【leetcode】127. Word Ladder 等長字符串集合的A到B的轉移最短長度

1. 題目

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:

Only one letter can be changed at a time
Each intermediate word must exist in the word list
For example,

Given:
beginWord = "hit"
endWord = "cog"
wordList = ["hot","dot","dog","lot","log"]
As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

Note:
Return 0 if there is no such transformation sequence.
All words have the same length.
All words contain only lowercase alphabetic characters.

2. 思路

註釋部分提出來：

// 26叉樹廣度優先遍歷，找到end，或者是樹沒法繼續延伸
// 樹的延伸是基於當前的字符串，每次變化一個字符[a-z]
// 若是新的字符串在wordlist裏就加入候選和延伸隊列裏
// 因爲是廣度遍歷，第一次找到的候選必定是最短路徑的所以
// 能夠將找到的候選從dict裏刪除,避免後續再找到的時候造成死循環
// 要計算是第幾層，用一個空字符串來做爲層之間的分隔符

這裏最坑的地方是標準答案不是求的轉移次數，而是求的轉移鏈的鏈長度，也就是轉移一次的狀況，長度是2，其餘相似的都是大了一。白白耗費了好多時間......

3. 代碼

耗時：263ms

class Solution {
public:
    // 26叉樹廣度優先遍歷，找到end，或者是樹沒法繼續延伸
    // 樹的延伸是基於當前的字符串，每次變化一個字符[a-z]
    // 若是新的字符串在wordlist裏就加入候選和延伸隊列裏
    // 因爲是廣度遍歷，第一次找到的候選必定是最短路徑的所以
    // 能夠將找到的候選從dict裏刪除,避免後續再找到的時候造成死循環
    // 要計算是第幾層，用一個空字符串來做爲層之間的分隔符
    int ladderLength(string beginWord, string endWord, unordered_set<string>& wordList) {
        if (beginWord == endWord) { return 0; }
        if (beginWord.length() == 0) { return 0; }
        wordList.erase(beginWord);
        deque<string> q;
        int step = 2;
        q.push_back(beginWord);
        q.push_back("");
        while (!q.empty()) {
            string s = q.front();
            q.pop_front();
            if (s == "") {
                if (q.empty()) { return 0; }
                q.push_back("");
                step++;
                continue;
            }
            //cout << "s=" << s << endl; 
            for (int w_i = 0; w_i < s.length(); w_i++) {
                char ch = s[w_i];
                for (char ch_i = 'a'; ch_i <= 'z'; ch_i++) {
                    if (ch_i == ch) { continue; }
                    s[w_i] = ch_i;
                    if (s == endWord) {
                        //cout << "find s=" << s << endl;
                        return step; 
                    }
                    if (wordList.find(s) != wordList.end()) {
                        q.push_back(s);
                        wordList.erase(s);
                        //cout << "new s=" << s << endl;
                    }
                    s[w_i] = ch;
                }
            }
        }
        return 0;
    }
};