JavaScript Regex to Java Regex - Replace and Lambdas

Wow, the original code is a bit over-verbose. I'd do:

return phoneStr.replace(/[a-zA-Z]/g, function(m) {
    var ix= m[0].toLowerCase().charCodeAt(0)-0x61; // ASCII 'a'
    return '22233344455566677778889999'.charAt(ix);
});

And consequently in Java, something like:

StringBuffer b= new StringBuffer();
Matcher m= Pattern.compile("[A-Za-z]").matcher(phoneStr);
while (m.find()) {
    int ix= (int) (m.group(0).toLowerCase().charAt(0)-'a');
    m.appendReplacement(b, "22233344455566677778889999".substring(ix, ix+1));
}
m.appendTail(b);
return b.toString();

Replacing with Java's Matcher is clumsy enough that you might just prefer to use an array for this case:

char[] c= phoneStr.toLowerCase().toCharArray();
for (int i= 0; i<c.length; i++)
    if (c[i]>='a' && c[i]<='z')
        c[i]= "22233344455566677778889999".charAt((int) (c[i]-'a'));
return new String(c);

A very basic way to do this would be :

String replaceCharsByNumbers(String stringToChange) {
    return stringToChange
            .replace('a', '2')
            .replace('b', '2')
            .replace('c', '2')
            .replace('d', '3')
            .replace('e', '3')
            .replace('f', '3')
            .replace('g', '4')
            .replace('h', '4')
            .replace('i', '4')
            .replace('j', '5')
            .replace('k', '5')
            .replace('l', '5')
            .replace('m', '6')
            .replace('n', '6')
            .replace('o', '6')
            .replace('p', '7')
            .replace('q', '7')
            .replace('r', '7')
            .replace('s', '7')
            .replace('t', '8')
            .replace('u', '8')
            .replace('v', '8')
            .replace('w', '9')
            .replace('x', '9')
            .replace('y', '9')
            .replace('z', '9');
}