題目1：根據第一二列，計算出第三列。即：求每組KH_VALUE狀態(1和0)變化的最小時間

--創建測試表
create table tmp as
select to_date(‘2017-04-21 16:22:00‘,‘yyyy-mm-dd hh24:mi:ss‘) dt, 1 kv from dual union all
select to_date(‘2017-04-21 16:23:00‘,‘yyyy-mm-dd hh24:mi:ss‘) dt, 1 kv from dual union all
select to_date(‘2017-04-21 16:24:00‘,‘yyyy-mm-dd hh24:mi:ss
 
‘) dt, 0 kv from dual union all
select to_date(‘2017-04-21 16:25:00‘,‘yyyy-mm-dd hh24:mi:ss‘) dt, 0 kv from dual union all
select to_date(‘2017-04-21 16:26:00‘,‘yyyy-mm-dd hh24:mi:ss‘) dt, 0 kv from dual union all
select to_date(‘2017-04-21 16:27:00‘,‘yyyy-mm-dd hh24:mi:ss‘) dt, 0 kv from dual union all
select to_date(‘
 
2017-04-21 16:28:00‘,‘yyyy-mm-dd hh24:mi:ss‘) dt, 1 kv from dual union all
select to_date(‘2017-04-21 16:29:00‘,‘yyyy-mm-dd hh24:mi:ss‘) dt, 1 kv from dual union all
select to_date(‘2017-04-21 16:30:00‘,‘yyyy-mm-dd hh24:mi:ss‘) dt, 0 kv from dual union all
select to_date(‘2017-04-21 16:31:00‘,‘yyyy-mm-dd hh24:mi:ss‘) dt, 0
 
 kv from dual;
--SQL實現:
select dt,kv,min(dt)over(partition by rn order by dt) new_dt
 from(select dt,kv,sum(kv2)over(order by dt) rn
        from(select dt,kv,
                    --case when lag(kv,1)over(order by dt) = kv then 0 else 1 end kv2
                    case when lag(kv,1)over(order by dt) = kv then 0 else row_number()over(order by dt) end kv2
               from tmp
             )
      )

題目2：按照c1的順序，求出c2狀態發生變化的開始及結束位置。

已知tmp表數據如下：
c1 c2
------
1 1
2 1
4 1
5 0
6 0
7 0
8 1
9 1
10 1
11 1
12 1
13 0
14 1
15 1
16 1
17 1
18 1
19 1

c1列為編號，c2為狀態(0,1),要求實現下面的效果：

開始位置,結束位置,狀態
1,4,1
5,7,0
8,12,1
13,13,0,
14,19,1

--創建測試表
create table tmp(c1 int ,c2 int );
insert into tmp values(1,1);
insert into tmp values(2,1);
insert into tmp values(4,1);
insert into tmp values(5,0);
insert into tmp values(6,0);
insert into tmp values(7,0);
insert into tmp values(8,1);
insert into tmp values(9,1);
insert into tmp values(10,1);
insert into tmp values(11,1);
insert into tmp values(12,1);
insert into tmp values(13,0);
insert into tmp values(14,1);
insert into tmp values(15,1);
insert into tmp values(16,1);
insert into tmp values(17,1);
insert into tmp values(18,1);
insert into tmp values(19,1);

--解法1：
select min(c1) start_c1,
max(c1) start_c2
c2
from(select c1,c2,
sum(rn)over(order by c1) rn
from(select c1,
c2,
decode(c2, lag(c2) over(order by c1), null, row_number() over(order by c1)) rn
from tmp
)
)
group by rn, c2;

--解法2：
select min(c1), max(c1), c2
from (select b.*,
row_number()over(partition by g order by c1) r1,
row_number()over(partition by g order by c1 desc) r2
from (select a.*, sum(t) over(order by c1) g
from (select t.*, decode(c2, lag(c2, 1, c2) over(order by c1), 0, 1) t
from tmp t
)a
) b
) c
where r1 = 1 or r2 = 1
group by g, c2
order by 1;

--解法3：
select min(c1) s, max(c1) e, c2
from (select c1, c2, sum(rn) over(order by c1) rn
from (select c1, c2,
case when lag(c2) over(order by c1) = c2 then 0 else 1 end rn
from tmp
)
)
group by c2, rn
order by s

根據狀態變化情況，求最大值和最小值