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.

class Solution {
    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
        Set<String> wordListSet = new HashSet<>(wordList); //将list转化为hashset
        if(!wordListSet.contains(endWord)) return 0;

        wordListSet.add(beginWord);

        Set<String> set = new HashSet<>();
        Queue<String> queue = new LinkedList<>();
        queue.offer(beginWord);
        set.add(beginWord);
        int level = 1;

        while(!queue.isEmpty()) {
            int size = queue.size(); //下面for循环是把每轮加进queue里的string处理完才让level+1
            for(int i = 0; i < size; i++) {
                String word = queue.poll();
                for(String nextWord : getNextWords(word, wordListSet)) {
                    if(set.contains(nextWord)) continue;
                    if(nextWord.equals(endWord)) return level + 1;

                    set.add(nextWord);
                    queue.offer(nextWord);
                }
            }
            level++;
        }
        return 0;
    }

    private List<String> getNextWords(String word, Set<String> wordList) {
        List<String> res = new ArrayList<>();
        for(char c = 'a'; c <= 'z'; c++) {
            for(int i = 0; i < word.length(); i++) {
                if(word.charAt(i) == c) continue;
                String nextWord = replace(word, c, i);
                if(wordList.contains(nextWord)) res.add(nextWord);
            }
        }
        return res;
    }

    private String replace(String word, char c, int position) {
        char[] chars = word.toCharArray(); //toCharArray将string转化为char的array
        chars[position] = c;
        return new String(chars); //将char的array转化为string
    }
}


第二次写
class Solution {
    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
        Set<String> wordListSet = new HashSet<>(wordList); //将list转化为hashset
        if(!wordListSet.contains(endWord)) return 0;

        //wordListSet.add(beginWord); //用不着

        Set<String> set = new HashSet<>();
        Queue<String> queue = new LinkedList<>();
        queue.offer(beginWord);
        set.add(beginWord);
        int level = 1;

        while(!queue.isEmpty()) {
            int size = queue.size(); //下面for循环是把每轮加进queue里的string处理完才让level+1
            for(int i = 0; i < size; i++) {
                String word = queue.poll();
                for(String nextWord : getNextWords(word, wordListSet)) {
                    if(nextWord.equals(endWord)) return level + 1;
                    if(!set.contains(nextWord)) {
                        set.add(nextWord);
                        queue.offer(nextWord);
                    } 
                }
            }
            level++;
        }
        return 0;
    }

    private List<String> getNextWords(String word, Set<String> wordList) {
        List<String> res = new ArrayList<>();
        for(char c = 'a'; c <= 'z'; c++) {
            for(int i = 0; i < word.length(); i++) {
                if(word.charAt(i) == c) continue;
                String nextWord = replace(word, c, i);
                if(wordList.contains(nextWord)) res.add(nextWord);
            }
        }
        return res;
    }

    private String replace(String word, char c, int position) {
        char[] chars = word.toCharArray(); //toCharArray将string转化为char的array
        chars[position] = c;
        return new String(chars); //将char的array转化为string
    }
}