How can I ignore spaces, punctuation marks, and all characters different than letters while checking a palindrome?

Question

I need to check a palindrome in a separate class but ignore non alphabet characters. So for example, radar would still qualify if it was written r,a,d,a,r

I believe I could use regex, but I don't know how.

Here is what I have so far,

 public static boolean isNonAlpha(char c) {
    return (c == '-' || c == '.' || c == ' ' || c == ')' || c == '(') || c == '<' || c == '>' || c == ',';
}

public static String checkInput(String test){
    int startChar = 0;
    int endChar = test.length() - 1;
    while (startChar < endChar) {
        if (test.charAt(startChar) != test.charAt(endChar)) {
            System.out.println("Your word is not a palindrome.");
            System.exit(0);
        } else {
            if (test.charAt(startChar) == test.charAt(endChar))
                startChar++;
                endChar--;
        }
    }
    System.out.println("Your word is indeed a palindrome.");        
    return test;

}

I'm stuck on how to incorporate my isNonAlpha method, or how to use regex

Standard regular expressions cannot detect palindromes as they require memory (of up to unlimited value) of what was previously accepted. http://stackoverflow.com/questions/233243/how-to-check-that-a-string-is-a-palindrome-using-regular-expressions However there are some implementations on that page — Joseph Young, Feb 28 '16 at 22:15
It might be possible to do in Javas regex implementation as shown in some languages though, but they technically aren't regular — Joseph Young, Feb 28 '16 at 22:18
[Remove all non-alphabet characters](http://stackoverflow.com/questions/11149759/remove-all-non-alphabetic-characters-from-a-string-array-in-java) and use the [regular palindrome check code](http://stackoverflow.com/a/4138856/3832970). See [this demo](http://ideone.com/ALr1Sl). Is it working as expected? — Wiktor Stribiżew, Feb 28 '16 at 22:21

Casimir et Hippolyte · Answer 1 · 2016-02-28T23:03:10.570

You can use this pattern with the matches method (add the case insensitive option if you want):

(?:[^a-z]*([a-z])(?=.*(\1[^a-z]*\2?+)$))+[^a-z]*[a-z]?[^a-z]*\2

If you want to match a single letter too, add |[^a-z]*[a-z][^a-z]* at the end.

demo regexplanet (Java)
demo regex101

details:

The idea is to capture one by one each letters from the start of the string in group 1 and to check each time in a lookahead if the same letter is present at the end. The capture group 2 is in the lookahead and capture at the end of the string its own content (from the previous repetition) and the new letter. At each repetition the capture group 2 grows with the new letter (and other characters that are not letters).

(?: # repeated non capturing group
    [^a-z]* # eventual other character before a letter
    ([a-z]) # the letter is captured in group 1
    (?=  # lookahead (to check the end of the string)
        .* 
        (
            \1      # backreference capture group1: the letter at the beginning
            [^a-z]* # other characters
            \2?+    # backreference capture group2: optional but possessive
                    # (that acts like a kind of conditional: if the group 2 already
                    # exists, it matches, otherwise not)
        )
        $  # anchor for the end of the string
    )
)+
[^a-z]*[a-z]?[^a-z]* # an eventual letter in the middle
\2 # backreference capture group 2

(With the matches method, anchors are implicit.)

How can I ignore spaces, punctuation marks, and all characters different than letters while checking a palindrome?

1 Answers1