Anagrams and good hash key

Given an array of strings, return all groups of strings that are anagrams.

Note: All inputs will be in lower-case.

Anagrams, are two collections of words that are consisted with same characters. It is not hard to check if two strings are anagrams (use a hashset to store all characters of one and check the other), however, how can we check a group of strings that contains different anagrams? Naturally we will think of using a map to store each group of anagrams with a hash key, then the next problem is, how to determine a good hash key?

At first, I thought that if two strings are consisted of same characters, the sum of all characters should be the same. That is true, however, it is possible that two strings composed of different words also have the same sum of characters. (e.g., "duh", "ill"). Here is a good one I found online: 

private int getKey(String s) {
        int key = 0;
        int a = 378551;
        int b = 63689;
        int[] count = new int[26];
        for (int i = 0; i < s.length(); i++) {
            int tmp = s.charAt(i) - 'a';
            count[tmp]++;
        }
        for (int num : count) {
            key = key * a + num;
            a = a * b;
        }
        return key;
    }

The rest of the method is trivial.

 public List anagrams(String[] strs) {
        if (strs == null)
            throw new NullPointerException("Null string array!");
        List rst = new ArrayList ();
        if (strs.length <= 1)
            return rst;
        Map> map = new HashMap> ();
        for (String s: strs) {
            int key = getKey(s);
            if (!map.containsKey(key)) {
                map.put(key, new ArrayList());
            }
            map.get(key).add(s);
        }
        for (ArrayList a : map.values()) {
            if (a.size() > 1)
                rst.addAll(a);
        }
        return rst;
        
    }
    private int getKey(String s) {
        int key = 0;
        int a = 378551;
        int b = 63689;
        int[] count = new int[26];
        for (int i = 0; i < s.length(); i++) {
            int tmp = s.charAt(i) - 'a';
            count[tmp]++;
        }
        for (int num : count) {
            key = key * a + num;
            a = a * b;
        }
        return key;
    }