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.

题目：给定一个链表，从尾部删除第n个节点并返回新的链表。

思路：使用两个指针，fast 和 slow，他们的距离是n，于是fast到尾的时候，n所在的节点就是须要删除的节点。

	public ListNode removeNthFromEnd(ListNode head, int n) {
		ListNode slow = head, fast = head;
		if (head.next == null)
			return null;
		for (int i = 1; i <= n; i++)
			slow = slow.next;
		if (slow == null) {
			head = head.next;
			return head;
		}
		while (slow.next != null) {
			slow = slow.next;
			fast = fast.next;
		}
		fast.next = fast.next.next;
		return head;
	}
    // Definition for singly-linked list.
    public class ListNode {
        int val;
        ListNode next;
        ListNode(int x) { val = x; }
    }