LeetCode 19: Remove Nth Node From End of List

Given a linked list, remove the n^th 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.

/**
 * 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 helper = new ListNode(0); // In case of removing the head
        helper.next = head;
        ListNode slower = helper;
        ListNode faster = helper;
        
        for (int i = 0; i < n; i++)
            faster = faster.next;
            
        while (faster.next != null)
        {
            slower = slower.next;
            faster = faster.next;
        }
        
        slower.next = slower.next.next;
        
        return helper.next;
    }
}