Leetcode 126. Word Ladder II[hard]

题目：
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 intermediate word must exist in the word list
For example,

Given:
beginWord = “hit”
endWord = “cog”
wordList = [“hot”,”dot”,”dog”,”lot”,”log”]
Return
[
[“hit”,”hot”,”dot”,”dog”,”cog”],
[“hit”,”hot”,”lot”,”log”,”cog”]
]
Note:
All words have the same length.
All words contain only lowercase alphabetic characters.

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这个题是真的烦，MLE,TLE,CLE真的是服了。
言归正传，经典题。将每个单词抽象成节点，相差一个字母的两个单词连边。
此题要求输出最短的词梯，所以使用bfs，找到最短词梯的长度。
题目还要输出所有的最短词梯，且考虑到所有最短词梯构成的并不一定是树，所以要使用dfs在bfs访问过的节点的基础上再次搜索。
为了方便dfs，在bfs的时候记录每个节点的step（从起始节点到当前节点的词梯长度），依据这个参数我们就可以还原bfs的搜索顺序了。

通过这个题，我还知道了，原来map中count比取值要快辣么多。。。。

class Solution {public:    vector< vector<string> > ans;    vector<string> word;    vector<bool> vis;    unordered_map<string, int> mp;    vector<int> q;    vector<string> tans;    vector<int> step;     int minstep;    vector< vector<string> > findLadders(string beginWord, string endWord, unordered_set<string> &wordList) {        minstep = -1;        prep(wordList);        bfs(mp[beginWord], mp[endWord]);        if (minstep != -1) {            tans.clear();            tans.resize(minstep + 1);            dfs(mp[endWord]);        }        return ans;    }    void prep(unordered_set<string> &wordList) {        mp.clear();        ans.clear();        word.clear();        word.push_back("null");        unordered_set<string>::iterator it;        int cnt = 0;        for (it = wordList.begin(); it != wordList.end(); it++) {            mp[*it] = ++cnt;            word.push_back(*it);        }        vis.resize(cnt + 1);        step.resize(cnt + 1);    }    void bfs(int ax, int ay) {        int h = 0, t = 1;        q.clear();        q.push_back(ax);        vis[ax] = true;        step[ax] = 0;        int len, num, sto;        char cc;        while (h < t) {            num = q[h];            if (num == ay) {                minstep = step[num];                return;            }            string str = word[num];            len = str.length();            for (int j = 0; j < len; j++) {                cc = str[j];                for (char k = 'a'; k <= 'z'; k++) {                    if (cc != k) {                        str[j] = k;                        if (mp.count(str) == 0) continue;                        sto = mp[str];                        if (!vis[sto]) {                            q.push_back(sto);                            vis[sto] = true;                            step[sto] = step[num] + 1;                            t++;                        }                    }                }                str[j] = cc;            }            h++;        }    }    void dfs(int x) {        tans[step[x]] = word[x];        if (step[x] == 0) {            ans.push_back(tans);            return;        }        string str = word[x];        int len = str.length();        for (int j = 0; j < len; j++) {            char cc = str[j];            for (char k = 'a'; k <= 'z'; k++) {                if (cc != k) {                    str[j] = k;                    if (mp.count(str) == 0) continue;                    int sto = mp[str];                    if (vis[sto] && step[sto] + 1 == step[x]) dfs(sto);                }            }            str[j] = cc;        }    }};