Word Ladder

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:

1. Only one letter can be changed at a time
2. Each intermediate word must exist in the dictionary

For example,

Given:
start = "hit"
end = "cog"
dict = ["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.

解题思路：

　　BFS, 类似于图的遍历， 将start一步内可以到达的单词加入到queue中，并将相应的长度放入另一个queue中，每次poll 出一个单词，看是否是目标单词，如果是，则返回相应的长度。

对每个单词的每一位进行'a' - 'z' 的变化，看是否在 dict中，如果在，则说明该单词是转换路径中的一个。将该词从字典中删除。（表示已经访问过了）

public class Solution {
    public int ladderLength(String start, String end, HashSet<String> dict) {
        int result = 0;
        
        if(dict.size() == 0){
            return result; 
        }
        
        dict.add(start);
        dict.add(end);
        
        result = BFS(start, end, dict);
        
        return result;
        
    }
    
    
    public int BFS(String start, String end, HashSet<String> dict){
        Queue<String> words = new LinkedList<String>();
        Queue<Integer> lengths = new LinkedList<Integer>();
        
        words.add(start);
        lengths.add(1);
        
        while(!words.isEmpty()){
            String word = words.poll();
            int len = lengths.poll();
            
            if(word.equals(end)){
                return len;
            }
            
            for(int i = 0; i< word.length(); i++){
                
                char[] arr = word.toCharArray();
                for (char c = 'a'; c <= 'z'; c++) {
                    
                    if(arr[i] == c){
                        continue;
                    }
                    
                    arr[i] = c;
                    
                    String str = String.valueOf(arr);
                    if(dict.contains(str)){
                        words.add(str);
                        lengths.add(len+1);
                        dict.remove(str);
                    }
                }
            }
        }
        return 0;
    }
    
}