249.Group Shifted Strings

    /*
     *249.Group Shifted Strings 
     *2016-6-18 by Mingyang
     *Given a string, we can "shift" each of its letter to its successive letter, 
     *for example: "abc" -> "bcd". We can keep "shifting" which forms the sequence:
     *"abc" -> "bcd" -> ... -> "xyz"
     *Given a list of strings which contains only lowercase alphabets, 
     *group all strings that belong to the same shifting sequence.
     *For example, given: ["abc", "bcd", "acef", "xyz", "az", "ba", "a", "z"], 
     *Return:
     *[
     *   ["abc","bcd","xyz"],
     *   ["az","ba"],
     *   ["acef"],
     *   ["a","z"]
     *]
     *这个题目一看就是HashMap，刚开始我说那String的长度来存作为key，但是发现相同长度的string可能不为同一阵营
     *所以后面该用一个新的string，就是把每两个字符之间的差存起来，写出一个string
     *["eqdf", "qcpr"]。
     *((‘q’ - 'e') + 26) % 26 = 12, ((‘d’ - 'q') + 26) % 26 = 13, ((‘f’ - 'd') + 26) % 26 = 2
     *((‘c’ - 'q') + 26) % 26 = 12, ((‘p’ - 'c') + 26) % 26 = 13, ((‘r’ - 'p') + 26) % 26 = 2
     *所以"eqdf"和"qcpr"是一组shifted strings。
     */
     public List<List<String>> groupStrings(String[] strings) {  
            List<List<String>> result = new ArrayList<List<String>>();  
            HashMap<String, List<String>> d = new HashMap<String, List<String>>();  
            for(int i = 0; i < strings.length; i++) {  
                StringBuffer sb = new StringBuffer();  
                for(int j = 0; j < strings[i].length(); j++) {  
                    sb.append(Integer.toString(((strings[i].charAt(j) - strings[i].charAt(0)) + 26) % 26));  
                    sb.append(" ");  
                }  
                String shift = sb.toString();  
                if(d.containsKey(shift)) {  
                    d.get(shift).add(strings[i]);  
                } else {  
                    List<String> l = new ArrayList<String>();  
                    l.add(strings[i]);  
                    d.put(shift, l);  
                }  
            }                
            for(String s : d.keySet()) {  
                Collections.sort(d.get(s));  
                result.add(d.get(s));  
            }   
            return result;  
        }