题目：
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.
代码：oj测试通过 Runtime: 53 ms
1 # Definition for singly-linked list. 2 # class ListNode: 3 # def __init__(self, x): 4 # self.val = x 5 # self.next = None 6 7 class Solution: 8 # @param head, a ListNode 9 # @param x, an integer 10 # @return a ListNode 11 def partition(self, head, x): 12 if head is None or head.next is None: 13 return head 14 15 h1 = ListNode(0) 16 dummyh1 = h1 17 h2 = ListNode(0) 18 dummyh2 = h2 19 20 p = head 21 while p is not None: 22 if p.val < x : 23 h1.next = p 24 h1 = h1.next 25 else: 26 h2.next = p 27 h2 = h2.next 28 p = p.next 29 30 h1.next = dummyh2.next 31 h2.next = None
思路：
这道题的意思就是：把比x小的单拎出来，把大于等于x的单拎出来，再把两个Linked List首尾接起来。
技巧是设定两个虚拟表头dummyh1和dummyh2：保留住两个新表头。
另外需要设定两个迭代变量h1和h2：用于迭代变量
leetcode 【 Partition List 】python 实现
原文：http://www.cnblogs.com/xbf9xbf/p/4214591.html