126. Word Ladder II

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

1 Only one letter can be changed at a time

2 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"]

  ]

 

vector<vector<string>> findLadders(string beginWord, string endWord, unordered_set<string> &wordList) {
        //very interesting problem
        //It can be solved with standard BFS. The tricky idea is doing BFS of paths instead of words!
        //Then the queue becomes a queue of paths.
        vector<vector<string>> ans;
        queue<vector<string>> paths;
        wordList.insert(endWord);
        paths.push({beginWord});
        int level = 1;
        int minLevel = INT_MAX;
        
        //"visited" records all the visited nodes on this level
        //these words will never be visited again after this level 
        //and should be removed from wordList. This is guaranteed
        // by the shortest path.
        unordered_set<string> visited; 
        
        while (!paths.empty()) {
            vector<string> path = paths.front();
            paths.pop();
            if (path.size() > level) {
                //reach a new level
                for (string w : visited) wordList.erase(w);
                visited.clear();
                if (path.size() > minLevel)
                    break;
                else
                    level = path.size();
            }
            string last = path.back();
            //find next words in wordList by changing
            //each element from 'a' to 'z'
            for (int i = 0; i < last.size(); ++i) {
                string news = last;
                for (char c = 'a'; c <= 'z'; ++c) {
                    news[i] = c;
                    if (wordList.find(news) != wordList.end()) {
                    //next word is in wordList
                    //append this word to path
                    //path will be reused in the loop
                    //so copy a new path
                        vector<string> newpath = path;
                        newpath.push_back(news);
                        visited.insert(news);
                        if (news == endWord) {
                            minLevel = level;
                            ans.push_back(newpath);
                        }
                        else
                            paths.push(newpath);
                    }
                }
            }
        }
        return ans;
    }

 

转载于:https://www.cnblogs.com/jxr041100/p/8435297.html