leetcode 【 Partition List 】python 实现

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

 # 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):
         if head is None or head.next is None:
             return head

         h1 = ListNode(0)
         dummyh1 = h1
         h2 = ListNode(0)
         dummyh2 = h2

         p = head
         while p is not None:
             if p.val < x :
                 h1.next = p
                 h1 = h1.next
             else:
                 h2.next = p
                 h2 = h2.next
             p = p.next

         h1.next = dummyh2.next
         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