[LeetCode]Word Ladder II

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 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.

class Solution {public:    //利用BFS构建图，然后利用DFS深度搜索解    vector<vector<string>> findLadders(string beginWord, string endWord, unordered_set<string> &wordList) {        unordered_set<string> current_step;        unordered_set<string> next_step;        unordered_map<string,unordered_set<string>> Graph;        unordered_set<string> unvisited = wordList;        current_step.insert(beginWord);        unvisited.erase(beginWord);        //每次删除一层的结构,同一层出现可能出现相同解要记住，所以每次只能一层层的删除        while(current_step.count(endWord)==0 && unvisited.size()>0){            for(auto pcu = current_step.begin(); pcu!=current_step.end(); pcu++){                string word = *pcu;                for(int i=0; i<beginWord.length(); ++i)                    for(int j=0; j<26; ++j){                        string temp = word;                        if(temp[i] == 'a'+j)                            continue;                        temp[i] = 'a'+j;                        if(unvisited.count(temp)>0){                            next_step.insert(temp);                            Graph[word].insert(temp);                        }                    }                            }                        if(next_step.empty()) break;            for(auto it = next_step.begin(); it!=next_step.end(); ++it){                unvisited.erase(*it);            }            current_step = next_step;            next_step.clear();        }        vector<vector<string>> ret;        vector<string> path;        DFS(Graph,beginWord,endWord,path,ret);        return ret;    }    void DFS(unordered_map<string,unordered_set<string>> &Graph,string beginWord,string endWord,vector<string> &path,vector<vector<string>> &ret){        path.push_back(beginWord);        if(beginWord == endWord){            //reverse(path.begin(),path.end());            ret.push_back(path);        }        unordered_set<string> adj = Graph[beginWord];        for(auto i=adj.begin(); i!=adj.end(); ++i){            DFS(Graph,*i,endWord,path,ret);            path.pop_back();        }    }};