[leetcode]127. Word Ladder单词接龙

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:

1. Only one letter can be changed at a time.
2. 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.

思路

BFS

 

代码

class Solution {
   public int ladderLength(String beginWord, String endWord, List<String> wordList) {
    // use dict to check duplicats 
    Set<String> dict = new HashSet<>(wordList);
    Queue<String> queue = new LinkedList<>();
    queue.add(beginWord);
    int level = 0;
    while(!queue.isEmpty()){
        int size = queue.size();
        for(int i = 0; i < size; i++){
            String cur = queue.remove();
            if(cur.equals(endWord)){ return level + 1;}
            for(int j = 0; j < cur.length(); j++){
                // hit -> {'h', 'i', 't'}
                char[] charArray = cur.toCharArray();
                for(char c = 'a'; c <='z'; c++){
                    // {'h', 'i', 't'} for'h', try checking 'a','b'...'z' which forms ait, bit...zit
                    charArray[j] = c;
                    String temp = new String(charArray);
                    if(dict.contains(temp)){
                        queue.add(temp);
                        // to avoid dead loop, like hit will find hit itself
                        dict.remove(temp);
                    }
                }
            }
        }
        level++;
    }
    return 0;
 }
}