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简单
数据库

603. 连续空余座位 🔒

English Version

题目描述

表: Cinema

+-------------+------+
| Column Name | Type |
+-------------+------+
| seat_id     | int  |
| free        | bool |
+-------------+------+
Seat_id 是该表的自动递增主键列。
在 PostgreSQL 中，free 存储为整数。请使用 ::boolean 将其转换为布尔格式。
该表的每一行表示第 i 个座位是否空闲。1 表示空闲，0 表示被占用。

 

查找电影院所有连续可用的座位。

返回按 seat_id 升序排序 的结果表。

测试用例的生成使得两个以上的座位连续可用。

结果表格式如下所示。

 

示例 1:

输入: 
Cinema 表:
+---------+------+
| seat_id | free |
+---------+------+
| 1       | 1    |
| 2       | 0    |
| 3       | 1    |
| 4       | 1    |
| 5       | 1    |
+---------+------+
输出: 
+---------+
| seat_id |
+---------+
| 3       |
| 4       |
| 5       |
+---------+

解法

方法一：自连接

我们可以使用自连接的方式，将相邻的两个座位连接起来，然后筛选出连续空余的座位并去重排序即可。

MySQL

# Write your MySQL query statement below
SELECT DISTINCT a.seat_id
FROM
    Cinema AS a
    JOIN Cinema AS b ON ABS(a.seat_id - b.seat_id) = 1 AND a.free AND b.free
ORDER BY 1;

方法二：窗口函数

我们也可以使用 LAG 和 LEAD 函数（或者 SUM() OVER(ROWS BETWEEN 1 PRECEDING AND 1 FOLLOWING)）来获取相邻的座位信息，然后筛选出连续空余的座位并去重排序即可。

MySQL

# Write your MySQL query statement below
WITH
    T AS (
        SELECT
            seat_id,
            (free + (LAG(free) OVER (ORDER BY seat_id))) AS a,
            (free + (LEAD(free) OVER (ORDER BY seat_id))) AS b
        FROM Cinema
    )
SELECT seat_id
FROM T
WHERE a = 2 OR b = 2;

方法三

MySQL

# Write your MySQL query statement below
WITH
    T AS (
        SELECT
            *,
            SUM(free = 1) OVER (
                ORDER BY seat_id
                ROWS BETWEEN 1 PRECEDING AND 1 FOLLOWING
            ) AS cnt
        FROM Cinema
    )
SELECT seat_id
FROM T
WHERE free = 1 AND cnt > 1
ORDER BY 1;