Program for Nth node from the end of a Linked List
Last Updated :
02 Apr, 2024
Given a Linked List and a number N, write a function that returns the value at the Nth node from the end of the Linked List.
Linked-List
Examples:
Input: 1 -> 2 -> 3 -> 4, N = 3
Output: 2
Input: 35 -> 15 -> 4 -> 20, N = 4
Output: 35 Â Â
Naive Approach: Follow the given steps to solve the problem using this approach:Â
- Calculate the length of the Linked List. Let the length be len.Â
- Print the (len – n + 1)th node from the beginning of the Linked List.Â
Below is the implementation of the above approach:
C
// C program to find N'th node from end
#include <stdio.h>
#include <stdlib.h>
/* Link list node */
typedef struct Node {
int data;
struct Node* next;
} Node;
/* Function to get the nth node from the last of a linked
* list*/
void printNthFromLast(Node* head, int N)
{
int len = 0, i;
Node* temp = head;
// Count the number of nodes in Linked List
while (temp != NULL) {
temp = temp->next;
len++;
}
// Check if value of N is not
// more than length of the linked list
if (len < N)
return;
temp = head;
// Get the (len-N+1)th node from the beginning
for (i = 1; i < len - N + 1; i++)
temp = temp->next;
printf("%d", temp->data);
return;
}
void push(struct 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 of the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
// Driver's Code
int main()
{
/* Start with the empty list */
struct Node* head = NULL;
// create linked 35->15->4->20
push(&head, 20);
push(&head, 4);
push(&head, 15);
push(&head, 35);
// Function call
printNthFromLast(head, 4);
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
// Java program to find N'th node from
// end of linked list
import java.io.*;
class LinkedList {
Node head; // head of the list
/* Linked List node */
class Node {
int data;
Node next;
Node(int d)
{
data = d;
next = null;
}
}
/* Function to get the Nth node from the last of a
linked list */
void printNthFromLast(int N)
{
int len = 0;
Node temp = head;
// 1) count the number of nodes in Linked List
while (temp != null) {
temp = temp.next;
len++;
}
// check if value of N is not more than length of
// the linked list
if (len < N)
return;
temp = head;
// 2) get the (len-N+1)th node from the beginning
for (int i = 1; i < len - N + 1; i++)
temp = temp.next;
System.out.println(temp.data);
}
/* 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;
}
// Driver's code
public static void main(String[] args)
{
LinkedList llist = new LinkedList();
llist.push(20);
llist.push(4);
llist.push(15);
llist.push(35);
// Function call
llist.printNthFromLast(4);
}
} // This code is contributed by Rajat Mishra
# Python3 program to find
# N'th node from end
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
# Function to get the nth node from
# the last of a linked list
def printNthFromLast(self, n):
temp = self.head # Used temp variable
length = 0
while temp is not None:
temp = temp.next
length += 1
# Print count
if n > length: # If entered location is greater
# than length of linked list
print('Location is greater than the' +
' length of LinkedList')
return
temp = self.head
for i in range(0, length - n):
temp = temp.next
print(temp.data)
# Driver's Code
if __name__ == "__main__":
llist = LinkedList()
llist.push(20)
llist.push(4)
llist.push(15)
llist.push(35)
# Function call
llist.printNthFromLast(4)
# This code is contributed by Yogesh Joshi
// C# program to find N'th node from end of linked list
using System;
public class LinkedList {
public Node head; // head of the list
/* Linked List node */
public class Node {
public int data;
public Node next;
public Node(int d)
{
data = d;
next = null;
}
}
/* Function to get the nth node from the last of a
linked list */
void printNthFromLast(int N)
{
int len = 0;
Node temp = head;
// 1) count the number of nodes in Linked List
while (temp != null) {
temp = temp.next;
len++;
}
// check if value of N is not more than length of
// the linked list
if (len < N)
return;
temp = head;
// 2) get the (len-N+1)th node from the beginning
for (int i = 1; i < len - N + 1; i++)
temp = temp.next;
Console.WriteLine(temp.data);
}
/* 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;
}
// Driver's code
public static void Main(String[] args)
{
LinkedList llist = new LinkedList();
llist.push(20);
llist.push(4);
llist.push(15);
llist.push(35);
// Function call
llist.printNthFromLast(4);
}
}
// This code is contributed by Rajput-Ji
// Simple Javascript program to find n'th node from end of linked list
/* Linked List node */
class Node {
constructor(d)
{
this.data = d;
this.next = null;
}
}
/* Function to get the nth node from the last of a
linked list */
class LinkedList
{
constructor(d){
this.head = d;
}
printNthFromLast(n)
{
let len = 0;
let temp = this.head;
// 1) count the number of nodes in Linked List
while (temp != null) {
temp = temp.next;
len++;
}
// check if value of n is not more than length of
// the linked list
if (len < n)
return;
temp = this.head;
// 2) get the (len-n+1)th node from the beginning
for (let i = 1; i < len - n + 1; i++)
temp = temp.next;
console.log(temp.data);
}
/* Inserts a new Node at front of the list. */
push(new_data)
{
/* 1 & 2: Allocate the Node &
Put in the data*/
let new_node = new Node(new_data);
/* 3. Make next of new Node as head */
new_node.next = this.head;
/* 4. Move the head to point to new Node */
this.head = new_node;
}
}
/*Driver program to test above methods */
let llist = new LinkedList();
llist.push(20);
llist.push(4);
llist.push(15);
llist.push(35);
llist.printNthFromLast(4);
// This code is contributed by Saurabh Jaiswal
// C++ program to find N'th node from end
#include <bits/stdc++.h>
using namespace std;
/* Link list node */
struct Node {
int data;
struct Node* next;
Node(int x) {
data = x;
next = NULL;
}
};
/* Function to get the Nth node from
the last of a linked list*/
void printNthFromLast(struct Node* head, int N)
{
int len = 0, i;
struct Node* temp = head;
// Count the number of nodes in Linked List
while (temp != NULL) {
temp = temp->next;
len++;
}
// Check if value of N is not
// more than length of the linked list
if (len < N)
return;
temp = head;
// Get the (len-N+1)th node from the beginning
for (i = 1; i < len - N + 1; i++)
temp = temp->next;
cout << temp->data;
return;
}
void push(struct Node** head_ref, int new_data)
{
/* allocate node */
struct Node* new_node = new Node(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;
}
// Driver's Code
int main()
{
/* Start with the empty list */
struct Node* head = NULL;
// create linked 35->15->4->20
push(&head, 20);
push(&head, 4);
push(&head, 15);
push(&head, 35);
// Function call
printNthFromLast(head, 4);
return 0;
}
Time complexity: O(M) where M is the size of the linked list
Auxiliary Space: O(1)
The given code implements a singly linked list in C and provides a function to print the Nth node from the end of the linked list.
Here is the algorithm of the code:
1.Define a struct called “Node” to represent each node of the linked list. It contains an integer data element and a pointer to the next node.
2.Define a function called “printNthFromLast” that takes two arguments: a pointer to the head of the linked list and an integer N representing the position of the node from the end.
3.Inside the “printNthFromLast” function, initialize a variable “len” to 0 and a temporary pointer “temp” to the head of the linked list.
4.Traverse the linked list using the temporary pointer “temp” and increment the “len” variable until “temp” becomes NULL.
5.Check if the value of N is less than or equal to the length of the linked list. If not, return from the function.
6.Reset the temporary pointer “temp” to the head of the linked list.
7.Traverse the linked list again, starting from the head, until the (len-N+1)th node is reached. Store the address of this node in the “temp” pointer.
8.Print the value of the “data” element of the “temp” node.
9.Define a function called “push” that takes two arguments: a double pointer to the head of the linked list and an integer representing the data to be inserted.
10.Inside the “push” function, allocate memory for a new node using the “malloc” function.
11.Set the “data” element of the new node to the input data value and set its “next” pointer to point to the current head of the linked list.
12.Update the head pointer to point to the new node.
13.In the main function, initialize the head pointer to NULL.
14.Insert elements 20, 4, 15, and 35 into the linked list using the “push” function.
15.Call the “printNthFromLast” function with the head pointer and the value 4 as arguments. This should print the value of the 4th node from the end of the linked list, which is 35.
Below is a recursive code for the same method. Thanks to Anuj Bansal for providing the following code.
C++
void printNthFromLast(struct Node* head, int N)
{
static int i = 0;
if (head == NULL)
return;
printNthFromLast(head->next, N);
if (++i == N)
cout<<head->data;
}
void printNthFromLast(struct Node* head, int N)
{
static int i = 0;
if (head == NULL)
return;
printNthFromLast(head->next, N);
if (++i == N)
printf("%d", head->data);
}
import java.io.*;
static void printNthFromLast(Node head, int N)
{
int i = 0;
if (head == null)
return;
printNthFromLast(head.next, N);
if (++i == N)
System.out.print(head.data);
}
// This code is contributed by rutvik_56.
static void printNthFromLast(Node head, int N)
{
static int i = 0;
if (head == null)
return;
printNthFromLast(head.next, N);
if (++i == N)
Console.Write(head.data);
}
// This code is contributed by pratham76.
function printNthFromLast(head , N)
{
function i = 0;
if (head == null)
return;
printNthFromLast(head.next, N);
if (++i == N)
document.write(head.data);
}
// This code is contributed by gauravrajput1
def printNthFromLast(head, N):
i = 0
if (head == None)
return
printNthFromLast(head.next, N);
i += 1
if (i == N):
print(head.data)
# This code is contributed by sunils0ni.
Time Complexity: O(M) where M is the length of the linked list.Â
Auxiliary Space: O(M) for call stack
Nth node from the end of a Linked List using two pointers:
As Nth node from the end equals to (Length – N + 1)th node from the start, so the idea is to Maintain two pointers starting from the head of the Linked-List and move one pointer to the Nth node from the start and then move both the pointers together until the pointer at the Nth position reaches the last node. Now the pointer which was moved later points at the Nth node from the end of the Linked-List
Below image is a dry run of the above approach:
Follow the given steps to solve the problem:
- Maintain two pointers main_ptr and ref_ptr
- Move ref_ptr to the Nth node from the start
- Now move both main_ptr and ref_ptr, until the ref_ptr reaches the last node
- Now print the data of the main_ptr, as it is at the Nth node from the end
Below is the implementation of the above approach:Â
C++
// C++ program to find n-th node
// from the end of the linked list.
#include <bits/stdc++.h>
using namespace std;
struct node {
int data;
node* next;
node(int val)
{
data = val;
next = NULL;
}
};
struct llist {
node* head;
llist() { head = NULL; }
// insert operation at the beginning of the list.
void insertAtBegin(int val)
{
node* newNode = new node(val);
newNode->next = head;
head = newNode;
}
// finding n-th node from the end.
void nthFromEnd(int n)
{
// create two pointers main_ptr and ref_ptr
// initially pointing to head.
node* main_ptr = head;
node* ref_ptr = head;
// if list is empty, return
if (head == NULL) {
cout << "List is empty" << endl;
return;
}
// move ref_ptr to the n-th node from beginning.
for (int i = 1; i < n; i++) {
ref_ptr = ref_ptr->next;
if (ref_ptr == NULL) {
cout << n
<< " is greater than no. of nodes in "
"the list"
<< endl;
return;
}
}
// move ref_ptr and main_ptr by one node until
// ref_ptr reaches end of the list.
while (ref_ptr != NULL && ref_ptr->next != NULL) {
ref_ptr = ref_ptr->next;
main_ptr = main_ptr->next;
}
cout << "Node no. " << n
<< " from end is: " << main_ptr->data << endl;
}
};
// Driver's code
int main()
{
llist ll;
ll.insertAtBegin(20);
ll.insertAtBegin(4);
ll.insertAtBegin(15);
ll.insertAtBegin(35);
// Function call
ll.nthFromEnd(4);
return 0;
}
// This code is contributed by sandeepkrsuman.
// Java program to find N'th
// node from end
import java.io.*;
class LinkedList {
Node head; // head of the list
/* Linked List node */
class Node {
int data;
Node next;
Node(int d)
{
data = d;
next = null;
}
}
/* Function to get the
Nth node from end of list */
void printNthFromLast(int N)
{
Node main_ptr = head;
Node ref_ptr = head;
int count = 0;
if (head != null) {
while (count < N) {
if (ref_ptr == null) {
System.out.println(
N + " is greater than the no "
+ " of nodes in the list");
return;
}
ref_ptr = ref_ptr.next;
count++;
}
if (ref_ptr == null) {
if (head != null)
System.out.println("Node no. " + N
+ " from last is "
+ head.data);
}
else {
while (ref_ptr != null) {
main_ptr = main_ptr.next;
ref_ptr = ref_ptr.next;
}
System.out.println("Node no. " + N
+ " from last is "
+ main_ptr.data);
}
}
}
/* 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;
}
// Driver's code
public static void main(String[] args)
{
LinkedList llist = new LinkedList();
llist.push(20);
llist.push(4);
llist.push(15);
llist.push(35);
// Function call
llist.printNthFromLast(4);
}
}
// This code is contributed by Rajat Mishra
// C# program to find N'th node from end
using System;
public class LinkedList {
Node head; // head of the list
/* Linked List node */
public class Node {
public int data;
public Node next;
public Node(int d)
{
data = d;
next = null;
}
}
/* Function to get the Nth node from end of list */
void printNthFromLast(int N)
{
Node main_ptr = head;
Node ref_ptr = head;
int count = 0;
if (head != null) {
while (count < N) {
if (ref_ptr == null) {
Console.WriteLine(
N + " is greater than the no "
+ " of nodes in the list");
return;
}
ref_ptr = ref_ptr.next;
count++;
}
if (ref_ptr == null) {
head = head.next;
if (head != null)
Console.WriteLine("Node no. " + N
+ " from last is "
+ main_ptr.data);
}
else {
while (ref_ptr != null) {
main_ptr = main_ptr.next;
ref_ptr = ref_ptr.next;
}
Console.WriteLine("Node no. " + N
+ " from last is "
+ main_ptr.data);
}
}
}
/* 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;
}
// Driver's code
public static void Main(String[] args)
{
LinkedList llist = new LinkedList();
llist.push(20);
llist.push(4);
llist.push(15);
llist.push(35);
// Function call
llist.printNthFromLast(4);
}
}
/* This code is contributed by PrinciRaj1992 */
// javascript program to find n'th
// node from end using slow and
// fast pointers
var head; // head of the list
/* Linked List node */
class Node {
constructor(val) {
this.data = val;
this.next = null;
}
}
/*
* Function to get the nth node from end of list
*/
function printNthFromLast(n)
{
var main_ptr = head;
var ref_ptr = head;
var count = 0;
if (head != null) {
while (count < n) {
if (ref_ptr == null) {
document.write(n + " is greater than the no " +
" of nodes in the list");
return;
}
ref_ptr = ref_ptr.next;
count++;
}
if (ref_ptr == null) {
if (head != null)
document.write("Node no. " + n + " from last is " + head.data);
} else {
while (ref_ptr != null) {
main_ptr = main_ptr.next;
ref_ptr = ref_ptr.next;
}
document.write("Node no. " + n + " from last is " + main_ptr.data);
}
}
}
/* 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;
}
/* Driver program to test above methods */
push(20);
push(4);
push(15);
push(35);
printNthFromLast(4);
// This code is contributed by Rajput-Ji
# Python3 program to find N'th node from end
# Node class
class Node:
# Constructor to initialize the node object
def __init__(self, data):
self.data = data
self.next = None
class LinkedList:
# Function 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
def printNthFromLast(self, N):
main_ptr = self.head
ref_ptr = self.head
count = 0
if(self.head is not None):
while(count < N):
if(ref_ptr is None):
print("%d is greater than the no. pf nodes in list" % (N))
return
ref_ptr = ref_ptr.next
count += 1
if(ref_ptr is None):
self.head = self.head.next
if(self.head is not None):
print("Node no. %d from last is % d "
% (N, main_ptr.data))
#For a single Node
if(self.head is None):
print("Node no. %d from last is % d "
% (N, main_ptr.data))
else:
while(ref_ptr is not None):
main_ptr = main_ptr.next
ref_ptr = ref_ptr.next
print("Node no. %d from last is %d"
% (N, main_ptr.data))
# Driver's code
if __name__ == '__main__':
llist = LinkedList()
llist.push(20)
llist.push(4)
llist.push(15)
llist.push(35)
# Function call
llist.printNthFromLast(4)
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)
Output35->15->4->20->NULL
Node no. 4 from end is: 35
Time Complexity: O(M) where M is the length of the linked list.
Auxiliary Space: O(1)
Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.
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