Java正则表达式 - 重叠匹配

Question

In the following code:

public static void main(String[] args) {
    List<String> allMatches = new ArrayList<String>();
    Matcher m = Pattern.compile("\\d+\\D+\\d+").matcher("2abc3abc4abc5");
    while (m.find()) {
        allMatches.add(m.group());
    }

    String[] res = allMatches.toArray(new String[0]);
    System.out.println(Arrays.toString(res));
}

The result is:

[2abc3, 4abc5]

[2abc3, 3abc4, 4abc5]

Answer 1

Make the matcher attempt to start its next scan from the latter \\d+ .

Matcher m = Pattern.compile("\\d+\\D+(\\d+)").matcher("2abc3abc4abc5");
if (m.find()) {
    do {
        allMatches.add(m.group());
    } while (m.find(m.start(1)));
}

Answer 2

Not sure if this is possible in Java, but in PCRE you could do the following:
(?=(\\d+\\D+\\d+)).

Explanation
The technique is to use a matching group in a lookahead, and then "eat" one character to move forward.

• (?= : start of positive lookahead
  • ( : start matching group 1
    • \\d+ : match a digit one or more times
    • \\D+ : match a non-digit character one or more times
    • \\d+ : match a digit one or more times
  • ) : end of group 1
• ) : end of lookahead
• . : match anything, this is to "move forward".

Online demo

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Thanks to Casimir et Hippolyte it really seems to work in Java. You just need to add backslashes and display the first capturing group: (?=(\\\\d+\\\\D+\\\\d+)). . Tested on www.regexplanet.com :

Answer 3

The above solution of HamZa works perfectly in Java. If you want to find a specific pattern in a text all you have to do is:

String regex = "\d+\D+\d+";

String updatedRegex = "(?=(" + regex + ")).";

Where the regex is the pattern you are looking for and to be overlapping you need to surround it with (?=(" at the start and ")). at the end.