【21】127. Word Ladder

127. Word Ladder

  • Total Accepted: 109845
  • Total Submissions: 571018
  • Difficulty: Medium
  • Contributors: Admin

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.

For example,

Given:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log","cog"]

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.
  • You may assume no duplicates in the word list.
  • You may assume beginWord and endWord are non-empty and are not the same.

UPDATE (2017/1/20):
The wordList parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.

DFS寻找所有路径，BFS最短路径。

 1 class Solution {
 2 public:
 3     int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
 4         if(beginWord == endWord) return 1;
 5         if(beginWord.size() < 1 || beginWord.size() != endWord.size()) return 0;
 6         queue<string> q;
 7         
 8         unordered_set<string> dict;//set
 9         for (int i = 0; i < wordList.size(); i++) {
10             dict.insert(wordList[i]);
11         }
12         
13         
14         q.push(beginWord);
15         dict.erase(beginWord);
16         int step = 2;
17         while(!q.empty()){
18             int size = q.size();
19             //string tmp = q.front();
20             //q.pop();
21             for(int i = 0; i < size; i++){
22                 
23                 string tmp = q.front();//pop在里层
24                 q.pop();
25                 
26                 for(int j = 0; j < beginWord.size(); j++){
27                     char old = tmp[j];
28                     for(char c = 'a'; c <= 'z'; c++){
29                         if(old == c) continue;
30                         tmp[j] = c;
31                         
32                         if(dict.find(tmp) != dict.end()){
33                             if(tmp == endWord){
34                                 return step;
35                             }
36                             q.push(tmp);
37                             dict.erase(tmp);
38                         }
39                         
40                     }
41                     tmp[j] = old;
42                 }
43                 //step ++;
44             }
45             step ++;//在while的最外层
46         }
47         //return step;
48         return 0;//否则不成立。只有一个可以return合理值的出口
49     }
50 };

 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
posted @ 2017-02-07 09:25  会咬人的兔子  阅读(129)  评论(0)  编辑  收藏  举报