Word Ladder

Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord to endWord, such that:

  1. Only one letter can be changed at a time
  2. Each intermediate word must exist in the word list

For example,

Given:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

Note:

    • Return 0 if there is no such transformation sequence.
    • All words have the same length.
    • All words contain only lowercase alphabetic characters. 
       1 public class Solution {
 2     public int ladderLength(String beginWord, String endWord, Set<String> wordList) {
 3         if (wordList.size () == 0) {
 4             return 0;
 5         }
 6         wordList.add(beginWord);
 7         wordList.add(endWord);
 8         HashSet<String> set = new HashSet<>();
 9         Queue<String> q = new LinkedList<>();
10         set.add(beginWord);
11         q.offer(beginWord);
12         int length = 1;
13         while (!q.isEmpty()) {
14             ++length;
15             int size = q.size();
16             for (int i = 0; i < size; i++) {
17                 String crt = q.poll();
18                 for (String next : getNextWords(crt, wordList)) {
19                     if (set.contains(next)) {
20                         continue;
21                     } 
22                     if (next.equals(endWord)) {
23                         return length;
24                     } 
25                     q.offer(next);
26                     set.add(next);
27                 }
28             }
29         }
30         return 0;
31     }
32      private String replace(String s, int index, char c) {
33         char[] chars = s.toCharArray();
34         chars[index] = c;
35         return new String(chars);
36     }
37     private ArrayList<String> getNextWords(String word, Set<String> dict) {
38         ArrayList<String> nextWords = new ArrayList<String>();
39         for (char c = 'a'; c <= 'z'; c++) {
40             for (int i = 0; i < word.length(); i++) {
41                 if (c == word.charAt(i)) {
42                     continue;
43                 }
44                 String nextWord = replace(word, i, c);
45                 if (dict.contains(nextWord)) {
46                     nextWords.add(nextWord);
47                 }
48             }
49         }
50         return nextWords;
51     }
52 
53 }

       

posted @ 2016-04-30 21:30  YuriFLAG  阅读(122)  评论(0)    收藏  举报