126. 单词接龙 II(C++)---BFS、DFS解题_c++单词接龙dfs

题目详情

给定两个单词（beginWord 和 endWord）和一个字典 wordList，找出所有从 beginWord 到 endWord 的最短转换序列。转换需遵循如下规则：

每次转换只能改变一个字母。
转换过程中的中间单词必须是字典中的单词。

说明:

• 如果不存在这样的转换序列，返回一个空列表。
• 所有单词具有相同的长度。
• 所有单词只由小写字母组成。
• 字典中不存在重复的单词。
• 你可以假设 beginWord 和 endWord 是非空的，且二者不相同。

示例 1:
输入:
beginWord = "hit",
endWord = "cog",
wordList = ["hot","dot","dog","lot","log","cog"]

输出:
[
  ["hit","hot","dot","dog","cog"],
 ["hit","hot","lot","log","cog"]
]

示例 2:
输入:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]

输出: []

解释: endWord "cog" 不在字典中，所以不存在符合要求的转换序列。

——题目难度:困难


 

这道力扣困难级别的题目很难，借鉴了大神的思路和代码，大致如下：

-下面代码解题 

class Solution {
public:
    vector<vector<string>> findLadders(string beginWord, string endWord, vector<string>& wordList) {
    	vector<vector<string>> ans;
		vector<string> path;
		if(std::find(wordList.begin(), wordList.end(), endWord) == wordList.end()) return ans;
		
		unordered_map<string,int> depth;
		unordered_map<string,vector<string>> neighbor;
		queue<string> que;
		unordered_set<string> wordSet(wordList.begin(), wordList.end());
		
		que.push(beginWord);
		depth[beginWord] = 1;
		
		while(!que.empty()) {
			string cur = que.front();
			que.pop();
			int wordLen = cur.size();
			for(int i=0;i<wordLen;i++)
			{
				string temp = cur;
				for(char c = 'a'; c <= 'z'; c++)
				{
					temp[i] = c;
					if(wordSet.count(temp)) {
						if(depth.count(temp) == 0) {
							depth[temp] = depth[cur] + 1;
							que.push(temp);
							neighbor[temp].push_back(cur);
						} else if(depth[temp] == depth[cur] + 1) { 
							 neighbor[temp].push_back(cur);
						}
					}
				}
			}
		}
		
		dfs(beginWord, endWord, path, neighbor, ans);
		return ans;
	}
	
	void dfs(string &beginWord, string &cur, vector<string> path,
	 		 unordered_map<string,vector<string>> &neighbor, vector<vector<string>> &ans)
	{
		if(cur == beginWord) {
			path.push_back(cur);
			std::reverse(path.begin(), path.end());
			ans.push_back(path);
			return ;
		}
		
		path.push_back(cur);
		
		 for(string word: neighbor[cur])
		{
			dfs(beginWord, word, path, neighbor, ans);
		}
 
	}
							
};

结果