126. 单词接龙 II（bfs）

 

难度困难521

按字典 wordList 完成从单词 beginWord 到单词 endWord 转化，一个表示此过程的 转换序列 是形式上像 beginWord -> s1 -> s2 -> ... -> sk 这样的单词序列，并满足：

• 每对相邻的单词之间仅有单个字母不同。
• 转换过程中的每个单词 si（1 <= i <= k）必须是字典 wordList 中的单词。注意，beginWord 不必是字典 wordList 中的单词。
• sk == endWord

给你两个单词 beginWord 和 endWord ，以及一个字典 wordList 。请你找出并返回所有从 beginWord 到 endWord 的 最短转换序列 ，如果不存在这样的转换序列，返回一个空列表。每个序列都应该以单词列表 [beginWord, s1, s2, ..., sk] 的形式返回。

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"]]
解释：存在 2 种最短的转换序列：
"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 wordList 中，所以不存在符合要求的转换序列。

提示：

• 1 <= beginWord.length <= 7
• endWord.length == beginWord.length
• 1 <= wordList.length <= 5000
• wordList[i].length == beginWord.length
• beginWord、endWord 和 wordList[i] 由小写英文字母组成
• beginWord != endWord
• wordList 中的所有单词 互不相同

class Solution {
public:
    vector<vector<string>> findLadders(string beginWord, string endWord, vector<string>& wordList) {
    vector<vector<string>> res;
    unordered_set<string> us(wordList.begin(), wordList.end());
        if (us.find(endWord) == us.end())
            return res;
        unordered_set<string> visited;
        queue<vector<string>> q;
        visited.insert(beginWord);
        q.push({beginWord});
        int step = 0;
        while (!q.empty()) {
            int sz = q.size();
            for(int i = 0 ; i< sz; i++){
                vector<string> cur = q.front(); q.pop();
                //cout << cur << endl;
                if (cur.back() == endWord) {
                    res.emplace_back(cur);
                    continue;
                }
                for (int j = 0; j < beginWord.size(); j++) {
                    string y = cur.back();
                    for (int k = 0 ; k < 26;k++) {
                        y[j] = 'a' + k;
                        if (y[j]==cur.back()[j]) {
                           continue;
                        }
                        if(us.find(y) != us.end() && visited.find(y) == visited.end()) {
                            cur.emplace_back(y);
                            q.push(cur);
                            visited.insert(y);
                            cur.pop_back();
                        }
                    }
                }
            }
            step ++;
        }
        return res;
    }
};