Word Ladder 2

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：

给定一个单词start, 问通过一系列变换是否能够得到单词end, 中间的单词都必须是给定的字典中的词。

其实就是 从最初的一个状态，能够找到一个path, 到达最终的一个状态。

Bingo! 图的搜索： 找到图中两个node之间的所有的最短路径。

图的搜索有 深度优先（DFS） 和 广度优先（BFS）两种。

BFS适用于解决寻找最短路径的问题，因此采用BFS

class Solution {public:vector<vector<string> > findLadders(string start, string end, unordered_set<string> &dict) {        //BFS   each node in graph has multipul fathers        vector<vector<string> > result;        vector<string> path;        bool found = false;        unordered_set<string> visited;        unordered_map<string, vector<string> > father;        unordered_set<string> current, next;                if(start.size() != end.size()) return result;                current.insert(start);        while(!current.empty())        {  for(auto i : current)     visited.insert(i);  for(auto str : current)  {            for(size_t i=0; i<str.size(); ++i)            {                string new_str(str);                for(char c = 'a'; c <= 'z'; ++c)                {                    if(c == new_str[i]) continue;                                        swap(c, new_str[i]);                    if(new_str == end || dict.count(new_str) > 0 && !visited.count(new_str)){                        //visited.insert(new_str);                        next.insert(new_str);                        father[new_str].push_back(str);                    }                    if(new_str == end){                        found = true;                        break;                    }                    swap(c, new_str[i]);                }            }  }// end for  current.clear();          if(!found){    swap(current, next);  }        }// end while                buildPath(father, path, start, end, result);        return result;    }private:    void buildPath(unordered_map<string, vector<string> > &father, vector<string> &path,                     const string &start, const string &word, vector<vector<string> > &result){        path.push_back(word);        if(word == start){            result.push_back(path);            reverse(result.back().begin(), result.back().end());        }        else{            for(auto f : father[word])              buildPath(father, path, start, f, result);        }        path.pop_back();    }};