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 transformed word must exist in the word list. Note that beginWord is not a transformed word.

Note:

• Return an empty list if there is no such transformation sequence.
• All words have the same length.
• All words contain only lowercase alphabetic characters.
• You may assume no duplicates in the word list.
• You may assume beginWord and endWord are non-empty and are not the same.

Example 1:

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

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

Example 2:

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

Output: []

Explanation: The endWord "cog" is not in wordList, therefore no possible transformation.

个人感觉这道题是相当有难度的一道题，它比之前那道 Word Ladder 要复杂很多，全场第四低的通过率12.9%正说明了这道题的难度，我也是研究了网上别人的解法很久才看懂，然后照葫芦画瓢的写了出来，下面这种解法的核心思想是BFS，大概思路如下：我们的目的是找出所有的路径，我们建立一个路径集paths，用以保存所有路径，然后是起始路径p，在p中先把起始单词放进去。然后定义两个整型变量level，和minLevel，其中level是记录循环中当前路径的长度，minLevel是记录最短路径的长度，这样的好处是，如果某条路径的长度超过了已有的最短路径的长度，那么舍弃，这样会提高运行速度，相当于一种剪枝。还要定义一个set变量words，用来记录已经循环过的路径中的词，然后就是BFS的核心了，循环路径集paths里的内容，取出队首路径，如果该路径长度大于level，说明字典中的有些词已经存入路径了，如果在路径中重复出现，则肯定不是最短路径，所以我们需要在字典中将这些词删去，然后将words清空，对循环对剪枝处理。然后我们取出当前路径的最后一个词，对每个字母进行替换并在字典中查找是否存在替换后的新词，这个过程在之前那道 Word Ladder 里面也有。如果替换后的新词在字典中存在，将其加入words中，并在原有路径的基础上加上这个新词生成一条新路径，如果这个新词就是结束词，则此新路径为一条完整的路径，加入结果中，并更新minLevel，若不是结束词，解将新路径加入路径集中继续循环。写了这么多，不知道你看晕了没有，还是看代码吧，这个最有效：

class Solution {
public:
    vector<vector<string>> findLadders(string beginWord, string endWord, vector<string>& wordList) {
        vector<vector<string>> res;
        unordered_set<string> dict(wordList.begin(), wordList.end());
        vector<string> p{beginWord};
        queue<vector<string>> paths;
        paths.push(p);
        int level = 1, minLevel = INT_MAX;
        unordered_set<string> 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<string> nextPath = t;
                    nextPath.push_back(newLast);
                    if (newLast == endWord) {
                        res.push_back(nextPath);
                        minLevel = level;
                    } else paths.push(nextPath);
                }
            }
        }
        return res;
    }
};

类似题目：

Word Ladder

参考资料：

https://leetcode.com/problems/word-ladder-ii/

https://leetcode.com/problems/word-ladder-ii/discuss/40487/Java-Solution-with-Iteration

http://yucoding.blogspot.com/2014/01/leetcode-question-word-ladder-ii.html

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