Group Shifted Strings(Easy)

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"]
]

解题思路

["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。

如此以来就很好理解了。

public class Solution {
   private String shiftStr(String str) {
        StringBuffer buffer = new StringBuffer();
        char[] char_array = str.toCharArray();
        int dist = str.charAt(0) - 'a';
        for (char c : char_array) 
            buffer.append((c - 'a' - dist + 26) % 26 + 'a');
        return buffer.toString();
    }

    public List<List<String>> groupStrings(String[] strings) {
        List<List<String>> result = new ArrayList<List<String>>();
        HashMap<String, List<String>> d = new HashMap<>();
        for(int i = 0; i < strings.length; i++) {
            String shift = shiftStr(strings[i]);

            if(d.containsKey(shift)) {
                d.get(shift).add(strings[i]);
            } else {
                List<String> l = new ArrayList<>();
                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;
    }
}