正则表达式匹配至少包含指定字符的字符串

Question

I have a huge dictionary which I'm trying to look through using a regex. What I would like to do is to find all the words in the dictionary which contain at least one occurrences of each character I provide in no particular order.

Right now I can find words which only contain the specified characters but like I said that is not exactly what I want.

Example:

I want at least one occurrence of each of the following characters {b, a, d}

astring.matches(regex)

I would expect words like:

badder, baddest, baffled

Notice they all contain at least one occurence of each character but in no particular order and other characters are present in the strings.

Anyone know how to do this? Other suggestions are also welcome!

Answer 1

You can use a lookahead to do this if it's available

(?=.*b)(?=.*a)(?=.*d)

However this is quite inefficient. Any reason you can't use multiple String.indexOf checks?

Answer 2

You need a series of look-aheads:

^(?=.*b)(?=.*a)(?=.*d).*

which is a pain to construct. However, you can ease the pain by using regex to build it:

String regex = "^" + "bad".replaceAll(".", "(?=.*$0)") + ".*";

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If using repeatedly with String.matches() , you would be better to use the following code, because every call to String.matches() compiles the regex again (there is no caching):

// do this once
Pattern pattern = Pattern.compile(regex); 

// reuse the pattern many times
if (pattern.matcher(input).matches())