249. Group Shifted Strings

The key point of this problem is to find "key" of every words. Time complexity: O(n)

How to find key of every string, we make every key as a string with 'a' as the first character.

1. get "shift": shift = cs[0]-'a';

2. minus every character by "shift"

3. convert the char[] to string

The string of step3 is the "key".

    Map<String, List<String>> map = new HashMap<>();
    public List<List<String>> groupStrings(String[] strings) {
        for(String s: strings){
            char[] cs = s.toCharArray();
            int shift = cs[0]-'a';
            for(int i=0;i<cs.length;i++){
                cs[i]=(char)(cs[i]-shift);
                if(cs[i]<'a')
                    cs[i]+=26;
                
            }
            String key = new String(cs);
            if(!map.containsKey(key)){
                map.put(key, new ArrayList<>());
            }
            map.get(key).add(s);
        }
        List<List<String>> res = new ArrayList<>();
        for(String key: map.keySet()){
            res.add(map.get(key));
        }
        return res;
    }