127. Word Ladder

最新推荐文章于 2021-10-31 08:20:30 发布

dici5041

最新推荐文章于 2021-10-31 08:20:30 发布

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原文链接：http://www.cnblogs.com/wentiliangkaihua/p/11546553.html

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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:

Only one letter can be changed at a time.
Each transformed word must exist in the word list. Note that beginWord is not a transformed word.

Note:

Return 0 if there is no such transformation sequence.
All words have the same length.
All words contain only lowercase alphabetic characters.
You may assume no duplicates in the word list.
You may assume beginWord and endWord are non-empty and are not the same.

Example 1:

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

Output: 5

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

Example 2:

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

Output: 0

Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.

// Author: Huahua
// Running time: 92 ms (<76.61%)
class Solution {
  public int ladderLength(String beginWord, String endWord, List<String> wordList) {
    Set<String> dict = new HashSet<>();
    for (String word : wordList) dict.add(word);
    
    if (!dict.contains(endWord)) return 0;  //不存在dict的endword直接返回0
    
    Queue<String> q = new ArrayDeque<>();
    q.offer(beginWord);
    
    int l = beginWord.length();
    int steps = 0;
    
    while (!q.isEmpty()) {
      ++steps;
      for (int s = q.size(); s > 0; --s) {
        String w = q.poll();       
        char[] chs = w.toCharArray();   //方便修改某一位
        for (int i = 0; i < l; ++i) {
          char ch = chs[i];
          for (char c = 'a'; c <= 'z'; ++c) {
            if (c == ch) continue;  //c == ch，重复，可以跳过
            chs[i] = c;
            String t = new String(chs);         
            if (t.equals(endWord)) return steps + 1;            
            if (!dict.contains(t)) continue;            //跳过无意义的case
            dict.remove(t);     //dict单词只能使用一次，否则路径一定会增加       
            q.offer(t);
          }
          chs[i] = ch;  //一位字符用完后要恢复原值
        }
      }
    }
    return 0;
  }
}