原题地址：https://oj.leetcode.com/problems/partition-list/
题意：
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.
解题思路：解决链表问题时，最好加一个头结点，问题会比较好解决。对这道题来说，创建两个头结点head1和head2，head1这条链表是小于x值的节点的链表，head2链表是大于等于x值的节点的链表，然后将head2链表链接到head链表的尾部即可。
代码：
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None class Solution:
# @param head, a ListNode
# @param x, an integer
# @return a ListNode
def partition(self, head, x):
head1 = ListNode(0)
head2 = ListNode(0)
Tmp = head
phead1 = head1
phead2 = head2
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