p180.java

/*Given a linked list consisting of L nodes and given a number N. The task is to find the Nth node from the end of the linked list.

Example 1:

Input:
N = 2
LinkedList: 1->2->3->4->5->6->7->8->9
Output: 8
Explanation: In the first example, there
are 9 nodes in linked list and we need
to find 2nd node from end. 2nd node
from end is 8.  
Example 2:

Input:
N = 5
LinkedList: 10->5->100->5
Output: -1
Explanation: In the second example, there
are 4 nodes in the linked list and we
need to find 5th from the end. Since 'n'
is more than the number of nodes in the
linked list, the output is -1.
Your Task:
The task is to complete the function getNthFromLast() which takes two arguments: reference to head and N and you need to return Nth from the end or -1 in case node doesn't exist.

Note:
Try to solve in a single traversal.

Expected Time Complexity: O(N).
Expected Auxiliary Space: O(1).

Constraints:
1 <= L <= 106
1 <= N <= 106*/

/* Structure of node
class Node
{
    int data;
    Node next;
    Node(int d) {data = d; next = null; }
}*/

class Solution{
    int getNthFromLast(Node head, int n){
    	Node first = head;
        Node second = head;
        for (int i = 1; i <= n; i++) {
            if (second == null) {
                return -1;
            }
            second = second.next;
        }
        while (second != null) {
            first = first.next;
            second = second.next;
        }
        return first.data; 
    }
}