[LeetCode 解題報告]127. Word Ladder

Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence 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 0 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: 5

Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

Example 2:

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

Output: 0

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

class Solution {
public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        unordered_set<string> dict(wordList.begin(), wordList.end());
        unordered_map<string, int> m;
        queue<string> que;
        m[beginWord] = 1;
        que.push(beginWord);
        
        while(!que.empty()) {
            string word = que.front();
            que.pop();
            
            for (int i = 0; i < word.size(); i ++) {
                string newWord = word;
                for (char ch='a'; ch <= 'z'; ch ++) { // 逐一枚舉
                    newWord[i] = ch;
                    if (dict.count(newWord) && newWord == endWord)
                        return m[word] +1;
                    if (dict.count(newWord) && !m.count(newWord)) {
                        que.push(newWord);
                        m[newWord] = m[word] + 1;
                    }
                }
            }
        }
        return 0;
    }
};