[LeetCode]题解（python）：086 - Partition List

题目来源

---

https://leetcode.com/problems/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.

---

题意分析

---

Input:

:type head: ListNode
:type x: int

Output:

:rtype: ListNode

Conditions:将一个list分成两个list，其中一个list的所有元素都小于x，另外一个list的所有元素都大于或等于x，注意要保持稳定性（即相等的元素，原来什么顺序就是什么顺序）

---

题目思路

---

设两个头节点，之后扩展这两个list就好，head1 list的所有元素都小于x，head2 list的所有元素都大于或等于x，然后将head2 list加到head1 list后面，然后返回需要的list 起点。

---

AC代码（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
         """
         head1 = ListNode(0)
         head2 = ListNode(0)

         p1 = head1
         p2 = head2

         tmp = head

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