// 解题思路: 快慢指针
// 快指针先走 K 步，然后两个指针一起，两者相差 K 步，当快指针走到末尾时，慢指针在倒数第 K 个节点的前一个。
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if (head == nullptr) 
            return nullptr;

        auto dummy = new ListNode(-1);
        dummy->next = head;

        auto fast = dummy, slow = dummy;
        while(n--) {
            fast = fast->next;
        }

        while(fast->next) {
            slow = slow->next;
            fast = fast->next;
        }

        slow->next = slow->next->next;
        return dummy->next;
    }
};