNthNodeFromEndHelper(dummy, n);
// Return new head i.e, dummy->next
return dummy->next;
}
void printLinkedList()
{
if (!this) {
cout << "Empty List\n";
return;
}
LinkedList* ptr = this;
while (ptr) {
cout << ptr->val << " ";
ptr = ptr->next;
}
cout << endl;
}
};
class TestCase {
private:
void printOutput(LinkedList* head)
{
// Output:
if (!head)
cout << "Empty Linked List\n";
else
head->printLinkedList();
}
void testCase1()
{
LinkedList* head = new LinkedList(1);
head = head->addNode(2);
head = head->addNode(3);
head = head->addNode(4);
head = head->addNode(5);
head->printLinkedList(); // Print: 1 2 3 4 5
head = head->removeNthNodeFromEnd(2);
printOutput(head); // Output: 1 2 3 5
}
void testCase2()
{
// Important Edge Case, where linkedList [1]
// and n=1,
LinkedList* head = new LinkedList(1);
head->printLinkedList(); // Print: 1
head = head->removeNthNodeFromEnd(2);
printOutput(head); // Output: Empty Linked List
}
void testCase3()
{
LinkedList* head = new LinkedList(1);
head = head->addNode(2);
head->printLinkedList(); // Print: 1 2
head = head->removeNthNodeFromEnd(1);
printOutput(head); // Output: 1
}
public:
void executeTestCases()
{
testCase1();
testCase2();
testCase3();
}
};
int main()
{
TestCase testCase;
testCase.executeTestCases();
return 0;
}
import java.util.*;
class LinkedList {
public int val;
public LinkedList next;
// Constructor to create the first node
public LinkedList(int val) {
this.val = val;
this.next = null;
}
// Method to add nodes to the linked list
public LinkedList addNode(int val) {
LinkedList node = new LinkedList(val);
if (this.next == null) {
this.next = node;
} else {
this.next.addNode(val);
}
return this;
}
// Method to print the linked list
public void printLinkedList() {
LinkedList node = this;
while (node != null) {
System.out.print(node.val + " ");
node = node.next;
}
System.out.println();
}
// Method to remove the nth node from the end of the linked list
public LinkedList removeNthNodeFromEnd(int n) {
LinkedList dummy = new LinkedList(0);
dummy.next = this;
LinkedList first = dummy;
LinkedList second = dummy;
for (int i = 0; i <= n; i++) {
first = first.next;
}
while (first != null) {
first = first.next;
second = second.next;
}
second.next = second.next.next;
return dummy.next;
}
}
class Main {
public static void main(String[] args) {
LinkedList list = new LinkedList(1);
list.addNode(2).addNode(3).addNode(4).addNode(5);
list.printLinkedList(); // Print: 1 2 3 4 5
list.removeNthNodeFromEnd(2);
list.printLinkedList(); // Output: 1 2 3 5
// Edge case where linked list has only one node
LinkedList list2 = new LinkedList(1);
list2.printLinkedList(); // Print: 1
list2.removeNthNodeFromEnd(1);
if (list2 == null || list2.next == null) {
System.out.println("Empty Linked List");
} else {
list2.printLinkedList();
}
LinkedList list3 = new LinkedList(1);
list3.addNode(2);
list3.printLinkedList(); // Print: 1 2
list3.removeNthNodeFromEnd(1);
list3.printLinkedList(); // Output: 1
}
}
# Python code addition
# class of linked list
class LinkedList:
# Constructor to create the first node
def __init__(self, val):
self.val = val
self.next = None
# Method to add nodes to the linked list
def addNode(self, val):
node = LinkedList(val)
if self.next is None:
self.next = node
else:
self.next.addNode(val)
return self
# Method to print the linked list
def printLinkedList(self):
node = self
while node is not None:
print(node.val, end=" ")
node = node.next
print()
# Method to remove the nth node from the end of the linked list
def removeNthNodeFromEnd(self, n):
dummy = LinkedList(0)
dummy.next = self
first = dummy
second = dummy
for i in range(n+1):
first = first.next
while first is not None:
first = first.next
second = second.next
second.next = second.next.next
return dummy.next
list = LinkedList(1)
list.addNode(2).addNode(3).addNode(4).addNode(5)
list.printLinkedList() # Print: 1 2 3 4 5
list.removeNthNodeFromEnd(2)
list.printLinkedList() # Output: 1 2 3 5
# Edge case where linked list has only one node
list2 = LinkedList(1)
list2.printLinkedList() # Print: 1
list2.removeNthNodeFromEnd(1)
if list2 is None or list2.next is None:
print("Empty Linked List")
else:
list2.printLinkedList()
list3 = LinkedList(1)
list3.addNode(2)
list3.printLinkedList() # Print: 1 2
list3.removeNthNodeFromEnd(1)
list3.printLinkedList() # Output: 1
# The code is contributed by Nidhi goel.
using System;
// Class of linked list
public class LinkedList {
// Constructor to create the first node
public int Val
{
get;
set;
}
public LinkedList Next
{
get;
set;
}
public LinkedList(int val)
{
Val = val;
Next = null;
}
// Method to add nodes to the linked list
public LinkedList AddNode(int val)
{
LinkedList node = new LinkedList(val);
if (Next == null) {
Next = node;
}
else {
Next.AddNode(val);
}
return this;
}
// Method to print the linked list
public void PrintLinkedList()
{
LinkedList node = this;
while (node != null) {
Console.Write(node.Val + " ");
node = node.Next;
}
Console.WriteLine();
}
// Method to remove the nth node from the end of the
// linked list
public LinkedList RemoveNthNodeFromEnd(int n)
{
LinkedList dummy = new LinkedList(0);
dummy.Next = this;
LinkedList first = dummy;
LinkedList second = dummy;
for (int i = 0; i <= n; i++) {
first = first.Next;
}
while (first != null) {
first = first.Next;
second = second.Next;
}
second.Next = second.Next.Next;
return dummy.Next;
}
}
class Program {
static void Main()
{
LinkedList list = new LinkedList(1);
list.AddNode(2).AddNode(3).AddNode(4).AddNode(5);
list.PrintLinkedList(); // Print: 1 2 3 4 5
list.RemoveNthNodeFromEnd(2);
list.PrintLinkedList(); // Output: 1 2 3 5
// Edge case where linked list has only one node
LinkedList list2 = new LinkedList(1);
list2.PrintLinkedList(); // Print: 1
LinkedList result = list2.RemoveNthNodeFromEnd(1);
if (result == null || result.Next == null) {
Console.WriteLine("Empty Linked List");
}
else {
result.PrintLinkedList();
}
LinkedList list3 = new LinkedList(1);
list3.AddNode(2);
list3.PrintLinkedList(); // Print: 1 2
list3.RemoveNthNodeFromEnd(1);
list3.PrintLinkedList(); // Output: 1
}
}
// Javascript code addition
// class of linked list
class LinkedList {
// Constructor to create the first node
constructor(val) {
this.val = val;
this.next = null;
}
// Method to add nodes to the linked list
addNode(val) {
const node = new LinkedList(val);
if (this.next === null) {
this.next = node;
} else {
this.next.addNode(val);
}
return this;
}
// Method to print the linked list
printLinkedList() {
let node = this;
while (node !== null) {
process.stdout.write(node.val + " ");
node = node.next;
}
console.log();
}
// Method to remove the nth node from the end of the linked list
removeNthNodeFromEnd(n) {
const dummy = new LinkedList(0);
dummy.next = this;
let first = dummy;
let second = dummy;
for (let i = 0; i <= n; i++) {
first = first.next;
}
while (first !== null) {
first = first.next;
second = second.next;
}
second.next = second.next.next;
return dummy.next;
}
}
const list = new LinkedList(1);
list.addNode(2).addNode(3).addNode(4).addNode(5);
list.printLinkedList(); // Print: 1 2 3 4 5
list.removeNthNodeFromEnd(2);
list.printLinkedList(); // Output: 1 2 3 5
// Edge case where linked list has only one node
const list2 = new LinkedList(1);
list2.printLinkedList(); // Print: 1
list2.removeNthNodeFromEnd(1);
if (list2 === null || list2.next === null) {
console.log("Empty Linked List");
} else {
list2.printLinkedList();
}
const list3 = new LinkedList(1);
list3.addNode(2);
list3.printLinkedList(); // Print: 1 2
list3.removeNthNodeFromEnd(1);
list3.printLinkedList(); // Output: 1
// The code is contributed by Nidhi goel.
Output
1 2 3 4 5 1 2 3 5 1 Empty Linked List 1 2 1
Two Pointer Approach - Slow and Fast Pointers
This problem can be solved by using two pointer approach as below:
- Take two pointers - fast and slow. And initialize their values as head node
- Iterate fast pointer till the value of n.
- Now, start iteration of fast pointer till the None value of the linked list. Also, iterate slow pointer.
- Hence, once the fast pointer will reach to the end the slow pointer will reach the node which you want to delete.
- Replace the next node of the slow pointer with the next to next node of the slow pointer.
// C++ code for the deleting a node from end using two
// pointer approach
#include <iostream>
using namespace std;
class LinkedList {
public:
// structure of a node
class Node {
public:
int data;
Node* next;
Node(int d)
{
data = d;
next = NULL;
}
};
// Head node
Node* head;
// Function for inserting a node at the beginning
void push(int data)
{
Node* new_node = new Node(data);
new_node->next = head;
head = new_node;
}
// Function to display the nodes in the list.
void display()
{
Node* temp = head;
while (temp != NULL) {
cout << temp->data << endl;
temp = temp->next;
}
}
// Function to delete the nth node from the end.
Node* deleteNthNodeFromEnd(Node* head, int n)
{
if(head==NULL) {
cout<<"Err : List is empty"<<endl;
return head;
}
if(n<=0) {
cout<<"Err : Number should be positive"<<endl;
return head;
}
Node* fast = head;
Node* slow = head;
for (int i = 1; i <=n; i++) {
if(fast == NULL && i!=n-1) {
cout<<"Err : Number is greater than length of List"<<endl;
return head;
}
fast = fast->next;
}
if (fast == NULL) {
return head->next;
}
while (fast->next != NULL) {
fast = fast->next;
slow = slow->next;
}
slow->next = slow->next->next;
return head;
}
};
int main()
{
LinkedList* l = new LinkedList();
// Create a list 1->2->3->4->5->NULL
l->push(5);
l->push(4);
l->push(3);
l->push(2);
l->push(1);
cout << "***** Linked List Before deletion *****"
<< endl;
l->display();
cout << "************** Delete nth Node from the End "
"*****"
<< endl;
l->head = l->deleteNthNodeFromEnd(l->head, 1);
cout << "*********** Linked List after Deletion *****"
<< endl;
l->display();
return 0;
}
// This code is contributed by lokesh (lokeshmvs21).
// Java code for deleting a node from the end using two
// Pointer Approach
class GFG {
class Node {
int data;
Node next;
Node(int data)
{
this.data = data;
this.next = null;
}
}
Node head;
// Function to insert node at the beginning of the list.
public void push(int data)
{
Node new_node = new Node(data);
new_node.next = head;
head = new_node;
}
// Function to print the nodes in the linked list.
public void display()
{
Node temp = head;
while (temp != null) {
System.out.println(temp.data);
temp = temp.next;
}
}
public void deleteNthNodeFromEnd(Node head, int n)
{
Node fast = head;
Node slow = head;
for (int i = 0; i < n; i++) {
fast = fast.next;
}
if (fast == null) {
head = head.next;
return;
}
while (fast.next != null) {
fast = fast.next;
slow = slow.next;
}
slow.next = slow.next.next;
return;
}
public static void main(String[] args)
{
GFG l = new GFG();
// Create a list 1->2->3->4->5->NULL
l.push(5);
l.push(4);
l.push(3);
l.push(2);
l.push(1);
System.out.println(
"***** Linked List Before deletion *****");
l.display();
System.out.println(
"************** Delete nth Node from the End *****");
l.deleteNthNodeFromEnd(l.head, 2);
System.out.println(
"*********** Linked List after Deletion *****");
l.display();
}
}
// This code is contributed by lokesh (lokeshmvs21).
class Node:
def __init__(self, data):
self.data = data
self.next = None
class LinkedList:
def __init__(self):
self.head = None
def push(self, data):
new_node = Node(data)
new_node.next = self.head
self.head = new_node
def display(self):
temp = self.head
while temp != None:
print(temp.data)
temp = temp.next
def deleteNthNodeFromEnd(self, head, n):
fast = head
slow = head
for _ in range(n):
fast = fast.next
if not fast:
return head.next
while fast.next:
fast = fast.next
slow = slow.next
slow.next = slow.next.next
return head
if __name__ == '__main__':
l = LinkedList()
l.push(5)
l.push(4)
l.push(3)
l.push(2)
l.push(1)
print('***** Linked List Before deletion *****')
l.display()
print('************** Delete nth Node from the End *****')
l.deleteNthNodeFromEnd(l.head, 2)
print('*********** Linked List after Deletion *****')
l.display()
// C# code for deleting a node from the end using two
// Pointer Approach
using System;
public class GFG {
public class Node {
public int data;
public Node next;
public Node(int data)
{
this.data = data;
this.next = null;
}
}
Node head;
void push(int data)
{
Node new_node = new Node(data);
new_node.next = head;
head = new_node;
}
void display()
{
Node temp = head;
while (temp != null) {
Console.WriteLine(temp.data);
temp = temp.next;
}
}
public void deleteNthNodeFromEnd(Node head, int n)
{
Node fast = head;
Node slow = head;
for (int i = 0; i < n; i++) {
fast = fast.next;
}
if (fast == null) {
head = head.next;
return;
}
while (fast.next != null) {
fast = fast.next;
slow = slow.next;
}
slow.next = slow.next.next;
return;
}
static public void Main()
{
// Code
GFG l = new GFG();
// Create a list 1->2->3->4->5->NULL
l.push(5);
l.push(4);
l.push(3);
l.push(2);
l.push(1);
Console.WriteLine(
"***** Linked List Before deletion *****");
l.display();
Console.WriteLine(
"************** Delete nth Node from the End *****");
l.deleteNthNodeFromEnd(l.head, 2);
Console.WriteLine(
"*********** Linked List after Deletion *****");
l.display();
}
}
// This code is contributed by lokesh(lokeshmvs21).
<script>
class Node{
constructor(data){
this.data = data
this.next = null
}
}
class LinkedList{
constructor(){
this.head = null
}
push(data){
let new_node = new Node(data)
new_node.next =this.head
this.head = new_node
}
display(){
let temp =this.head
while(temp != null){
document.write(temp.data,"</br>")
temp = temp.next
}
}
deleteNthNodeFromEnd(head, n){
let fast = head
let slow = head
for(let i=0;i<n;i++){
fast = fast.next
}
if(!fast)
return head.next
while(fast.next){
fast = fast.next
slow = slow.next
}
slow.next = slow.next.next
return head
}
}
// driver code
let l = new LinkedList()
l.push(5)
l.push(4)
l.push(3)
l.push(2)
l.push(1)
document.write('***** Linked List Before deletion *****',"</br>")
l.display()
document.write('************** Delete nth Node from the End *****',"</br>")
l.deleteNthNodeFromEnd(l.head, 2)
document.write('*********** Linked List after Deletion *****',"</br>")
l.display()
// This code is contributed by shinjanpatra
</script>
Output
***** Linked List Before deletion ***** 1 2 3 4 5 ************** Delete nth Node from the End ***** *********** Linked List after Deletion ***** 1 2 3 5
Time complexity: O(n)
Space complexity: O(1) using constant space