# [Solved] 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. ## Question

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:

```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 Solution

```# 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: ListNode, x: int) -> ListNode:
l1 = ListNode(0)
y = l1
l2 = ListNode(0)
z = l2