Remove Nth Node From End of List

Given the head of a linked list, remove the nth node from the end of the list and return its head.

Follow up: Could you do this in one pass? 

Example 1:

Input: head = [1,2,3,4,5], n = 2
Output: [1,2,3,5]

Example 2:

Input: head = [1], n = 1
Output: []

Example 3:

Input: head = [1,2], n = 1
Output: [1]

Solution:

public ListNode removeNthFromEnd(ListNode head, int n) {
        if(head == null)
            return head;        
        ListNode dummy = new ListNode(0);
        dummy.next = head;        
        ListNode first, second;
        first=second=dummy;
        for(int i=0; i<n+1; i++) {
            first=first.next;
        }
        while(first!=null){
            first=first.next;
            second=second.next;
        }
        // Remove nth node from end
        second.next = second.next.next;
        return dummy.next;
    }

Remove Nth Node From End of List

Given a linked list, remove the nth node from the end of list and return its head.

For example,

   Given linked list: 1->2->3->4->5, and n = 2.

   After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:
Given n will always be valid.
Try to do this in one pass.

Sol:

 public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode start = new ListNode(0);
        ListNode first=start, second=start;
        start.next = head;
     
        // Move n nodes
        for (int i=0; i<=n; i++) {
            first = first.next;
        }
        // Move both pointers one node at a time until end of list
        while(first!=null) {
           first = first.next;
           second = second.next;
        }
       
       // remove nth node from end
       second.next = second.next.next;
       return start.next;
    }