> For the complete documentation index, see [llms.txt](https://dfine.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://dfine.gitbook.io/leetcode/876.middle_of_the_linked_list.md).

# 876.Middle of the Linked List

**876.Middle of the Linked List**

难度:Easy

> 给定一个带有头结点 head 的非空单链表，返回链表的中间结点。

如果有两个中间结点，则返回第二个中间结点。

```
示例 1：

输入：[1,2,3,4,5]
输出：此列表中的结点 3 (序列化形式：[3,4,5])
返回的结点值为 3 。 (测评系统对该结点序列化表述是 [3,4,5])。
注意，我们返回了一个 ListNode 类型的对象 ans，这样：
ans.val = 3, ans.next.val = 4, ans.next.next.val = 5, 以及 ans.next.next.next = NULL.
示例 2：

输入：[1,2,3,4,5,6]
输出：此列表中的结点 4 (序列化形式：[4,5,6])
由于该列表有两个中间结点，值分别为 3 和 4，我们返回第二个结点。
 

提示：

给定链表的结点数介于 1 和 100 之间。
```

简单的快慢指针：

```
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* middleNode(ListNode* head) {
        if(!head->next) return head;
        if(!head->next->next) return head->next;
        ListNode* slow=head->next;
        ListNode* fast=head->next->next;
        while(fast->next && fast->next->next)
        {
            slow=slow->next;
            fast=fast->next->next;
        }
        
        return fast->next ? slow->next : slow;
        
        
    }
};
```


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