> For the complete documentation index, see [llms.txt](https://protegejj.gitbook.io/oj-practices/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://protegejj.gitbook.io/oj-practices/chapter1/linked-list/876-middle-of-the-linked-list.md).

# 876     Middle of the Linked List

## 876. [Middle of the Linked List](https://leetcode.com/problems/middle-of-the-linked-list/description/)

## 1. Question

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`.

## 2. Implementation

思路: 其实就是Linked List里找中点的基础操作

```java
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode middleNode(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }

        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode fast = dummy, slow = dummy;

        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }

        return fast == null ? slow : slow.next;
    }
}
```

## 3. Time & Space Complexity

时间复杂度O(n), 空间复杂度O(1)