126. 单词接龙 II

给定两个单词（beginWord 和 endWord）和一个字典 wordList，找出所有从 beginWord 到 endWord 的最短转换序列。转换需遵循如下规则：

每次转换只能改变一个字母。

转换过程中的中间单词必须是字典中的单词。

说明:

如果不存在这样的转换序列，返回一个空列表。

所有单词具有相同的长度。

所有单词只由小写字母组成。

字典中不存在重复的单词。

你可以假设 beginWord 和 endWord 是非空的，且二者不相同。

示例 1:

输入:

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


输出:

[
  ["hit","hot","dot","dog","cog"],
  ["hit","hot","lot","log","cog"]
]


示例 2:

输入:

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


输出: []

解释: endWord “cog” 不在字典中，所以不存在符合要求的转换序列。

class Solution {
    public  List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {
        //结果
        List<List<String>> res = new ArrayList<>();
        if(wordList == null) return res;
        //bfs搜索所用的字典
        Set<String> dicts = new HashSet<>(wordList);
        if(!dicts.contains(endWord)) return res;
        if(dicts.contains(beginWord)) dicts.remove(beginWord);
        //bfs搜索最短路径所用的开始和结束的字典
        Set<String> endList = new HashSet<>(),
                    beginList = new HashSet<>();
        //每个点所对应的邻接点，list
        Map<String, List<String>> map = new HashMap<>();
        beginList.add(beginWord);
        endList.add(endWord);
        bfs(map, beginList, endList, beginWord, endWord,dicts, false);   
        //dfs的前进路线保存list
        List<String> subList = new ArrayList<>();
        subList.add(beginWord); 
        dfs(map, res, subList, beginWord, endWord);
        return res;
    }
    void dfs (Map<String, List<String>> map, 
        List<List<String>> result, List<String> subList, 
        String beginWord, String endWord) {
        if(beginWord.equals(endWord)) {
          result.add(new ArrayList<>(subList));
          return; 
        }
        if (!map.containsKey(beginWord)) {
          return; 
        }
        for (String word : map.get(beginWord)) {
          subList.add(word);
          dfs(map, result, subList, word, endWord);
          subList.remove(subList.size() - 1);
        }
    }
    //reverse是双端bfs的一个优化
    void bfs(Map<String, List<String>> map, Set<String> beginList, Set<String> endList, String beginWord, String endWord,Set<String> wordList, boolean reverse){
        if(beginList.size() == 0) return;
        wordList.removeAll(beginList);
        boolean finish = false;
        Set<String> temp = new HashSet<>();
        for(String str : beginList){
            char[] charr = str.toCharArray();
            for(int chI = 0; chI < charr.length; chI++){
                char old = charr[chI];
                for(char ch = 'a'; ch <= 'z'; ch++){
                    if(ch == old)
                        continue;
                    charr[chI] = ch;
                    String newstr = new String(charr);
                    if(!wordList.contains(newstr)){
                        continue;
                    }
                    //若是在某一层找到了最后的节点，就直接标记找到了，即一票决定。这里因为要找所有的最短路径，所以循环还是要继续的。
                    if(endList.contains(newstr)){
                        finish = true;
                    }else{
                        temp.add(newstr);
                    }
                    //无论怎么变换方向，永远用开始方向的字符做key，是为了后面的dfs，单一方向搜索
                    String key = reverse? newstr:str;
                    String value = reverse ? str : newstr;
                    if(!map.containsKey(key)){
                        map.put(key, new ArrayList<>());
                    }
                    map.get(key).add(value);

                }
                charr[chI] = old;
            }
        }
        if(!finish) {
            if(temp.size() > endList.size()){
                bfs(map, endList, temp, beginWord, endWord,wordList, !reverse);
            }else{
                bfs(map, temp, endList, beginWord, endWord, wordList, reverse);
            }
        }
    }
}