Given 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.


Input: head = 1->4->3->2->5->2, x = 3
Output: 1->2->2->4->3->5


# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
# = None

class Solution:
    def partition(self, head, x):
        :type head: ListNode
        :type x: int
        :rtype: ListNode
        if head is None or is None:
            return head

        shead, scurr, lhead, lcurr = None, None, None, None

        curr = head
        while curr is not None:
            if curr.val < x:
                if shead is None:
                    shead = curr
           = curr
                scurr = curr
                if lhead is None:
                    lhead = curr
           = curr
                lcurr = curr

            curr =

        if scurr is not None:
   = lhead
            shead = lhead

        if lcurr is not None:
   = None

        return shead