0127.单词接龙.java

import java.util.HashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
import java.util.Set;

/*
 * @lc app=leetcode.cn id=127 lang=java
 *
 * [127] 单词接龙
 *
 * https://leetcode.cn/problems/word-ladder/description/
 *
 * algorithms
 * Hard (47.67%)
 * Likes:    1067
 * Dislikes: 0
 * Total Accepted:    156.1K
 * Total Submissions: 327K
 * Testcase Example:  '"hit"\n"cog"\n["hot","dot","dog","lot","log","cog"]'
 *
 * 字典 wordList 中从单词 beginWord 和 endWord 的 转换序列 是一个按下述规格形成的序列 beginWord -> s1 ->
 * s2 -> ... -> sk：
 * 
 * 
 * 每一对相邻的单词只差一个字母。
 * 对于 1 <= i <= k 时，每个 si 都在 wordList 中。注意， beginWord 不需要在 wordList 中。
 * sk == endWord
 * 
 * 
 * 给你两个单词 beginWord 和 endWord 和一个字典 wordList ，返回 从 beginWord 到 endWord 的 最短转换序列
 * 中的 单词数目 。如果不存在这样的转换序列，返回 0 。
 * 
 * 
 * 示例 1：
 * 
 * 
 * 输入：beginWord = "hit", endWord = "cog", wordList =
 * ["hot","dot","dog","lot","log","cog"]
 * 输出：5
 * 解释：一个最短转换序列是 "hit" -> "hot" -> "dot" -> "dog" -> "cog", 返回它的长度 5。
 * 
 * 
 * 示例 2：
 * 
 * 
 * 输入：beginWord = "hit", endWord = "cog", wordList =
 * ["hot","dot","dog","lot","log"]
 * 输出：0
 * 解释：endWord "cog" 不在字典中，所以无法进行转换。
 * 
 * 
 * 
 * 提示：
 * 
 * 
 * 1 <= beginWord.length <= 10
 * endWord.length == beginWord.length
 * 1 <= wordList.length <= 5000
 * wordList[i].length == beginWord.length
 * beginWord、endWord 和 wordList[i] 由小写英文字母组成
 * beginWord != endWord
 * wordList 中的所有字符串 互不相同
 * 
 * 
 */

// @lc code=start
class Solution {
    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
        Set<String> set = new HashSet<>(wordList);
        if (set.size() == 0 || !set.contains(endWord))
            return 0;

        set.remove(beginWord);
        Queue<String> queue = new LinkedList<>();
        Set<String> visited = new HashSet<>();
        queue.offer(beginWord);
        visited.add(beginWord);

        int step = 1;
        while (!queue.isEmpty()) {
            int count = queue.size();
            for (int i = 0; i < count; i++) {
                String currentWord = queue.poll();
                if (changeWordEveryOneLetter(currentWord, endWord, queue, set, visited))
                    return ++step;
            }

            step++;
        }

        return 0;
    }

    private boolean changeWordEveryOneLetter(String currentWord, String endWord, Queue<String> queue, Set<String> words,
            Set<String> visited) {
        char[] cs = currentWord.toCharArray();
        for (int i = 0; i < endWord.length(); i++) {
            char c = cs[i];
            for (char j = 'a'; j <= 'z'; j++) {
                if (c == j)
                    continue;

                cs[i] = j;
                String nextWord = String.valueOf(cs);
                if (words.contains(nextWord)) {
                    if (endWord.equals(nextWord))
                        return true;
                    if (!visited.contains(nextWord)) {
                        queue.add(nextWord);
                        visited.add(nextWord);
                    }
                }
            }

            cs[i] = c;
        }

        return false;
    }
}
// @lc code=end