# 删除链表的倒数第 N 个结点
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 <= 300 <= Node.val <= 1001 <= n <= sz
## template
```python
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
class LinkList:
def __init__(self):
self.head=None
def initList(self, data):
self.head = ListNode(data[0])
r=self.head
p = self.head
for i in data[1:]:
node = ListNode(i)
p.next = node
p = p.next
return r
def convert_list(self,head):
ret = []
if head == None:
return
node = head
while node != None:
ret.append(node.val)
node = node.next
return ret
class Solution:
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
v = ListNode(0, head)
handle = v
index = []
while v is not None:
index.append(v)
v = v.next
pre = len(index)-n-1
next = len(index)-n+1
index[pre].next = index[next] if next >= 0 and next < len(
index) else None
return handle.next
# %%
l = LinkList()
list1 = [1,2,3,4,5]
head = l.initList(list1)
n = 2
s = Solution()
print(l.convert_list(s.removeNthFromEnd(head, n)))
```
## 答案
```python
```
## 选项
### A
```python
```
### B
```python
```
### C
```python
```