One approach is to do that using the String class itself. Let's say that your string is something like that:

String s = "some text";
boolean hasNonAlpha = s.matches("^.*[^a-zA-Z0-9 ].*$");

one other is to use an external library, such as Apache commons:

String s = "some text";
boolean hasNonAlpha = !StringUtils.isAlphanumeric(s);

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;
    }

}