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.思路
BFS
代码
1 class Solution { 2 public int ladderLength(String beginWord, String endWord, List<String> wordList) { 3 // use dict to check duplicats 4 Set<String> dict = new HashSet<>(wordList); 5 Queue<String> queue = new LinkedList<>(); 6 queue.add(beginWord); 7 int level = 0; 8 while(!queue.isEmpty()){ 9 int size = queue.size(); 10 for(int i = 0; i < size; i++){ 11 String cur = queue.remove(); 12 if(cur.equals(endWord)){ return level + 1;} 13 for(int j = 0; j < cur.length(); j++){ 14 // hit -> {'h', 'i', 't'} 15 char[] charArray = cur.toCharArray(); 16 for(char c = 'a'; c <='z'; c++){ 17 // {'h', 'i', 't'} for'h', try checking 'a','b'...'z' which forms ait, bit...zit 18 charArray[j] = c; 19 String temp = new String(charArray); 20 if(dict.contains(temp)){ 21 queue.add(temp); 22 // to avoid dead loop, like hit will find hit itself 23 dict.remove(temp); 24 } 25 } 26 } 27 } 28 level++; 29 } 30 return 0; 31 } 32 }
转载于:https://www.cnblogs.com/liuliu5151/p/9823991.html
