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->5.1
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26# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def partition(self, head, x):
"""
:type head: ListNode
:type x: int
:rtype: ListNode
"""
head1 = list1 = ListNode(0)
head2 = list2 = ListNode(0)
while head:
if head.val < x:
list1.next = head
list1 = list1.next
else:
list2.next = head
list2 = list2.next
head = head.next
list2.next = None
list1.next = head2.next
return head1.next