给你一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。

输入：head = [1,2,3,4,5], n = 2
输出：[1,2,3,5]

示例 2：
输入：head = [1], n = 1
输出：[]

示例 3：
输入：head = [1,2], n = 1
输出：[1]

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int k) {
         auto dummy = new ListNode(-1);//弄一个头结点
        dummy->next = head;

        int n = 0;
        for (auto p = dummy->next; p; p = p->next) n ++ ;

        auto p = dummy;
        for (int i = 0; i < n - k ; i ++ ) p = p->next;
        p->next = p->next->next;

        return dummy->next;
    }
};