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.

由于现做的126题，所以这道题其实只用BFS就可以了。

用126的答案。

public class Solution {

    public int ladderLength(String beginWord, String endWord, Set<String> wordList) {
        if( beginWord == null || beginWord.length()  == 0 || wordList.size() == 0 || beginWord.length() != endWord.length() )
            return 0;
        return BFS(beginWord,endWord,wordList);
    }

    public int BFS(String beginWord,String endWord,Set<String> wordList){


        Queue queue = new LinkedList<String>();
        queue.add(beginWord);
        int result = 1;
        while( !queue.isEmpty() ){
            String str = (String) queue.poll();
            if( str.equals(endWord) )
                continue;
            for( int i = 0 ;i <beginWord.length();i++){
                char[] word = str.toCharArray();
                for( char ch = 'a';ch<='z';ch++) {
                    word[i] = ch;
                    String Nword = new String(word);
                    if ( wordList.contains(Nword)) {
                        if (!map.containsKey(Nword)) {
                            map.put(Nword, (int) map.get(str) + 1);
                            queue.add(Nword);
                        }
                    }
                    if(  Nword.equals(endWord)  )
                        return (int) map.get(str) + 1;
                }
            }
        }
        return 0;
    }
}

 

 

去掉map，会快一些。

public class Solution {

    public int ladderLength(String beginWord, String endWord, Set<String> wordList) {
        if( beginWord == null || beginWord.length()  == 0 || wordList.size() == 0 || beginWord.length() != endWord.length() )
            return 0;

        Queue queue = new LinkedList<String>();
        queue.add(beginWord);
        int result = 1;
        while( ! queue.isEmpty() ){
            int len = queue.size();
            for( int i = 0;i<len;i++){
                String str = (String) queue.poll();
                for( int ii = 0; ii < str.length();ii++){
                    char[] word = str.toCharArray();
                    for( char ch = 'a'; ch<='z';ch++){
                        word[ii] = ch;
                        String newWord = new String(word);
                        if( wordList.contains(newWord) ){
                            wordList.remove(newWord);
                            queue.add(newWord);
                        }
                        if( newWord.equals(endWord) )
                            return result+1;
                    }
                }
            }
            result++;
        }
        return 0;
    }
}

还有更快的做法，一般是前后一起建立队列来做，会快很多。