[Leetcode] Anagrams

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

Note: All inputs will be in lower-case.

We sort each string in alphabet order, then each anagrams would have the same result. We could use this as the key to group all the anagrams in a hashmap. The time complexity of this method would be O(n*mlogm) where m is the average length of all strings. A better way is to use count sort of the strings since there are only 26 characters of them. The time complexity would reduce to O(n*m).

    public ArrayList<String> anagrams(String[] strs) {
        HashMap<String, List<String>> map = new HashMap<String, List<String>>();
        for(int i = 0; i < strs.length; i++) {
            char[] chs = strs[i].toCharArray();
            Arrays.sort(chs);
            String str = new String(chs);
            if(map.containsKey(str) == true) {
                map.get(str).add(strs[i]);
            }
            else {
                List<String> list = new ArrayList<String>();
                list.add(strs[i]);
                map.put(str, list);
            }
        }
        ArrayList<String> res = new ArrayList<String>();
        for(String key : map.keySet()) {
            if(map.get(key).size() > 1) {
                for(String str : map.get(key)) 
                    res.add(str);
            }
        }
        
        return res;
    }