127. Word Ladder

Hard (Khó) C++ 🔗 Xem trên LeetCode

📋 Đề Bài

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.

🧠 Thuật Toán & Kỹ Thuật

Hash Table (Bảng băm)Union Find (Tập hợp rời rạc)String (Chuỗi)
⏱️ Thời gian O(n²)
💾 Không gian O(n)

💻 Lời Giải

C++ 0127-word-ladder.cpp 
class Solution {
public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        bool flag = false;
        for (string word : wordList) {
            if (word == endWord) {
                flag = true;
                break;
            }
        }
        if (!flag) {
            return 0;
        }
        queue<string> mq;
        int step = 0;
        mq.push(beginWord);
        unordered_set<string> us(wordList.begin(), wordList.end());
        unordered_set<string> visited;
        visited.insert(beginWord);
        
        while (!mq.empty()) {
            int n = mq.size();
            while (n--) {
                string word = mq.front();
                mq.pop();
                if (word == endWord) {
                    return step + 1;
                }
                int m = word.size();
                for (int i = 0; i < m; ++i) {
                    char tmp = word[i];
                    for (char c = 'a'; c <= 'z'; c++) {
                        if (tmp == c) {
                            continue;
                        }
                        word[i] = c;
                        if (visited.find(word) == visited.end() and us.find(word) != us.end()) {
                            mq.push(word);
                            visited.insert(word);
                        }
                    }
                    word[i] = tmp;
                }
            }
            step += 1;
        }
        return 0;
    }
};