java问题：如何match两个正规表达式 - 未名空间MITBBS历史存档

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java问题：如何match两个正规表达式

java问题：如何match两个正规表达式# Java - 爪哇娇娃

r*t2005-06-09 07:06

1 楼

想知道两个正规表达式有没有交集（只要知道有没有，不要知道交集是什么）
例如，正规表达式
^abc/.*\\z （abc开始的string）
和
^.*ef\\z （ef结尾的string）
有交集；
和
^ef/.*\\z （ef开始的string）就没有
应该怎么做呢？

S*O2005-06-09 07:06

2 楼

this has nothing to do w/ java bah. more like an algorithm...

【在 r**t 的大作中提到】

: 想知道两个正规表达式有没有交集（只要知道有没有，不要知道交集是什么）
: 例如，正规表达式
: ^abc/.*\\z （abc开始的string）
: 和
: ^.*ef\\z （ef结尾的string）
: 有交集；
: 和
: ^ef/.*\\z （ef开始的string）就没有
: 应该怎么做呢？

g*s2005-06-09 07:06

3 楼

OP meant how to do this with Java reg exp package ya(if Java reg exp package
implemented this).

【在 S*O 的大作中提到】

: this has nothing to do w/ java bah. more like an algorithm...

S*O2005-06-09 07:06

4 楼

if so, then use reg1 to match the objects, w/ the matched result,
run reg2, if there are any matches, that means these 2 reg has intersection.
otherwise no.
that would work bah?

【在 g*s 的大作中提到】

: OP meant how to do this with Java reg exp package ya(if Java reg exp package
: implemented this).

R*r2005-06-09 07:06

5 楼

How do you create the strings? you only have reg expressions.

【在 S*O 的大作中提到】

: if so, then use reg1 to match the objects, w/ the matched result,
: run reg2, if there are any matches, that means these 2 reg has intersection.
: otherwise no.
: that would work bah?

S*O2005-06-09 07:06

6 楼

sth like this:
public static boolean hasIntersection(string pattern1, String pattern2,
String[] strings) {
RE reg1 = new RE(pattern1);
RE reg2 - new RE(pattern2);
for (int i=0;iif(reg1.isMatch(strings[i]))
if(reg2.isMatch(strings[i])
return true;
return false;
}

【在 S*O 的大作中提到】

: if so, then use reg1 to match the objects, w/ the matched result,
: run reg2, if there are any matches, that means these 2 reg has intersection.
: otherwise no.
: that would work bah?

S*O2005-06-09 07:06

7 楼

what strings?

【在 R*******r 的大作中提到】

: How do you create the strings? you only have reg expressions.

g*s2005-06-09 07:06

8 楼

reg1-matched strings are infinite ya...

【在 S*O 的大作中提到】

: if so, then use reg1 to match the objects, w/ the matched result,
: run reg2, if there are any matches, that means these 2 reg has intersection.
: otherwise no.
: that would work bah?

g*s2005-06-09 07:06

9 楼

not check whether to have intersection on some string set, but universal
string set ya.

【在 S*O 的大作中提到】

: sth like this:
: public static boolean hasIntersection(string pattern1, String pattern2,
: String[] strings) {
: RE reg1 = new RE(pattern1);
: RE reg2 - new RE(pattern2);
: for (int i=0;i: if(reg1.isMatch(strings[i]))
: if(reg2.isMatch(strings[i])
: return true;
: return false;

S*O2005-06-09 07:06

10 楼

oh, i see...
when there is no target-to-be-matched, how to find out 2 expressions
have intersection or not...
hmmmmm

【在 g*s 的大作中提到】

: reg1-matched strings are infinite ya...

S*O2005-06-09 07:06

11 楼

then the question is if there is a way to
join multiple regular expression patterns...
if u can join them together, then u know if they intersect or not.

【在 S*O 的大作中提到】

: oh, i see...
: when there is no target-to-be-matched, how to find out 2 expressions
: have intersection or not...
: hmmmmm

w*e2005-06-09 07:06

12 楼

Regex can be turned into finite state (If I am wrong, please point out). You
can turn them into 2 state machines (FSM), and try to minimize those states.
If Minimize FSM is same, then they are match, otherwise not.
You can find algorithm by search google about Minimize FSM.
best regard

【在 S*O 的大作中提到】

: then the question is if there is a way to
: join multiple regular expression patterns...
: if u can join them together, then u know if they intersect or not.