Leetcode: 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.

Solution:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode fast = dummy;
        for (int i=0; i<n; i++)
        {
            fast = fast != null ? fast.next : null;
        }
        
        if(fast == null || n < 0)
            return head;
            
        ListNode slow = dummy;
        while(fast.next != null)
        {
            fast = fast.next;
            slow = slow.next;
        }
        
        slow.next = slow.next.next;
        return dummy.next;
    }
}