LeetCode: Word Ladder II [127]

【题目】

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.

【题意】

    给定两个单词start和end, 一个词典，找到全部的最短转换序列。

几个注意事项：
    1. 每次变换仅仅能改变一个字符
    2. 变换的中间单词必须在词典中
    3. 全部单词长度同样
    4. 全部单词字符都小写

【思路】

     思路和word Ladder是同样的，仅仅只是本题须要把左右的序列输出。

     为了恢复转换序列在搜索的过程中，我们须要记录每一个可达单词的前继单词（所谓单词可达，就是start通过若干次字符变换后能够转换成当前单词）。

     一旦我们找到end, 我们就能够通过前继恢复路径。这跟用dijkstra找最短路径的方法事实上非常相似。

    

【代码】

class Solution {
public:
    void getSequences(vector<vector<string> >&result, vector<string>&sequence, string&start, string end, map<string, vector<string> >&percursors){
        sequence.push_back(end);
        if(start==end){
            vector<string> v=sequence;
            reverse(v.begin(), v.end());
            result.push_back(v);
            return;
        }
        
        //找end的前驱
        vector<string> pres = percursors[end];
        for(int i=0; i<pres.size(); i++){
            getSequences(result, sequence, start, pres[i], percursors);
            sequence.pop_back();
        }
    }

    vector<vector<string> > findLadders(string start, string end, unordered_set<string> &dict) {
        vector<vector<string> > result;
        if(start==end)return result;
        
        //记录前驱的map
        map<string, vector<string> > percursors;
        //标记是否已经找到最短序列
        bool isFind=false;
        //交替存储相邻
        queue<string> q1;
        queue<string> q2;
        q1.push(start);
        //找前驱
        while(!q1.empty() || !q2.empty()){
            //存放当前层单词
            set<string> words;
            if(!q1.empty()){
                while(!q1.empty()){
                    string curword=q1.front(); q1.pop();
                    for(int i=0; i<curword.length(); i++){
                        string tword=curword;
                        for(char c='a'; c<='z'; c++){
                            if(c!=curword[i]){
                                tword[i]=c;
                                //推断是否是end
                                if(tword==end){
                                    isFind=true;
                                    //保存前驱
                                    percursors[tword].push_back(curword);
                                    //保存当前层单词
                                    words.insert(tword);
                                }
                                else if(dict.find(tword)!=dict.end()){
                                    //假设tword在词典中。则保存它的前驱
                                    percursors[tword].push_back(curword);
                                    //保存当前层单词
                                    words.insert(tword);
                                }
                            }
                        }
                    }
                }
                //将当前层的单词保存到q2
                for(set<string>::iterator it=words.begin(); it!=words.end(); it++){
                    q2.push(*it);
                    dict.erase(*it);
                }
            }
            else{
                while(!q2.empty()){
                    string curword=q2.front(); q2.pop();
                    for(int i=0; i<curword.length(); i++){
                        string tword=curword;
                        for(char c='a'; c<='z'; c++){
                            if(c!=curword[i]){
                                tword[i]=c;
                                //推断是否是end
                                if(tword==end){
                                    isFind=true;
                                    //保存前驱
                                    percursors[tword].push_back(curword);
                                    //保存当前层单词
                                    words.insert(tword);
                                }
                                else if(dict.find(tword)!=dict.end()){
                                    //假设tword在词典中，则保存它的前驱
                                    percursors[tword].push_back(curword);
                                    //保存当前层单词
                                    words.insert(tword);
                                }
                            }
                        }
                    }
                }
                //将当前层的单词保存到q1
                for(set<string>::iterator it=words.begin(); it!=words.end(); it++){
                    q1.push(*it);
                    dict.erase(*it);
                }
            }
            if(isFind)break;
        }
        //生成全部序列
        vector<string>sequence;
        getSequences(result, sequence, start, end, percursors);
        return result;
    }
};