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 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;
}
}