diff --git a/codes/leetcode/README.md b/codes/leetcode/README.md
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+# Leetcode 数据库篇题解
+
+> [题库地址](https://leetcode-cn.com/problemset/database/)
diff --git a/codes/leetcode/big-countries.sql b/codes/leetcode/big-countries.sql
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+--  这里有张 World 表
+--
+--  +-----------------+------------+------------+--------------+---------------+
+--  | name            | continent  | area       | population   | gdp           |
+--  +-----------------+------------+------------+--------------+---------------+
+--  | Afghanistan     | Asia       | 652230     | 25500100     | 20343000      |
+--  | Albania         | Europe     | 28748      | 2831741      | 12960000      |
+--  | Algeria         | Africa     | 2381741    | 37100000     | 188681000     |
+--  | Andorra         | Europe     | 468        | 78115        | 3712000       |
+--  | Angola          | Africa     | 1246700    | 20609294     | 100990000     |
+--  +-----------------+------------+------------+--------------+---------------+
+--  如果一个国家的面积超过300万平方公里，或者人口超过2500万，那么这个国家就是大国家。
+--
+--  编写一个SQL查询，输出表中所有大国家的名称、人口和面积。
+--
+--  例如，根据上表，我们应该输出:
+--
+--  +--------------+-------------+--------------+
+--  | name         | population  | area         |
+--  +--------------+-------------+--------------+
+--  | Afghanistan  | 25500100    | 652230       |
+--  | Algeria      | 37100000    | 2381741      |
+--  +--------------+-------------+--------------+
+
+SELECT name, population, area FROM World
+WHERE area > 3000000 OR population > 25000000;
+
diff --git a/codes/leetcode/combine-two-tables.sql b/codes/leetcode/combine-two-tables.sql
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diff --git a/codes/leetcode/duplicate-emails.sql b/codes/leetcode/duplicate-emails.sql
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+--    编写一个 SQL 查询，查找 Person 表中所有重复的电子邮箱。
+--
+--    示例：
+--
+--    +----+---------+
+--    | Id | Email   |
+--    +----+---------+
+--    | 1  | a@b.com |
+--    | 2  | c@d.com |
+--    | 3  | a@b.com |
+--    +----+---------+
+--    根据以上输入，你的查询应返回以下结果：
+--
+--    +---------+
+--    | Email   |
+--    +---------+
+--    | a@b.com |
+--    +---------+
+--    说明：所有电子邮箱都是小写字母。
+
+SELECT Email FROM Person GROUP BY Email HAVING COUNT(*) > 1;
diff --git a/codes/leetcode/not-boring-movies.sql b/codes/leetcode/not-boring-movies.sql
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+--    某城市开了一家新的电影院，吸引了很多人过来看电影。该电影院特别注意用户体验，专门有个 LED显示板做电影推荐，上面公布着影评和相关电影描述。
+--
+--    作为该电影院的信息部主管，您需要编写一个 SQL查询，找出所有影片描述为非 boring (不无聊) 的并且 id 为奇数 的影片，结果请按等级 rating 排列。
+--
+--
+--
+--    例如，下表 cinema:
+--
+--    +---------+-----------+--------------+-----------+
+--    |   id    | movie     |  description |  rating   |
+--    +---------+-----------+--------------+-----------+
+--    |   1     | War       |   great 3D   |   8.9     |
+--    |   2     | Science   |   fiction    |   8.5     |
+--    |   3     | irish     |   boring     |   6.2     |
+--    |   4     | Ice song  |   Fantacy    |   8.6     |
+--    |   5     | House card|   Interesting|   9.1     |
+--    +---------+-----------+--------------+-----------+
+--    对于上面的例子，则正确的输出是为：
+--
+--    +---------+-----------+--------------+-----------+
+--    |   id    | movie     |  description |  rating   |
+--    +---------+-----------+--------------+-----------+
+--    |   5     | House card|   Interesting|   9.1     |
+--    |   1     | War       |   great 3D   |   8.9     |
+--    +---------+-----------+--------------+-----------+
+
+SELECT * FROM cinema
+WHERE description != 'boring' AND id % 2 = 1
+ORDER BY rating DESC;
diff --git a/codes/leetcode/swap-salary.sql b/codes/leetcode/swap-salary.sql
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+--    给定一个 salary表，如下所示，有m=男性 和 f=女性的值 。
+--    交换所有的 f 和 m 值(例如，将所有 f 值更改为 m，反之亦然)。要求使用一个更新查询，并且没有中间临时表。
+--
+--    例如:
+--
+--    | id | name | sex | salary |
+--    |----|------|-----|--------|
+--    | 1  | A    | m   | 2500   |
+--    | 2  | B    | f   | 1500   |
+--    | 3  | C    | m   | 5500   |
+--    | 4  | D    | f   | 500    |
+--    运行你所编写的查询语句之后，将会得到以下表:
+--
+--    | id | name | sex | salary |
+--    |----|------|-----|--------|
+--    | 1  | A    | f   | 2500   |
+--    | 2  | B    | m   | 1500   |
+--    | 3  | C    | f   | 5500   |
+--    | 4  | D    | m   | 500    |
+
+UPDATE salary
+SET sex =
+    CASE sex
+        WHEN 'm'
+        THEN 'f'
+        ELSE 'm'
+    END;