diff --git a/leetcode/0382.Linked-List-Random-Node/382. Linked List Random Node.go b/leetcode/0382.Linked-List-Random-Node/382. Linked List Random Node.go
new file mode 100644
index 00000000..0e54f754
--- /dev/null
+++ b/leetcode/0382.Linked-List-Random-Node/382. Linked List Random Node.go	
@@ -0,0 +1,46 @@
+package leetcode
+
+import (
+	"math/rand"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+type Solution struct {
+	head *ListNode
+}
+
+/** @param head The linked list's head.
+  Note that the head is guaranteed to be not null, so it contains at least one node. */
+func Constructor(head *ListNode) Solution {
+	return Solution{head: head}
+}
+
+/** Returns a random node's value. */
+func (this *Solution) GetRandom() int {
+	scope, selectPoint, curr := 1, 0, this.head
+	for curr != nil {
+		if rand.Float64() < 1.0/float64(scope) {
+			selectPoint = curr.Val
+		}
+		scope += 1
+		curr = curr.Next
+	}
+	return selectPoint
+}
+
+/**
+ * Your Solution object will be instantiated and called as such:
+ * obj := Constructor(head);
+ * param_1 := obj.GetRandom();
+ */
diff --git a/leetcode/0382.Linked-List-Random-Node/382. Linked List Random Node_test.go b/leetcode/0382.Linked-List-Random-Node/382. Linked List Random Node_test.go
new file mode 100644
index 00000000..f3371344
--- /dev/null
+++ b/leetcode/0382.Linked-List-Random-Node/382. Linked List Random Node_test.go	
@@ -0,0 +1,28 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+func Test_Problem382(t *testing.T) {
+	header := structures.Ints2List([]int{1, 2, 3, 4, 5, 6, 7, 8, 9, 0})
+	obj := Constructor(header)
+	fmt.Printf("obj = %v\n", structures.List2Ints(header))
+	param1 := obj.GetRandom()
+	fmt.Printf("param_1 = %v\n", param1)
+	param1 = obj.GetRandom()
+	fmt.Printf("param_1 = %v\n", param1)
+	param1 = obj.GetRandom()
+	fmt.Printf("param_1 = %v\n", param1)
+	param1 = obj.GetRandom()
+	fmt.Printf("param_1 = %v\n", param1)
+	param1 = obj.GetRandom()
+	fmt.Printf("param_1 = %v\n", param1)
+	param1 = obj.GetRandom()
+	fmt.Printf("param_1 = %v\n", param1)
+
+	fmt.Printf("\n\n\n")
+}
diff --git a/leetcode/0382.Linked-List-Random-Node/README.md b/leetcode/0382.Linked-List-Random-Node/README.md
new file mode 100644
index 00000000..db5ecc06
--- /dev/null
+++ b/leetcode/0382.Linked-List-Random-Node/README.md
@@ -0,0 +1,105 @@
+# [382. Linked List Random Node](https://leetcode.com/problems/linked-list-random-node/)
+
+
+## 题目
+
+Given a singly linked list, return a random node's value from the linked list. Each node must have the **same probability** of being chosen.
+
+Implement the `Solution` class:
+
+- `Solution(ListNode head)` Initializes the object with the integer array nums.
+- `int getRandom()` Chooses a node randomly from the list and returns its value. All the nodes of the list should be equally likely to be choosen.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/03/16/getrand-linked-list.jpg](https://assets.leetcode.com/uploads/2021/03/16/getrand-linked-list.jpg)
+
+```
+Input
+["Solution", "getRandom", "getRandom", "getRandom", "getRandom", "getRandom"]
+[[[1, 2, 3]], [], [], [], [], []]
+Output
+[null, 1, 3, 2, 2, 3]
+
+Explanation
+Solution solution = new Solution([1, 2, 3]);
+solution.getRandom(); // return 1
+solution.getRandom(); // return 3
+solution.getRandom(); // return 2
+solution.getRandom(); // return 2
+solution.getRandom(); // return 3
+// getRandom() should return either 1, 2, or 3 randomly. Each element should have equal probability of returning.
+
+```
+
+**Constraints:**
+
+- The number of nodes in the linked list will be in the range `[1, 104]`.
+- `-10^4 <= Node.val <= 10^4`
+- At most `10^4` calls will be made to `getRandom`.
+
+**Follow up:**
+
+- What if the linked list is extremely large and its length is unknown to you?
+- Could you solve this efficiently without using extra space?
+
+## 题目大意
+
+给定一个单链表，随机选择链表的一个节点，并返回相应的节点值。保证每个节点被选的概率一样。
+
+进阶: 如果链表十分大且长度未知，如何解决这个问题？你能否使用常数级空间复杂度实现？
+
+## 解题思路
+
+- rand.Float64() 可以返回 [0.0,1.0) 之间的随机数。利用这个函数完成我们的随机化取节点的过程。
+
+## 代码
+
+```go
+package leetcode
+
+import (
+	"math/rand"
+
+	"github.com/halfrost/LeetCode-Go/structures"
+)
+
+// ListNode define
+type ListNode = structures.ListNode
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+type Solution struct {
+	head *ListNode
+}
+
+/** @param head The linked list's head.
+  Note that the head is guaranteed to be not null, so it contains at least one node. */
+func Constructor(head *ListNode) Solution {
+	return Solution{head: head}
+}
+
+/** Returns a random node's value. */
+func (this *Solution) GetRandom() int {
+	scope, selectPoint, curr := 1, 0, this.head
+	for curr != nil {
+		if rand.Float64() < 1.0/float64(scope) {
+			selectPoint = curr.Val
+		}
+		scope += 1
+		curr = curr.Next
+	}
+	return selectPoint
+}
+
+/**
+ * Your Solution object will be instantiated and called as such:
+ * obj := Constructor(head);
+ * param_1 := obj.GetRandom();
+ */
+```
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