【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 tox.

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 = Noneclass 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