Fastest way to check a string is alphanumeric in Java

Question

What is the fastest way to check that a String contains only alphanumeric characters.

I've got some code that is going to chew up a lot of CPU and I wonder if there is going to be a quicker way than using pre-compiled regular expressions.

Answer 1

Use String.matches(), like:

String myString = "qwerty123456";
System.out.println(myString.matches("[A-Za-z0-9]+"));

That may not be the absolute "fastest" possible approach. But in general there's not much point in trying to compete with the people who write the language's "standard library" in terms of performance.

Answer 2

I've written the tests that compare using regular expressions (as per other answers) against not using regular expressions. Tests done on a quad core OSX10.8 machine running Java 1.6

Interestingly using regular expressions turns out to be about 5-10 times slower than manually iterating over a string. Furthermore the isAlphanumeric2() function is marginally faster than isAlphanumeric(). One supports the case where extended Unicode numbers are allowed, and the other is for when only standard ASCII numbers are allowed.

public class QuickTest extends TestCase {

    private final int reps = 1000000;

    public void testRegexp() {
        for(int i = 0; i < reps; i++)
            ("ab4r3rgf"+i).matches("[a-zA-Z0-9]");
    }

public void testIsAlphanumeric() {
    for(int i = 0; i < reps; i++)
        isAlphanumeric("ab4r3rgf"+i);
}

public void testIsAlphanumeric2() {
    for(int i = 0; i < reps; i++)
        isAlphanumeric2("ab4r3rgf"+i);
}

    public boolean isAlphanumeric(String str) {
        for (int i=0; i<str.length(); i++) {
            char c = str.charAt(i);
            if (!Character.isLetterOrDigit(c))
                return false;
        }

        return true;
    }

    public boolean isAlphanumeric2(String str) {
        for (int i=0; i<str.length(); i++) {
            char c = str.charAt(i);
            if (c < 0x30 || (c >= 0x3a && c <= 0x40) || (c > 0x5a && c <= 0x60) || c > 0x7a)
                return false;
        }
        return true;
    }

}

Answer 3

A regex will probably be quite efficient, because you would specify ranges: [0-9a-zA-Z]. Assuming the implementation code for regexes is efficient, this would simply require an upper and lower bound comparison for each range. Here's basically what a compiled regex should do:

boolean isAlphanumeric(String str) {
    for (int i=0; i<str.length(); i++) {
        char c = str.charAt(i);
        if (c < 0x30 || (c >= 0x3a && c <= 0x40) || (c > 0x5a && c <= 0x60) || c > 0x7a)
            return false;
    }

    return true;
}

I don't see how your code could be more efficient than this, because every character will need to be checked, and the comparisons couldn't really be any simpler.