Print the last k nodes of the linked list in reverse order | Iterative Approaches
Last Updated :
11 Jul, 2025
Given a linked list containing N nodes and a positive integer K where K should be less than or equal to N. The task is to print the last K nodes of the list in reverse order.
Examples:
Input : list: 1->2->3->4->5, K = 2
Output : 5 4
Input : list: 3->10->6->9->12->2->8, K = 4
Output : 8 2 12 9
The solution discussed in previous post uses recursive approach. The following article discusses three iterative approaches to solve the above problem.
Approach 1: The idea is to use stack data structure. Push all the linked list nodes data value to stack and pop first K elements and print them.
Below is the implementation of above approach:
C++
// C++ implementation to print the last k nodes
// of linked list in reverse order
#include <bits/stdc++.h>
using namespace std;
// Structure of a node
struct Node {
int data;
Node* next;
};
// Function to get a new node
Node* getNode(int data)
{
// allocate space
Node* newNode = new Node;
// put in data
newNode->data = data;
newNode->next = NULL;
return newNode;
}
// Function to print the last k nodes
// of linked list in reverse order
void printLastKRev(Node* head, int k)
{
// if list is empty
if (!head)
return;
// Stack to store data value of nodes.
stack<int> st;
// Push data value of nodes to stack
while (head) {
st.push(head->data);
head = head->next;
}
int cnt = 0;
// Pop first k elements of stack and
// print them.
while (cnt < k) {
cout << st.top() << " ";
st.pop();
cnt++;
}
}
// Driver code
int main()
{
// Create list: 1->2->3->4->5
Node* head = getNode(1);
head->next = getNode(2);
head->next->next = getNode(3);
head->next->next->next = getNode(4);
head->next->next->next->next = getNode(5);
int k = 4;
// print the last k nodes
printLastKRev(head, k);
return 0;
}
// Java implementation to print the last k nodes
// of linked list in reverse order
import java.util.*;
class GFG
{
// Structure of a node
static class Node
{
int data;
Node next;
};
// Function to get a new node
static Node getNode(int data)
{
// allocate space
Node newNode = new Node();
// put in data
newNode.data = data;
newNode.next = null;
return newNode;
}
// Function to print the last k nodes
// of linked list in reverse order
static void printLastKRev(Node head, int k)
{
// if list is empty
if (head == null)
return;
// Stack to store data value of nodes.
Stack<Integer> st = new Stack<Integer>();
// Push data value of nodes to stack
while (head != null)
{
st.push(head.data);
head = head.next;
}
int cnt = 0;
// Pop first k elements of stack and
// print them.
while (cnt < k)
{
System.out.print(st.peek() + " ");
st.pop();
cnt++;
}
}
// Driver code
public static void main(String[] args)
{
// Create list: 1->2->3->4->5
Node head = getNode(1);
head.next = getNode(2);
head.next.next = getNode(3);
head.next.next.next = getNode(4);
head.next.next.next.next = getNode(5);
int k = 4;
// print the last k nodes
printLastKRev(head, k);
}
}
// This code is contributed by PrinciRaj1992
# Python3 implementation to print the last k nodes
# of linked list in reverse order
import sys
import math
# Structure of a node
class Node:
def __init__(self,data):
self.data = data
self.next = None
# Function to get a new node
def getNode(data):
# allocate space and return new node
return Node(data)
# Function to print the last k nodes
# of linked list in reverse order
def printLastKRev(head,k):
# if list is empty
if not head:
return
# Stack to store data value of nodes.
stack = []
# Push data value of nodes to stack
while(head):
stack.append(head.data)
head = head.next
cnt = 0
# Pop first k elements of stack and
# print them.
while(cnt < k):
print("{} ".format(stack[-1]),end="")
stack.pop()
cnt += 1
# Driver code
if __name__=='__main__':
# Create list: 1->2->3->4->5
head = getNode(1)
head.next = getNode(2)
head.next.next = getNode(3)
head.next.next.next = getNode(4)
head.next.next.next.next = getNode(5)
k = 4
# print the last k nodes
printLastKRev(head,k)
# This Code is Contributed by Vikash Kumar 37
// C# implementation to print the last k nodes
// of linked list in reverse order
using System;
using System.Collections.Generic;
class GFG
{
// Structure of a node
public class Node
{
public int data;
public Node next;
};
// Function to get a new node
static Node getNode(int data)
{
// allocate space
Node newNode = new Node();
// put in data
newNode.data = data;
newNode.next = null;
return newNode;
}
// Function to print the last k nodes
// of linked list in reverse order
static void printLastKRev(Node head, int k)
{
// if list is empty
if (head == null)
return;
// Stack to store data value of nodes.
Stack<int> st = new Stack<int>();
// Push data value of nodes to stack
while (head != null)
{
st.Push(head.data);
head = head.next;
}
int cnt = 0;
// Pop first k elements of stack and
// print them.
while (cnt < k)
{
Console.Write(st.Peek() + " ");
st.Pop();
cnt++;
}
}
// Driver code
public static void Main(String[] args)
{
// Create list: 1->2->3->4->5
Node head = getNode(1);
head.next = getNode(2);
head.next.next = getNode(3);
head.next.next.next = getNode(4);
head.next.next.next.next = getNode(5);
int k = 4;
// print the last k nodes
printLastKRev(head, k);
}
}
// This code contributed by Rajput-Ji
<script>
// JavaScript implementation to print the last k nodes
// of linked list in reverse order
// Structure of a node
class Node {
constructor(val) {
this.data = val;
this.next = null;
}
}
// Function to get a new node
function getNode(data) {
// allocate space
var newNode = new Node();
// put in data
newNode.data = data;
newNode.next = null;
return newNode;
}
// Function to print the last k nodes
// of linked list in reverse order
function printLastKRev(head , k) {
// if list is empty
if (head == null)
return;
// Stack to store data value of nodes.
var st = [];
// Push data value of nodes to stack
while (head != null) {
st.push(head.data);
head = head.next;
}
var cnt = 0;
// Pop first k elements of stack and
// print them.
while (cnt < k) {
document.write(st.pop() + " ");
cnt++;
}
}
// Driver code
// Create list: 1->2->3->4->5
var head = getNode(1);
head.next = getNode(2);
head.next.next = getNode(3);
head.next.next.next = getNode(4);
head.next.next.next.next = getNode(5);
var k = 4;
// print the last k nodes
printLastKRev(head, k);
// This code contributed by aashish1995
</script>
Time Complexity: O(N)
Auxiliary Space: O(N)
The auxiliary space of the above approach can be reduced to O(k). The idea is to use two pointers. Place first pointer to beginning of the list and move second pointer to k-th node form beginning. Then find k-th node from end using approach discussed in this article: Find kth node from end of linked list. After finding kth node from end push all the remaining nodes in the stack. Pop all elements one by one from stack and print them.
Below is the implementation of the above efficient approach:
C++
// C++ implementation to print the last k nodes
// of linked list in reverse order
#include <bits/stdc++.h>
using namespace std;
// Structure of a node
struct Node {
int data;
Node* next;
};
// Function to get a new node
Node* getNode(int data)
{
// allocate space
Node* newNode = new Node;
// put in data
newNode->data = data;
newNode->next = NULL;
return newNode;
}
// Function to print the last k nodes
// of linked list in reverse order
void printLastKRev(Node* head, int k)
{
// if list is empty
if (!head)
return;
// Stack to store data value of nodes.
stack<int> st;
// Declare two pointers.
Node *first = head, *sec = head;
int cnt = 0;
// Move second pointer to kth node.
while (cnt < k) {
sec = sec->next;
cnt++;
}
// Move first pointer to kth node from end
while (sec) {
first = first->next;
sec = sec->next;
}
// Push last k nodes in stack
while (first) {
st.push(first->data);
first = first->next;
}
// Last k nodes are reversed when pushed
// in stack. Pop all k elements of stack
// and print them.
while (!st.empty()) {
cout << st.top() << " ";
st.pop();
}
}
// Driver code
int main()
{
// Create list: 1->2->3->4->5
Node* head = getNode(1);
head->next = getNode(2);
head->next->next = getNode(3);
head->next->next->next = getNode(4);
head->next->next->next->next = getNode(5);
int k = 4;
// print the last k nodes
printLastKRev(head, k);
return 0;
}
// Java implementation to print
// the last k nodes of linked list
// in reverse order
import java.util.*;
class GFG
{
// Structure of a node
static class Node
{
int data;
Node next;
};
// Function to get a new node
static Node getNode(int data)
{
// allocate space
Node newNode = new Node();
// put in data
newNode.data = data;
newNode.next = null;
return newNode;
}
// Function to print the last k nodes
// of linked list in reverse order
static void printLastKRev(Node head, int k)
{
// if list is empty
if (head == null)
return;
// Stack to store data value of nodes.
Stack<Integer> st = new Stack<Integer>();
// Declare two pointers.
Node first = head, sec = head;
int cnt = 0;
// Move second pointer to kth node.
while (cnt < k)
{
sec = sec.next;
cnt++;
}
// Move first pointer to kth node from end
while (sec != null)
{
first = first.next;
sec = sec.next;
}
// Push last k nodes in stack
while (first != null)
{
st.push(first.data);
first = first.next;
}
// Last k nodes are reversed when pushed
// in stack. Pop all k elements of stack
// and print them.
while (!st.empty())
{
System.out.print(st.peek() + " ");
st.pop();
}
}
// Driver code
public static void main(String[] args)
{
// Create list: 1->2->3->4->5
Node head = getNode(1);
head.next = getNode(2);
head.next.next = getNode(3);
head.next.next.next = getNode(4);
head.next.next.next.next = getNode(5);
int k = 4;
// print the last k nodes
printLastKRev(head, k);
}
}
// This code is contributed by Princi Singh
# Python3 implementation to print the last k nodes
# of linked list in reverse order
# Node class
class Node:
# Function to initialise the node object
def __init__(self, data):
self.data = data # Assign data
self.next = None
# Function to get a new node
def getNode(data):
# allocate space
newNode = Node(0)
# put in data
newNode.data = data
newNode.next = None
return newNode
# Function to print the last k nodes
# of linked list in reverse order
def printLastKRev( head, k):
# if list is empty
if (head == None):
return
# Stack to store data value of nodes.
st = []
# Declare two pointers.
first = head
sec = head
cnt = 0
# Move second pointer to kth node.
while (cnt < k) :
sec = sec.next
cnt = cnt + 1
# Move first pointer to kth node from end
while (sec != None):
first = first.next
sec = sec.next
# Push last k nodes in stack
while (first != None):
st.append(first.data)
first = first.next
# Last k nodes are reversed when pushed
# in stack. Pop all k elements of stack
# and print them.
while (len(st)):
print( st[-1], end= " ")
st.pop()
# Driver code
# Create list: 1.2.3.4.5
head = getNode(1)
head.next = getNode(2)
head.next.next = getNode(3)
head.next.next.next = getNode(4)
head.next.next.next.next = getNode(5)
k = 4
# print the last k nodes
printLastKRev(head, k)
# This code is contributed by Arnab Kundu
// C# implementation to print
// the last k nodes of linked list
// in reverse order
using System;
using System.Collections.Generic;
class GFG
{
// Structure of a node
class Node
{
public int data;
public Node next;
};
// Function to get a new node
static Node getNode(int data)
{
// allocate space
Node newNode = new Node();
// put in data
newNode.data = data;
newNode.next = null;
return newNode;
}
// Function to print the last k nodes
// of linked list in reverse order
static void printLastKRev(Node head, int k)
{
// if list is empty
if (head == null)
return;
// Stack to store data value of nodes.
Stack<int> st = new Stack<int>();
// Declare two pointers.
Node first = head, sec = head;
int cnt = 0;
// Move second pointer to kth node.
while (cnt < k)
{
sec = sec.next;
cnt++;
}
// Move first pointer to kth node from end
while (sec != null)
{
first = first.next;
sec = sec.next;
}
// Push last k nodes in stack
while (first != null)
{
st.Push(first.data);
first = first.next;
}
// Last k nodes are reversed when pushed
// in stack. Pop all k elements of stack
// and print them.
while (st.Count != 0)
{
Console.Write(st.Peek() + " ");
st.Pop();
}
}
// Driver code
public static void Main(String[] args)
{
// Create list: 1->2->3->4->5
Node head = getNode(1);
head.next = getNode(2);
head.next.next = getNode(3);
head.next.next.next = getNode(4);
head.next.next.next.next = getNode(5);
int k = 4;
// print the last k nodes
printLastKRev(head, k);
}
}
// This code is contributed by 29AjayKumar
<script>
// JavaScript implementation to print
// the last k nodes of linked list
// in reverse order
// Structure of a node
class Node {
constructor() {
this.data = 0;
this.next = null;
}
}
// Function to get a new node
function getNode(data) {
// allocate space
var newNode = new Node();
// put in data
newNode.data = data;
newNode.next = null;
return newNode;
}
// Function to print the last k nodes
// of linked list in reverse order
function printLastKRev(head, k) {
// if list is empty
if (head == null) return;
// Stack to store data value of nodes.
var st = [];
// Declare two pointers.
var first = head,
sec = head;
var cnt = 0;
// Move second pointer to kth node.
while (cnt < k) {
sec = sec.next;
cnt++;
}
// Move first pointer to kth node from end
while (sec != null) {
first = first.next;
sec = sec.next;
}
// Push last k nodes in stack
while (first != null) {
st.push(first.data);
first = first.next;
}
// Last k nodes are reversed when pushed
// in stack. Pop all k elements of stack
// and print them.
while (st.length != 0) {
document.write(st[st.length - 1] + " ");
st.pop();
}
}
// Driver code
// Create list: 1->2->3->4->5
var head = getNode(1);
head.next = getNode(2);
head.next.next = getNode(3);
head.next.next.next = getNode(4);
head.next.next.next.next = getNode(5);
var k = 4;
// print the last k nodes
printLastKRev(head, k);
</script>
Time Complexity: O(N)
Auxiliary Space: O(k)
Approach-2:
- Count the number of nodes in the linked list.
- Declare an array with the number of nodes as its size.
- Start storing the value of nodes of the linked list from the end of the array i.e. reverse manner.
- Print k values from starting of the array.
C++
#include <iostream>
using namespace std;
// Structure of a node
struct Node {
int data;
Node* next;
};
// Function to get a new node
Node* getNode(int data){
// allocate space
Node* newNode = new Node;
// put in data
newNode->data = data;
newNode->next = NULL;
return newNode;
}
// Function to print the last k nodes
// of linked list in reverse order
void printLastKRev(Node* head,
int& count, int k) {
struct Node* cur = head;
while(cur != NULL){
count++;
cur = cur->next;
}
int arr[count], temp = count;
cur = head;
while(cur != NULL){
arr[--temp] = cur->data;
cur = cur->next;
}
for(int i = 0; i < k; i++)
cout << arr[i] << " ";
}
//
// Driver code
int main()
{
// Create list: 1->2->3->4->5
Node* head = getNode(1);
head->next = getNode(2);
head->next->next = getNode(3);
head->next->next->next = getNode(4);
head->next->next->next->next = getNode(5);
head->next->next->next->next->next = getNode(10);
int k = 4, count = 0;
// print the last k nodes
printLastKRev(head, count, k);
return 0;
}
// Java code implementation for above approach
class GFG
{
// Structure of a node
static class Node
{
int data;
Node next;
};
// Function to get a new node
static Node getNode(int data)
{
// allocate space
Node newNode = new Node();
// put in data
newNode.data = data;
newNode.next = null;
return newNode;
}
// Function to print the last k nodes
// of linked list in reverse order
static void printLastKRev(Node head,
int count, int k)
{
Node cur = head;
while(cur != null)
{
count++;
cur = cur.next;
}
int []arr = new int[count];
int temp = count;
cur = head;
while(cur != null)
{
arr[--temp] = cur.data;
cur = cur.next;
}
for(int i = 0; i < k; i++)
System.out.print(arr[i] + " ");
}
// Driver code
public static void main(String[] args)
{
// Create list: 1.2.3.4.5
Node head = getNode(1);
head.next = getNode(2);
head.next.next = getNode(3);
head.next.next.next = getNode(4);
head.next.next.next.next = getNode(5);
head.next.next.next.next.next = getNode(10);
int k = 4, count = 0;
// print the last k nodes
printLastKRev(head, count, k);
}
}
// This code is contributed by 29AjayKumar
# Python3 code implementation for above approach
# Structure of a node
class Node:
def __init__(self, data):
self.data = data
self.next = None
# Function to get a new node
def getNode(data):
# allocate space
newNode = Node(data)
return newNode
# Function to print the last k nodes
# of linked list in reverse order
def printLastKRev(head, count,k):
cur = head;
while(cur != None):
count += 1
cur = cur.next;
arr = [0 for i in range(count)]
temp = count;
cur = head;
while(cur != None):
temp -= 1
arr[temp] = cur.data;
cur = cur.next;
for i in range(k):
print(arr[i], end = ' ')
# Driver code
if __name__=='__main__':
# Create list: 1.2.3.4.5
head = getNode(1);
head.next = getNode(2);
head.next.next = getNode(3);
head.next.next.next = getNode(4);
head.next.next.next.next = getNode(5);
head.next.next.next.next.next = getNode(10);
k = 4
count = 0;
# print the last k nodes
printLastKRev(head, count, k);
# This code is contributed by rutvik_56
// C# code implementation for above approach
using System;
class GFG
{
// Structure of a node
class Node
{
public int data;
public Node next;
};
// Function to get a new node
static Node getNode(int data)
{
// allocate space
Node newNode = new Node();
// put in data
newNode.data = data;
newNode.next = null;
return newNode;
}
// Function to print the last k nodes
// of linked list in reverse order
static void printLastKRev(Node head,
int count, int k)
{
Node cur = head;
while(cur != null)
{
count++;
cur = cur.next;
}
int []arr = new int[count];
int temp = count;
cur = head;
while(cur != null)
{
arr[--temp] = cur.data;
cur = cur.next;
}
for(int i = 0; i < k; i++)
Console.Write(arr[i] + " ");
}
// Driver code
public static void Main(String[] args)
{
// Create list: 1.2.3.4.5
Node head = getNode(1);
head.next = getNode(2);
head.next.next = getNode(3);
head.next.next.next = getNode(4);
head.next.next.next.next = getNode(5);
head.next.next.next.next.next = getNode(10);
int k = 4, count = 0;
// print the last k nodes
printLastKRev(head, count, k);
}
}
// This code is contributed by PrinciRaj1992
<script>
// Javascript implementation of the approach
// Structure of a node
class Node {
constructor() {
this.data = 0;
this.next = null;
}
}
// Function to get a new node
function getNode( data)
{
// allocate space
var newNode = new Node();
// put in data
newNode.data = data;
newNode.next = null;
return newNode;
}
// Function to print the last k nodes
// of linked list in reverse order
function printLastKRev( head, count, k)
{
var cur = head;
while(cur != null)
{
count++;
cur = cur.next;
}
let arr = new Array(count);
let temp = count;
cur = head;
while(cur != null)
{
arr[--temp] = cur.data;
cur = cur.next;
}
for(let i = 0; i < k; i++)
document.write(arr[i] + " ");
}
// Driver Code
// Create list: 1.2.3.4.5
var head = getNode(1);
head.next = getNode(2);
head.next.next = getNode(3);
head.next.next.next = getNode(4);
head.next.next.next.next = getNode(5);
head.next.next.next.next.next = getNode(10);
let k = 4, count = 0;
// print the last k nodes
printLastKRev(head, count, k);
// This code is contributed b jana_sayantan
</script>
Time Complexity: O(N)
Auxiliary Space: O(N)
Approach-3: The idea is to first reverse the linked list iteratively as discussed in following post: Reverse a linked list. After reversing print first k nodes of the reversed list. After printing restore the list by reversing the list again.
Below is the implementation of above approach:
C++
// C++ implementation to print the last k nodes
// of linked list in reverse order
#include <bits/stdc++.h>
using namespace std;
// Structure of a node
struct Node {
int data;
Node* next;
};
// Function to get a new node
Node* getNode(int data)
{
// allocate space
Node* newNode = new Node;
// put in data
newNode->data = data;
newNode->next = NULL;
return newNode;
}
// Function to reverse the linked list.
Node* reverseLL(Node* head)
{
if (!head || !head->next)
return head;
Node *prev = NULL, *next = NULL, *curr = head;
while (curr) {
next = curr->next;
curr->next = prev;
prev = curr;
curr = next;
}
return prev;
}
// Function to print the last k nodes
// of linked list in reverse order
void printLastKRev(Node* head, int k)
{
// if list is empty
if (!head)
return;
// Reverse linked list.
head = reverseLL(head);
Node* curr = head;
int cnt = 0;
// Print first k nodes of linked list.
while (cnt < k) {
cout << curr->data << " ";
cnt++;
curr = curr->next;
}
// Restore the list.
head = reverseLL(head);
}
// Driver code
int main()
{
// Create list: 1->2->3->4->5
Node* head = getNode(1);
head->next = getNode(2);
head->next->next = getNode(3);
head->next->next->next = getNode(4);
head->next->next->next->next = getNode(5);
int k = 4;
// print the last k nodes
printLastKRev(head, k);
return 0;
}
// Java implementation to print the last k nodes
// of linked list in reverse order
import java.util.*;
class GFG
{
// Structure of a node
static class Node
{
int data;
Node next;
};
// Function to get a new node
static Node getNode(int data)
{
// allocate space
Node newNode = new Node();
// put in data
newNode.data = data;
newNode.next = null;
return newNode;
}
// Function to reverse the linked list.
static Node reverseLL(Node head)
{
if (head == null || head.next == null)
return head;
Node prev = null, next = null, curr = head;
while (curr != null)
{
next = curr.next;
curr.next = prev;
prev = curr;
curr = next;
}
return prev;
}
// Function to print the last k nodes
// of linked list in reverse order
static void printLastKRev(Node head, int k)
{
// if list is empty
if (head == null)
return;
// Reverse linked list.
head = reverseLL(head);
Node curr = head;
int cnt = 0;
// Print first k nodes of linked list.
while (cnt < k)
{
System.out.print(curr.data + " ");
cnt++;
curr = curr.next;
}
// Restore the list.
head = reverseLL(head);
}
// Driver code
public static void main(String[] args)
{
// Create list: 1->2->3->4->5
Node head = getNode(1);
head.next = getNode(2);
head.next.next = getNode(3);
head.next.next.next = getNode(4);
head.next.next.next.next = getNode(5);
int k = 4;
// print the last k nodes
printLastKRev(head, k);
}
}
// This code is contributed by 29AjayKumar
# Python3 implementation to print the
# last k nodes of linked list in
# reverse order
# Structure of a node
class Node:
def __init__(self, data):
self.data = data
self.next = None
# Function to get a new node
def getNode(data):
# Allocate space
newNode = Node(data)
return newNode
# Function to reverse the linked list.
def reverseLL(head):
if (not head or not head.next):
return head
prev = None
next = None
curr = head;
while (curr):
next = curr.next
curr.next = prev
prev = curr
curr = next
return prev
# Function to print the last k nodes
# of linked list in reverse order
def printLastKRev(head, k):
# If list is empty
if (not head):
return
# Reverse linked list.
head = reverseLL(head)
curr = head
cnt = 0
# Print first k nodes of linked list.
while (cnt < k):
print(curr.data, end = ' ')
cnt += 1
curr = curr.next
# Restore the list.
head = reverseLL(head)
# Driver code
if __name__=='__main__':
# Create list: 1.2.3.4.5
head = getNode(1)
head.next = getNode(2)
head.next.next = getNode(3)
head.next.next.next = getNode(4)
head.next.next.next.next = getNode(5)
k = 4
# Print the last k nodes
printLastKRev(head, k)
# This code is contributed by pratham76
// C# implementation to print the last k nodes
// of linked list in reverse order
using System;
class GFG
{
// Structure of a node
public class Node
{
public int data;
public Node next;
};
// Function to get a new node
static Node getNode(int data)
{
// allocate space
Node newNode = new Node();
// put in data
newNode.data = data;
newNode.next = null;
return newNode;
}
// Function to reverse the linked list.
static Node reverseLL(Node head)
{
if (head == null || head.next == null)
return head;
Node prev = null, next = null, curr = head;
while (curr != null)
{
next = curr.next;
curr.next = prev;
prev = curr;
curr = next;
}
return prev;
}
// Function to print the last k nodes
// of linked list in reverse order
static void printLastKRev(Node head, int k)
{
// if list is empty
if (head == null)
return;
// Reverse linked list.
head = reverseLL(head);
Node curr = head;
int cnt = 0;
// Print first k nodes of linked list.
while (cnt < k)
{
Console.Write(curr.data + " ");
cnt++;
curr = curr.next;
}
// Restore the list.
head = reverseLL(head);
}
// Driver code
public static void Main(String[] args)
{
// Create list: 1->2->3->4->5
Node head = getNode(1);
head.next = getNode(2);
head.next.next = getNode(3);
head.next.next.next = getNode(4);
head.next.next.next.next = getNode(5);
int k = 4;
// print the last k nodes
printLastKRev(head, k);
}
}
// This code is contributed by Princi Singh
<script>
// Javascript implementation to print the last k nodes
// of linked list in reverse order
// Structure of a node
class Node
{
constructor()
{
this.data = 0;
this.next = null;
}
}
// Function to get a new node
function getNode(data)
{
// allocate space
let newNode = new Node();
// put in data
newNode.data = data;
newNode.next = null;
return newNode;
}
// Function to reverse the linked list.
function reverseLL(head)
{
if (head == null || head.next == null)
return head;
let prev = null, next = null, curr = head;
while (curr != null)
{
next = curr.next;
curr.next = prev;
prev = curr;
curr = next;
}
return prev;
}
// Function to print the last k nodes
// of linked list in reverse order
function printLastKRev(head,k)
{
// if list is empty
if (head == null)
return;
// Reverse linked list.
head = reverseLL(head);
let curr = head;
let cnt = 0;
// Print first k nodes of linked list.
while (cnt < k)
{
document.write(curr.data + " ");
cnt++;
curr = curr.next;
}
// Restore the list.
head = reverseLL(head);
}
// Driver code
// Create list: 1->2->3->4->5
let head = getNode(1);
head.next = getNode(2);
head.next.next = getNode(3);
head.next.next.next = getNode(4);
head.next.next.next.next = getNode(5);
let k = 4;
// print the last k nodes
printLastKRev(head, k);
// This code is contributed by avanitrachhadiya2155
</script>
Time Complexity: O(N)
Auxiliary Space: O(1)
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