【LeetCode】127. Word Ladder 解題報告

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127. Word Ladder / Hard

A transformation sequence from word beginWord to word endWord using a dictionary wordList is a sequence of words beginWord -> s1 -> s2 -> ... -> sk such that:

  • Every adjacent pair of words differs by a single letter.
  • Every si for 1 <= i <= k is in wordList. Note that beginWord does not need to be in wordList.
  • sk == endWord

Given two words, beginWord and endWord, and a dictionary wordList, return the number of words in the shortest transformation sequence from beginWord to endWord, or 0 if no such sequence exists.

Example 1:

Input: beginWord = “hit”, endWord = “cog”, wordList = [“hot”,”dot”,”dog”,”lot”,”log”,”cog”]
Output: 5
Explanation: One shortest transformation sequence is “hit” -> “hot” -> “dot” -> “dog” -> cog”, which is 5 words long.

Example 2:

Input: beginWord = “hit”, endWord = “cog”, wordList = [“hot”,”dot”,”dog”,”lot”,”log”]
Output: 0
Explanation: The endWord “cog” is not in wordList, therefore there is no valid transformation sequence.

Constraints:

  • 1 <= beginWord.length <= 10
  • endWord.length == beginWord.length
  • 1 <= wordList.length <= 5000
  • wordList[i].length == beginWord.length
  • beginWord, endWord, and wordList[i] consist of lowercase English letters.
  • beginWord != endWord
  • All the words in wordList are unique.

Solution: BFS

思路

這題我們必須以詞語之間的關係去思考，以範例來說，不同詞之間的關係可以畫成這張圖。
我們可以利用 BFS 來找出最短路徑。

那麼如何建立出詞語之間的關係呢？
首先將各個詞做前處理，也就是依序將每個字元改為 * 替代，並放入 map 中。
舉例來說，*ot 所對應的 string vector 會存放 “hot”, “pot”, “rot” 等詞彙。

建出這樣的 map 後，我們從 beginWord 開始，一樣先做前處理，這時就能找到下一個字，進而可以使用 BFS 來解決問題了。

例如 hit 的其中一個變形是 h*t，在 map 中會找到 hit, hot，為了避免走回頭路或是迴圈，
還需要再另外宣告一個 visited 陣列來檢查是否走過。而 hot 的變形 *ot 又可以找到 lot 與 dot。

為了知道 BFS 走了多遠，放進 queue 的不只有 string，還搭配一個 int 來記錄廣度。
一旦下一個變體的陣列中有 endWord，便直接將廣度加一後回傳。

效能

Complexity

  • Time Complexity: O(M^2xN), where M is the length of each word and N is the total number of words in the input word list.
  • Space Complexity: O(M^2xN)

LeetCode Result

Code

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class Solution {
public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        int sz = beginWord.size();
        unordered_map<string, vector<string>> hmap;
        unordered_map<string, bool> visited;
        
        for(const auto& word: wordList) {
            visited[word] = false;
            for(int i = 0; i < sz; ++i) {
                string newWord = word.substr(0, i) + "*" + word.substr(i+1, sz-i-1); 
                if(hmap.count(newWord)) {
                    hmap[newWord].emplace_back(word);
                } else {
                    hmap[newWord] = vector<string>{word};
                }
            }
        }

        queue<pair<string, int>> q;
        q.push(pair<string, int>{beginWord, 1});
        while(!q.empty()) {
            string word = q.front().first; 
            int level = q.front().second;
            q.pop();
            
            for(int i = 0; i < sz; ++i) {
                string newWord = word.substr(0, i) + "*" + word.substr(i+1, sz-i-1); 
                if(hmap.count(newWord)) {
                    for(const auto& nextWord: hmap[newWord]) {
                        if(nextWord == endWord) {
                            return level + 1;
                        } else if(!visited[nextWord]) {
                            visited[nextWord] = true;
                            q.push(pair<string, int>{nextWord, level+1});
                        }
                    }
                }
            }
        }
        return 0;
    }
};