diff --git a/content/posts/leetcode.md b/content/posts/leetcode.md
index ef12ec1..d783371 100644
--- a/content/posts/leetcode.md
+++ b/content/posts/leetcode.md
@@ -650,3 +650,46 @@ func hasCycle(head *ListNode) bool {
 	return false
 }
 ```
+
+## day10 2024-03-07
+
+### 876. Middle of the Linked List
+
+Given the head of a singly linked list, return the middle node of the linked list.
+
+If there are two middle nodes, return the second middle node.
+
+![0307rm4x7SjNblpu](https://testingcf.jsdelivr.net/gh/game-loader/picbase@master/uPic/0307rm4x7SjNblpu.png)
+
+### 题解
+
+本题寻找链表中间位置的元素, 是一个经典快慢指针的题目. 只需要让快指针前进的速度为2, 慢指针为1, 则快指针到达链表末尾时慢指针正好指向中间位置. 要注意链表元素个数为奇数和偶数时的处理方法的不同.
+
+### 代码
+
+```go
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func middleNode(head *ListNode) *ListNode {
+    fast := head
+    slow := head
+    for fast.Next != nil && fast.Next.Next != nil{
+        fast = fast.Next.Next
+        slow = slow.Next
+    }
+    // consider list number is even
+    if fast.Next != nil{
+        slow = slow.Next
+    }
+    return slow
+}
+```
+
+### 总结
+
+本题是一道经典的快慢指针的简单题目, 进一步深化了快慢指针的应用.