[LeetCode]Word Ladder II

最新推荐文章于 2018-05-20 11:46:33 发布

CiaoLiang

最新推荐文章于 2018-05-20 11:46:33 发布

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本文详细介绍了WordLadder II问题的解决方法，包括使用BFS构建图和DFS深度搜索来找到从beginWord到endWord的最短转换路径。通过实例演示了如何实现这一过程，并提供了具体的代码实现。

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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:

Only one letter can be changed at a time
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();
        }
    }
};