feat: add sql solutions: No.1045 & No.1050

Author: yanglbme

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

@@ -72,7 +72,20 @@ Result 表：
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    customer_id
+FROM
+    Customer
+GROUP BY
+    customer_id
+HAVING
+    COUNT(DISTINCT(product_key)) = (
+        SELECT
+            COUNT(1)
+        FROM
+            Product
+    );
 ```
 
 <!-- tabs:end -->

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

@@ -77,7 +77,20 @@ The customers who bought all the products (5 and 6) are customers with id 1 and
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    customer_id
+FROM
+    Customer
+GROUP BY
+    customer_id
+HAVING
+    COUNT(DISTINCT(product_key)) = (
+        SELECT
+            COUNT(1)
+        FROM
+            Product
+    );
 ```
 
 <!-- tabs:end -->

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

@@ -0,0 +1,14 @@
+# Write your MySQL query statement below
+SELECT
+    customer_id
+FROM
+    Customer
+GROUP BY
+    customer_id
+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

@@ -52,12 +52,20 @@ Result 表：
 
 <!-- 这里可写通用的实现逻辑 -->
 
+`GROUP BY` + `HAVING` 解决。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    actor_id, director_id
+FROM
+    ActorDirector
+GROUP BY actor_id, director_id
+HAVING count(1) >= 3;
 ```
 
 <!-- tabs:end -->

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

@@ -90,12 +90,20 @@ The only pair is (1, 1) where they cooperated exactly 3 times.
 
 ## Solutions
 
+Use `GROUP BY` & `HAVING`.
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    actor_id, director_id
+FROM
+    ActorDirector
+GROUP BY actor_id, director_id
+HAVING count(1) >= 3;
 ```
 
 <!-- tabs:end -->

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

@@ -0,0 +1,7 @@
+# Write your MySQL query statement below
+SELECT
+    actor_id, director_id
+FROM
+    ActorDirector
+GROUP BY actor_id, director_id
+HAVING count(1) >= 3;