[LeetCode] 126. Word Ladder II 詞語階梯之二

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.

個人感覺這道題是相當有難度的一道題，它比之前那道 Word Ladder 要復雜很多，全場第四低的通過率 12.9% 正說明了這道題的難度，博主也是研究了網(wǎng)上別人的解法很久才看懂，然后照葫蘆畫瓢的寫了出來，下面這種解法的核心思想是 BFS，大概思路如下：目的是找出所有的路徑，這里建立一個路徑集 paths，用以保存所有路徑，然后是起始路徑p，在p中先把起始單詞放進去。然后定義兩個整型變量 level，和 minLevel，其中 level 是記錄循環(huán)中當前路徑的長度，minLevel 是記錄最短路徑的長度，這樣的好處是，如果某條路徑的長度超過了已有的最短路徑的長度，那么舍棄，這樣會提高運行速度，相當于一種剪枝。還要定義一個 HashSet 變量 words，用來記錄已經(jīng)循環(huán)過的路徑中的詞，然后就是 BFS 的核心了，循環(huán)路徑集 paths 里的內(nèi)容，取出隊首路徑，如果該路徑長度大于 level，說明字典中的有些詞已經(jīng)存入路徑了，如果在路徑中重復出現(xiàn)，則肯定不是最短路徑，所以需要在字典中將這些詞刪去，然后將 words 清空，對循環(huán)對剪枝處理。然后取出當前路徑的最后一個詞，對每個字母進行替換并在字典中查找是否存在替換后的新詞，這個過程在之前那道 Word Ladder 里面也有。如果替換后的新詞在字典中存在，將其加入 words 中，并在原有路徑的基礎上加上這個新詞生成一條新路徑，如果這個新詞就是結(jié)束詞，則此新路徑為一條完整的路徑，加入結(jié)果中，并更新 minLevel，若不是結(jié)束詞，則將新路徑加入路徑集中繼續(xù)循環(huán)。寫了這么多，不知道你看暈了沒有，還是看代碼吧，這個最有效：

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;
    }
};

Github 同步地址：

https://github.com/grandyang/leetcode/issues/126

類似題目：

Word Ladder

參考資料：

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

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

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

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