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 1Input: 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++
// 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) |
Java
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 |
Python3
# Python code for the deleting a node from end# in two traversalclass Node: def __init__(self, value): self.data = value self.next = Nonedef length(head): temp = head count = 0 while(temp != None): count += 1 temp = temp.next return countdef 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 headif __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) |
C#
// 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); }} |
Javascript
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++
// 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 codeint 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 |
C
/* 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 thelinked 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 programint 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
// Java implementation of the approachclass 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
# Python3 implementation of the approachclass Node: def __init__(self, new_data): self.data = new_data self.next = Noneclass 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 Codellist = 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#
// C# implementation of the approachusing 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 |
Javascript
<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> |
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++
// 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;} |
Java
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 }} |
Javascript
// Javascript code addition// class of linked listclass 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 5list.removeNthNodeFromEnd(2);list.printLinkedList(); // Output: 1 2 3 5// Edge case where linked list has only one nodeconst list2 = new LinkedList(1);list2.printLinkedList(); // Print: 1list2.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 2list3.removeNthNodeFromEnd(1);list3.printLinkedList(); // Output: 1// The code is contributed by Nidhi goel. |
Python3
# Python code addition# class of linked listclass 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.nextlist = LinkedList(1)list.addNode(2).addNode(3).addNode(4).addNode(5)list.printLinkedList() # Print: 1 2 3 4 5list.removeNthNodeFromEnd(2)list.printLinkedList() # Output: 1 2 3 5# Edge case where linked list has only one nodelist2 = LinkedList(1)list2.printLinkedList() # Print: 1list2.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 2list3.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++
// 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. 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; }};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->deleteNthNodeFromEnd(l->head, 2); cout << "*********** Linked List after Deletion *****" << endl; l->display(); return 0;}// This code is contributed by lokesh (lokeshmvs21). |
Java
// Java code for deleting a node from the end using two// Pointer Approachclass 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). |
Python
class Node: def __init__(self, data): self.data = data self.next = Noneclass 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 headif __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#
// C# code for deleting a node from the end using two// Pointer Approachusing 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). |
Javascript
<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 codelet 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
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