feat: add solutions to lc problems: No.0579,0585

* No.0579.Find Cumulative Salary of an Employee
* No.0585.Investments in 2016

Author: yanglbme

solution/0500-0599/0579.Find Cumulative Salary of an Employee/README.md

@@ -88,7 +88,30 @@
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+select
+    id,
+    month,
+    sum(salary) over (
+        partition by id
+        order by
+            month range 2 preceding
+    ) as Salary
+from
+    employee
+where
+    (id, month) not in (
+        select
+            id,
+            max(month)
+        from
+            Employee
+        group by
+            id
+    )
+order by
+    id,
+    month desc
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0579.Find Cumulative Salary of an Employee/README_EN.md

@@ -113,7 +113,30 @@ So the cumulative salary summary for this employee is:
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+select
+    id,
+    month,
+    sum(salary) over (
+        partition by id
+        order by
+            month range 2 preceding
+    ) as Salary
+from
+    employee
+where
+    (id, month) not in (
+        select
+            id,
+            max(month)
+        from
+            Employee
+        group by
+            id
+    )
+order by
+    id,
+    month desc
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0579.Find Cumulative Salary of an Employee/Solution.sql

@@ -0,0 +1,24 @@
+# Write your MySQL query statement below
+select
+    id,
+    month,
+    sum(salary) over (
+        partition by id
+        order by
+            month range 2 preceding
+    ) as Salary
+from
+    employee
+where
+    (id, month) not in (
+        select
+            id,
+            max(month)
+        from
+            Employee
+        group by
+            id
+    )
+order by
+    id,
+    month desc

solution/0500-0599/0585.Investments in 2016/README.md

@@ -66,7 +66,22 @@
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+with t as (
+    select
+        tiv_2016,
+        count(pid) over(partition by tiv_2015) as cnt1,
+        count(pid) over(partition by concat(lat, lon)) as cnt2
+    from
+        Insurance
+)
+select
+    round(sum(TIV_2016), 2) tiv_2016
+from
+    t
+where
+    cnt1 != 1
+    and cnt2 = 1;
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0585.Investments in 2016/README_EN.md

@@ -72,7 +72,22 @@ So, the result is the sum of tiv_2016 of the first and last record, which is 45.
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+with t as (
+    select
+        tiv_2016,
+        count(pid) over(partition by tiv_2015) as cnt1,
+        count(pid) over(partition by concat(lat, lon)) as cnt2
+    from
+        Insurance
+)
+select
+    round(sum(TIV_2016), 2) tiv_2016
+from
+    t
+where
+    cnt1 != 1
+    and cnt2 = 1;
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0585.Investments in 2016/Solution.sql

@@ -0,0 +1,16 @@
+# Write your MySQL query statement below
+with t as (
+    select
+        tiv_2016,
+        count(pid) over(partition by tiv_2015) as cnt1,
+        count(pid) over(partition by concat(lat, lon)) as cnt2
+    from
+        Insurance
+)
+select
+    round(sum(TIV_2016), 2) tiv_2016
+from
+    t
+where
+    cnt1 != 1
+    and cnt2 = 1;