LeetCode----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();
    }
};