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* Added the file * Added word-ladder program * Update README.md
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/* | ||
Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord to endWord, such that: | ||
Only one letter can be changed at a time. | ||
Each transformed word must exist in the word list. | ||
Note: | ||
Return 0 if there is no such transformation sequence. | ||
All words have the same length. | ||
All words contain only lowercase alphabetic characters. | ||
You may assume no duplicates in the word list. | ||
You may assume beginWord and endWord are non-empty and are not the same. | ||
Problem Link: https://leetcode.com/problems/word-ladder/ | ||
It is a popular interview program which is based on BFS | ||
*/ | ||
import java.util.*; | ||
class Solution { | ||
public int ladderLength(String beginWord, String endWord, List<String> wordList) { | ||
HashSet<String> hs=new HashSet<>(wordList); | ||
if(!hs.contains(endWord)) | ||
{ | ||
return 0; | ||
} | ||
int steps=1; | ||
Queue<String> q=new LinkedList<>(); | ||
q.add(beginWord); | ||
while(!q.isEmpty()) | ||
{ | ||
int count=q.size(); | ||
for(int i=0;i<count;i++) | ||
{ | ||
String curr=q.poll(); | ||
char a[]=curr.toCharArray(); | ||
for(int j=0;j<a.length;j++) | ||
{ | ||
char temp=a[j]; | ||
for(char c='a';c<='z';c++) | ||
{ | ||
if(a[j]==c) continue; | ||
a[j]=c; | ||
String test=new String(a); | ||
if(test.equals(endWord)) return steps+1; | ||
if(hs.contains(test)) | ||
{ | ||
q.add(test); | ||
hs.remove(test); | ||
} | ||
} | ||
a[j]=temp; | ||
} | ||
} | ||
steps++; | ||
} | ||
return 0; | ||
} | ||
} |
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