127. Word Ladder（BFS）

题目：

Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence frombeginWord to endWord, such that:

1. Only one letter can be changed at a time
2. 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.

求最短路径长度，相当于找字典树最小深度，用bfs

代码：

public class Solution {    public int ladderLength(String beginWord, String endWord, Set<String> wordList) {        Map<String, Integer> map = new HashMap<String, Integer>();        LinkedList<String> queue = new LinkedList<String>();        queue.add(beginWord);        wordList.remove(beginWord);        map.put(beginWord, 1);        while(!queue.isEmpty())        {        StringBuilder builder;        String cur = queue.pop();        int level = map.get(cur);        for(int i=0;i<cur.length();i++)        {        builder = new StringBuilder(cur);        for(int j='a';j<='z';j++)        {        builder.setCharAt(i, (char)j);        String temp = builder.toString();                if(temp.equals(cur))        {        continue;        }        if(temp.equals(endWord))        {        return level+1;        }        if(wordList.contains(temp))        {        wordList.remove(temp);        queue.add(temp);        map.put(temp, level+1);        }        }        }        }                return 0;    }}