I think something like this is what you want.
For alternating characters:
(?=(.)(?!\1)(.))(?:\1\2){2,}
\0 will be the entire alternating sequence, \1 and \2 are the two (distinct) alternating characters.
For run of N and M characters, possibly separated by other characters (replace N and M with numbers here):
(?=(.))\1{N}.*?(?=(?!\1)(.))\2{M}
\0 will be entire match, including infix. \1 is the character repeated (at least) N times, \2 is the character repeated (at least) M times.
Here's a test harness in Java.
import java.util.regex.*;
public class Regex3 {
static String runNrunM(int N, int M) {
return "(?=(.))\\1{N}.*?(?=(?!\\1)(.))\\2{M}"
.replace("N", String.valueOf(N))
.replace("M", String.valueOf(M));
}
static void dumpMatches(String text, String pattern) {
Matcher m = Pattern.compile(pattern).matcher(text);
System.out.println(text + " <- " + pattern);
while (m.find()) {
System.out.println(" match");
for (int g = 0; g <= m.groupCount(); g++) {
System.out.format(" %d: [%s]%n", g, m.group(g));
}
}
}
public static void main(String[] args) {
String[] tests = {
"foobababababaf foobaafoobaaaooo",
"xxyyyy axxayyyya zzzzzzzzzzzzzz"
};
for (String test : tests) {
dumpMatches(test, "(?=(.)(?!\\1)(.))(?:\\1\\2){2,}");
}
for (String test : tests) {
dumpMatches(test, runNrunM(3, 3));
}
for (String test : tests) {
dumpMatches(test, runNrunM(2, 4));
}
}
}
This produces the following output:
foobababababaf foobaafoobaaaooo <- (?=(.)(?!\1)(.))(?:\1\2){2,}
match
0: [bababababa]
1: [b]
2: [a]
xxyyyy axxayyyya zzzzzzzzzzzzzz <- (?=(.)(?!\1)(.))(?:\1\2){2,}
foobababababaf foobaafoobaaaooo <- (?=(.))\1{3}.*?(?=(?!\1)(.))\2{3}
match
0: [aaaooo]
1: [a]
2: [o]
xxyyyy axxayyyya zzzzzzzzzzzzzz <- (?=(.))\1{3}.*?(?=(?!\1)(.))\2{3}
match
0: [yyyy axxayyyya zzz]
1: [y]
2: [z]
foobababababaf foobaafoobaaaooo <- (?=(.))\1{2}.*?(?=(?!\1)(.))\2{4}
xxyyyy axxayyyya zzzzzzzzzzzzzz <- (?=(.))\1{2}.*?(?=(?!\1)(.))\2{4}
match
0: [xxyyyy]
1: [x]
2: [y]
match
0: [xxayyyy]
1: [x]
2: [y]
Explanation
(?=(.)(?!\1)(.))(?:\1\2){2,} has two parts
(?=(.)(?!\1)(.)) establishes \1 and \2 using lookahead
- Nested negative lookahead ensures that
\1 != \2
- Using lookahead to capture lets
\0 have the entire match (instead of just the "tail" end)
(?:\1\2){2,} captures the \1\2 sequence, which must repeat at least twice.
(?=(.))\1{N}.*?(?=(?!\1)(.))\2{M} has three parts
(?=(.))\1{N} captures \1 in a lookahead, and then match it N times
- Using lookahead to capture means the repetition can be
N instead of N-1
.*? allows an infix to separate the two runs, reluctant to keep it as short as possible(?=(?!\1)(.))\2{M}
- Similar to first part
- Nested negative lookahead ensures that
\1 != \2
The run regex will match longer runs, e.g. run(2,2) matches "xxxyyy":
xxxyyy <- (?=(.))\1{2}.*?(?=(?!\1)(.))\2{2}
match
0: [xxxyy]
1: [x]
2: [y]
Also, it does not allow overlapping matches. That is, there is only one run(2,3) in "xx11yyy222".
xx11yyy222 <- (?=(.))\1{2}.*?(?=(?!\1)(.))\2{3}
match
0: [xx11yyy]
1: [x]
2: [y]