Word Ladder

127. Word Ladder

Difficulty: Medium

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 transformed word must exist in the word list. Note that beginWord is not a transformed word.

每次只能改变一个字母，改变之后的单词必须在wordlist中,类似于数组的编辑距离，但是对结果进行的限制。

For example,

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

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.
• You may assume no duplicates in the word list.
• You may assume beginWord and endWord are non-empty and are not the same.

UPDATE (2017/1/20):
The wordList parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.

Solution

使用层次遍历 bfs即可，因为Note里面有限定条件，因此不必自己构建无向图，按26个英文字母遍历即可。

class Solution {
    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
        Set<String> wordSet = new HashSet<>(wordList); // 把list转成set,防止超时 contains由O(n)变为O(1)
        if(!wordSet.contains(endWord))return 0;
        int res = 1;
        LinkedList<String> q = new LinkedList<>();
        q.addLast(beginWord);      
        while(!q.isEmpty()){     
            int qSize = q.size();
            while(qSize-->0){
                String s = q.getFirst();
                for(int j=0;j<s.length();j++){
                    char[] temp = s.toCharArray();
                    for(int i=0;i<27;i++){
                        char c = (char)('a'+i);
                        temp[j] = c;
                        String word = String.valueOf(temp);
                        if(word.equals(endWord))return res+1;
                        if(wordSet.contains(word)){
                            q.addLast(word);
                            wordSet.remove(word);
                        }
                    }
                }
                q.removeFirst();
            }
            res++;
        }
        return 0;
    }
}