[LeetCode] 86. Partition List 分隔链表 @python

Description

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.

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给定一个链表和一个特定值 x，对链表进行分隔，使得所有小于 x 的节点都在大于或等于 x 的节点之前。

你应当保留两个分区中每个节点的初始相对位置。

例如，

给定1->4->3->2->5->2 和 x = 3,

返回1->2->2->4->3->5.

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Solution

构建两个新链表，小于x的放在链表1，大于或等于x的放在链表2，最后将链表1的表尾指向链表2的表头即可。

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# -*- coding: utf-8 -*-
"""
Created on Sun Mar 18 15:35:41 2018

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

class Solution:
    def partition(self, head, x):
        """
        :type head: ListNode
        :type x: int
        :rtype: ListNode
        """
        head1 = ListNode(0)
        head2 = ListNode(0)
        phead1 = head1
        phead2 = head2
        tmp = head
        while tmp:
            if tmp.val < x:
                phead1.next = tmp
                tmp = tmp.next
                phead1 = phead1.next
                phead1.next = None
            else:
                phead2.next = tmp
                tmp = tmp.next
                phead2 = phead2.next
                phead2.next = None
        phead1.next = head2.next
        head = head1.next
        return head