feat: add sql solution to lc problem: No.1949

No.1949.Strong Friendship

Author: yanglbme

solution/1900-1999/1949.Strong Friendship/README.md

@@ -79,7 +79,36 @@
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+# Write your MySQL query statement below
+with t as (
+    select
+        *
+    from
+        Friendship
+    union
+    all
+    select
+        user2_id,
+        user1_id
+    from
+        Friendship
+)
+select
+    t1.user1_id,
+    t1.user2_id,
+    count(1) common_friend
+from
+    t t1
+    join t t2 on t1.user2_id = t2.user1_id
+    join t t3 on t1.user1_id = t3.user1_id
+where
+    t3.user2_id = t2.user2_id
+    and t1.user1_id < t1.user2_id
+group by
+    t1.user1_id,
+    t1.user2_id
+having
+    count(1) >= 3;
 ```
 
 <!-- tabs:end -->

solution/1900-1999/1949.Strong Friendship/README_EN.md

@@ -72,7 +72,36 @@ We did not include the friendship of users 2 and 3 because they only have two co
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+with t as (
+    select
+        *
+    from
+        Friendship
+    union
+    all
+    select
+        user2_id,
+        user1_id
+    from
+        Friendship
+)
+select
+    t1.user1_id,
+    t1.user2_id,
+    count(1) common_friend
+from
+    t t1
+    join t t2 on t1.user2_id = t2.user1_id
+    join t t3 on t1.user1_id = t3.user1_id
+where
+    t3.user2_id = t2.user2_id
+    and t1.user1_id < t1.user2_id
+group by
+    t1.user1_id,
+    t1.user2_id
+having
+    count(1) >= 3;
 ```
 
 <!-- tabs:end -->

solution/1900-1999/1949.Strong Friendship/Solution.sql

@@ -0,0 +1,30 @@
+# Write your MySQL query statement below
+with t as (
+    select
+        *
+    from
+        Friendship
+    union
+    all
+    select
+        user2_id,
+        user1_id
+    from
+        Friendship
+)
+select
+    t1.user1_id,
+    t1.user2_id,
+    count(1) common_friend
+from
+    t t1
+    join t t2 on t1.user2_id = t2.user1_id
+    join t t3 on t1.user1_id = t3.user1_id
+where
+    t3.user2_id = t2.user2_id
+    and t1.user1_id < t1.user2_id
+group by
+    t1.user1_id,
+    t1.user2_id
+having
+    count(1) >= 3;