bobo老师，这道题这两种BFS时间复杂度在我看来是一样的，但是最终运行的时间差距很大，具体原因是什么呢？
1.528ms

class Solution {
public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        
        if( find( wordList.begin(), wordList.end(), endWord) == wordList.end())
            return 0;
        unordered_set<string> unSet( wordList.begin(), wordList.end());
        queue< pair<string,int> > q; // < word in List, the depth till this word>
        q.push( make_pair( beginWord, 1));
        unSet.erase( beginWord);
        vector<string> visited;
        
        while( !q.empty() && !unSet.empty()) {
            string s = q.front().first;
            int depth = q.front().second;
            q.pop();
            visited.clear();
            auto iter = unSet.begin();
            for( ; iter != unSet.end(); ++iter) {
                if( stringDefference( s, *iter) == 1) {
                    if( *iter == endWord)
                        return depth + 1;
                    q.push( make_pair( *iter, depth + 1));
                    visited.push_back( *iter);
                    // unSet.erase( iter); // wrong to erase directly
                }
            }
            for( auto it = visited.begin(); 
                it != visited.end(); ++it) 
                unSet.erase( *it);
        }
        
        return 0;
    }
    
private:
    int stringDefference( const string& s1, const string& s2) {
        
        assert( s1.size() == s2.size());
        assert( s1 != s2);
        int res = 0;
        for( int i=0; i<s1.size(); ++i) {
            if( s1[i] != s2[i])
                res ++;
        }
        
        return res;
    }
};

2. 80ms

class Solution {
public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
   
        if( find( wordList.begin(), wordList.end(), endWord) == wordList.end()) 
            return 0;     
        unordered_set<string> dict( wordList.begin(), wordList.end());
        queue< pair< string, int> > que;
        que.push(make_pair(beginWord, 1));
        dict.erase(beginWord); 
        while( !que.empty() && dict.size()>0) {
            string str = que.front().first;
            int ladder = que.front().second;
            que.pop();

            for( int i=0; i<str.size(); ++i) {
                string tmpstr = str;
                for(int j=0; j<26; ++j) {
                    tmpstr[i] = 'a'+j;
                    if(dict.find(tmpstr) != dict.end()) {
                        if( tmpstr == endWord) return ladder+1;
                        que.push( make_pair(tmpstr, ladder+1));
                        dict.erase(tmpstr);
                    }
                }
            }
        }
        return 0;
    }
};