feat: add solutions to lc problems

solution/0600-0699/0601.Human Traffic of Stadium/README.md

@@ -73,19 +73,18 @@ id</strong> 为 5、6、7、8 的四行 id 连续，并且每行都有 &gt;= 100
 ```sql
 # Write your MySQL query statement below
 WITH
-    s AS (
-        SELECT *, id - row_number() OVER (ORDER BY id) AS rk
+    S AS (
+        SELECT
+            *,
+            id - (row_number() OVER (ORDER BY id)) AS rk
         FROM Stadium
         WHERE people >= 100
     ),
-    t AS (
-        SELECT *, count(*) OVER (PARTITION BY rk) AS cnt
-        FROM s
-    )
+    T AS (SELECT *, count(1) OVER (PARTITION BY rk) AS cnt FROM S)
 SELECT id, visit_date, people
-FROM t
+FROM T
 WHERE cnt >= 3
-ORDER BY visit_date;
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

solution/0600-0699/0601.Human Traffic of Stadium/README_EN.md

@@ -68,19 +68,18 @@ The rows with ids 2 and 3 are not included because we need at least three consec
 ```sql
 # Write your MySQL query statement below
 WITH
-    s AS (
-        SELECT *, id - row_number() OVER (ORDER BY id) AS rk
+    S AS (
+        SELECT
+            *,
+            id - (row_number() OVER (ORDER BY id)) AS rk
         FROM Stadium
         WHERE people >= 100
     ),
-    t AS (
-        SELECT *, count(*) OVER (PARTITION BY rk) AS cnt
-        FROM s
-    )
+    T AS (SELECT *, count(1) OVER (PARTITION BY rk) AS cnt FROM S)
 SELECT id, visit_date, people
-FROM t
+FROM T
 WHERE cnt >= 3
-ORDER BY visit_date;
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

solution/0600-0699/0601.Human Traffic of Stadium/Solution.sql

@@ -1,15 +1,14 @@
 # Write your MySQL query statement below
 WITH
-    s AS (
-        SELECT *, id - row_number() OVER (ORDER BY id) AS rk
+    S AS (
+        SELECT
+            *,
+            id - (row_number() OVER (ORDER BY id)) AS rk
         FROM Stadium
         WHERE people >= 100
     ),
-    t AS (
-        SELECT *, count(*) OVER (PARTITION BY rk) AS cnt
-        FROM s
-    )
+    T AS (SELECT *, count(1) OVER (PARTITION BY rk) AS cnt FROM S)
 SELECT id, visit_date, people
-FROM t
+FROM T
 WHERE cnt >= 3
-ORDER BY visit_date;
+ORDER BY 1;

solution/0600-0699/0602.Friend Requests II Who Has the Most Friends/README.md

@@ -71,21 +71,17 @@ RequestAccepted 表：
 ### **SQL**
 
 ```sql
-SELECT
-    ids AS id,
-    COUNT(*) AS num
-FROM
-    (
-        SELECT
-            requester_id AS ids
-        FROM RequestAccepted
-        UNION ALL
-        SELECT
-            accepter_id
-        FROM RequestAccepted
-    ) AS t
-GROUP BY ids
-ORDER BY num DESC
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT requester_id, accepter_id FROM RequestAccepted
+        UNION
+        SELECT accepter_id, requester_id FROM RequestAccepted
+    )
+SELECT requester_id AS id, count(accepter_id) AS num
+FROM T
+GROUP BY 1
+ORDER BY 2 DESC
 LIMIT 1;
 ```
 

solution/0600-0699/0602.Friend Requests II Who Has the Most Friends/README_EN.md

@@ -60,21 +60,17 @@ The person with id 3 is a friend of people 1, 2, and 4, so he has three friends
 ### **SQL**
 
 ```sql
-SELECT
-    ids AS id,
-    COUNT(*) AS num
-FROM
-    (
-        SELECT
-            requester_id AS ids
-        FROM RequestAccepted
-        UNION ALL
-        SELECT
-            accepter_id
-        FROM RequestAccepted
-    ) AS t
-GROUP BY ids
-ORDER BY num DESC
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT requester_id, accepter_id FROM RequestAccepted
+        UNION
+        SELECT accepter_id, requester_id FROM RequestAccepted
+    )
+SELECT requester_id AS id, count(accepter_id) AS num
+FROM T
+GROUP BY 1
+ORDER BY 2 DESC
 LIMIT 1;
 ```
 

solution/0600-0699/0602.Friend Requests II Who Has the Most Friends/Solution.sql

@@ -1,16 +1,12 @@
-SELECT
-    ids AS id,
-    COUNT(*) AS num
-FROM
-    (
-        SELECT
-            requester_id AS ids
-        FROM RequestAccepted
-        UNION ALL
-        SELECT
-            accepter_id
-        FROM RequestAccepted
-    ) AS t
-GROUP BY ids
-ORDER BY num DESC
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT requester_id, accepter_id FROM RequestAccepted
+        UNION
+        SELECT accepter_id, requester_id FROM RequestAccepted
+    )
+SELECT requester_id AS id, count(accepter_id) AS num
+FROM T
+GROUP BY 1
+ORDER BY 2 DESC
 LIMIT 1;

solution/0600-0699/0603.Consecutive Available Seats/README.md

@@ -66,9 +66,23 @@ Cinema 表:
 SELECT DISTINCT a.seat_id
 FROM
     Cinema AS a
-    JOIN Cinema AS b
-        ON ABS(a.seat_id - b.seat_id) = 1 AND a.free = TRUE AND b.free = TRUE
-ORDER BY a.seat_id;
+    JOIN Cinema AS b ON abs(a.seat_id - b.seat_id) = 1 AND a.free AND b.free
+ORDER BY 1;
+```
+
+```sql
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            seat_id,
+            (free + (lag(free) OVER (ORDER BY seat_id))) AS a,
+            (free + (lead(free) OVER (ORDER BY seat_id))) AS b
+        FROM Cinema
+    )
+SELECT seat_id
+FROM T
+WHERE a = 2 OR b = 2;
 ```
 
 <!-- tabs:end -->

solution/0600-0699/0603.Consecutive Available Seats/README_EN.md

@@ -63,9 +63,23 @@ Cinema table:
 SELECT DISTINCT a.seat_id
 FROM
     Cinema AS a
-    JOIN Cinema AS b
-        ON ABS(a.seat_id - b.seat_id) = 1 AND a.free = TRUE AND b.free = TRUE
-ORDER BY a.seat_id;
+    JOIN Cinema AS b ON abs(a.seat_id - b.seat_id) = 1 AND a.free AND b.free
+ORDER BY 1;
+```
+
+```sql
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            seat_id,
+            (free + (lag(free) OVER (ORDER BY seat_id))) AS a,
+            (free + (lead(free) OVER (ORDER BY seat_id))) AS b
+        FROM Cinema
+    )
+SELECT seat_id
+FROM T
+WHERE a = 2 OR b = 2;
 ```
 
 <!-- tabs:end -->

solution/0600-0699/0603.Consecutive Available Seats/Solution.sql

@@ -1,7 +1,12 @@
 # Write your MySQL query statement below
-SELECT DISTINCT a.seat_id
-FROM
-    Cinema AS a
-    JOIN Cinema AS b
-        ON ABS(a.seat_id - b.seat_id) = 1 AND a.free = TRUE AND b.free = TRUE
-ORDER BY a.seat_id;
+WITH
+    T AS (
+        SELECT
+            seat_id,
+            (free + (lag(free) OVER (ORDER BY seat_id))) AS a,
+            (free + (lead(free) OVER (ORDER BY seat_id))) AS b
+        FROM Cinema
+    )
+SELECT seat_id
+FROM T
+WHERE a = 2 OR b = 2;

solution/0600-0699/0607.Sales Person/README.md

@@ -149,4 +149,19 @@ WHERE
     );
 ```
 
+```sql
+# Write your MySQL query statement below
+SELECT name
+FROM SalesPerson
+WHERE
+    sales_id NOT IN (
+        SELECT sales_id
+        FROM
+            Company AS c
+            JOIN Orders AS o USING (com_id)
+        GROUP BY sales_id
+        HAVING sum(name = 'RED') > 0
+    );
+```
+
 <!-- tabs:end -->

solution/0600-0699/0607.Sales Person/README_EN.md

@@ -144,4 +144,19 @@ WHERE
     );
 ```
 
+```sql
+# Write your MySQL query statement below
+SELECT name
+FROM SalesPerson
+WHERE
+    sales_id NOT IN (
+        SELECT sales_id
+        FROM
+            Company AS c
+            JOIN Orders AS o USING (com_id)
+        GROUP BY sales_id
+        HAVING sum(name = 'RED') > 0
+    );
+```
+
 <!-- tabs:end -->

solution/0600-0699/0608.Tree Node/README.md

@@ -79,36 +79,16 @@
 
 ### **SQL**
 
-```sql
-SELECT
-    id,
-    (
-        CASE
-            WHEN p_id IS NULL THEN 'Root'
-            WHEN id IN (
-                SELECT p_id
-                FROM tree
-            ) THEN 'Inner'
-            ELSE 'Leaf'
-        END
-    ) AS type
-FROM tree;
-```
-
 ```sql
 # Write your MySQL query statement below
 SELECT
     id,
     CASE
         WHEN p_id IS NULL THEN 'Root'
-        WHEN id IN (
-            SELECT
-                p_id
-            FROM tree
-        ) THEN 'Inner'
+        WHEN id IN (SELECT p_id FROM Tree) THEN 'Inner'
         ELSE 'Leaf'
-    END AS Type
-FROM tree;
+    END AS type
+FROM Tree;
 ```
 
 <!-- tabs:end -->

solution/0600-0699/0608.Tree Node/README_EN.md

@@ -90,36 +90,16 @@ Tree table:
 
 ### **SQL**
 
-```sql
-SELECT
-    id,
-    (
-        CASE
-            WHEN p_id IS NULL THEN 'Root'
-            WHEN id IN (
-                SELECT p_id
-                FROM tree
-            ) THEN 'Inner'
-            ELSE 'Leaf'
-        END
-    ) AS type
-FROM tree;
-```
-
 ```sql
 # Write your MySQL query statement below
 SELECT
     id,
     CASE
         WHEN p_id IS NULL THEN 'Root'
-        WHEN id IN (
-            SELECT
-                p_id
-            FROM tree
-        ) THEN 'Inner'
+        WHEN id IN (SELECT p_id FROM Tree) THEN 'Inner'
         ELSE 'Leaf'
-    END AS Type
-FROM tree;
+    END AS type
+FROM Tree;
 ```
 
 <!-- tabs:end -->

solution/0600-0699/0608.Tree Node/Solution.sql

@@ -1,13 +1,9 @@
+# Write your MySQL query statement below
 SELECT
     id,
-    (
-        CASE
-            WHEN p_id IS NULL THEN 'Root'
-            WHEN id IN (
-                SELECT p_id
-                FROM tree
-            ) THEN 'Inner'
-            ELSE 'Leaf'
-        END
-    ) AS type
-FROM tree;
+    CASE
+        WHEN p_id IS NULL THEN 'Root'
+        WHEN id IN (SELECT p_id FROM Tree) THEN 'Inner'
+        ELSE 'Leaf'
+    END AS type
+FROM Tree;

solution/0600-0699/0613.Shortest Distance in a Line/README.md

@@ -59,4 +59,12 @@ FROM
     JOIN Point AS p2 ON p1.x != p2.x;
 ```
 
+```sql
+# Write your MySQL query statement below
+SELECT x - lag(x) OVER (ORDER BY x) AS shortest
+FROM Point
+ORDER BY 1
+LIMIT 1, 1;
+```
+
 <!-- tabs:end -->