leetcode JAVA Word Ladder 3.44 难度系数3

Question：

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:
Only one letter can be changed at a time
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.

public class Solution {
public int ladderLength(String start, String end, HashSet<String> dict) {
int path = 0;

Queue<String> queue = new LinkedList<String>();
queue.offer(start);
HashSet<String> hs = new HashSet<>();
hs.add(start);
int length1 = 1;
int length2 = 0;
while(!queue.isEmpty()){
String s = queue.poll();
length1--;
HashSet<String> hsstr = nextStr(s,dict,hs);
length2 +=hsstr.size();
for(String str:hsstr){
if(str.equals(end)){
return path+2;
}else{
queue.offer(str);
}
}
if(length1==0){
path++;
length1=length2;
length2 = 0;
}
}
return 0;
}

private HashSet<String> nextStr(String s, HashSet<String> dict,
HashSet<String> hs) {
HashSet<String> hsstr = new HashSet<>();
for(int i=0;i<s.length();i++){
StringBuffer sb = new StringBuffer(s);
for(char c='a';c<='z';c++){
sb.setCharAt(i,c);
String temp = sb.toString();
if(dict.contains(temp)&&!hs.contains(temp)){
hs.add(temp);
hsstr.add(temp);
}
}
}
return hsstr;
}
}