Use [^A-Za-z0-9].

Note: removed the space since that is not typically considered alphanumeric.

Change your regex to:

.*\\W+.*

The Java Pattern class, which is Java's implementation of regex, supports Unicode Categories, e.g. \p{Lu}. Since you want alphanumeric, that would be Categories L (Letter) and N (Number).

Since your example shows you also want to keep spaces, you need to include that. Let's use the Predefined Character Class \s, so you also get to keep newlines and tabs.

To find anything but the specified characters, use a Negation Character Class: [^abc]

All-in-all, that means [^\s\p{L}\p{N}]:

String output = input.replaceAll("[^\\s\\p{L}\\p{N}]+", "");

Where What is that an animal No It is a plane
Dónde Qué es eso un animal No Es un avión
Onde O que é isso um animal Não É um avião

Or see regex101.com for demo.

Of course, there are multiple ways to do it.

You could alternatively use the POSIX Character Class \p{Alnum}, and then enable UNICODE_CHARACTER_CLASS, using (?U).

String output = input.replaceAll("(?U)[^\\s\\p{Alnum}]+", "");

Where What is that an animal No It is a plane
Dónde Qué es eso un animal No Es un avión
Onde O que é isso um animal Não É um avião

Now, if you didn't want spaces, that could be simplified by using \P{xx} instead:

String output = input.replaceAll("(?U)\\P{Alnum}+", "");

WhereWhatisthatananimalNoItisaplane
DóndeQuéesesounanimalNoEsunavión
OndeOqueéissoumanimalNãoÉumavião

Considering you want to check for ASCII Alphanumeric characters, Try this: "^[a-zA-Z0-9]*$". Use this RegEx in String.matches(Regex), it will return true if the string is alphanumeric, else it will return false.

public boolean isAlphaNumeric(String s){
    String pattern= "^[a-zA-Z0-9]*$";
    return s.matches(pattern);
}

If it will help, read this for more details about regex: http://www.vogella.com/articles/JavaRegularExpressions/article.html