127. Word Ladder

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单向bfs
方法也是想到和写了出来，就是没有用find这个方法导致超时！！！用了iterator
其实就是bfs不断判断是否是那就ok了，数据结构的使用方法还是有点弱！

class Solution {public:    void addnext(string s, queue<string>& qu, unordered_set<string>& wordList){        wordList.erase(s);        for(int i = 0; i < s.length(); ++ i){            char temp = s[i];            for(int j = 0; j < 26; ++ j){                s[i] = 'a' + j;                if(wordList.find(s) != wordList.end()){                    qu.push(s);                    wordList.erase(s);                }            }            s[i] = temp;        }    }    int ladderLength(string beginWord, string endWord, unordered_set<string>& wordList) {        wordList.insert(endWord);        queue<string>qu;        addnext(beginWord, qu, wordList);        int sum = 2;        while(!qu.empty()){            int n = qu.size();            for(int i = 1; i <= n; ++ i){                string t = qu.front();                qu.pop();                if(t == endWord) return sum;                addnext(t, qu, wordList);            }            sum++;        }        return 0;    }};

双向bfs
双向bfs，比单向更快，没有用queue，用了unorder set一样可以做，每次结束就装换，开两个，一头一尾，这题十分考验打码能力，非思维能力，2刷值得继续练习！

class Solution {public:    int ladderLength(string beginWord, string endWord, unordered_set<string>& wordList) {        unordered_set<string> head, tail, *phead, *ptail;        head.insert(beginWord);        tail.insert(endWord);        int sum = 2;        while(!head.empty() && !tail.empty()){            if(head.size() < tail.size()){                phead = &head;                ptail = &tail;            }            else{                phead = &tail;                ptail = &head;            }            unordered_set<string> temp;            for(auto it = phead -> begin(); it != phead -> end(); it++){                string word = *it;                wordList.erase(word);                for(int i = 0; i < word.length(); ++ i){                    char t = word[i];                    for(int j = 0; j < 26; ++ j){                        word[i] = 'a' + j;                        if(ptail -> find(word) != ptail -> end())                            return sum;                        if(wordList.find(word) != wordList.end()){                            temp.insert(word);                            wordList.erase(word);                        }                    }                    word[i] = t;                }            }            sum++;            swap(*phead, temp);        }        return 0;    }};

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