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.

原题链接：https://oj.leetcode.com/problems/remove-nth-node-from-end-of-list/

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; }    }