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。
示例 1:

| 12
 
 | 输入:head = [1,2,3,4,5], n = 2输出:[1,2,3,5]
 
 | 
示例 2:
示例 3:
提示:
- 链表中结点的数目为 sz
- 1 <= sz <= 30
- 0 <= Node.val <= 100
- 1 <= n <= sz
进阶:你能尝试使用一趟扫描实现吗?
 
| 12
 3
 4
 5
 6
 7
 8
 9
 10
 11
 12
 13
 14
 15
 16
 17
 18
 19
 20
 21
 
 | class Solution {public ListNode removeNthFromEnd(ListNode head, int n) {
 ListNode dummy = new ListNode(0);
 dummy.next = head;
 ListNode fast = dummy;
 ListNode slow = dummy;
 while(n-- != 0 && fast != null){
 fast = fast.next;
 }
 if(fast == null) return head;
 
 while(fast.next != null){
 fast = fast.next;
 slow = slow.next;
 }
 
 slow.next = slow.next.next;
 return dummy.next;
 
 }
 }
 
 |