C# Program For Writing A Function To Get Nth Node In A Linked List

Write a GetNth() function that takes a linked list and an integer index and returns the data value stored in the node at that index position.

Example:

Input:  1->10->30->14,  index = 2
Output: 30  
The node at index 2 is 30

Algorithm:

1. Initialize count = 0
2. Loop through the link list
     a. If count is equal to the passed index then return current
         node
     b. Increment count
     c. Change current to point to next of the current.

Implementation:

C#

// C# program to find n'th node
// in linked list
using System;
using System.Diagnostics;

public class Node
{
    public int data;
    public Node next;
    public Node(int d)
    {
        data = d;
        next = null;
    }
}

class LinkedList
{
    // Head of list
    Node head;

    // Takes index as argument and
    // return data at index
    public int GetNth(int index)
    {
        Node current = head;

        // Index of Node we are
        // currently looking at
        int count = 0;
        while (current != null)
        {
            if (count == index)
                return current.data;
            count++;
            current = current.next;
        }

        /* If we get to this line, the caller
           was asking for a non-existent element
           so we assert fail */
        Debug.Assert(false);
        return 0;
    }

    /* Given a reference to the head of a list
       and an int, inserts a new Node on the
       front of the list. */
    public void push(int new_data)
    {
        // 1. Alloc the Node and put data
        Node new_Node = new Node(new_data);

        // 2. Make next of new Node as head
        new_Node.next = head;

        // 3. Move the head to point to new Node
        head = new_Node;
    }

    // Driver code
    public static void Main(String[] args)
    {
        // Start with empty list
        LinkedList llist = new LinkedList();

        // Use push() to construct list
        // 1->12->1->4->1
        llist.push(1);
        llist.push(4);
        llist.push(1);
        llist.push(12);
        llist.push(1);

        // Check the count function
        Console.WriteLine("Element at index 3 is " +
                           llist.GetNth(3));
    }
}

// This code is contributed by Arnab Kundu

Output:

Element at index 3 is 4

Time Complexity: O(n)

Space complexity: O(1) using constant space

Method 2- With Recursion:

This method is contributed by MY_DOOM.

Algorithm: 

getnth(node, n)
1. Initialize count = 0
2. if count==n
     return node->data
3. else
    return getnth(node->next, n-1)

Implementation:

C#

// C# program to find n'th node in
// linked list using recursion
using System;

class GFG
{
    // Link list node
    public class Node
    {
        public int data;
        public Node next;
        public Node(int data)
        {
            this.data = data;
        }
    }

    /* Given a reference (pointer to pointer)
       to the head of a list and an int, push
       a new node on the front of the list. */
    static Node push(Node head, int new_data)
    {
        // Allocate node
        Node new_node = new Node(new_data);

        // Put in the data
        new_node.data = new_data;

        new_node.next = head;
        head = new_node;
        return head;
    }

    /* Takes head pointer of the linked list
       and index as arguments and return data
       at index*/
    static int GetNth(Node head, int n)
    {
        // Base Condition
        if (head == null)
            return -1;

        int count = 0;

        // If count equal too n return
        // node.data
        // Test Condition
        if (count == n)
            return head.data;

        // Recursively decrease n and
        // increase head to next pointer
        return GetNth(head.next, n - 1);
    }

    // Driver code
    public static void Main()
    {
        // Start with the empty list
        Node head = null;

        // Use push() to construct list
        // 1.12.1.4.1
        head = push(head, 1);
        head = push(head, 4);
        head = push(head, 1);
        head = push(head, 12);
        head = push(head, 1);

        // Check the count function
        Console.Write("Element at index 3 is {0}",
                       GetNth(head, 3));
    }
}
// Code improvement by Aishwarya Mittal
// This code contributed by PrinciRaj1992

Output:

Element at index 3 is 4

Time Complexity: O(n) 

Space Complexity : O(n) for call stack since using recursion.

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