## Description

Given a non-empty, singly linked list with head node `head`

, return a middle node of linked list.

If there 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`

.

## Solutions

找到链表的中间值。

### 1. Two Pointers

使用快慢指针即可。

```
# Time: O(n)
# Space: O(1)
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def middleNode(self, head: ListNode) -> ListNode:
if not head or (head and not head.next):
return head
fast, slow = head, head
while fast:
if not fast.next:
break
fast = fast.next.next
slow = slow.next
return slow
# Runtime: 32 ms, faster than 35.49% of Python3
# Memory Usage: 13.6 MB, less than 7.14% of Python3
```