## 题目描述

Given two words (beginWord and endWord), and a dictionary’s word list, find all shortest transformation sequence(s) from beginWord to endWord, such that:

Only one letter can be changed at a time
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"]​`​​
Return

`[    ["hit","hot","dot","dog","cog"],    ["hit","hot","lot","log","cog"]  ]`

Note:
Return an empty list 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.

## AC 代码

`class Solution {private:    // 记录广度搜索状态的结构    struct LadderState    {        vector<string> frontState;  // 记录当前状态是由哪些状态转换而来        int level;                  // 记录当前状态的层数    };private:    // 判断两个单词是否只相差一个字母    bool isNear(string lhv, string rhv)    {        if(lhv.size() != rhv.size()) return false;        int cnt = 0;        for(int i = 0; i < lhv.size() && cnt < 2; ++i)        {            if(lhv[i] != rhv[i])            {                ++cnt;            }        }        return (cnt == 1);    }        // 返回到达 endWord 的所有最短路径    set<vector<string>> recoverPaths(string beginWord, string endWord, map<string, LadderState>& wordStates)    {        if(beginWord == endWord) return {{beginWord}};        set<vector<string>> paths;        for(size_t i = 0; i < wordStates[endWord].frontState.size(); ++i)        {            set<vector<string>> curPaths = recoverPaths(beginWord, wordStates[endWord].frontState[i], wordStates);            for(auto path : curPaths)            {                path.push_back(endWord);                paths.insert(path);            }        }        return paths;    }public:    vector<vector<string>> findLadders(string beginWord, string endWord, vector<string>& wordList) {        multimap<string, string> wordDir; //wordDir[i] = j 表示 i 到 j 只有一个字母不同        // 计算距离，将相邻的节点存入 wordDir 中        for(int i = 0; i < wordList.size(); ++i)        {            for(int j = i+1; j < wordList.size(); ++j)            {                if(isNear(wordList[i], wordList[j]))                {                    wordDir.insert({wordList[i], wordList[j]});                    wordDir.insert({wordList[j], wordList[i]});                }            }        }        for(int i = 0; i < wordList.size(); ++i)        {            if(isNear(wordList[i], beginWord))            {                wordDir.insert({wordList[i], beginWord});                wordDir.insert({beginWord, wordList[i]});            }        }        // 初始化各个节点的状态        map<string, LadderState> wordStates;        for(size_t i = 0; i < wordList.size(); ++i)        {            wordStates[wordList[i]] = { {}, 0};        }        wordStates[beginWord] = { {}, 0 };        // 计算最短路径        queue<string> que;        unordered_set<string> flags;        que.push(beginWord);        que.push("#");        flags.insert(beginWord);        int stepNum = 0;        while(!que.empty())        {            string curWord = que.front();            que.pop();            if(curWord == "#")             {                ++stepNum;                if(!que.empty()) que.push("#");                continue;            }            if(curWord == endWord)            {                continue;            }            for(auto iter = wordDir.find(curWord); iter != wordDir.end() && iter->first == curWord; ++iter)            {                if(flags.find(iter->second) == flags.end())                {                    flags.insert(iter->second);                    que.push(iter->second);                    wordStates[iter->second].level = wordStates[curWord].level + 1;                }                if(wordStates[iter->second].level == wordStates[curWord].level + 1)                {                    wordStates[iter->second].frontState.push_back(curWord);                }            }        }        // 根据 wordStates 恢复各个路径        set<vector<string>> paths = recoverPaths(beginWord, endWord, wordStates);        vector<vector<string>> ans(paths.begin(), paths.end());        return ans;    }};`