关键是逆向思维：当快指针遍历完，我们希望慢指针在哪里？

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode *dummy = new ListNode(0); ////// 虚拟头节点防止访问空指针(->next->next), 凡是有可能修改头节点的都加一个虚拟头节点
        dummy -> next = head;
        ListNode *p = dummy, *q = dummy;

        for(int i = 0; i < n+1; i ++){ // 当快指针遍历完，我们希望慢指针在前n+1个位置
            q = q->next;
        }

        while(q){
            q = q->next, p = p->next;
        }

        p->next = p->next->next;

        return dummy->next;
    }
};