feat: add solutions to lc problem: No.1142

No.1142. User Activity for the Past 30 Days II

Author: yanglbme

solution/1000-1099/1093.Statistics from a Large Sample/README.md

@@ -83,7 +83,9 @@
 -   更新 $cnt = cnt + count[k]$；
 -   如果 $count[k] \gt count[mode]$，那么更新 $mode = k$。
 
-遍历结束后，我们再根据 $cnt$ 的奇偶性来更新中位数 $median$，如果 $cnt$ 是奇数，那么中位数就是第 $\lfloor \frac{cnt}{2} \rfloor + 1$ 个数字，如果 $cnt$ 是偶数，那么中位数就是第 $\lfloor \frac{cnt}{2} \rfloor$ 和第 $\lfloor \frac{cnt}{2} \rfloor + 1$ 个数字的平均值。
+遍历结束后，我们再根据 $cnt$ 的奇偶性来计算中位数 $median$，如果 $cnt$ 是奇数，那么中位数就是第 $\lfloor \frac{cnt}{2} \rfloor + 1$ 个数字，如果 $cnt$ 是偶数，那么中位数就是第 $\lfloor \frac{cnt}{2} \rfloor$ 和第 $\lfloor \frac{cnt}{2} \rfloor + 1$ 个数字的平均值。
+
+> 这里我们通过一个简单的辅助函数 $find(i)$ 来找到第 $i$ 个数字，具体实现可以参考下面的代码。
 
 最后，我们将 $mi, mx, \frac{s}{cnt}, median, mode$ 放入答案数组中返回即可。
 

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

@@ -72,15 +72,15 @@ Activity table:
 
 ```sql
 SELECT
-	activity_date AS DAY,
-	COUNT( DISTINCT user_id ) AS active_users 
+    activity_date AS day,
+    COUNT(DISTINCT user_id) AS active_users
 FROM
-	Activity 
+    Activity
 WHERE
-	DATEDIFF( '2019-07-27', activity_date ) > 0 
-	AND DATEDIFF( '2019-07-27', activity_date ) < 30 
+    DATEDIFF('2019-07-27', activity_date) >= 0 and
+    DATEDIFF('2019-07-27', activity_date) < 30
 GROUP BY
-	activity_date;
+    activity_date;
 ```
 
 <!-- tabs:end -->

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

@@ -68,15 +68,15 @@ Activity table:
 
 ```sql
 SELECT
-	activity_date AS DAY,
-	COUNT( DISTINCT user_id ) AS active_users 
+    activity_date AS day,
+    COUNT(DISTINCT user_id) AS active_users
 FROM
-	Activity 
+    Activity
 WHERE
-	DATEDIFF( '2019-07-27', activity_date ) > 0 
-	AND DATEDIFF( '2019-07-27', activity_date ) < 30 
+    DATEDIFF('2019-07-27', activity_date) >= 0 and
+    DATEDIFF('2019-07-27', activity_date) < 30
 GROUP BY
-	activity_date;
+    activity_date;
 ```
 
 <!-- tabs:end -->

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

@@ -1,10 +1,10 @@
 SELECT
-	activity_date AS DAY,
-	COUNT( DISTINCT user_id ) AS active_users 
+    activity_date AS day,
+    COUNT(DISTINCT user_id) AS active_users
 FROM
-	Activity 
+    Activity
 WHERE
-	DATEDIFF( '2019-07-27', activity_date ) > 0 
-	AND DATEDIFF( '2019-07-27', activity_date ) < 30 
+    DATEDIFF('2019-07-27', activity_date) >= 0 and
+    DATEDIFF('2019-07-27', activity_date) < 30
 GROUP BY
-	activity_date;
+    activity_date;

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

@@ -71,7 +71,14 @@ Activity 表：
 ### **SQL**
 
 ```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
 ```
 
 <!-- tabs:end -->

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

@@ -67,7 +67,14 @@ Activity table:
 ### **SQL**
 
 ```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
 ```
 
 <!-- tabs:end -->

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

@@ -0,0 +1,8 @@
+# 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