Group Shifted Strings

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"]
]

Note: For the return value, each inner list's elements must follow the lexicographic order.

题解：

既然是group，那么我们需要有一个规则，把符合规则的放在一起。那么，规则是什么呢？

根据题目的意思，对于两串字符串，如果字符串的相邻字符的差值是一致的，那么我们就可以把它们放在一起。

但是对于az, ba这样的，字符之间的差值不一样啊。但是为何它们是一组的呢？

az字符之间的差值是25，ba之间字符的差值是-1，但是字符是每隔26就一循环，所以，对于-1，你一旦加上26就是25.

 public class Solution {
     public List<List<String>> groupStrings(String[] strings) {
         List<List<String>> result = new ArrayList<List<String>>();
         HashMap<String, List<String>> map = new HashMap<String, List<String>>();
         for (int i = ; i < strings.length; i++) {
             StringBuffer sb = new StringBuffer();
             for (int j = ; j < strings[i].length(); j++) {
                 sb.append(Integer.toString(((strings[i].charAt(j) - strings[i].charAt()) + ) % ));
                 sb.append(" ");
             }
             String shift = sb.toString();
 
             if (map.containsKey(shift)) {
                 map.get(shift).add(strings[i]);
             } else {
                 List<String> list = new ArrayList<String>();
                 list.add(strings[i]);
                 map.put(shift, list);
             }
         }
 
         for (String s : map.keySet()) {
             Collections.sort(map.get(s));
             result.add(map.get(s));
         }
         return result;
     }
 }