LeetCode OJ - Word Ladder II

Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, 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"]

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.

分析：一道给力的题，有人说是五星题。采用BFS遍历，并使用vector<string, vector<string> >邻接表来构建图模型，最后使用DFS深度搜索图来得到所有结果。如果遍历图有重复元素，可以使用一个标记为来记录visited。

class Solution {    vector<vector<string> > ret;    string END;public:    vector<vector<string>> findLadders(string start, string end, unordered_set<string> &dict) {        END = end;                queue<string> que, next;        que.push(start);        dict.erase(start);                unordered_set<string> del;        del.insert(start);        unordered_map<string, vector<string> > path;                bool found = false;        while(!que.empty()) {            for(auto x : del) dict.erase(x);            del.clear();            while(!que.empty()) {                string x = que.front();  que.pop();                                //根据cur查找，并push到next队列                for(int i = 0; i < x.size(); i++) {                    string word = x;                         for(char ch = 'a'; ch <= 'z'; ch++) {                                                if(word[i] == ch) continue;                        word[i] = ch;                                                                        //找到可达字符集合，记录到path中                        if(dict.count(word)) {                            if(word == end) found = true;                            if(!del.count(word))next.push(word);                            del.insert(word);                            path[x].push_back(word);                         }                    }                }            }                        if(found) break;            swap(que, next);        }                vector<string> item;        item.push_back(start);        DFS(path, start, item);        return ret;    }private:        void DFS(unordered_map<string, vector<string> > &path, string last, vector<string> &item) {        if(last == END) {            ret.push_back(item);            return ;        }                for(int i = 0; i < path[last].size(); i++) {            item.push_back(path[last][i]);            DFS(path, path[last][i], item);            item.pop_back();        }         }};