Delete Nth node from the end of the given linked list
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.
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 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 |
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 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
// 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
# 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#
// 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 |
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> |
Created Linked list is: 2 3 1 7 Linked List after Deletion is: 2 3 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; } |
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.
Python
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() |
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 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> |
***** 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)