task: #626

README.md

@@ -75,6 +75,7 @@ Have a good contributing!
     - [184. Department Highest Salary](./leetcode/medium/184.%20Department%20Highest%20Salary.sql)
     - [570. Managers with at Least 5 Direct Reports](./leetcode/medium/570.%20Managers%20with%20at%20Least%205%20Direct%20Reports.sql)
     - [602. Friend Requests II: Who Has the Most Friends](./leetcode/medium/602.%20Friend%20Requests%20II:%20Who%20Has%20the%20Most%20Friends.sql)
+    - [626. Exchange Seats](./leetcode/medium/626.%20Exchange%20Seats.sql)
     - [1045. Customers Who Bought All Products](./leetcode/medium/1045.%20Customers%20Who%20Bought%20All%20Products.sql)
     - [1070. Product Sales Analysis III](./leetcode/medium/1070.%20Product%20Sales%20Analysis%203.sql)
     - [1158. Market Analysis 1](./leetcode/medium/1158.%20Market%20Analysis%201.sql)

leetcode/medium/626. Exchange Seats.sql

@@ -0,0 +1,43 @@
+/*
+Question 626. Exchange Seats
+Link: https://leetcode.com/problems/exchange-seats/description/?envType=study-plan-v2&envId=top-sql-50
+
+Table: Seat
+
++-------------+---------+
+| Column Name | Type    |
++-------------+---------+
+| id          | int     |
+| student     | varchar |
++-------------+---------+
+id is the primary key (unique value) column for this table.
+Each row of this table indicates the name and the ID of a student.
+The ID sequence always starts from 1 and increments continuously.
+
+
+Write a solution to swap the seat id of every two consecutive students. If the number of students is odd, the id of the last student is not swapped.
+
+Return the result table ordered by id in ascending order.
+*/
+
+SELECT 
+    s.id,
+    (CASE
+        WHEN s.id % 2 = 0 THEN s1.student
+        WHEN s3.student IS NULL THEN s.student
+        WHEN s.id % 2 = 1 THEN s3.student
+    END) AS student
+FROM Seat AS s
+LEFT JOIN 
+    Seat AS s1
+    ON s.id = (
+        SELECT MIN(s2.id) FROM Seat AS s2
+        WHERE s1.id < s2.id
+    )
+LEFT JOIN 
+    Seat AS s3
+    ON s.id = (
+        SELECT MAX(s4.id) FROM Seat AS s4
+        WHERE s3.id > s4.id
+    )
+