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
      
       & wordList) { 4         if(beginWord == endWord) return 1; 5         if(beginWord.size() < 1 || beginWord.size() != endWord.size()) return 0; 6         queue
       
         q; 7          8         unordered_set
        
          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 };