Leecode consecutive numbers

Original title

Write a SQL query to find all numbers that appear at least three consecutive times.

+----+-----+
| Id | Num |
+----+-----+
| 1  |  1  |
| 2  |  1  |
| 3  |  1  |
| 4  |  2  |
| 5  |  1  |
| 6  |  2  |
| 7  |  2  |
+----+-----+

For example, given the above Logstable, only 1 consecutive number of at least three times.

+-----------------+
| ConsecutiveNums |
+-----------------+
| 1               |
+-----------------+

Source: LeetCode
Link: https://leetcode-cn.com/problems/consecutive-numbers
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Solution 1

Methods: DISTINCTand WHEREstatements

Consecutive identical numbers mean Id is attached, since this question is asked at least 3 consecutive times, we use Logsand check for three consecutive identical digits.

SELECT *
FROM
    Logs l1,
    Logs l2,
    Logs l3
WHERE
    l1.Id = l2.Id - 1
    AND l2.Id = l3.Id - 1
    AND l1.Num = l2.Num
    AND l2.Num = l3.Num
;

Then we choose any Num from the above table to get the answer we want. At the same time, we need to add keywords DISTINCT, because if a number appears more than 3 times in a row, repeated elements will be returned.

SELECT DISTINCT
    l1.Num AS ConsecutiveNums
FROM
    Logs l1,
    Logs l2,
    Logs l3
WHERE
    l1.Id = l2.Id - 1
    AND l2.Id = l3.Id - 1
    AND l1.Num = l2.Num
    AND l2.Num = l3.Num
;

Solution 2

The window function lead()solves perfectly

The deviation function using the window function is perfectly implemented. It can be understood like this: copy num into the two columns num1 and num2, and then move num1 up by one row as a whole, and move num2 up by two rows as a whole, as follows:
Insert picture description here
so as long as num=num1=num2 is enough.

# Write your MySQL query statement below
select distinct num as ConsecutiveNums  from
(
    select num,lead(num,1)over()as num1,lead(num,2)over()as num2 
    from logs
) as c
where c.num = c.num1 and c.num1 = c.num2

Solution 3

Use variables instead of connecting multiple tables to reduce the time and space complexity.
Pre record the value of the previous number, count, nums record the number of occurrences of the previous number

IF() expression usage

select 
    distinct num as ConsecutiveNums
from
(
    select
    num,
    if(@pre=num,@count:=@count+1,@count:=1) as nums,
    @pre:=num
    from logs as l,
        (select @pre:=null,@count:=1)as pc
) as n
where nums>=3