leetcode 127. Word ladder

Seawolf · 发表于 2019-12-24 11:22:40

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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 that beginWord is not a transformed word.
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.
Example 1:
Input:
beginWord = "hit",
endWord = "cog",
wordList = ["hot","dot","dog","lot","log","cog"]
Output: 5
Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.
Example 2:
Input:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]
Output: 0
Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.

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BFS

class Solution {
public int ladderLength(String beginWord, String endWord, List<String> wordAsList) {
if(!wordAsList.contains(endWord)){
return 0;
}
Queue<String> beginSet = new LinkedList<String>();
HashSet<String> wordList = new HashSet<String>(wordAsList);
beginSet.add(beginWord);
int level = 0;
while(!beginSet.isEmpty()){
int size = beginSet.size();
for(int i = 0 ; i< size; i++){
String word = beginSet.poll();
if(word.equals(endWord)) return level+1;
char[] wordset = word.toCharArray();
for(int k = 0 ; k< wordset.length; k++){
char temp = wordset[k];
for(char j = 'a' ; j<= 'z'; j++){
wordset[k] = j;
String str = new String(wordset);
if(wordList.contains(str)){
wordList.remove(str);
beginSet.add(str);
}
}
wordset[k] = temp;
}
}
level ++;
}
return 0;
}
}

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Bidirection BFS

class Solution {
public int ladderLength(String beginWord, String endWord, List<String> wordAsList) {
if(!wordAsList.contains(endWord)){
return 0;
}
Set<String> beginSet = new HashSet<String>();
Set<String> endSet = new HashSet<String>();
HashSet<String> wordList = new HashSet<String>(wordAsList);
beginSet.add(beginWord);
endSet.add(endWord);
int level = 1;
while(!beginSet.isEmpty() && !endSet.isEmpty()){
if(beginSet.size() > endSet.size()){
Set <String> temp = endSet;
endSet = beginSet;
beginSet = temp;
}
Set <String> temp = new HashSet<String>();
for(String word : beginSet){
char[] wordset = word.toCharArray();
int len = beginWord.length();
for(int k = 0 ; k< len; k++){
char temp1 = wordset[k];
for(char j = 'a' ; j<= 'z'; j++){
wordset[k] = j;
String str = new String(wordset);
if(endSet.contains(str)) return level+1;
if(!wordList.contains(str)) continue;
wordList.remove(str);
temp.add(str);
}
wordset[k] = temp1;
}
}
Set <String> temp2 = beginSet;
beginSet = temp;
temp = temp2;
level ++;
}
return 0;
}
}

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[学习笔记] leetcode 127. Word ladder