Word Ladder

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:

• 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.

import java.util.HashMap;

class Solution {
    public int ladderLength(String start, String end, List<String> dictionary) {
        // Boolean parts works as visited
        HashMap<String, Boolean> dict = new HashMap<>();
        for (String x : dictionary)
            dict.put(x, false);
        if (!dict.containsKey(end))
            return 0;
        Queue<String> q = new LinkedList<String>();
        q.add(start);
        int level = 2;
        while (!q.isEmpty()) {
            int size = q.size();
            for (int i = 0; i < size; i++) {
                String tmp = q.poll();
                for (int j = 0; j < tmp.length(); j++) {
                    char[] chars = tmp.toCharArray();
                    for (char c = 'a'; c <= 'z'; c++) {
                        chars[j] = c;
                        String tmp2 = new String(chars);
                        if (tmp2.equals(end))
                            return level;
                        if (dict.containsKey(tmp2) && dict.get(tmp2) == false) {
                            q.add(tmp2);
                            dict.put(tmp2, true);
                        }
                    }
                }
            }
            level++;
        }
        return 0;
    }
}