127. Word Ladder

Set会快很多
替换各个位置的字符 在wordDict里面判断存不存在
类似BFS
https://blog.csdn.net/u014532901/article/details/78820124

 

 

 

 1 class Solution {
 2     public int ladderLength(String beginWord, String endWord, List<String> wordList) {
 3         if(!wordList.contains(endWord)) return 0;
 4         Set<String> wordSet=new HashSet<>(wordList);
 5         Queue<String> visited = new LinkedList<>();
 6         visited.offer(beginWord);
 7         wordList.remove(beginWord);
 8         int distance = 1;
 9         int size = visited.size();
10         while(!visited.isEmpty()) {
11             for(int j = 0 ; j < size; j++) {
12                 String str = visited.poll();
13                 for(int k = 0; k < str.length(); k++) {
14                     char[] arr = str.toCharArray();
15                     for(int i = 0; i < 26; i++) {
16                         arr[k] = (char)('a' + i);
17                         String str1 = new String(arr);
18                         if(wordSet.contains(str1)) {
19                             visited.offer(str1);
20                             wordSet.remove(str1);
21                         }
22                     }
23                 }
24             }
25             distance++;
26             size = visited.size();
27             if(visited.contains(endWord)) {
28                 break;
29             }else if(visited.isEmpty()) {
30                 return 0;
31             }
32         }
33         return distance;
34         
35     }
36 }

 

posted @ 2018-09-23 10:58  jasoncool1  阅读(155)  评论(0)  编辑  收藏  举报