LeetCode-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 length5.

Note:

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

//思路：双端HashSet,将其分为起始集合和结束集合，当起始集合中的单词变换一次后得到结束集合中的某词时可以结束
//当起始、结束集合中有任意一个为空时，总的也为空
//由于只有26个字母，因此，采用字母替换法所能达到的26*n在n较大时远远优于n*n
//init
import java.util.*;
public class Solution {
    public int ladderLength(String start, String end, HashSet<String> dict) {
        HashSet<String> beginSet=new HashSet<String>();
        HashSet<String> endSet=new HashSet<String>();
        HashSet<String> visited=new HashSet<String>();
         
        beginSet.add(start);
        endSet.add(end);
        visited.add(start);
        int res=1;
         
        while(!beginSet.isEmpty()&&!endSet.isEmpty()){
            //保证每次对较少数目的endSet做处理
            if(beginSet.size()>endSet.size()){
                HashSet<String> temp=beginSet;
                beginSet=endSet;
                endSet=temp;
            }
             
            HashSet<String> temp=new HashSet<String>();
             
            for(String str:beginSet){
                char[] chs=str.toCharArray();
                for(int i=0;i<chs.length;i++){
                    char old=chs[i];
                    for(char ch='a';ch<='z';ch++){
                        chs[i]=ch;
                        String cur=String.valueOf(chs);
                        if(endSet.contains(cur)){
                            return res+1;
                        }
                        if(!visited.contains(cur)&&dict.contains(cur)){
                            temp.add(cur);
                            visited.add(cur);
                        }
                    }
                    chs[i]=old;
                }
                 
            }
            beginSet=temp;
            res++;
        }
        return 0;
    }
}