[LeetCode] Word Ladder

Given two words (beginWord and endWord), and a dictionary, 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 intermediate word must exist in the dictionary

For example,

Given:
start = "hit"
end = "cog"
dict = ["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.

题目这么长感觉会被吓到，其实这题是典型的广度搜索，为什么用广搜，因为他要求“最短”变换长。每次变一个字母，也就是说每个字母的变换方式决定了那一层的节点数，我们可以一层一层的搜，直到搜到某一层的某个变换刚好是endWord 那么流程结束，因为是一层一层扩占所以保证了结果的最小性。 广搜是最合适的方案。

int ladderLength(string beginWord, string endWord, unordered_set<string>& wordDict) {
    int length = 1;
    if (beginWord == endWord) {
        return length;
    }
    queue<string> q;
    queue<string> t;
    unordered_set<string> checked;
    q.push(beginWord);
    
    while (!q.empty()) {
        string topWord = q.front();
        q.pop();
        
        for (int i = 0; i < topWord.length(); i++) {
            for (char c = 'a'; c <= 'z'; c++) {
                string newWord = topWord;
                newWord.replace(i, 1, string(1, c));
                if (newWord == endWord) {
                    return length + 1;
                }
                if (checked.find(newWord) == checked.end() && wordDict.find(newWord) != wordDict.end()) {
                    t.push(newWord);
                    checked.insert(newWord);
                }
            }
        }
        
        if (q.empty()) {
            length++;
            swap(q, t);
        }
    }
    
    return 0;
}

这里值得一提的trick 是交替使用两个队列，每当一个为空的时候，说明结束了一层的搜索，所以length需要加1，然后使用swap 来将两个队列交换。