Problem description:

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]

Solution:

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# 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:
        if not head:
            return head
        l1 = h1 = ListNode(0)
        l2 = h2 = ListNode(0)
        
        while head:
            if head.val < x:
                l1.next = head
                l1 = l1.next
            else:
                l2.next = head
                l2 = l2.next
            head = head.next
        l1.next = h2.next
        l2.next = None
        return h1.next

time complexity: $O(n)$
space complexity: $O(1)$
reference:
related problem: