feat: update solutions to lc problems: No.0176~0178 (doocs#1711)

* No.0176.Second Highest Salary
* No.0177.Nth Highest Salary
* No.0178.Rank Scores

Author: yanglbme

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

@@ -79,11 +79,11 @@ Employee 表：
 
 **方法二：使用 MAX() 函数和子查询**
 
-使用 `MAX()` 函数，从小于 `MAX()` 的 Salary 中挑选最大值 `MAX()` 即可。
+我们也可以使用 `MAX()` 函数，从小于 `MAX()` 的薪水中挑选一个最大的薪水即可。
 
-**方法三：使用 IFNULL() 和窗口函数**
+**方法三：使用 DISTINCT 和窗口函数**
 
-我们也可以先通过 `dense_rank()` 函数计算出每个员工的薪水排名，然后再筛选出排名为 $2$ 的员工即可，如果不存在第二高的薪水，那么就返回 `null`。
+我们还可以先通过 `DENSE_RANK()` 函数计算出每个员工的薪水排名，然后再筛选出排名为 $2$ 的员工薪水即可。
 
 <!-- tabs:start -->
 
@@ -93,38 +93,24 @@ Employee 表：
 # Write your MySQL query statement below
 SELECT
     (
-        SELECT DISTINCT Salary
+        SELECT DISTINCT salary
         FROM Employee
-        ORDER BY Salary DESC
-        LIMIT 1 OFFSET 1
+        ORDER BY salary DESC
+        LIMIT 1, 1
     ) AS SecondHighestSalary;
 ```
 
 ```sql
 # Write your MySQL query statement below
-SELECT MAX(Salary) AS SecondHighestSalary
+SELECT MAX(salary) AS SecondHighestSalary
 FROM Employee
-WHERE
-    Salary < (
-        SELECT MAX(Salary)
-        FROM Employee
-    );
+WHERE salary < (SELECT MAX(salary) FROM Employee);
 ```
 
 ```sql
 # Write your MySQL query statement below
-WITH
-    S AS (
-        SELECT salary, dense_rank() OVER (ORDER BY salary DESC) AS rk
-        FROM Employee
-    )
-SELECT
-    ifnull(
-        SELECT salary
-        FROM S
-        WHERE rk = 2
-        LIMIT 1, NULL
-    ) AS SecondHighestSalary;
+WITH T AS (SELECT salary, DENSE_RANK() OVER (ORDER BY salary DESC) AS rk FROM Employee)
+SELECT (SELECT DISTINCT salary FROM T WHERE rk = 2) AS SecondHighestSalary;
 ```
 
 <!-- tabs:end -->

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

@@ -78,38 +78,24 @@ Employee table:
 # Write your MySQL query statement below
 SELECT
     (
-        SELECT DISTINCT Salary
+        SELECT DISTINCT salary
         FROM Employee
-        ORDER BY Salary DESC
-        LIMIT 1 OFFSET 1
+        ORDER BY salary DESC
+        LIMIT 1, 1
     ) AS SecondHighestSalary;
 ```
 
 ```sql
 # Write your MySQL query statement below
-SELECT MAX(Salary) AS SecondHighestSalary
+SELECT MAX(salary) AS SecondHighestSalary
 FROM Employee
-WHERE
-    Salary < (
-        SELECT MAX(Salary)
-        FROM Employee
-    );
+WHERE salary < (SELECT MAX(salary) FROM Employee);
 ```
 
 ```sql
 # Write your MySQL query statement below
-WITH
-    S AS (
-        SELECT salary, dense_rank() OVER (ORDER BY salary DESC) AS rk
-        FROM Employee
-    )
-SELECT
-    ifnull(
-        SELECT salary
-        FROM S
-        WHERE rk = 2
-        LIMIT 1, NULL
-    ) AS SecondHighestSalary;
+WITH T AS (SELECT salary, DENSE_RANK() OVER (ORDER BY salary DESC) AS rk FROM Employee)
+SELECT (SELECT DISTINCT salary FROM T WHERE rk = 2) AS SecondHighestSalary;
 ```
 
 <!-- tabs:end -->

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

@@ -1,8 +1,3 @@
 # Write your MySQL query statement below
-SELECT
-    (
-        SELECT DISTINCT Salary
-        FROM Employee
-        ORDER BY Salary DESC
-        LIMIT 1 OFFSET 1
-    ) AS SecondHighestSalary;
+WITH T AS (SELECT salary, DENSE_RANK() OVER (ORDER BY salary DESC) AS rk FROM Employee)
+SELECT (SELECT DISTINCT salary FROM T WHERE rk = 2) AS SecondHighestSalary;

solution/0100-0199/0177.Nth Highest Salary/README.md

@@ -70,6 +70,10 @@ n = 2
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：排序 + LIMIT**
+
+我们可以先对 `salary` 进行降序排序，然后使用 `LIMIT` 语句获取第 $n$ 高的工资。
+
 <!-- tabs:start -->
 
 ### **SQL**
@@ -81,9 +85,9 @@ BEGIN
   RETURN (
       # Write your MySQL query statement below.
       SELECT (
-          SELECT DISTINCT Salary
+          SELECT DISTINCT salary
           FROM Employee
-          ORDER BY Salary DESC
+          ORDER BY salary DESC
           LIMIT 1 OFFSET N
       )
   );

solution/0100-0199/0177.Nth Highest Salary/README_EN.md

@@ -77,9 +77,9 @@ BEGIN
   RETURN (
       # Write your MySQL query statement below.
       SELECT (
-          SELECT DISTINCT Salary
+          SELECT DISTINCT salary
           FROM Employee
-          ORDER BY Salary DESC
+          ORDER BY salary DESC
           LIMIT 1 OFFSET N
       )
   );

solution/0100-0199/0177.Nth Highest Salary/Solution.sql

@@ -4,9 +4,9 @@ BEGIN
   RETURN (
       # Write your MySQL query statement below.
       SELECT (
-          SELECT DISTINCT Salary
+          SELECT DISTINCT salary
           FROM Employee
-          ORDER BY Salary DESC
+          ORDER BY salary DESC
           LIMIT 1 OFFSET N
       )
   );

solution/0100-0199/0178.Rank Scores/README.md

@@ -66,9 +66,7 @@ Scores 表:
 
 <!-- 这里可写通用的实现逻辑 -->
 
-<!-- tabs:start -->
-
-### **MySQL8**
+**方法一：使用窗口函数 `DENSE_RANK()`**
 
 使用 `DENSE_RANK()` 函数，语法如下：
 
@@ -86,20 +84,22 @@ DENSE_RANK() OVER (
 
 与 `RANK()` 函数不同，`DENSE_RANK()` 函数始终返回连续的排名值。
 
-题解如下：
+**方法二：变量**
+
+MySQL 8 开始才提供了 `ROW_NUMBER()`，`RANK()`，`DENSE_RANK()` 等[窗口函数](https://dev.mysql.com/doc/refman/8.0/en/window-function-descriptions.html)，在之前的版本，可以使用变量实现类似的功能。
+
+<!-- tabs:start -->
+
+### **SQL**
 
 ```sql
 # Write your MySQL query statement below
 SELECT
-    Score,
-    DENSE_RANK() OVER (ORDER BY Score DESC) AS 'Rank'
+    score,
+    DENSE_RANK() OVER (ORDER BY score DESC) AS 'rank'
 FROM Scores;
 ```
 
-### **MySQL5**
-
-MySQL 8 开始才提供了 `ROW_NUMBER()`，`RANK()`，`DENSE_RANK()` 等[窗口函数](https://dev.mysql.com/doc/refman/8.0/en/window-function-descriptions.html)，在之前的版本，可以使用变量实现类似的功能：
-
 ```sql
 SELECT
     Score,

solution/0100-0199/0178.Rank Scores/README_EN.md

@@ -64,7 +64,7 @@ Scores table:
 
 <!-- tabs:start -->
 
-### **MySQL8**
+**Solution 1: Use `DENSE_RANK()`**
 
 Use `DENSE_RANK()` to solve this problem.
 
@@ -75,20 +75,22 @@ DENSE_RANK() OVER (
 )
 ```
 
-Solution:
+**Solution 2: Use variables**
+
+MySQL only provides [window function](https://dev.mysql.com/doc/refman/8.0/en/window-function-descriptions.html) after version 8. In previous versions, variables can be used to achieve similar functions.
+
+<!-- tabs:start -->
+
+### **SQL**
 
 ```sql
 # Write your MySQL query statement below
 SELECT
-    Score,
-    DENSE_RANK() OVER (ORDER BY Score DESC) AS 'Rank'
+    score,
+    DENSE_RANK() OVER (ORDER BY score DESC) AS 'rank'
 FROM Scores;
 ```
 
-### **MySQL5**
-
-MySQL only provides [window function](https://dev.mysql.com/doc/refman/8.0/en/window-function-descriptions.html) after version 8. In previous versions, variables can be used to achieve similar functions:
-
 ```sql
 SELECT
     Score,

solution/0100-0199/0178.Rank Scores/Solution.sql

@@ -1,5 +1,5 @@
 # Write your MySQL query statement below
 SELECT
-    Score,
-    DENSE_RANK() OVER (ORDER BY Score DESC) AS 'Rank'
+    score,
+    DENSE_RANK() OVER (ORDER BY score DESC) AS 'rank'
 FROM Scores;