LeetCode题解：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:

Only one letter can be changed at a time
Each intermediate word must exist in the word list
For example,

Given:
beginWord = “hit”
endWord = “cog”
wordList = [“hot”,”dot”,”dog”,”lot”,”log”]
As one shortest transformation is “hit” -> “hot” -> “dot” -> “dog” -> “cog”,
return its length 5.

Note:
Return 0 if there is no such transformation sequence.
All words have the same length.
All words contain only lowercase alphabetic characters.

题意：给定两个单词和字典，判断单词A是否能通过变换字典中的的单词，并不断向下变换为单词B

解决思路：修改单词中某个单词，然后看字典里是否存在该单词，存在则提取

代码：

public class Solution {
    public int ladderLength(String start, String end, Set<String> dict) {
        LinkedList<String> queue = new LinkedList<String>();
        queue.add(start);
        dict.add(end);
        int step = 0;

        while (!queue.isEmpty()) {
            LinkedList<String> level = new LinkedList<String>();
            step++;
            while (!queue.isEmpty()) {
                String q = queue.poll();
                if(q.equals(end))
                    return step;

                char[] t = q.toCharArray();
                for(int i = 0; i < start.length(); i++){
                    for(char c = 'a'; c <= 'z'; c++){
                        char temp = t[i];
                        t[i] = c;
                        String s = String.copyValueOf(t);
                        t[i] = temp;
                        if(dict.contains(s)){
                            level.add(s);
                            dict.remove(s);
                        }
                    }
                }
            }
            queue = level;
        }

        return 0;
    }
}