feat: add sql solutions to leetcode problems: No.0175,0176

* https://leetcode-cn.com/problems/combine-two-tables/
* https://leetcode-cn.com/problems/second-highest-salary/

Author: yanglbme

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

@@ -30,25 +30,27 @@ PersonId 是上表主键
 AddressId 是上表主键
 </pre>
 
-<p>&nbsp;</p>
-
 <p>编写一个 SQL 查询，满足条件：无论 person 是否有地址信息，都需要基于上述两表提供&nbsp;person 的以下信息：</p>
 
-<p>&nbsp;</p>
-
 <pre>FirstName, LastName, City, State
 </pre>
 
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
 
+左连接。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
-```
-select p.FirstName, p.LastName, a.City, a.State from Person p left join Address a on p.PersonId = a.PersonId;
+```sql
+# Write your MySQL query statement below
+SELECT p.FirstName, p.LastName, a.City, a.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

@@ -66,8 +66,12 @@ FirstName, LastName, City, State
 
 ### **SQL**
 
-```
-select p.FirstName, p.LastName, a.City, a.State from Person p left join Address a on p.PersonId = a.PersonId;
+```sql
+# Write your MySQL query statement below
+SELECT p.FirstName, p.LastName, a.City, a.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 +1,5 @@
-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 p.FirstName, p.LastName, a.City, a.State
+FROM Person p
+LEFT JOIN Address a
+ON p.PersonId = a.PersonId;

solution/0100-0199/0176.Second Highest Salary/README.md

@@ -33,9 +33,29 @@
 
 ### **SQL**
 
+解法 1：使用 LIMIT 语句和子查询。
+
+```sql
+# Write your MySQL query statement below
+SELECT
+(
+    SELECT DISTINCT Salary
+    FROM Employee
+    ORDER BY Salary DESC
+    LIMIT 1 OFFSET 1
+) AS SecondHighestSalary;
 ```
-select (select distinct Salary from Employee order by Salary desc limit 1 offset 1) as
-SecondHighestSalary;
+
+解法 2：使用 `MAX()` 函数，从小于 `MAX()` 的 Salary 中挑选最大值 `MAX()` 即可。
+
+```sql
+# Write your MySQL query statement below
+SELECT MAX(Salary) AS SecondHighestSalary
+FROM Employee
+WHERE Salary < (
+    SELECT MAX(Salary)
+    FROM Employee
+);
 ```
 
 <!-- tabs:end -->

solution/0100-0199/0176.Second Highest Salary/README_EN.md

@@ -46,9 +46,29 @@
 
 ### **SQL**
 
+Solution 1: Use Sub Query and LIMIT.
+
+```sql
+# Write your MySQL query statement below
+SELECT
+(
+    SELECT DISTINCT Salary
+    FROM Employee
+    ORDER BY Salary DESC
+    LIMIT 1 OFFSET 1
+) AS SecondHighestSalary;
 ```
-select (select distinct Salary from Employee order by Salary desc limit 1 offset 1) as
-SecondHighestSalary;
+
+Solution 2: Use `MAX()` function.
+
+```sql
+# Write your MySQL query statement below
+SELECT MAX(Salary) AS SecondHighestSalary
+FROM Employee
+WHERE Salary < (
+    SELECT MAX(Salary)
+    FROM Employee
+);
 ```
 
 <!-- tabs:end -->

solution/0100-0199/0176.Second Highest Salary/Solution.sql

@@ -1,2 +1,8 @@
-select (select distinct Salary from Employee order by Salary desc limit 1 offset 1) as 
-SecondHighestSalary;
+# Write your MySQL query statement below
+SELECT
+(
+    SELECT DISTINCT Salary
+    FROM Employee
+    ORDER BY Salary DESC
+    LIMIT 1 OFFSET 1
+) AS SecondHighestSalary;