feat: add sql solutions to lc problems

solution/1100-1199/1127.User Purchase Platform/README.md

@@ -65,29 +65,29 @@ Result table:
 ```sql
 # Write your MySQL query statement below
 WITH
-    s AS (
+    P AS (
         SELECT DISTINCT spend_date, 'desktop' AS platform FROM Spending
         UNION
-        SELECT DISTINCT spend_date, 'mobile' AS platform FROM Spending
+        SELECT DISTINCT spend_date, 'mobile' FROM Spending
         UNION
-        SELECT DISTINCT spend_date, 'both' AS platform FROM Spending
+        SELECT DISTINCT spend_date, 'both' FROM Spending
     ),
-    t AS (
+    T AS (
         SELECT
             user_id,
             spend_date,
-            if(count(platform) = 2, 'both', platform) AS platform,
-            sum(amount) AS amount
+            sum(amount) AS amount,
+            if(count(platform) = 1, platform, 'both') AS platform
         FROM Spending
         GROUP BY 1, 2
     )
 SELECT
-    t1.*,
+    p.*,
     ifnull(sum(amount), 0) AS total_amount,
-    count(t2.user_id) AS total_users
+    count(t.user_id) AS total_users
 FROM
-    s AS t1
-    LEFT JOIN t AS t2 USING (spend_date, platform)
+    P AS p
+    LEFT JOIN T AS t USING (spend_date, platform)
 GROUP BY 1, 2;
 ```
 

solution/1100-1199/1127.User Purchase Platform/README_EN.md

@@ -69,29 +69,29 @@ On 2019-07-02, user 2 purchased using mobile <strong>only</strong>, user 3 purch
 ```sql
 # Write your MySQL query statement below
 WITH
-    s AS (
+    P AS (
         SELECT DISTINCT spend_date, 'desktop' AS platform FROM Spending
         UNION
-        SELECT DISTINCT spend_date, 'mobile' AS platform FROM Spending
+        SELECT DISTINCT spend_date, 'mobile' FROM Spending
         UNION
-        SELECT DISTINCT spend_date, 'both' AS platform FROM Spending
+        SELECT DISTINCT spend_date, 'both' FROM Spending
     ),
-    t AS (
+    T AS (
         SELECT
             user_id,
             spend_date,
-            if(count(platform) = 2, 'both', platform) AS platform,
-            sum(amount) AS amount
+            sum(amount) AS amount,
+            if(count(platform) = 1, platform, 'both') AS platform
         FROM Spending
         GROUP BY 1, 2
     )
 SELECT
-    t1.*,
+    p.*,
     ifnull(sum(amount), 0) AS total_amount,
-    count(t2.user_id) AS total_users
+    count(t.user_id) AS total_users
 FROM
-    s AS t1
-    LEFT JOIN t AS t2 USING (spend_date, platform)
+    P AS p
+    LEFT JOIN T AS t USING (spend_date, platform)
 GROUP BY 1, 2;
 ```
 

solution/1100-1199/1127.User Purchase Platform/Solution.sql

@@ -1,26 +1,26 @@
 # Write your MySQL query statement below
 WITH
-    s AS (
+    P AS (
         SELECT DISTINCT spend_date, 'desktop' AS platform FROM Spending
         UNION
-        SELECT DISTINCT spend_date, 'mobile' AS platform FROM Spending
+        SELECT DISTINCT spend_date, 'mobile' FROM Spending
         UNION
-        SELECT DISTINCT spend_date, 'both' AS platform FROM Spending
+        SELECT DISTINCT spend_date, 'both' FROM Spending
     ),
-    t AS (
+    T AS (
         SELECT
             user_id,
             spend_date,
-            if(count(platform) = 2, 'both', platform) AS platform,
-            sum(amount) AS amount
+            sum(amount) AS amount,
+            if(count(platform) = 1, platform, 'both') AS platform
         FROM Spending
         GROUP BY 1, 2
     )
 SELECT
-    t1.*,
+    p.*,
     ifnull(sum(amount), 0) AS total_amount,
-    count(t2.user_id) AS total_users
+    count(t.user_id) AS total_users
 FROM
-    s AS t1
-    LEFT JOIN t AS t2 USING (spend_date, platform)
+    P AS p
+    LEFT JOIN T AS t USING (spend_date, platform)
 GROUP BY 1, 2;

solution/1100-1199/1132.Reported Posts II/README.md

@@ -98,18 +98,20 @@ Removals table:
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    round(avg(avg_per * 100), 2) AS average_daily_percent
-FROM
-    (
+WITH
+    T AS (
         SELECT
-            count(DISTINCT t2.post_id) / count(DISTINCT t1.post_id) AS avg_per
+            count(DISTINCT t2.post_id) / count(
+                DISTINCT t1.post_id
+            ) * 100 AS percent
         FROM
             Actions AS t1
             LEFT JOIN Removals AS t2 ON t1.post_id = t2.post_id
-        WHERE t1.extra = 'spam'
+        WHERE extra = 'spam'
         GROUP BY action_date
-    ) AS t3;
+    )
+SELECT round(avg(percent), 2) AS average_daily_percent
+FROM T;
 ```
 
 <!-- tabs:end -->

solution/1100-1199/1132.Reported Posts II/README_EN.md

@@ -93,18 +93,20 @@ Note that the output is only one number and that we do not care about the remove
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    round(avg(avg_per * 100), 2) AS average_daily_percent
-FROM
-    (
+WITH
+    T AS (
         SELECT
-            count(DISTINCT t2.post_id) / count(DISTINCT t1.post_id) AS avg_per
+            count(DISTINCT t2.post_id) / count(
+                DISTINCT t1.post_id
+            ) * 100 AS percent
         FROM
             Actions AS t1
             LEFT JOIN Removals AS t2 ON t1.post_id = t2.post_id
-        WHERE t1.extra = 'spam'
+        WHERE extra = 'spam'
         GROUP BY action_date
-    ) AS t3;
+    )
+SELECT round(avg(percent), 2) AS average_daily_percent
+FROM T;
 ```
 
 <!-- tabs:end -->

solution/1100-1199/1132.Reported Posts II/Solution.sql

@@ -1,13 +1,15 @@
 # Write your MySQL query statement below
-SELECT
-    round(avg(avg_per * 100), 2) AS average_daily_percent
-FROM
-    (
+WITH
+    T AS (
         SELECT
-            count(DISTINCT t2.post_id) / count(DISTINCT t1.post_id) AS avg_per
+            count(DISTINCT t2.post_id) / count(
+                DISTINCT t1.post_id
+            ) * 100 AS percent
         FROM
             Actions AS t1
             LEFT JOIN Removals AS t2 ON t1.post_id = t2.post_id
-        WHERE t1.extra = 'spam'
+        WHERE extra = 'spam'
         GROUP BY action_date
-    ) AS t3;
+    )
+SELECT round(avg(percent), 2) AS average_daily_percent
+FROM T;

solution/1100-1199/1141.User Activity for the Past 30 Days I/README.md

@@ -66,19 +66,22 @@ Activity table:
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：GROUP BY + HAVING**
+
+我们查询出所有在 `2019-07-27` 且在 $30$ 天内的所有活动记录，然后按照日期分组，统计每天的去重活跃用户数。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
-SELECT
-    activity_date AS day,
-    COUNT(DISTINCT user_id) AS active_users
+# Write your MySQL query statement below
+SELECT activity_date AS day, count(DISTINCT user_id) AS active_users
 FROM Activity
 WHERE
-    DATEDIFF('2019-07-27', activity_date) >= 0
-    AND DATEDIFF('2019-07-27', activity_date) < 30
-GROUP BY activity_date;
+    activity_date <= '2019-07-27'
+    AND datediff('2019-07-27', activity_date) < 30
+GROUP BY 1;
 ```
 
 <!-- tabs:end -->

solution/1100-1199/1141.User Activity for the Past 30 Days I/README_EN.md

@@ -67,14 +67,13 @@ Activity table:
 ### **SQL**
 
 ```sql
-SELECT
-    activity_date AS day,
-    COUNT(DISTINCT user_id) AS active_users
+# Write your MySQL query statement below
+SELECT activity_date AS day, count(DISTINCT user_id) AS active_users
 FROM Activity
 WHERE
-    DATEDIFF('2019-07-27', activity_date) >= 0
-    AND DATEDIFF('2019-07-27', activity_date) < 30
-GROUP BY activity_date;
+    activity_date <= '2019-07-27'
+    AND datediff('2019-07-27', activity_date) < 30
+GROUP BY 1;
 ```
 
 <!-- tabs:end -->

solution/1100-1199/1141.User Activity for the Past 30 Days I/Solution.sql

@@ -1,8 +1,7 @@
-SELECT
-    activity_date AS day,
-    COUNT(DISTINCT user_id) AS active_users
+# Write your MySQL query statement below
+SELECT activity_date AS day, count(DISTINCT user_id) AS active_users
 FROM Activity
 WHERE
-    DATEDIFF('2019-07-27', activity_date) >= 0
-    AND DATEDIFF('2019-07-27', activity_date) < 30
-GROUP BY activity_date;
+    activity_date <= '2019-07-27'
+    AND datediff('2019-07-27', activity_date) < 30
+GROUP BY 1;

solution/1100-1199/1142.User Activity for the Past 30 Days II/README.md

@@ -72,15 +72,18 @@ Activity 表：
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    IFNULL(
-        ROUND(COUNT(DISTINCT session_id) / COUNT(DISTINCT user_id), 2),
-        0
-    ) AS average_sessions_per_user
-FROM Activity
-WHERE
-    DATEDIFF('2019-07-27', activity_date) >= 0
-    AND DATEDIFF('2019-07-27', activity_date) < 30;
+WITH
+    T AS (
+        SELECT
+            count(DISTINCT session_id) AS sessions
+        FROM Activity
+        WHERE
+            activity_date <= '2019-07-27'
+            AND datediff('2019-07-27', activity_date) < 30
+        GROUP BY user_id
+    )
+SELECT ifnull(round(avg(sessions), 2), 0) AS average_sessions_per_user
+FROM T;
 ```
 
 <!-- tabs:end -->

solution/1100-1199/1142.User Activity for the Past 30 Days II/README_EN.md

@@ -68,15 +68,18 @@ Activity table:
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    IFNULL(
-        ROUND(COUNT(DISTINCT session_id) / COUNT(DISTINCT user_id), 2),
-        0
-    ) AS average_sessions_per_user
-FROM Activity
-WHERE
-    DATEDIFF('2019-07-27', activity_date) >= 0
-    AND DATEDIFF('2019-07-27', activity_date) < 30;
+WITH
+    T AS (
+        SELECT
+            count(DISTINCT session_id) AS sessions
+        FROM Activity
+        WHERE
+            activity_date <= '2019-07-27'
+            AND datediff('2019-07-27', activity_date) < 30
+        GROUP BY user_id
+    )
+SELECT ifnull(round(avg(sessions), 2), 0) AS average_sessions_per_user
+FROM T;
 ```
 
 <!-- tabs:end -->

solution/1100-1199/1142.User Activity for the Past 30 Days II/Solution.sql

@@ -1,10 +1,13 @@
 # Write your MySQL query statement below
-SELECT
-    IFNULL(
-        ROUND(COUNT(DISTINCT session_id) / COUNT(DISTINCT user_id), 2),
-        0
-    ) AS average_sessions_per_user
-FROM Activity
-WHERE
-    DATEDIFF('2019-07-27', activity_date) >= 0
-    AND DATEDIFF('2019-07-27', activity_date) < 30;
+WITH
+    T AS (
+        SELECT
+            count(DISTINCT session_id) AS sessions
+        FROM Activity
+        WHERE
+            activity_date <= '2019-07-27'
+            AND datediff('2019-07-27', activity_date) < 30
+        GROUP BY user_id
+    )
+SELECT ifnull(round(avg(sessions), 2), 0) AS average_sessions_per_user
+FROM T;

solution/1100-1199/1148.Article Views I/README.md

@@ -55,17 +55,20 @@ Views 表：
 
 <!-- 这里可写通用的实现逻辑 -->
 
-“`DISTINCT` + `ORDER BY`” 实现。
+**方法一：DISTINCT + WHERE**
+
+我们利用 `WHERE` 子句来筛选出 `author_id` 和 `viewer_id` 相等的记录，然后利用 `DISTINCT` 来去重，最后按照 `id` 排序即可。
 
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
-SELECT DISTINCT (author_id) AS id
+# Write your MySQL query statement below
+SELECT DISTINCT author_id AS id
 FROM Views
 WHERE author_id = viewer_id
-ORDER BY id;
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

solution/1100-1199/1148.Article Views I/README_EN.md

@@ -61,10 +61,11 @@ Views table:
 ### **SQL**
 
 ```sql
-SELECT DISTINCT (author_id) AS id
+# Write your MySQL query statement below
+SELECT DISTINCT author_id AS id
 FROM Views
 WHERE author_id = viewer_id
-ORDER BY id;
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

solution/1100-1199/1148.Article Views I/Solution.sql

@@ -1,4 +1,5 @@
-SELECT DISTINCT (author_id) AS id
+# Write your MySQL query statement below
+SELECT DISTINCT author_id AS id
 FROM Views
 WHERE author_id = viewer_id
-ORDER BY id;
+ORDER BY 1;