使用lag（）函数获取前一天，如果前一天>1，则计算新的\u trip \u标志，然后计数（新的\u trip \u标志）。
演示：

with table1 as (
select 'ABC' as touristid, 1  as day union all
select 'ABC' as touristid, 1  as day union all
select 'ABC' as touristid, 2  as day union all
select 'ABC' as touristid, 4  as day union all
select 'ABC' as touristid, 5  as day union all
select 'ABC' as touristid, 6  as day union all
select 'ABC' as touristid, 8  as day union all
select 'ABC' as touristid, 10 as day 
)
select touristid, count(new_trip_flag) trip_cnt
  from 
       ( -- calculate new_trip_flag
         select touristid,
                case when (day-prev_day) > 1 or prev_day is NULL then true end  new_trip_flag
           from       
                ( -- get prev_day
                  select touristid, day, 
                         lag(day) over(partition by touristid order by day) prev_day
                    from table1
                )s
        )s
 group by touristid;

结果：

touristid   trip_cnt    
ABC         4

同样的方法也适用于Hive。