[LeetCode]Remove Nth Node From End of List

Question:
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.

本题难度easy，常规方法就不介绍了。我在discuss看到华人写的另一种解法，方法很不错。帖子是：​​My java solution, using a fast pointer​​​，作者：传说选手。
方法：
1、fast pointer move n steps in advance
2、cur pointer and fast pointer move together
3、do the reconnection
我对其代码进行了改进：

public class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        //require
        if(head==null)
            return head;
        ListNode fast=head;
        ListNode fake=new ListNode(0);
        fake.next=head;
        //invariant
        while(n>0){//move fast n steps in advance
            fast=fast.next;
            n--;
        }
        ListNode cur=fake;
        while(fast!=null){
            fast=fast.next;
            cur=cur.next;
        }
        cur.next=cur.next.next;
        //ensure
        return fake.next;
    }
}