An O(n) solution to anagrams in constant space using XOR

xorAnagram.java

boolean isAnagram( String s1, String s2) {
   if (s1.length() != s2.length()) return false;
   int x = 0;
   for (int i = 0; i<s1.length(); i++) {
      x ^= s1.charAt(i) ^ s2.charAt(i);
   }
   return x == 0;
}