力扣【SQL连续问题】

180. 连续出现的数字

SELECT DISTINCT if(a.num = b.num AND b.num = c.num,a.num,null) AS ConsecutiveNums
FROM Logs a
LEFT OUTER JOIN Logs b
ON a.id+1 = b.id
LEFT OUTER JOIN Logs c
ON a.id+2 = c.id
WHERE if(a.num = b.num AND b.num = c.num,a.num,null) IS NOT NULL

603. 连续空余座位

SELECT a.seat_id AS seat_id
FROM Cinema a
LEFT OUTER JOIN Cinema b
ON a.seat_id + 1 = b.seat_id
LEFT OUTER JOIN Cinema c
ON a.seat_id - 1 = c.seat_id
WHERE a.free = 1 AND (b.free = 1 OR c.free = 1) 
ORDER BY seat_id

-- 执行速度慢
SELECT DISTINCT a.seat_id AS seat_id
FROM Cinema a
LEFT OUTER JOIN Cinema b
ON ABS(a.seat_id - b.seat_id) = 1
WHERE a.free = 1 AND b.free = 1
ORDER BY seat_id

613. 直线上的最近距离

-- 方法一：
SELECT MIN(ABS(a.x - b.x)) AS shortest
FROM Point a
LEFT OUTER JOIN Point b
ON a.x <> b.x

-- 方法二：
select  x - lag(x) over(order by x) as shortest
  from point
 order by shortest
 limit 1
 offset 1

-- 方法二中的中间输出 
select  x, 
 		lag(x) over(order by x) AS lag_x,    -- 取比每个 x 小的值中，但又最靠近x（数值上最大的）的那一个值 
  		x - lag(x) over(order by x) as shortest  -- 计算每个 x 与其 lag_x 之间的距离
  from point 

1285. 找到连续区间的开始和结束数字

①先给每个数进行排名

②用这些数减去自己的排名，如果减了之后的结果是一样的，说明这几个数是连续的

（原理：等差数列的值，减去等差数列的值，结果才能是一样的。）

③用logid减去排名得出来的数进行group by，也就是把连续的数全都放在一个一个小组里面，求出每个小组的最大值和最小值就可以了

SELECT MIN(a.log_id) AS START_ID, MAX(a.log_id) AS END_ID
FROM (SELECT log_id, 
        ROW_NUMBER() OVER (ORDER BY log_id ASC) rn,
        log_id - ROW_NUMBER() OVER (ORDER BY log_id ASC) reference
      FROM Logs) a 
GROUP BY a.reference
ORDER BY start_id