feat: update solutions to lc problems: No.0175,0570

* No.0175.Combine Two Tables
* No.0570.Managers with at Least 5 Direct Reports

Author: yanglbme

solution/0100-0199/0175.Combine Two Tables/README.md

@@ -78,19 +78,22 @@ addressId = 1 包含了 personId = 2 的地址信息。</pre>
 
 <!-- 这里可写通用的实现逻辑 -->
 
-左连接。
+**方法一：左连接**
 
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
-SELECT p.FirstName,
-    p.LastName,
-    a.City,
-    a.State
-FROM Person p
-    LEFT JOIN Address a ON p.PersonId = a.PersonId;
+# Write your MySQL query statement below
+select
+    firstName,
+    lastName,
+    city,
+    state
+from
+    Person p
+    left join Address a on p.personId = a.personId
 ```
 
 <!-- tabs:end -->

solution/0100-0199/0175.Combine Two Tables/README_EN.md

@@ -81,12 +81,15 @@ addressId = 1 contains information about the address of personId = 2.
 ### **SQL**
 
 ```sql
-SELECT p.FirstName,
-    p.LastName,
-    a.City,
-    a.State
-FROM Person p
-    LEFT JOIN Address a ON p.PersonId = a.PersonId;
+# Write your MySQL query statement below
+select
+    firstName,
+    lastName,
+    city,
+    state
+from
+    Person p
+    left join Address a on p.personId = a.personId
 ```
 
 <!-- tabs:end -->

solution/0100-0199/0175.Combine Two Tables/Solution.sql

@@ -1,6 +1,9 @@
-SELECT p.FirstName,
-    p.LastName,
-    a.City,
-    a.State
-FROM Person p
-    LEFT JOIN Address a ON p.PersonId = a.PersonId;
+# Write your MySQL query statement below
+select
+    firstName,
+    lastName,
+    city,
+    state
+from
+    Person p
+    left join Address a on p.personId = a.personId

solution/0500-0599/0570.Managers with at Least 5 Direct Reports/README.md

@@ -64,7 +64,23 @@ Employee 表:
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+select
+    name
+from
+    Employee e1
+    join (
+        select
+            managerId
+        from
+            Employee
+        where
+            managerId is not null
+        group by
+            managerId
+        having
+            count(1) >= 5
+    ) e2 on e1.id = e2.managerId;
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0570.Managers with at Least 5 Direct Reports/README_EN.md

@@ -60,7 +60,23 @@ Employee table:
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+select
+    name
+from
+    Employee e1
+    join (
+        select
+            managerId
+        from
+            Employee
+        where
+            managerId is not null
+        group by
+            managerId
+        having
+            count(1) >= 5
+    ) e2 on e1.id = e2.managerId;
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0570.Managers with at Least 5 Direct Reports/Solution.sql

@@ -0,0 +1,17 @@
+# Write your MySQL query statement below
+select
+    name
+from
+    Employee e1
+    join (
+        select
+            managerId
+        from
+            Employee
+        where
+            managerId is not null
+        group by
+            managerId
+        having
+            count(1) >= 5
+    ) e2 on e1.id = e2.managerId;