Question:

Finding Nth node from end of linked list: Java

Summary:

To find the Nth node from the end of the linked list, first we have to break down the getNthFromLast method:

For this, we initially set two pointers to the head of the linked list, fast and slow.
A loop runs n times, moving the fast pointer forward by n nodes. If the value of n is greater than the size of the linked list, it returns -1.
After the first loop, the fast pointer is n nodes ahead of the slow pointer.
Another loop runs until the fast pointer reaches the end of the linked list. In each iteration, both the fast and slow pointers move one node at a time.
When the fast pointer reaches the end, the slow pointer will be at the nth node from the end.
The data of the nth node from the end is returned.

Solution:

/* Structure of node

class Node

{

int data;

Node next;

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

}

class Solution

{

//Function to find the data of nth node from the end of a linked list.

int getNthFromLast(Node head, int n)

{

// Your code here

Node fast = head;

Node slow = head;

for(int i=0;i<n;i++)

{

// if the nth position is greater than the size of the linked list, -1.

if(fast == null)

{

return -1;

}

fast=fast.next;

}

while(fast!=null)

{

slow=slow.next;

fast=fast.next;

}

return slow.data;

}

Explanation:

This code defines a solution class with a method getNthFromLast that finds the data of the nth node from the end of a linked list. The linked list is represented using a Node class, which has an integer data field (int data) and a reference to the next node (Node next). The Node class also has a constructor that initializes the data and sets the next reference to null.

The first loop with the fast pointer moves it n nodes ahead, checking for the end of the linked list. If n is greater than the size of the linked list, it returns -1.