问如何使平衡组捕获？

EN

(?=([A-Z](?:(?:(?'O'{)[^{}]*)+(?:(?'-O'})[^{}]*?)+)+(?(O)(?!))))

1: R{R{abc}aD{mnoR{xyz}}}
1: R{abc}
1: D{mnoR{xyz}}
1: R{xyz}

(?=                         # Opening positive lookahead
    ([A-Z]                  # Opening capture group and any uppercase letter (to match R & D)
        (?:                 # First non-capture group opening
            (?:             # Second non-capture group opening
                (?'O'{)     # Get the named opening brace
                [^{}]*      # Any non-brace
            )+              # Close of second non-capture group and repeat over as many times as necessary
            (?:             # Third non-capture group opening
                (?'-O'})    # Removal of named opening brace when encountered
                [^{}]*?     # Any other non-brace characters in case there are more nested braces
            )+              # Close of third non-capture group and repeat over as many times as necessary
        )+                  # Close of first non-capture group and repeat as many times as necessary for multiple side by side nested braces
        (?(O)(?!))          # Condition to prevent unbalanced braces
    )                       # Close capture group
)                           # Close positive lookahead

(?=([A-Z]{(?:[^{}]|(?1))+}))

(?=                    # Opening positive lookahead
    ([A-Z]             # Opening capture group and any uppercase letter (to match R & D)
        {              # Opening brace
            (?:        # Opening non-capture group
                [^{}]  # Matches non braces
            |          # OR
                (?1)   # Recurse first capture group
            )+         # Close non-capture group and repeat as many times as necessary
        }              # Closing brace
    )                  # Close of capture group
)                      # Close of positive lookahead

/**
 * Finds all matches (i.e. including sub/nested matches) of the regex in the input string.
 * 
 * @param input
 *          The input string.
 * @param regex
 *          The regex pattern. It has to target the most nested substrings. For example, given the following input string
 *          <code>A{01B{23}45C{67}89}</code>, if you want to catch every <code>X{*}</code> substrings (where <code>X</code> is a capital letter),
 *          you have to use <code>[A-Z][{][^{]+?[}]</code> or <code>[A-Z][{][^{}]+[}]</code> instead of <code>[A-Z][{].+?[}]</code>.
 * @param format
 *          The format must follow the <a href= "http://docs.oracle.com/javase/7/docs/api/java/util/Formatter.html#syntax" >format string
 *          syntax</a>. It will be given one single integer as argument, so it has to contain (and to contain only) a <code>%d</code> flag. The
 *          format must not be foundable anywhere in the input string. If <code>null</code>, <code>ééé%dèèè</code> will be used.
 * @return The list of all the matches of the regex in the input string.
 */
public static List<String> findAllMatches(String input, String regex, String format) {

    if (format == null) {
        format = "ééé%dèèè";
    }
    int counter = 0;
    Map<String, String> matches = new LinkedHashMap<String, String>();
    Pattern pattern = Pattern.compile(regex);
    Matcher matcher = pattern.matcher(input);

    // if a substring has been found
    while (matcher.find()) {
        // create a unique replacement string using the counter
        String replace = String.format(format, counter++);
        // store the relation "replacement string --> initial substring" in a queue
        matches.put(replace, matcher.group());
        String end = input.substring(matcher.end(), input.length());
        String start = input.substring(0, matcher.start());
        // replace the found substring by the created unique replacement string
        input = start + replace + end;
        // reiterate on the new input string (faking the original matcher.find() implementation)
        matcher = pattern.matcher(input);
    }

    List<Entry<String, String>> entries = new LinkedList<Entry<String, String>>(matches.entrySet());

    // for each relation "replacement string --> initial substring" of the queue
    for (int i = 0; i < entries.size(); i++) {
        Entry<String, String> current = entries.get(i);
        // for each relation that could have been found before the current one (i.e. more nested)
        for (int j = 0; j < i; j++) {
            Entry<String, String> previous = entries.get(j);
            // if the current initial substring contains the previous replacement string
            if (current.getValue().contains(previous.getKey())) {
                // replace the previous replacement string by the previous initial substring in the current initial substring
                current.setValue(current.getValue().replace(previous.getKey(), previous.getValue()));
            }
        }
    }

    return new LinkedList<String>(matches.values());
}

String input = "tes{}tR{R{abc}aD{mnoR{xyz}}}";
String regex = "[A-Z][{][^{}]+[}]";
findAllMatches(input, regex, null);

R{abc}
R{xyz}
D{mnoR{xyz}}
R{R{abc}aD{mnoR{xyz}}}

(?>
    (?<O>)\p{Lu}\{   # Push to the O stack, and match an upper-case letter and {
    |                # OR
    \}(?<-O>)        # Match } and pop from the stack
    |                # OR
    \p{Ll}           # Match a lower-case letter
)+
(?(O)(?!))        # Make sure the stack is empty

(?>(?<O>)\p{Lu}\{|\}(?<-O>)|\p{Ll})+(?(O)(?!))

(?>(?<O>)\p{Lu}\{|\}(?<Target-O>)|\p{Ll})+(?(O)(?!))