feat: add sql solutions to lc problems (doocs#1195)

* No.0619.Biggest Single Number
* No.0620.Not Boring Movies
* No.1045.Customers Who Bought All Products
* No.1068.Product Sales Analysis I
* No.1069.Product Sales Analysis II
* No.1070.Product Sales Analysis III
* No.1075.Project Employees I

Author: yanglbme

solution/0600-0699/0619.Biggest Single Number/README.md

@@ -96,14 +96,28 @@ MyNumbers table:
 ### **SQL**
 
 ```sql
-SELECT MAX(a.num) AS num
+# Write your MySQL query statement below
+SELECT max(num) AS num
 FROM
     (
         SELECT num
         FROM MyNumbers
         GROUP BY num
-        HAVING count(*) = 1
-    ) AS a;
+        HAVING count(1) = 1
+    ) AS t;
+```
+
+```sql
+# Write your MySQL query statement below
+SELECT
+    CASE
+        WHEN count(1) = 1 THEN num
+        ELSE NULL
+    END AS num
+FROM MyNumbers
+GROUP BY num
+ORDER BY 1 DESC
+LIMIT 1;
 ```
 
 <!-- tabs:end -->

solution/0600-0699/0619.Biggest Single Number/README_EN.md

@@ -84,14 +84,28 @@ MyNumbers table:
 ### **SQL**
 
 ```sql
-SELECT MAX(a.num) AS num
+# Write your MySQL query statement below
+SELECT max(num) AS num
 FROM
     (
         SELECT num
         FROM MyNumbers
         GROUP BY num
-        HAVING count(*) = 1
-    ) AS a;
+        HAVING count(1) = 1
+    ) AS t;
+```
+
+```sql
+# Write your MySQL query statement below
+SELECT
+    CASE
+        WHEN count(1) = 1 THEN num
+        ELSE NULL
+    END AS num
+FROM MyNumbers
+GROUP BY num
+ORDER BY 1 DESC
+LIMIT 1;
 ```
 
 <!-- tabs:end -->

solution/0600-0699/0619.Biggest Single Number/Solution.sql

@@ -1,8 +1,10 @@
-SELECT MAX(a.num) AS num
-FROM
-    (
-        SELECT num
-        FROM MyNumbers
-        GROUP BY num
-        HAVING count(*) = 1
-    ) AS a;
+# Write your MySQL query statement below
+SELECT
+    CASE
+        WHEN count(1) = 1 THEN num
+        ELSE NULL
+    END AS num
+FROM MyNumbers
+GROUP BY num
+ORDER BY 1 DESC
+LIMIT 1;

solution/0600-0699/0620.Not Boring Movies/README.md

@@ -48,9 +48,10 @@
 ### **SQL**
 
 ```sql
+# Write your MySQL query statement below
 SELECT *
-FROM cinema
-WHERE description NOT LIKE '%boring%' AND mod(id, 2) = 1
+FROM Cinema
+WHERE description != 'boring' AND id % 2 = 1
 ORDER BY rating DESC;
 ```
 

solution/0600-0699/0620.Not Boring Movies/README_EN.md

@@ -61,9 +61,10 @@ We have three movies with odd-numbered IDs: 1, 3, and 5. The movie with ID = 3 i
 ### **SQL**
 
 ```sql
+# Write your MySQL query statement below
 SELECT *
-FROM cinema
-WHERE description NOT LIKE '%boring%' AND mod(id, 2) = 1
+FROM Cinema
+WHERE description != 'boring' AND id % 2 = 1
 ORDER BY rating DESC;
 ```
 

solution/0600-0699/0620.Not Boring Movies/Solution.sql

@@ -1,4 +1,5 @@
+# Write your MySQL query statement below
 SELECT *
-FROM cinema
-WHERE description NOT LIKE '%boring%' AND mod(id, 2) = 1
+FROM Cinema
+WHERE description != 'boring' AND id % 2 = 1
 ORDER BY rating DESC;

solution/0600-0699/0626.Exchange Seats/README.md

@@ -103,4 +103,17 @@ from
     seat
 ```
 
+```sql
+# Write your MySQL query statement below
+SELECT
+    CASE
+        WHEN id & 1 = 0 THEN id - 1
+        WHEN row_number() OVER (ORDER BY id) != count(id) OVER () THEN id + 1
+        ELSE id
+    END AS id,
+    student
+FROM Seat
+ORDER BY 1;
+```
+
 <!-- tabs:end -->

solution/0600-0699/0626.Exchange Seats/README_EN.md

@@ -99,4 +99,17 @@ from
     seat
 ```
 
+```sql
+# Write your MySQL query statement below
+SELECT
+    CASE
+        WHEN id & 1 = 0 THEN id - 1
+        WHEN row_number() OVER (ORDER BY id) != count(id) OVER () THEN id + 1
+        ELSE id
+    END AS id,
+    student
+FROM Seat
+ORDER BY 1;
+```
+
 <!-- tabs:end -->

solution/1000-1099/1045.Customers Who Bought All Products/README.md

@@ -66,22 +66,20 @@ Result 表：
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：GROUP BY + HAVING**
+
+我们将 `Customer` 表按照 `customer_id` 进行分组，然后使用 `HAVING` 子句筛选出购买了所有产品的客户。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    customer_id
+SELECT customer_id
 FROM Customer
-GROUP BY customer_id
-HAVING
-    COUNT(DISTINCT product_key) = (
-        SELECT
-            COUNT(1)
-        FROM Product
-    );
+GROUP BY 1
+HAVING count(DISTINCT product_key) = (SELECT count(1) FROM Product);
 ```
 
 <!-- tabs:end -->

solution/1000-1099/1045.Customers Who Bought All Products/README_EN.md

@@ -79,16 +79,10 @@ The customers who bought all the products (5 and 6) are customers with IDs 1 and
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    customer_id
+SELECT customer_id
 FROM Customer
-GROUP BY customer_id
-HAVING
-    COUNT(DISTINCT product_key) = (
-        SELECT
-            COUNT(1)
-        FROM Product
-    );
+GROUP BY 1
+HAVING count(DISTINCT product_key) = (SELECT count(1) FROM Product);
 ```
 
 <!-- tabs:end -->

solution/1000-1099/1045.Customers Who Bought All Products/Solution.sql

@@ -1,11 +1,5 @@
 # Write your MySQL query statement below
-SELECT
-    customer_id
+SELECT customer_id
 FROM Customer
-GROUP BY customer_id
-HAVING
-    COUNT(DISTINCT product_key) = (
-        SELECT
-            COUNT(1)
-        FROM Product
-    );
+GROUP BY 1
+HAVING count(DISTINCT product_key) = (SELECT count(1) FROM Product);

solution/1000-1099/1050.Actors and Directors Who Cooperated At Least Three Times/README.md

@@ -51,18 +51,19 @@ Result 表：
 
 <!-- 这里可写通用的实现逻辑 -->
 
-`GROUP BY` + `HAVING` 解决。
+**方法一：使用 `GROUP BY` + `HAVING`**
+
+我们将 `ActorDirector` 表按照 `actor_id` 和 `director_id` 进行分组，然后使用 `HAVING` 过滤出合作次数大于等于 $3$ 次的组。
 
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
-SELECT
-    actor_id,
-    director_id
+# Write your MySQL query statement below
+SELECT actor_id, director_id
 FROM ActorDirector
-GROUP BY actor_id, director_id
+GROUP BY 1, 2
 HAVING count(1) >= 3;
 ```
 

solution/1000-1099/1050.Actors and Directors Who Cooperated At Least Three Times/README_EN.md

@@ -60,11 +60,10 @@ Use `GROUP BY` & `HAVING`.
 ### **SQL**
 
 ```sql
-SELECT
-    actor_id,
-    director_id
+# Write your MySQL query statement below
+SELECT actor_id, director_id
 FROM ActorDirector
-GROUP BY actor_id, director_id
+GROUP BY 1, 2
 HAVING count(1) >= 3;
 ```
 

solution/1000-1099/1050.Actors and Directors Who Cooperated At Least Three Times/Solution.sql

@@ -1,6 +1,5 @@
-SELECT
-    actor_id,
-    director_id
+# Write your MySQL query statement below
+SELECT actor_id, director_id
 FROM ActorDirector
-GROUP BY actor_id, director_id
+GROUP BY 1, 2
 HAVING count(1) >= 3;

solution/1000-1099/1068.Product Sales Analysis I/README.md

@@ -78,19 +78,20 @@ Result 表：
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：使用 `JOIN` 内连接**
+
+我们直接使用 `JOIN` 连接 `Sales` 和 `Product` 两张表，连接字段为 `product_id`，然后选择需要的字段即可。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    p.product_name AS product_name,
-    s.year AS year,
-    s.price AS price
+SELECT product_name, year, price
 FROM
-    Sales AS s
-    LEFT JOIN Product AS p ON s.product_id = p.product_id;
+    Sales
+    JOIN Product USING (product_id);
 ```
 
 <!-- tabs:end -->

solution/1000-1099/1068.Product Sales Analysis I/README_EN.md

@@ -88,13 +88,10 @@ From sale_id = 7, we can conclude that Apple was sold for 9000 in the year 2011.
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    p.product_name AS product_name,
-    s.year AS year,
-    s.price AS price
+SELECT product_name, year, price
 FROM
-    Sales AS s
-    LEFT JOIN Product AS p ON s.product_id = p.product_id;
+    Sales
+    JOIN Product USING (product_id);
 ```
 
 <!-- tabs:end -->

solution/1000-1099/1068.Product Sales Analysis I/Solution.sql

@@ -1,8 +1,5 @@
 # Write your MySQL query statement below
-SELECT
-    p.product_name AS product_name,
-    s.year AS year,
-    s.price AS price
+SELECT product_name, year, price
 FROM
-    Sales AS s
-    LEFT JOIN Product AS p ON s.product_id = p.product_id;
+    Sales
+    JOIN Product USING (product_id);

solution/1000-1099/1069.Product Sales Analysis II/README.md

@@ -71,17 +71,19 @@ Result 表：
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：使用 `GROUP BY`**
+
+我们可以使用 `GROUP BY`，按照 `product_id` 分组，然后每一组对 `quantity` 求和。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    product_id,
-    sum(quantity) AS total_quantity
+SELECT product_id, sum(quantity) AS total_quantity
 FROM Sales
-GROUP BY product_id;
+GROUP BY 1;
 ```
 
 <!-- tabs:end -->

solution/1000-1099/1069.Product Sales Analysis II/README_EN.md

@@ -83,11 +83,9 @@ Product table:
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    product_id,
-    sum(quantity) AS total_quantity
+SELECT product_id, sum(quantity) AS total_quantity
 FROM Sales
-GROUP BY product_id;
+GROUP BY 1;
 ```
 
 <!-- tabs:end -->

solution/1000-1099/1069.Product Sales Analysis II/Solution.sql

@@ -1,6 +1,4 @@
 # Write your MySQL query statement below
-SELECT
-    product_id,
-    sum(quantity) AS total_quantity
+SELECT product_id, sum(quantity) AS total_quantity
 FROM Sales
-GROUP BY product_id;
+GROUP BY 1;

solution/1000-1099/1070.Product Sales Analysis III/README.md

@@ -102,4 +102,21 @@ WHERE
     );
 ```
 
+```sql
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            *,
+            rank() OVER (
+                PARTITION BY product_id
+                ORDER BY year
+            ) AS rk
+        FROM Sales
+    )
+SELECT product_id, year AS first_year, quantity, price
+FROM T
+WHERE rk = 1;
+```
+
 <!-- tabs:end -->