LeetCode-Alien Dictionary

There is a new alien language which uses the latin alphabet. However, the order among letters are unknown to you. You receive a list of words from the dictionary, where words are sorted lexicographically by the rules of this new language. Derive the order of letters in this language.

For example,
Given the following words in dictionary, 

[
  "wrt",
  "wrf",
  "er",
  "ett",
  "rftt"
]

The correct order is: "wertf".

Note:

    1. You may assume all letters are in lowercase.
    2. If the order is invalid, return an empty string.
    3. There may be multiple valid order of letters, return any one of them is fine.

Solution:

　　

 1 public class Solution {
 2     public void addressTwoWords(List<List<Character>> relSet, int[] preList, String word1, String word2) {
 3         int len = Math.min(word1.length(), word2.length());
 4         boolean relFound = false;
 5         for (int i = 0; i < len; i++) {
 6             char c1 = word1.charAt(i);
 7             char c2 = word2.charAt(i);
 8             if (c1 != c2 && !relFound) {
 9                 if (relSet.get(c1 - 'a') == null) {
10                     relSet.set(c1 - 'a', new ArrayList<Character>());
11                 }
12                 relSet.get(c1 - 'a').add(c2);
13                 preList[c2 - 'a'] = (preList[c2 - 'a'] == -1) ? 1 : preList[c2 - 'a'] + 1;
14                 relFound = true;
15             } else {
16                 if (preList[c2 - 'a'] == -1)
17                     preList[c2 - 'a'] = 0;
18             }
19         }
20         int len2 = Math.max(word1.length(), word2.length());
21         for (int i = len; i < len2; i++) {
22             if (i < word2.length() && preList[word2.charAt(i) - 'a'] == -1)
23                 preList[word2.charAt(i) - 'a'] = 0;
24         }
25     }
26 
27     public String alienOrder(String[] words) {
28         if (words.length == 0)
29             return "";
30         if (words.length == 1)
31             return words[0];
32 
33         List<List<Character>> relSet = new ArrayList<List<Character>>();
34         int[] preList = new int[26];
35         for (int i = 0; i < 26; i++) {
36             relSet.add(null);
37             preList[i] = -1;
38         }
39         StringBuilder builder = new StringBuilder();
40 
41         addressTwoWords(relSet, preList, "", words[0]);
42         for (int i = 0; i < words.length - 1; i++) {
43             addressTwoWords(relSet, preList, words[i], words[i + 1]);
44         }
45 
46         // Topological sort
47         HashSet<Character> set = new HashSet<Character>();
48         for (int i = 0; i < 26; i++)
49             if (preList[i] == 0) {
50                 char c = (char)('a' + i);
51                 set.add(c);
52             }
53 
54         while (!set.isEmpty()) {
55             Character c = set.iterator().next();
56             builder.append(c);
57             List<Character> sList = relSet.get(c - 'a');
58             if (sList != null)
59                 for (int i = 0; i < sList.size(); i++) {
60                     preList[sList.get(i) - 'a']--;
61                     if (preList[sList.get(i) - 'a'] == 0)
62                         set.add(sList.get(i));
63                 }
64             set.remove(c);
65         }
66         for (int i = 0; i < 26; i++)
67             if (preList[i] > 0)
68                 return "";
69         return builder.toString();
70     }
71 }

 

posted @ 2016-07-18 14:18  LiBlog  阅读(135)  评论(0)  编辑  收藏  举报