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.

思路1：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。注意：删除节点时需要删除指针的前驱pre；增加dummy head处理删除头节点的特殊情况。

public class Solution {	public ListNode removeNthFromEnd(ListNode head, int n) {		if (head == null)			return null;		ListNode dummyHead = new ListNode(-1);		dummyHead.next = head;		ListNode s = head;		ListNode t = head;		n = n - 1;		for (int i = 0; i < n; i++) {			t = t.next;		}		ListNode preS = dummyHead;		while (t.next != null) {			preS = preS.next;			s = s.next;			t = t.next;		}		// remove s		preS.next = s.next;		s.next = null;		return dummyHead.next;	}	public static void main(String[] args) {		ListNode head = new ListNode(1);		head.next = new ListNode(2);		head.next.next = new ListNode(3);		head.next.next.next = new ListNode(4);		head.next.next.next.next = new ListNode(5);		ListNode newHead = new Solution().removeNthFromEnd(head, 1);	}}