<p>Given the <code>head</code> of a linked list and a value <code>x</code>, partition it such that all nodes <strong>less than</strong> <code>x</code> come before nodes <strong>greater than or equal</strong> to <code>x</code>.</p>

<p>You should <strong>preserve</strong> the original relative order of the nodes in each of the two partitions.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/04/partition.jpg" style="width: 662px; height: 222px;" />
<pre>
<strong>Input:</strong> head = [1,4,3,2,5,2], x = 3
<strong>Output:</strong> [1,2,2,4,3,5]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> head = [2,1], x = 2
<strong>Output:</strong> [1,2]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the list is in the range <code>[0, 200]</code>.</li>
	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
	<li><code>-200 &lt;= x &lt;= 200</code></li>
</ul>

# 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]:
        lower_x = ListNode(0, None)
        higher_x = ListNode(0, None)

        curr_low = lower_x
        curr_high = higher_x

        curr = head 
        while curr:
            if curr.val < x:
                curr_low.next = curr
                curr_low = curr
            else:
                curr_high.next = curr
                curr_high = curr
            curr = curr.next
            
        curr_low.next = higher_x.next
        curr_high.next = None

        return lower_x.next
        


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