# 86. Partition List

#### Medium

***

Given the `head` of a linked list and a value `x`, partition it such that all nodes **less than** `x` come before nodes **greater than or equal** to `x`.

You should **preserve** the original relative order of the nodes in each of the two partitions.

 

**Example 1:**

![](https://assets.leetcode.com/uploads/2021/01/04/partition.jpg)

```
Input: head = [1,4,3,2,5,2], x = 3
Output: [1,2,2,4,3,5]
```

**Example 2:**

```
Input: head = [2,1], x = 2
Output: [1,2]
```

 

**Constraints:**

* The number of nodes in the list is in the range `[0, 200]`.
* `-100 <= Node.val <= 100`
* `-200 <= x <= 200`

```python
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def partition(self, head: Optional[ListNode], x: int) -> Optional[ListNode]:
        p1,less = None, None
        p2,more = None, None
        while head:
            if head.val < x:
                if less is None:
                    p1, less = head, head
                else:
                    less.next = head
                    less = head
            else:
                if more is None:
                    p2, more = head, head
                else:
                    more.next = head
                    more = head
            head = head.next
        if more:
            more.next = None
        if less:
            less.next = p2
        return p1 or p2
```