# 分隔链表
给你一个链表的头节点 head 和一个特定值 x ，请你对链表进行分隔，使得所有 小于 x 的节点都出现在 大于或等于 x 的节点之前。
你应当 保留 两个分区中每个节点的初始相对位置。
示例 1：
输入：head = [1,4,3,2,5,2], x = 3
输出：[1,2,2,4,3,5]
示例 2：
输入：head = [2,1], x = 2
输出：[1,2]
提示：
- 链表中节点的数目在范围
[0, 200] 内 -100 <= Node.val <= 100-200 <= x <= 200
## 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 partition(self, head, x):
"""
:type head: ListNode
:type x: int
:rtype: ListNode
"""
if head is None:
return None
less = lesshead = None
last = pos = head
while pos is not None:
if pos.val < x:
if lesshead is None:
lesshead = pos
else:
less.next = pos
less = pos
if head == pos:
last = head = pos.next
else:
last.next = pos.next
else:
last = pos
pos = pos.next
if lesshead is not None:
less.next = head
else:
lesshead = head
return lesshead
# %%
l = LinkList()
list1 = [1,4,3,2,5,2]
l1 = l.initList(list1)
x = 3
s = Solution()
print(l.convert_list(s.partition(l1, x)))
```
## 答案
```python
```
## 选项
### A
```python
```
### B
```python
```
### C
```python
```