leetcode@ [126] Word Ladder II (BFS + 层次遍历 + DFS)

https://leetcode.com/problems/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:
    void dfs(vector<vector<string> >& res, unordered_map<string, vector<string> >& fa, vector<string> load, string beginWord, string curWord) {
        if(curWord == beginWord) {
            reverse(load.begin(), load.end());
            res.push_back(load);
            reverse(load.begin(), load.end());
            return;
        }
        
        for(int i=0; i<fa[curWord].size(); ++i) {
            load.push_back(fa[curWord][i]);
            dfs(res, fa, load, beginWord, fa[curWord][i]);
            load.pop_back();
        }
    }

    vector<vector<string>> findLadders(string beginWord, string endWord, unordered_set<string> &wordList) {
        vector<vector<string> > res;
        if(beginWord.compare(endWord) == 0)  return res;
        
        wordList.insert(endWord);
        
        unordered_map<string, vector<string> > fa;
        unordered_set<string> vis;
        unordered_set<string> lev;
        unordered_set<string> next_lev;

        lev.insert(beginWord);
        bool found = false;
        
        while(!lev.empty() && !found) {
            if(lev.find(endWord) != lev.end())  found = true;
            
            for(auto str: lev)  vis.insert(str);
            
            for(auto str : lev) {
                for(int i=0; i<str.length(); ++i) {
                    for(char ch = 'a'; ch <= 'z'; ++ch) {
                        if(str[i] != ch) {
                            string tmp = str;
                            tmp[i] = ch;
                            if(wordList.find(tmp) != wordList.end() && vis.find(tmp) == vis.end()) {
                                next_lev.insert(tmp);
                                fa[tmp].push_back(str);
                            }
                        }
                    }
                }   
            }
            
            lev.clear();
            swap(lev, next_lev);
        }
        
        if(found) {
            vector<string> load;
            load.push_back(endWord);
            dfs(res, fa, load, beginWord, endWord);
        }
        
        return res;
    }
};
原文地址:https://www.cnblogs.com/fu11211129/p/5281928.html