# 反转链表 II

给你单链表的头指针 <code>head</code> 和两个整数 <code>left</code> 和 <code>right</code> ，其中 <code>left <= right</code> 。请你反转从位置 <code>left</code> 到位置 <code>right</code> 的链表节点，返回 <strong>反转后的链表</strong> 。<p> </p><p><strong>示例 1：</strong></p><img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0092.Reverse%20Linked%20List%20II/images/rev2ex2.jpg" style="width: 542px; height: 222px;" /><pre><strong>输入：</strong>head = [1,2,3,4,5], left = 2, right = 4<strong><br />输出：</strong>[1,4,3,2,5]</pre><p><strong>示例 2：</strong></p><pre><strong>输入：</strong>head = [5], left = 1, right = 1<strong><br />输出：</strong>[5]</pre><p> </p><p><strong>提示：</strong></p><ul>	<li>链表中节点数目为 <code>n</code></li>	<li><code>1 <= n <= 500</code></li>	<li><code>-500 <= Node.val <= 500</code></li>	<li><code>1 <= left <= right <= n</code></li></ul><p> </p><p><strong>进阶：</strong> 你可以使用一趟扫描完成反转吗？</p>

## template

```python
class ListNode(object):
	def __init__(self, x):
		self.val = x
		self.next = None
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(object):
	def reverseBetween(self, head, m, n):
		"""
		:type head: ListNode
		:type m: int
		:type n: int
		:rtype: ListNode
		"""
		if m == n:
			return head
		split_node, prev, curr = None, None, head
		count = 1
		while count <= m and curr is not None:
			if count == m:
				split_node = prev
			prev = curr
			curr = curr.next
			count += 1
		tail, next_node = prev, None
		while curr is not None and count <= n:
			next_temp = curr.next
			curr.next = prev
			prev = curr
			curr = next_temp
			count += 1
		if split_node is not None:
			split_node.next = prev
		if tail is not None:
			tail.next = curr
		if m == 1:
			return prev
		return head
# %%
l = LinkList()
list1 = [1,2,3,4,5]
l1 = l.initList(list1)
left = 2
right = 4
s = Solution()
print(l.convert_list(s.reverseBetween(l1, left, right)))
```

## 答案

```python

```

## 选项

### A

```python

```

### B

```python

```

### C

```python

```