LeetCode Group Shifted Strings

原題連接在這裏：https://leetcode.com/problems/group-shifted-strings/

題目：

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"], 
A solution is:

[
  ["abc","bcd","xyz"],
  ["az","ba"],
  ["acef"],
  ["a","z"]
]

題解：

與Group Anagrams相似. 維護一個HashMap, key是每一個string 的 base型.

Time Complexity: O(n * strings.length), n 是每一個string的平均長度.

Space: O(hm.size()), HashMap size.

AC Java:

 1 public class Solution {
 2     public List<List<String>> groupStrings(String[] strings) {
 3         if(strings == null || strings.length == 0){
 4             return new ArrayList<List<String>>();
 5         }
 6         
 7         HashMap<String, List<String>> hm = new HashMap<String, List<String>>();
 8         for(String str : strings){
 9             String base  = getBase(str);
10             if(!hm.containsKey(base)){
11                 List<String> ls = new ArrayList<String>();
12                 hm.put(base, ls);
13             }
14             hm.get(base).add(str);
15         }
16         
17         return new ArrayList<List<String>>(hm.values());
18     }
19     
20     private String getBase(String s){
21         if(s == null || s.length() == 0){
22             return s;
23         }
24         
25         StringBuilder sb = new StringBuilder();
26         int offset = s.charAt(0) - 'a';
27         for(int i = 0; i<s.length(); i++){
28             char c = (char)(s.charAt(i) - offset);
29             if(c < 'a'){
30                 c += 26;
31             }
32             sb.append(c);
33         }
34         return sb.toString();
35     }
36 }