给定一个链表，删除链表的倒数第 n 个节点，并且返回链表的头结点。

示例：

给定一个链表: 1->2->3->4->5, 和 n = 2.

当删除了倒数第二个节点后，链表变为 1->2->3->5.
说明：

给定的 n 保证是有效的。

进阶：

你能尝试使用一趟扫描实现吗？

思路：双指针，一个先走n步，然后两个一起走

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
  ListNode dummy = new ListNode(0);
    dummy.next = head;
    ListNode first = dummy;
    ListNode second = dummy;
    // Advances first pointer so that the gap between first and second is n nodes apart
    for (int i = 1; i <= n + 1; i++) {
        first = first.next;
    }
    // Move first to the end, maintaining the gap
    while (first != null) {
        first = first.next;
        second = second.next;
    }
    second.next = second.next.next;
    return dummy.next;
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode l = new ListNode(0);
        l.next = head;
        ListNode node1 = l;
        ListNode node2 = l;
        for(int i = 0;i <= n;i++){
            node1 = node1.next;
        }
        while(node1 != null){
            node1 = node1.next;
            node2 = node2.next;
        }   
        node2.next = node2.next.next;
        return l.next;
    }
}