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Java Program For Inserting A Node After The N-th Node From The End

  • Last Updated : 11 Jan, 2022

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. 

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

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.

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.

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.

Please refer complete article on Insert a node after the n-th node from the end for more details!


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