Insert a node after the n-th node from the end
Insert a node x after the nth node from the end in the given singly linked list. It is guaranteed that the list contains the nth node from the end. Also 1 <= n.
Examples:
Input : list: 1->3->4->5
n = 4, x = 2
Output : 1->2->3->4->5
4th node from the end is 1 and
insertion has been done after this node.
Input : list: 10->8->3->12->5->18
n = 2, x = 11
Output : 10->8->3->12->5->11->18
Method 1 (Using length of the list):
Find the length of the linked list, i.e, the number of nodes in the list. Let it be len. Now traverse the list from the 1st node upto the (len-n+1)th node from the beginning and insert the new node after this node. This method requires two traversals of the list.
C++
// C++ implementation to insert a node after// the n-th node from the end#include <bits/stdc++.h>using namespace std;// structure of a nodestruct Node { int data; Node* next;};// function to get a new nodeNode* getNode(int data){ // allocate memory for the node Node* newNode = (Node*)malloc(sizeof(Node)); // put in the data newNode->data = data; newNode->next = NULL; return newNode;}// function to insert a node after the// nth node from the endvoid insertAfterNthNode(Node* head, int n, int x){ // if list is empty if (head == NULL) return; // get a new node for the value 'x' Node* newNode = getNode(x); Node* ptr = head; int len = 0, i; // find length of the list, i.e, the // number of nodes in the list while (ptr != NULL) { len++; ptr = ptr->next; } // traverse up to the nth node from the end ptr = head; for (i = 1; i <= (len - n); i++) ptr = ptr->next; // insert the 'newNode' by making the // necessary adjustment in the links newNode->next = ptr->next; ptr->next = newNode;}// function to print the listvoid printList(Node* head){ while (head != NULL) { cout << head->data << " "; head = head->next; }}// Driver program to test aboveint main(){ // Creating list 1->3->4->5 Node* head = getNode(1); head->next = getNode(3); head->next->next = getNode(4); head->next->next->next = getNode(5); int n = 4, x = 2; cout << "Original Linked List: "; printList(head); insertAfterNthNode(head, n, x); cout << "\nLinked List After Insertion: "; printList(head); return 0;} |
Java
// Java implementation to insert a node after // the n-th node from the end 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 memory for the node Node newNode = new Node(); // put in the data newNode.data = data; newNode.next = null; return newNode; } // function to insert a node after the // nth node from the end static void insertAfterNthNode(Node head, int n, int x) { // if list is empty if (head == null) return; // get a new node for the value 'x' Node newNode = getNode(x); Node ptr = head; int len = 0, i; // find length of the list, i.e, the // number of nodes in the list while (ptr != null) { len++; ptr = ptr.next; } // traverse up to the nth node from the end ptr = head; for (i = 1; i <= (len - n); i++) ptr = ptr.next; // insert the 'newNode' by making the // necessary adjustment in the links newNode.next = ptr.next; ptr.next = newNode; } // function to print the list static void printList(Node head) { while (head != null) { System.out.print(head.data + " "); head = head.next; } } // Driver code public static void main(String[] args) { // Creating list 1->3->4->5 Node head = getNode(1); head.next = getNode(3); head.next.next = getNode(4); head.next.next.next = getNode(5); int n = 4, x = 2; System.out.print("Original Linked List: "); printList(head); insertAfterNthNode(head, n, x); System.out.println(); System.out.print("Linked List After Insertion: "); printList(head); }} // This code is contributed by prerna saini |
Python3
# Python implementation to insert a node after # the n-th node from the end # Linked List node class Node: def __init__(self, data): self.data = data self.next = None# function to get a new node def getNode(data) : # allocate memory for the node newNode = Node(0) # put in the data newNode.data = data newNode.next = None return newNode # function to insert a node after the # nth node from the end def insertAfterNthNode(head, n, x) : # if list is empty if (head == None) : return # get a new node for the value 'x' newNode = getNode(x) ptr = head len = 0 i = 0 # find length of the list, i.e, the # number of nodes in the list while (ptr != None) : len = len + 1 ptr = ptr.next # traverse up to the nth node from the end ptr = head i = 1 while ( i <= (len - n) ) : ptr = ptr.next i = i + 1 # insert the 'newNode' by making the # necessary adjustment in the links newNode.next = ptr.next ptr.next = newNode # function to print the list def printList( head) : while (head != None): print(head.data ,end = " ") head = head.next # Driver code # Creating list 1->3->4->5 head = getNode(1) head.next = getNode(3) head.next.next = getNode(4) head.next.next.next = getNode(5) n = 4x = 2print("Original Linked List: ") printList(head) insertAfterNthNode(head, n, x) print()print("Linked List After Insertion: ") printList(head) # This code is contributed by Arnab Kundu |
C#
// C# implementation to insert a node after // the n-th node from the end 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 memory for the node Node newNode = new Node(); // put in the data newNode.data = data; newNode.next = null; return newNode; } // function to insert a node after the // nth node from the end static void insertAfterNthNode(Node head, int n, int x) { // if list is empty if (head == null) return; // get a new node for the value 'x' Node newNode = getNode(x); Node ptr = head; int len = 0, i; // find length of the list, i.e, the // number of nodes in the list while (ptr != null) { len++; ptr = ptr.next; } // traverse up to the nth node from the end ptr = head; for (i = 1; i <= (len - n); i++) ptr = ptr.next; // insert the 'newNode' by making the // necessary adjustment in the links newNode.next = ptr.next; ptr.next = newNode; } // function to print the list static void printList(Node head) { while (head != null) { Console.Write(head.data + " "); head = head.next; } } // Driver code public static void Main(String[] args) { // Creating list 1->3->4->5 Node head = getNode(1); head.next = getNode(3); head.next.next = getNode(4); head.next.next.next = getNode(5); int n = 4, x = 2; Console.Write("Original Linked List: "); printList(head); insertAfterNthNode(head, n, x); Console.WriteLine(); Console.Write("Linked List After Insertion: "); printList(head); }} // This code has been contributed by 29AjayKumar |
Output:
Original Linked List: 1 3 4 5 Linked List After Insertion: 1 2 3 4 5
Time Complexity: O(n), where n is the number of nodes in the list.
Method 2 (Single traversal):
This method uses two pointers, one is slow_ptr and the other is fast_ptr. First move the fast_ptr up to the nth node from the beginning. Make the slow_ptr point to the 1st node of the list. Now, simultaneously move both the pointers until fast_ptr points to the last node. At this point the slow_ptr will be pointing to the nth node from the end. Insert the new node after this node. This method requires single traversal of the list.
C++
// C++ implementation to insert a node after the// nth node from the end#include <bits/stdc++.h>using namespace std;// structure of a nodestruct Node { int data; Node* next;};// function to get a new nodeNode* getNode(int data){ // allocate memory for the node Node* newNode = (Node*)malloc(sizeof(Node)); // put in the data newNode->data = data; newNode->next = NULL; return newNode;}// function to insert a node after the// nth node from the endvoid insertAfterNthNode(Node* head, int n, int x){ // if list is empty if (head == NULL) return; // get a new node for the value 'x' Node* newNode = getNode(x); // Initializing the slow and fast pointers Node* slow_ptr = head; Node* fast_ptr = head; // move 'fast_ptr' to point to the nth node // from the beginning for (int i = 1; i <= n - 1; i++) fast_ptr = fast_ptr->next; // iterate until 'fast_ptr' points to the // last node while (fast_ptr->next != NULL) { // move both the pointers to the // respective next nodes slow_ptr = slow_ptr->next; fast_ptr = fast_ptr->next; } // insert the 'newNode' by making the // necessary adjustment in the links newNode->next = slow_ptr->next; slow_ptr->next = newNode;}// function to print the listvoid printList(Node* head){ while (head != NULL) { cout << head->data << " "; head = head->next; }}// Driver program to test aboveint main(){ // Creating list 1->3->4->5 Node* head = getNode(1); head->next = getNode(3); head->next->next = getNode(4); head->next->next->next = getNode(5); int n = 4, x = 2; cout << "Original Linked List: "; printList(head); insertAfterNthNode(head, n, x); cout << "\nLinked List After Insertion: "; printList(head); return 0;} |
Java
// Java implementation to // insert a node after the // nth node from the end 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 memory for the node Node newNode = new Node(); // put in the data newNode.data = data; newNode.next = null; return newNode; } // function to insert a node after // the nth node from the end static void insertAfterNthNode(Node head, int n, int x) { // if list is empty if (head == null) return; // get a new node for the value 'x' Node newNode = getNode(x); // Initializing the slow // and fast pointers Node slow_ptr = head; Node fast_ptr = head; // move 'fast_ptr' to point to the // nth node from the beginning for (int i = 1; i <= n - 1; i++) fast_ptr = fast_ptr.next; // iterate until 'fast_ptr' points // to the last node while (fast_ptr.next != null) { // move both the pointers to the // respective next nodes slow_ptr = slow_ptr.next; fast_ptr = fast_ptr.next; } // insert the 'newNode' by making the // necessary adjustment in the links newNode.next = slow_ptr.next; slow_ptr.next = newNode; } // function to print the list static void printList(Node head) { while (head != null) { System.out.print(head.data + " "); head = head.next; } } // Driver code public static void main(String[] args) { // Creating list 1->3->4->5 Node head = getNode(1); head.next = getNode(3); head.next.next = getNode(4); head.next.next.next = getNode(5); int n = 4, x = 2; System.out.println("Original Linked List: "); printList(head); insertAfterNthNode(head, n, x); System.out.println(); System.out.println("Linked List After Insertion: "); printList(head); }} // This code is contributed by// Prerna Saini. |
Python3
# Python3 implementation to insert a# node after the nth node from the end # Structure of a nodeclass Node: def __init__(self, data): self.data = data self.next = None # Function to get a new nodedef getNode(data): # Allocate memory for the node newNode = Node(data) return newNode# Function to insert a node after the# nth node from the enddef insertAfterNthNode(head, n, x): # If list is empty if (head == None): return # Get a new node for the value 'x' newNode = getNode(x) # Initializing the slow and fast pointers slow_ptr = head fast_ptr = head # Move 'fast_ptr' to point to the nth # node from the beginning for i in range(1, n): fast_ptr = fast_ptr.next # Iterate until 'fast_ptr' points to the # last node while (fast_ptr.next != None): # Move both the pointers to the # respective next nodes slow_ptr = slow_ptr.next fast_ptr = fast_ptr.next # Insert the 'newNode' by making the # necessary adjustment in the links newNode.next = slow_ptr.next slow_ptr.next = newNode# Function to print the listdef printList(head): while (head != None): print(head.data, end = ' ') head = head.next # Driver codeif __name__=='__main__': # Creating list 1.3.4.5 head = getNode(1) head.next = getNode(3) head.next.next = getNode(4) head.next.next.next = getNode(5) n = 4 x = 2 print("Original Linked List: ", end = '') printList(head) insertAfterNthNode(head, n, x) print("\nLinked List After Insertion: ", end = '') printList(head) # This code is contributed by rutvik_56 |
C#
// C# implementation to // insert a node after the // nth node from the end 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 memory for the node Node newNode = new Node(); // put in the data newNode.data = data; newNode.next = null; return newNode; } // function to insert a node after // the nth node from the end static void insertAfterNthNode(Node head, int n, int x) { // if list is empty if (head == null) return; // get a new node for the value 'x' Node newNode = getNode(x); // Initializing the slow // and fast pointers Node slow_ptr = head; Node fast_ptr = head; // move 'fast_ptr' to point to the // nth node from the beginning for (int i = 1; i <= n - 1; i++) fast_ptr = fast_ptr.next; // iterate until 'fast_ptr' points // to the last node while (fast_ptr.next != null) { // move both the pointers to the // respective next nodes slow_ptr = slow_ptr.next; fast_ptr = fast_ptr.next; } // insert the 'newNode' by making the // necessary adjustment in the links newNode.next = slow_ptr.next; slow_ptr.next = newNode; } // function to print the list static void printList(Node head) { while (head != null) { Console.Write(head.data + " "); head = head.next; } } // Driver code public static void Main() { // Creating list 1->3->4->5 Node head = getNode(1); head.next = getNode(3); head.next.next = getNode(4); head.next.next.next = getNode(5); int n = 4, x = 2; Console.WriteLine("Original Linked List: "); printList(head); insertAfterNthNode(head, n, x); Console.WriteLine(); Console.WriteLine("Linked List After Insertion: "); printList(head); }}/* This code contributed by PrinciRaj1992 */ |
Output:
Original Linked List: 1 3 4 5 Linked List After Insertion: 1 2 3 4 5
Time Complexity: O(n), where n is the number of nodes in the list.
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