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