Solution_126.java

/*
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。
 */

import java.util.*;
import java.util.List;

public class Solution {

    public static void main(String[] args) {
        String beginword = "hit";
        String endword = "cog";
        List<String> wordlist = new ArrayList<>();
        wordlist.add("hot");
        wordlist.add("dot");
        wordlist.add("dog");
        wordlist.add("lot");
        wordlist.add("log");
        wordlist.add("cog");
        Solution solution = new Solution();
        System.out.println(solution.findLadders(beginword, endword, wordlist));
    }

    public List<List<String>> findLadders(String beginWord, String endWord, List<String> wordList) {
        List<List<String>> ans = new ArrayList<>();
        // 如果不含有结束单词，直接结束，不然后边会造成死循环
        if (!wordList.contains(endWord)) {
            return ans;
        }
        // 利用 BFS 得到所有的邻居节点,以及每个节点的所在层数
        HashMap<String, Integer> distance = new HashMap<>();
        HashMap<String, ArrayList<String>> map = new HashMap<>();
        bfs(beginWord, endWord, wordList, map, distance);

        ArrayList<String> temp = new ArrayList<String>();
        // temp 用来保存当前的路径
        temp.add(beginWord);
        findLaddersHelper(beginWord, endWord, map, distance, temp, ans);

        return ans;
    }

    private void findLaddersHelper(String beginWord, String endWord, HashMap<String, ArrayList<String>> map,
                                   HashMap<String, Integer> distance, ArrayList<String> temp, List<List<String>> ans) {
        if (beginWord.equals(endWord)) {
            ans.add(new ArrayList<String>(temp));
            return;
        }
        // 得到所有的下一个的节点
    /*
      "a"
      "c"
      ["a","b","c"]*/
        //之所以是 map.getOrDefault 而不是 get，就是上边的情况 get 会出错
        ArrayList<String> neighbors = map.getOrDefault(beginWord, new ArrayList<String>());
        for (String neighbor : neighbors) {
            //判断层数是否符合
            if (distance.get(beginWord) + 1 == distance.get(neighbor)) {
                temp.add(neighbor);
                findLaddersHelper(neighbor, endWord, map, distance, temp, ans);
                temp.remove(temp.size() - 1);
            }
        }
    }

    public void bfs(String beginWord, String endWord, List<String> wordList, HashMap<String, ArrayList<String>> map,
                    HashMap<String, Integer> distance) {
        Queue<String> queue = new LinkedList<>();
        queue.offer(beginWord);
        distance.put(beginWord, 0);
        boolean isFound = false;
        int depth = 0;
        Set<String> dict = new HashSet<>(wordList);
        while (!queue.isEmpty()) {
            int size = queue.size();
            depth++;
            for (int j = 0; j < size; j++) {
                String temp = queue.poll();
                // 一次性得到所有的下一个的节点
                ArrayList<String> neighbors = getNeighbors(temp, dict);
                map.put(temp, neighbors);
                for (String neighbor : neighbors) {
                    if (!distance.containsKey(neighbor)) {
                        distance.put(neighbor, depth);
                        if (neighbor.equals(endWord)) {
                            isFound = true;
                        }
                        queue.offer(neighbor);
                    }

                }
            }
            if (isFound) {
                break;
            }
        }
    }

    private ArrayList<String> getNeighbors(String node, Set<String> dict) {
        ArrayList<String> res = new ArrayList<String>();
        char chs[] = node.toCharArray();

        for (char ch = 'a'; ch <= 'z'; ch++) {
            for (int i = 0; i < chs.length; i++) {
                if (chs[i] == ch)
                    continue;
                char old_ch = chs[i];
                chs[i] = ch;
                if (dict.contains(String.valueOf(chs))) {
                    res.add(String.valueOf(chs));
                }
                chs[i] = old_ch;
            }

        }
        //System.out.println(node+" :  "+res.toString());
        return res;
    }

}