题目地址：https://leetcode.com/problems/middle-of-the-linked-list/description/
题目描述
Given a non-empty, singly linked list with head node head, return a middle node of linked list.
Ifthere are two middle nodes, return the second middle node.
Example 1:
Input: [1,2,3,4,5]
Output: Node 3 from this list (Serialization: [3,4,5])
The returned node has value 3. (The judge's serialization of this node is [3,4,5]).
Note that we returned a ListNode object ans, such that:
ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, and ans.next.next.next = NULL.
Example 2:
Input: [1,2,3,4,5,6]
Output: Node 4 from this list (Serialization: [4,5,6])
Since the list has two middle nodes with values 3 and 4, we return the second one.
Note:
- The number of nodes in the given list will be between 1 and 100.
题目大意
给出链表中的中间节点。如果链表长度为偶数，返回的应该是中间的第二个节点。
解题方法
使用哑结点
这个题很简单，就是快慢指针两个移动即可。
需要注意的是，如果链表长度为偶数，返回的应该是中间的第二个节点。这个做的方法，是判断fast指针是指向了最末尾的None还是链表结尾的node。如果是链表结尾说明是偶数个点，如果是none说明是奇数个点。
代码如下：
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def middleNode(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
dummy = ListNode(0)
dummy.next = head
slow, fast = dummy, dummy
while fast and fast.next:
slow = slow.next
fast = fast.next.next
return slow.next if fast else slow
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不使用哑结点
如果不使用哑结点，初始时直接将两个指针指向head能获得更快的速度。这么做的前提是题目已经告诉了节点的数量最少是1，否则如果初始化是空的话fast.next会报错。
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def middleNode(self, head):
"""
:type head: ListNode
:rtype: ListNode
"""
slow, fast = head, head
while fast and fast.next:
fast = fast.next.next
slow = slow.next
return slow
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