leetcode: 86. Partition List

# 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.
#
# For example,
# Given 1->4->3->2->5->2 and x = 3,
# return 1->2->2->4->3->5.

class ListNode():
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution():
    def partition(self, head, x):
        h1 = t1 = ListNode(-1)
        h2 = t2 = ListNode(-1)
        cur = head
        while cur:
            if cur.val < x:
                t1.next = cur
                t1 = t1.next
            else:
                t2.next = cur
                t2 = t2.next
            cur = cur.next
        t1.next, t2.next = h2.next, None
        return h1.next


if __name__ == "__main__":
    head, head.next, head.next.next, head.next.next.next, head.next.next.next.next, head.next.next.next.next.next \
        = ListNode(1), ListNode(4), ListNode(3), ListNode(2), ListNode(5), ListNode(2)
    print(Solution().partition(head, 3))