diff --git a/README.md b/README.md
index 1f8c7e5..8514e55 100644
--- a/README.md
+++ b/README.md
@@ -86,7 +86,7 @@ LeetCode 与 LintCode 解题记录。此为个人练习仓库，代码中对重
   - LeetCode 13. 罗马数字转整数 <https://leetcode-cn.com/problems/roman-to-integer/>
   - LintCode 419. 罗马数字转整数 <https://www.lintcode.com/problem/roman-to-integer/description>
 
-## 数组
+## 数组/队列/集合/映射
 
 - [电话号码的字母组合](src/array/letter-combinations-of-a-phone-number.js)
 
@@ -283,6 +283,10 @@ LeetCode 与 LintCode 解题记录。此为个人练习仓库，代码中对重
   - LeetCode 221. 最大正方形 <https://leetcode-cn.com/problems/maximal-square/>
   - LintCode 436. 最大正方形 <https://www.lintcode.com/problem/maximal-square/description>
 
+- [数据流中第一个唯一的数字](src/array/first-unique-number-in-data-stream.js)
+
+  - LintCode 685. 数据流中第一个唯一的数字 <https://www.lintcode.com/problem/first-unique-number-in-data-stream/description>
+
 ## 栈
 
 - [最大矩阵](src/stack/maximal-rectangle.js)
diff --git a/src/array/first-unique-number-in-data-stream.js b/src/array/first-unique-number-in-data-stream.js
new file mode 100644
index 0000000..7e508d3
--- /dev/null
+++ b/src/array/first-unique-number-in-data-stream.js
@@ -0,0 +1,23 @@
+/**
+ * @param nums: a continuous stream of numbers
+ * @param number: a number
+ * @return: returns the first unique number
+ */
+export const firstUniqueNumber = function (nums, number) {
+  const due = new Set()
+  const queue = new Set()
+  for (const n of nums) {
+    if (queue.has(n)) {
+      queue.delete(n)
+      due.add(n)
+    } else if (!due.has(n)) {
+      queue.add(n)
+    }
+
+    if (n === number) break
+  }
+
+  if (!due.has(number) && !queue.has(number)) return -1
+
+  return Array.from(queue)[0] || -1
+}
diff --git a/test/array/first-unique-number-in-data-stream.test.js b/test/array/first-unique-number-in-data-stream.test.js
new file mode 100644
index 0000000..5f093f0
--- /dev/null
+++ b/test/array/first-unique-number-in-data-stream.test.js
@@ -0,0 +1,7 @@
+import { firstUniqueNumber } from '../../src/array/first-unique-number-in-data-stream'
+
+test('数据流中第一个唯一的数字', () => {
+  expect(firstUniqueNumber([1, 2, 2, 1, 3, 4, 4, 5, 6], 5)).toBe(3)
+  expect(firstUniqueNumber([1, 2, 2, 1, 3, 4, 4, 5, 6], 7)).toBe(-1)
+  expect(firstUniqueNumber([1, 2, 2, 1, 3, 4], 3)).toBe(3)
+})