Word Ladder 未完成

Given two words (beginWord and endWord), and a dictionary, 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 intermediate word must exist in the dictionary

For example,

Given:
start = "hit"
end = "cog"
dict = ["hot","dot","dog","lot","log"]

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.

每次只能改变一个字符，求最短路径。

开始想法为列出二维矩阵，找出变化一次，变化两次，知道变化为end，从而求最短路径。然而发现需要内存过多，同时超时。

改为采用BFS，这样首先找到的肯定是最短路径。但是同样超时。看到网上都是用java实现的，不知道是什么问题。

 class Solution {
 private:
     int isOneDiff(string beginWord, string endWord)
     {
         int n=beginWord.size();
         int m=endWord.size();
         if(n!=m) return -;
         int count=;
         for(int i=;i<n;i++)
         {
             if(beginWord[i]!=endWord[i])
                 count++;

         }
         if(count>) return -;
         return count;
     }
 public:
     int ladderLength(string beginWord, string endWord, unordered_set<string>& wordDict) {
         int n=wordDict.size();
         if(beginWord.empty()||endWord.empty()||n<||beginWord.size()!=endWord.size()||isOneDiff(beginWord,endWord)==)
             return ;
         if(isOneDiff(beginWord,endWord)==)
             return ;
         if((wordDict.find(beginWord)!=wordDict.end())&&(wordDict.find(endWord)!=wordDict.end())&&(n==))
             return ;
         queue<string> q;
         map<string,int> wordmap;
         int wordlength=beginWord.size();
         int count=;
         q.push(beginWord);
         wordmap.insert(pair<string,int>(beginWord,count));
         while(!q.empty())
         {
             string tmpword=q.front();
             count=wordmap[tmpword];
             q.pop();
             for(int i=;i<wordlength;i++)
             {

                 for(char j='a';j<='z';j++)
                 {
                     if(j==tmpword[i]) continue;
                     tmpword[i]=j;
                     if(tmpword==endWord) return count+;
                     if(wordDict.find(tmpword)!=wordDict.end())
                     {
                         q.push(tmpword);
                         wordmap.insert(pair<string,int>(tmpword,count+));
                     }
                 }
             }
         }

         return ;
     }
 };