【LEETCODE】86- Partition List [Python]

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.

题意：
给个list，给个值，把list分为两部分，小于x的放在前，大于等于x的放在后，两部分各自保持原有顺序

思路：
use p to move from head to end and compare each value with x
create two dummy point, one is used to link points that are <x, another is to link points that are >=x



Python：

# 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
"""

if head is None or head.next is None or x is None:
return head

p1=head1=ListNode(0)
p2=head2=ListNode(0)
p=head

while p:
if p.val<x:
p1.next=p
p1=p1.next
else:
p2.next=p
p2=p2.next
p=p.next
p1.next=head2.next
p2.next=None
return head1.next