//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。 
//
// 进阶:你能尝试使用一趟扫描实现吗? 
//
// 
//
// 示例 1: 
//
// 
//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]
// 
//
// 
//
// 提示: 
//
// 
// 链表中结点的数目为 sz 
// 1 <= sz <= 30 
// 0 <= Node.val <= 100 
// 1 <= n <= sz 
// 
// Related Topics 链表 双指针 
// 👍 1226 👎 0

/*
* 19 删除链表的倒数第 N 个结点
* 2021-02-22 17:18:53
* @author oxygenbytes
*/ 
#include "leetcode.h" 
//leetcode submit region begin(Prohibit modification and deletion)
/**
 * 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) {

        if(!head->next && n == 1){
            return nullptr;
        }

        auto fast = head, slow = head;

        while(n--)  if(fast) fast = fast->next;

        if(!fast) return head->next;
        auto keep = slow;
        while(fast){
            fast = fast->next;
            keep = slow;
            slow = slow->next;
        }
        keep->next = slow->next;

        return head;
    }
};
//leetcode submit region end(Prohibit modification and deletion)