feat: add sql solutions to lc problems

.prettierignore

@@ -15,8 +15,6 @@ node_modules/
 /solution/bash_problem_readme_template_en.md
 /solution/0100-0199/0177.Nth Highest Salary/Solution.sql
 /solution/0600-0699/0627.Swap Salary/Solution.sql
-/solution/1000-1099/1076.Project Employees II/Solution.sql
-/solution/1000-1099/1082.Sales Analysis I/Solution.sql
 /solution/1100-1199/1173.Immediate Food Delivery I/Solution.sql
 /solution/1400-1499/1454.Active Users/Solution.sql
 /solution/1400-1499/1484.Group Sold Products By The Date/Solution.sql

solution/1000-1099/1076.Project Employees II/README.md

@@ -74,26 +74,29 @@ Result table:
 
 ```sql
 # Write your MySQL query statement below
-SELECT  project_id
-FROM Project p
-GROUP BY  project_id
-HAVING COUNT(employee_id) >= all(
-SELECT  COUNT(employee_id)
+SELECT project_id
 FROM Project
-GROUP BY  project_id )
+GROUP BY 1
+HAVING
+    count(1) >= all(
+        SELECT count(1)
+        FROM Project
+        GROUP BY project_id
+    );
 ```
 
 ```sql
 # Write your MySQL query statement below
-SELECT project_id
-FROM
-    (
+WITH
+    T AS (
         SELECT
             project_id,
-            dense_rank() OVER (ORDER BY COUNT(employee_id) DESC) AS rk
+            rank() OVER (ORDER BY count(employee_id) DESC) AS rk
         FROM Project
-        GROUP BY project_id
-    ) AS t
+        GROUP BY 1
+    )
+SELECT project_id
+FROM T
 WHERE rk = 1;
 ```
 

solution/1000-1099/1076.Project Employees II/README_EN.md

@@ -83,26 +83,29 @@ Employee table:
 
 ```sql
 # Write your MySQL query statement below
-SELECT  project_id
-FROM Project p
-GROUP BY  project_id
-HAVING COUNT(employee_id) >= all(
-SELECT  COUNT(employee_id)
+SELECT project_id
 FROM Project
-GROUP BY  project_id )
+GROUP BY 1
+HAVING
+    count(1) >= all(
+        SELECT count(1)
+        FROM Project
+        GROUP BY project_id
+    );
 ```
 
 ```sql
 # Write your MySQL query statement below
-SELECT project_id
-FROM
-    (
+WITH
+    T AS (
         SELECT
             project_id,
-            dense_rank() OVER (ORDER BY COUNT(employee_id) DESC) AS rk
+            rank() OVER (ORDER BY count(employee_id) DESC) AS rk
         FROM Project
-        GROUP BY project_id
-    ) AS t
+        GROUP BY 1
+    )
+SELECT project_id
+FROM T
 WHERE rk = 1;
 ```
 

solution/1000-1099/1076.Project Employees II/Solution.sql

@@ -1,8 +1,12 @@
 # Write your MySQL query statement below
-SELECT  project_id
-FROM Project p
-GROUP BY  project_id
-HAVING COUNT(employee_id) >= all(
-SELECT  COUNT(employee_id)
-FROM Project
-GROUP BY  project_id )
+WITH
+    T AS (
+        SELECT
+            project_id,
+            rank() OVER (ORDER BY count(employee_id) DESC) AS rk
+        FROM Project
+        GROUP BY 1
+    )
+SELECT project_id
+FROM T
+WHERE rk = 1;

solution/1000-1099/1077.Project Employees III/README.md

@@ -74,28 +74,31 @@ employee_id 为 1 和 3 的员工在 project_id 为 1 的项目中拥有最丰
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：内连接 + 窗口函数**
+
+我们先将 `Project` 表和 `Employee` 表进行内连接，然后使用窗口函数 `rank()` 对 `Project` 表进行分组，按照 `experience_years` 降序排列，最后取出每个项目中经验最丰富的雇员。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    project_id,
-    employee_id
-FROM
-    (
+WITH
+    T AS (
         SELECT
-            p.project_id,
-            p.employee_id,
+            project_id,
+            employee_id,
             rank() OVER (
-                PARTITION BY p.project_id
-                ORDER BY e.experience_years DESC
+                PARTITION BY project_id
+                ORDER BY experience_years DESC
             ) AS rk
         FROM
-            Project AS p
-            LEFT JOIN Employee AS e ON p.employee_id = e.employee_id
-    ) AS t
+            Project
+            JOIN Employee USING (employee_id)
+    )
+SELECT project_id, employee_id
+FROM T
 WHERE rk = 1;
 ```
 

solution/1000-1099/1077.Project Employees III/README_EN.md

@@ -85,22 +85,21 @@ Employee table:
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    project_id,
-    employee_id
-FROM
-    (
+WITH
+    T AS (
         SELECT
-            p.project_id,
-            p.employee_id,
+            project_id,
+            employee_id,
             rank() OVER (
-                PARTITION BY p.project_id
-                ORDER BY e.experience_years DESC
+                PARTITION BY project_id
+                ORDER BY experience_years DESC
             ) AS rk
         FROM
-            Project AS p
-            LEFT JOIN Employee AS e ON p.employee_id = e.employee_id
-    ) AS t
+            Project
+            JOIN Employee USING (employee_id)
+    )
+SELECT project_id, employee_id
+FROM T
 WHERE rk = 1;
 ```
 

solution/1000-1099/1077.Project Employees III/Solution.sql

@@ -1,18 +1,17 @@
 # Write your MySQL query statement below
-SELECT
-    project_id,
-    employee_id
-FROM
-    (
+WITH
+    T AS (
         SELECT
-            p.project_id,
-            p.employee_id,
+            project_id,
+            employee_id,
             rank() OVER (
-                PARTITION BY p.project_id
-                ORDER BY e.experience_years DESC
+                PARTITION BY project_id
+                ORDER BY experience_years DESC
             ) AS rk
         FROM
-            Project AS p
-            LEFT JOIN Employee AS e ON p.employee_id = e.employee_id
-    ) AS t
+            Project
+            JOIN Employee USING (employee_id)
+    )
+SELECT project_id, employee_id
+FROM T
 WHERE rk = 1;

solution/1000-1099/1082.Sales Analysis I/README.md

@@ -88,13 +88,31 @@ Product 表：
 
 ```sql
 # Write your MySQL query statement below
-SELECT  seller_id
+SELECT seller_id
 FROM Sales
-GROUP BY  seller_id
-HAVING SUM(price) >= ALL (
-SELECT  SUM(price)
-FROM Sales
-GROUP BY  seller_id )
+GROUP BY seller_id
+HAVING
+    SUM(price) >= ALL(
+        SELECT SUM(price)
+        FROM Sales
+        GROUP BY seller_id
+    );
+```
+
+```sql
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            seller_id,
+            sum(price) AS tot,
+            rank() OVER (ORDER BY sum(price) DESC) AS rk
+        FROM Sales
+        GROUP BY seller_id
+    )
+SELECT seller_id
+FROM T
+WHERE rk = 1;
 ```
 
 <!-- tabs:end -->

solution/1000-1099/1082.Sales Analysis I/README_EN.md

@@ -84,13 +84,31 @@ Sales table:
 
 ```sql
 # Write your MySQL query statement below
-SELECT  seller_id
+SELECT seller_id
 FROM Sales
-GROUP BY  seller_id
-HAVING SUM(price) >= ALL (
-SELECT  SUM(price)
-FROM Sales
-GROUP BY  seller_id )
+GROUP BY seller_id
+HAVING
+    SUM(price) >= ALL(
+        SELECT SUM(price)
+        FROM Sales
+        GROUP BY seller_id
+    );
+```
+
+```sql
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            seller_id,
+            sum(price) AS tot,
+            rank() OVER (ORDER BY sum(price) DESC) AS rk
+        FROM Sales
+        GROUP BY seller_id
+    )
+SELECT seller_id
+FROM T
+WHERE rk = 1;
 ```
 
 <!-- tabs:end -->

solution/1000-1099/1082.Sales Analysis I/Solution.sql

@@ -1,8 +1,12 @@
-# Write your MySQL query statement below
-SELECT  seller_id
-FROM Sales
-GROUP BY  seller_id
-HAVING SUM(price) >= ALL (
-SELECT  SUM(price)
-FROM Sales
-GROUP BY  seller_id )
+WITH
+    T AS (
+        SELECT
+            seller_id,
+            sum(price) AS tot,
+            rank() OVER (ORDER BY sum(price) DESC) AS rk
+        FROM Sales
+        GROUP BY seller_id
+    )
+SELECT seller_id
+FROM T
+WHERE rk = 1;

solution/2500-2599/2544.Alternating Digit Sum/README.md

@@ -56,7 +56,9 @@
 
 **方法一：模拟**
 
-从最高有效位开始，每次取出一位数字，根据其相邻数字的符号，决定当前数字的符号，然后将当前数字加入答案。
+直接根据题目描述模拟即可。
+
+我们定义一个初始符号 $sign=1$，然后从最高有效位开始，每次取出一位数字 $x$，与 $sign$ 相乘，将结果加到答案中，然后将 $sign$ 取反，继续处理下一位数字，直到处理完所有数字。
 
 时间复杂度 $O(\log n)$，空间复杂度 $O(\log n)$。其中 $n$ 为给定数字。
 
@@ -66,6 +68,12 @@
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
+```python
+class Solution:
+    def alternateDigitSum(self, n: int) -> int:
+        return sum((-1) ** i * int(x) for i, x in enumerate(str(n)))
+```
+
 ```python
 class Solution:
     def alternateDigitSum(self, n: int) -> int:

solution/2500-2599/2544.Alternating Digit Sum/README_EN.md

@@ -60,6 +60,12 @@
 
 ### **Python3**
 
+```python
+class Solution:
+    def alternateDigitSum(self, n: int) -> int:
+        return sum((-1) ** i * int(x) for i, x in enumerate(str(n)))
+```
+
 ```python
 class Solution:
     def alternateDigitSum(self, n: int) -> int:

solution/2500-2599/2544.Alternating Digit Sum/Solution.py

@@ -1,8 +1,3 @@
-class Solution:
-    def alternateDigitSum(self, n: int) -> int:
-        ans, sign = 0, 1
-        for c in str(n):
-            x = int(c)
-            ans += sign * x
-            sign *= -1
-        return ans
+class Solution:
+    def alternateDigitSum(self, n: int) -> int:
+        return sum((-1) ** i * int(x) for i, x in enumerate(str(n)))