If you're using guava on your project (and if you're not, I believe you should consider it), the CharMatcher class handles this very nicely:

Your first example might be:

result = CharMatcher.WHITESPACE.removeFrom(dirtyString);

while your second might be:

result = CharMatcher.anyOf(" *#&").removeFrom(dirtyString);
// or alternatively
result = CharMatcher.noneOf(" *#&").retainFrom(dirtyString);

or if you want to be more flexible with whitespace (tabs etc), you can combine them rather than writing your own:

CharMatcher illegal = CharMatcher.WHITESPACE.or(CharMatcher.anyOf("*#&"));
result = illegal.removeFrom(dirtyString);

or you might instead specify legal characters, which depending on your requirements might be:

CharMatcher legal = CharMatcher.JAVA_LETTER; // based on Unicode char class
CharMatcher legal = CharMatcher.ASCII.and(CharMatcher.JAVA_LETTER); // only letters which are also ASCII, as your examples
CharMatcher legal = CharMatcher.inRange('a', 'z'); // lowercase only
CharMatcher legal = CharMatcher.inRange('a', 'z').or(CharMatcher.inRange('A', 'Z')); // either case

followed by retainFrom(dirtyString) as above.

Very nice, powerful API.

Use replaceAll.

This will do it:

String dirtyString = "This contains spaces which are not allowed";
String result = dirtyString.replaceAll("\\s", "");

and works by replacing all whitespace with 'nothing'.

String resultString = subjectString.replaceAll("\\P{L}+", "");

will replace any non-letter characters with nothing.

I also prefer the whitelisting-approach. You'll never know what comes around. There seem to be more encodings in than characters. This way you can control it all:

public String convert(String s) {
  s = StringUtils.removePattern(s, "[^A-Za-zäöüÄÖÜß?!$,. 0-9\\-\\+\\*\\?=&%\\$§\"\\!\\^#:;,_²³°\\[\\]\\{\\}<>\\|~]'`'");
  return s.trim();
}

This contains all german umlauts and french accents and ... you know - just look at your keyboard. I think I picked them all. Feel free to omit special chars like < > to prevent code-injection...

Filter code points

Regex is not the only avenue to your goal. You can get the code point integer number for each character in your string, then filter out those not considered a letter in Unicode.

The String#codePoints method returns an IntStream, a stream of int primitive values, one per character.

The Character class can tell us if the character assigned to each of those code point numbers in Unicode is considered a letter, as opposed to whitespace, digits, punctuation, and so on.

Those code points passing our test are converted back to a String by way of the StringBuilder class.

String input = " this*is#a*&very_dirty&String" ; 
String onlyLetters = 
        input 
        .codePoints()
        .filter(
            codePoint -> Character.isLetter( codePoint ) 
        )
        .collect(               
            StringBuilder :: new ,        
            StringBuilder :: appendCodePoint , 
            StringBuilder :: append    
        )        
        .toString() 
;

See this code run live at Ideone.com.

thisisaverydirtyString