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ConsecutiveNumbers
2015-11-21 01:49:06 来源: 作者: 【 】 浏览:0
Tags:ConsecutiveNumbers
    Write a SQL query to find all numbers that appear at least three times consecutively.

    +----+-----+
    | Id | Num |
    +----+-----+
    | 1  |  1  |
    | 2  |  1  |
    | 3  |  1  |
    | 4  |  2  |
    | 5  |  1  |
    | 6  |  2  |
    | 7  |  2  |
    +----+-----+
    For example, given the above Logs table, 1 is the only number that appears consecutively for at least three times.
    解法一:

    从上往下对齐排名,如果相等,则排名相等,不相等则排名加一
    代码如下:
    select num,@curRank := @curRank+IF(@preNum = num,0,1) as rank ,@preNum:=num from logs s,(select @curRank := 0) r,(select @preNum := null) p;
    +------+------+--------------+
    | num  | rank | @preNum:=num |
    +------+------+--------------+
    |    1 |    1 |            1 |
    |    1 |    1 |            1 |
    |    1 |    1 |            1 |
    |    2 |    2 |            2 |
    |    1 |    3 |            1 |
    |    2 |    4 |            2 |
    |    2 |    4 |            2 |
    +------+------+--------------+
    如上所示,如果一个num的连续排名超过3则符合题意。
    select num,count(rank) as count from (
        select num,@curRank := @curRank+IF(@preNum = num,0,1) as rank ,@preNum:=num from logs s,(select @curRank := 0) r,(select @preNum := null) p ) t 
        group by rank;
    +------+-------+
    | num  | count |
    +------+-------+
    |    1 |     3 |
    |    2 |     1 |
    |    1 |     1 |
    |    2 |     2 |
    +------+-------+
    得到上述的表之后,再使用having条件选择。
    最终结果:
    select distinct num from(
        select num,count(rank) as count from (
            select num,@curRank := @curRank+IF(@preNum = num,0,1) as rank ,@preNum:=num from logs s,(select @curRank := 0) r,(select @preNum := null) p ) t 
            group by rank having count >= 3;
        ) tmp;
    解法二:
    1、从第一条记录搜索,前后相同,count++,不相等,count=1;
    2、判断,若rank大于3则符合题意
    select distinct num 
        from ( 
            select num,@curRank := @preRank+IF(@preNum = num,1,0),@preRank :=IF(@preNum = num,@curRank,1) as rank,@preNum := num 
                from Logs s,(select @preRank := 1) r,(select @preNum := null) p
        ) t 
        where rank >= 3;
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