leetcode 127. Word Ladder ----- java
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 intermediate word must exist in the word list
For example,
Given:
beginWord = "hit"
endWord = "cog"
wordList = ["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.
由于现做的126题,所以这道题其实只用BFS就可以了。
用126的答案。
public class Solution { public int ladderLength(String beginWord, String endWord, Set<String> wordList) {
if( beginWord == null || beginWord.length() == 0 || wordList.size() == 0 || beginWord.length() != endWord.length() )
return 0;
return BFS(beginWord,endWord,wordList);
} public int BFS(String beginWord,String endWord,Set<String> wordList){ Queue queue = new LinkedList<String>();
queue.add(beginWord);
int result = 1;
while( !queue.isEmpty() ){
String str = (String) queue.poll();
if( str.equals(endWord) )
continue;
for( int i = 0 ;i <beginWord.length();i++){
char[] word = str.toCharArray();
for( char ch = 'a';ch<='z';ch++) {
word[i] = ch;
String Nword = new String(word);
if ( wordList.contains(Nword)) {
if (!map.containsKey(Nword)) {
map.put(Nword, (int) map.get(str) + 1);
queue.add(Nword);
}
}
if( Nword.equals(endWord) )
return (int) map.get(str) + 1;
}
}
}
return 0;
}
}
去掉map,会快一些。
public class Solution { public int ladderLength(String beginWord, String endWord, Set<String> wordList) {
if( beginWord == null || beginWord.length() == 0 || wordList.size() == 0 || beginWord.length() != endWord.length() )
return 0; Queue queue = new LinkedList<String>();
queue.add(beginWord);
int result = 1;
while( ! queue.isEmpty() ){
int len = queue.size();
for( int i = 0;i<len;i++){
String str = (String) queue.poll();
for( int ii = 0; ii < str.length();ii++){
char[] word = str.toCharArray();
for( char ch = 'a'; ch<='z';ch++){
word[ii] = ch;
String newWord = new String(word);
if( wordList.contains(newWord) ){
wordList.remove(newWord);
queue.add(newWord);
}
if( newWord.equals(endWord) )
return result+1;
}
}
}
result++;
}
return 0;
}
}
还有更快的做法,一般是前后一起建立队列来做,会快很多。
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