leetcode126 Word Ladder II

思路：

宽搜过程中分层记录路径，递归还原。
实现：

 class Solution
 {
 public:
     void getPath(string now, string beginWord, string endWord, vector<string>& buf, unordered_map<string, unordered_set<string>>& par, vector<vector<string>>& ret)
     {
         if (now == beginWord)
         {
             vector<string> tmp(, endWord);
             for (auto it : buf) tmp.push_back(it);
             ret.push_back(tmp);
             reverse(ret.back().begin(), ret.back().end());
             return;
         }
         for (auto it : par[now])
         {
             buf.push_back(it);
             getPath(it, beginWord, endWord, buf, par, ret);
             buf.pop_back();
         }
     }
     vector<vector<string>> findLadders(string beginWord, string endWord, vector<string>& wordList)
     {
         unordered_set<string> tmp;
         for (auto it : wordList) tmp.insert(it);
         unordered_map<string, unordered_set<string>> par;
         unordered_set<string> q;
         q.insert(beginWord);
         bool flg = false;
         while (!q.empty())
         {
             unordered_set<string> next;
             for (auto it : q)
             {
                  for (int i = ; i < it.length(); i++)
                 {
                     for (char c = 'a'; c <= 'z'; c++)
                     {
                         string buf = it;
                         if (buf[i] == c) continue;
                         buf[i] = c;
                         if (!tmp.count(buf)) continue;
                         if (!q.count(buf))
                         {
                             next.insert(buf);
                             par[buf].insert(it);
                         }
                         if (buf == endWord) flg = true;
                     }
                 }
             }
             for (auto it : q) { tmp.erase(it); }
             q = next;
             if (flg) break;
         }
         vector<vector<string>> ret;
         if (flg)
         {
             vector<string> buf;
             getPath(endWord, beginWord, endWord, buf, par, ret);
         }
         return ret;
     }
 };