//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)