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 -1;
        int count=0;
        for(int i=0;i<n;i++)
        {
            if(beginWord[i]!=endWord[i])
                count++;
            
        }
        if(count>1) return -1;
        return count;
    }
public:
    int ladderLength(string beginWord, string endWord, unordered_set<string>& wordDict) {
        int n=wordDict.size();
        if(beginWord.empty()||endWord.empty()||n<1||beginWord.size()!=endWord.size()||isOneDiff(beginWord,endWord)==0)
            return 0;
        if(isOneDiff(beginWord,endWord)==1)
            return 2;
        if((wordDict.find(beginWord)!=wordDict.end())&&(wordDict.find(endWord)!=wordDict.end())&&(n==2))
            return 0;
        queue<string> q;
        map<string,int> wordmap;
        int wordlength=beginWord.size();
        int count=1;
        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=0;i<wordlength;i++)
            {
                
                for(char j='a';j<='z';j++)
                {
                    if(j==tmpword[i]) continue;
                    tmpword[i]=j;
                    if(tmpword==endWord) return count+1;
                    if(wordDict.find(tmpword)!=wordDict.end()) 
                    {
                        q.push(tmpword);
                        wordmap.insert(pair<string,int>(tmpword,count+1));
                    }
                }
            }
        }
        
        return 0;
    }
};