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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 node 
struct Node { 
    int data; 
    Node* next; 
}; 
  
// function to get a new node 
Node* 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 end 
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 
void printList(Node* head) 
{ 
    while (head != NULL) { 
        cout << head->data << " "; 
        head = head->next; 
    } 
} 
  
// Driver program to test above 
int 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 

Python

# 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 = 4
x = 2
  
print("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 node 
struct Node { 
    int data; 
    Node* next; 
}; 
  
// function to get a new node 
Node* 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 end 
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 
void printList(Node* head) 
{ 
    while (head != NULL) { 
        cout << head->data << " "; 
        head = head->next; 
    } 
} 
  
// Driver program to test above 
int 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. 

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.

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python

C#

C++

Java

C#

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