【广度优先遍历】leetcode - Word Ladder II

题目：leetcode

Word Ladder II

 Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) fromstart 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.

class Solution {public:    vector<vector<string>> findLadders(string start, string end, unordered_set<string> &dict) {        vector<vector<string>> res;        if(start==end && dict.count(start))        {            res.push_back(vector<string>(1,start));            return res;        }                unordered_set<string> cur,next,used;        unordered_map<string,unordered_set<string>> father;//键是单词，值是父亲，记录单词的父子关系        cur.emplace(start);        bool should_continue=true;                while(should_continue && !cur.empty())        {            //mark临时记录父子关系            //之所以设置mark记录本轮产生的父子关系，是因为同一个单词，可能有多个父亲，不能只记录一个，而且只能记录距离最近的父亲            unordered_map<string,unordered_set<string>> mark;            for(auto &i:next)            {                used.emplace(i);            }            //记得清空next！            next.clear();            for(auto word:cur)            {                set<string> r=NewWords(word,dict,used);                for(auto i:r)                {                    if(i==end)                        should_continue=false;                    next.emplace(i);                    if(father.count(i)==0)                    {                        mark[i].emplace(word);                    }                }            }            //把mark的内容导入father中            for(auto &j:mark)            {                father[j.first]=j.second;            }            swap(cur,next);        }        //根据父子关系，重建路径        find_path(father,start,end,res);        return res;    }            void find_path(unordered_map<string, unordered_set<string>> &father, string start, string end, vector<vector<string>> &res){vector<string> path;path.push_back(end);find_path_core(father, start, end, res, path);}void find_path_core(unordered_map<string, unordered_set<string>> &father, string start, string end, vector<vector<string>> &res, vector<string> &path){if (end == start){reverse(path.begin(), path.end());res.push_back(path);reverse(path.begin(), path.end());//注意！再反转回来！！！！return;}for (auto &i : father[end]){    if(find(path.begin(),path.end(),i)!=path.end())        continue;path.push_back(i);find_path_core(father, start, i, res, path);path.pop_back();}}        set<string> NewWords(string &word,unordered_set<string> &dict,unordered_set<string> &used)    {         set<string> res;        for(int i=0;i<word.size();++i)        {            char c=word[i];            for(int j=0;j<26;++j)            {                char tmp='a'+j;                if(tmp==c)                    continue;                word[i]=tmp;                if(dict.count(word)>0 && used.count(word)==0)                    res.emplace(word);            }             word[i]=c;//注意还原单词        }        return res;    }    };