这个匹配包含重复项的字符串。这基本上是你要找的内容的反向。

.*(.)+.*\1.*

如果合适的话，你可以选择与这个表达式 * 不匹配 * 的字符串。或者检查regex negative match来改进它。

我有一个非常丑陋的方式来做这个使用regex。
它使用已命名的捕获组(?<name>)、反向引用\k<name>和负前瞻Assert(?!)。

@Test
public void testForNonRepeatedCharactersUsingRegex() {
    Pattern p = Pattern.compile("^(?<c1>[a-zA-Z])(?!.*\\k<c1>)" +
            "(?<c3>[a-zA-Z])(?!.*\\k<c3>)" +
            "(?<c2>[a-zA-Z])(?!.*\\k<c2>)" +
            "(?<c4>[a-zA-Z])(?!.*\\k<c4>)" +
            "(?<c5>[a-zA-Z])(?!.*\\k<c5>)" +
            "(?<c6>[a-zA-Z])(?!.*\\k<c6>)" +
            "(?<c7>[a-zA-Z])(?!.*\\k<c7>)" +
            ".$");
    Assertions.assertTrue(p.matcher("abcdefgh").matches());
    Assertions.assertTrue(p.matcher("ABCDEFGH").matches());
    Assertions.assertFalse(p.matcher("12345678").matches());
    Assertions.assertFalse(p.matcher("abefgh").matches());
    Assertions.assertFalse(p.matcher("").matches());
    Assertions.assertFalse(p.matcher("abcdefga").matches());
    Assertions.assertFalse(p.matcher("aaaabbbb").matches());
}

如果您可以不使用regex，那么一个更干净、更清晰的解决方案是：

@Test
public void testForNonRepeatedCharactersUsingStreams() {
    Assertions.assertTrue(uniqueCharacterSequence("abcdefgh", 8));
    Assertions.assertTrue(uniqueCharacterSequence("ABCDEFGH", 8));
    Assertions.assertFalse(uniqueCharacterSequence("12345678", 8));
    Assertions.assertFalse(uniqueCharacterSequence("abefgh", 8));
    Assertions.assertFalse(uniqueCharacterSequence("", 8));
    Assertions.assertFalse(uniqueCharacterSequence("abcdefga", 8));
    Assertions.assertFalse(uniqueCharacterSequence("aaaabbbb", 8));
}

private static boolean uniqueCharacterSequence(String s, int length) {
    return  s.length() == length &&
            s.codePoints().allMatch(Character::isLetter) &&
            s.codePoints().distinct().count() == length;
}