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.

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) {
 7  *         val = x;
 8  *         next = null;
 9  *     }
10  * }
11  */
12 public class Solution {
13     public ListNode removeNthFromEnd(ListNode head, int n) {
14         List<ListNode> list = new ArrayList<ListNode>();
15         ListNode temp = head;
16         if(null == head.next)
17             return null;
18         while(null != temp){
19             list.add(temp);
20             temp = temp.next;
21         }//只要一趟，把所有节点保存起来
22         
23         if(list.size() - n - 1 >= 0){
24             temp = list.get(list.size() - n - 1);
25             temp.next = temp.next.next;
26             return head;
27         }
28         head = head.next;
29         return head;
30     }
31 }

ps:依然是双指针思想，两个指针相隔n-1，每次两个指针向后一步，当后面一个指针没有后继了，前面一个指针就是要删除的节点。博客园上面看到的，就没有实现了