126. 单词接龙 II

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：
1.每次转换只能改变一个字母。
2.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。

代码

class Solution {
public:
    vector> findLadders(string beginWord, string endWord, vector& wordList) {
        vector> res;
        unordered_set dict(wordList.begin(), wordList.end());
        vector p{beginWord};
        queue> paths;
        paths.push(p);
        int level = 1, minLevel = INT_MAX;
        unordered_set words;
        while (!paths.empty()) {
            auto t = paths.front(); paths.pop();
            if (t.size() > level) {
                for (string w : words) dict.erase(w);
                words.clear();
                level = t.size();
                if (level > minLevel) break;
            }
            string last = t.back();
            for (int i = 0; i < last.size(); ++i) {
                string newLast = last;
                for (char ch = 'a'; ch <= 'z'; ++ch) {
                    newLast[i] = ch;
                    if (!dict.count(newLast)) continue;
                    words.insert(newLast);
                    vector nextPath = t;
                    nextPath.push_back(newLast);
                    if (newLast == endWord) {
                        res.push_back(nextPath);
                        minLevel = level;
                    } else paths.push(nextPath);
                }
            }
        }
        return res;
    }
};