我想在Oracle sql中完成以下任务：
仅返回Table1中开始日期和结束日期跨越Table2中开始日期和结束日期的行。

Table 2:
 ------------------------------------
| uid | startdate |enddate | typenum |
 ------------------------------------
| 123 | 09/1/22   | 09/3/22 | 25     |

应返回表1中与表2的开始日期和结束日期重叠的任何行。例如：

Table 1 - qualified rows:
 ------------------------------------
| uid | startdate |enddate | typenum |
 ------------------------------------
| 123 | 09/1/22   | 09/3/22 | 10    |        
| 123 | 09/2/22   | 09/6/22 | 30    |  
| 123 | 08/28/22   | 08/30/22 | 25    |
| 123 | 09/4/22   | 09/5/22 | 25    |

Table 1 - disqualified rows:
 ------------------------------------
| uid | startdate |enddate | typenum |
 ------------------------------------
| 123 | 09/1/22   | 09/3/22 | 25     |        
| 123 | 09/2/22   | 09/6/22 | 25     |  
| 123 | 08/30/22   | 09/2/22 | 25    |
| 123 | 08/30/22   | 09/7/22 | 25    |

此查询可以工作，但是否可以进一步优化和清理？

Select * From Table1 T1
INNER JOIN Table2  T2 ON T1.uid = T2.uid
AND T1.typenum = 25
AND ( 
    ( T1.startdate BETWEEN T2.startdate AND T2.endate OR T1.endate BETWEEN T2.startdate AND 
    T2.endate )
    OR ( T1.startdate >  T2.startdate AND T1.endate < T2.endate )
    OR ( T1.startdate <  T2.startdate AND T1.endate > T2.endate )
    )