feat: add solutions to lc problems: No.1148, No.1149

Author: yanglbme

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

@@ -51,17 +51,26 @@ Views 表：
 +------+
 </pre>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
 
+“`DISTINCT` + `ORDER BY`” 实现。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    DISTINCT(author_id) as id
+FROM
+    Views
+WHERE
+    author_id = viewer_id
+ORDER BY
+    id;
 ```
 
-<!-- tabs:end -->
+<!-- tabs:end -->

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

@@ -48,15 +48,22 @@ Result table:
 +------+
 </pre>
 
-
 ## Solutions
 
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    DISTINCT(author_id) as id
+FROM
+    Views
+WHERE
+    author_id = viewer_id
+ORDER BY
+    id;
 ```
 
-<!-- tabs:end -->
+<!-- tabs:end -->

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

@@ -0,0 +1,9 @@
+# Write your MySQL query statement below
+SELECT
+    DISTINCT(author_id) as id
+FROM
+    Views
+WHERE
+    author_id = viewer_id
+ORDER BY
+    id;

solution/1100-1199/1149.Article Views II/README.md

@@ -49,17 +49,28 @@ Result table:
 | 6    |
 +------+</pre>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
 
+“`DISTINCT` + `GROUP BY`” 实现。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    DISTINCT(viewer_id) as id
+FROM
+    Views
+GROUP BY
+    view_date, viewer_id
+HAVING
+    COUNT(DISTINCT(article_id)) > 1
+ORDER BY
+    id;
 ```
 
-<!-- tabs:end -->
+<!-- tabs:end -->

solution/1100-1199/1149.Article Views II/README_EN.md

@@ -56,7 +56,17 @@ Result table:
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    DISTINCT(viewer_id) as id
+FROM
+    Views
+GROUP BY
+    view_date, viewer_id
+HAVING
+    COUNT(DISTINCT(article_id)) > 1
+ORDER BY
+    id;
 ```
 
 <!-- tabs:end -->

solution/1100-1199/1149.Article Views II/Solution.sql

@@ -0,0 +1,11 @@
+# Write your MySQL query statement below
+SELECT
+    DISTINCT(viewer_id) as id
+FROM
+    Views
+GROUP BY
+    view_date, viewer_id
+HAVING
+    COUNT(DISTINCT(article_id)) > 1
+ORDER BY
+    id;