This is an alternate solution when verifying if a String is an anagram.

The algorithm:

• For both Strings: Remove special characters (!,’.)
• For both Strings: Split the String into array of characters
• Iterate both arrays and check if they have same size and characters are the same

To remove the special characters we can use regex:

public class SpecialCharRemover {

    public static String removeSpecial(String s1){
        if(isNull(s1)){
            return null;
        }
        return s1.replaceAll("[^\\w\\s]","");
    }
}

Creating character map

public class CharMap {

    public static Map<String, Integer> getCharMap(String s){
        if(isNull(s)){
            return new HashMap<>();
        }
        Map<String, Integer> charMap = new HashMap<>();
        for(String curr : asList(s.split(""))){
            if(charMap.containsKey(curr)){
                charMap.put(curr, charMap.get(curr) + 1);
            }else{
                charMap.put(curr, 1);
            }
        }

        return charMap;
    }
}

wrapping it all up

public class Anagram {

    public static boolean isAnagram(String s1, String s2){
        if(isNull(s1) || isNull(s2)){
            return false;
        }
        Map<String, Integer> charMap1 = CharMap.getCharMap(SpecialCharRemover.removeSpecial(s1));
        Map<String, Integer> charMap2 = CharMap.getCharMap(SpecialCharRemover.removeSpecial(s2));

        if(charMap1.keySet().size() != charMap2.keySet().size()){
            return false;
        }

        for(String curr : charMap1.keySet()){
            if(!charMap1.get(curr).equals(charMap2.get(curr))){
                return false;
            }
        }
        return true;
    }
}

Source code: Here