Implement a magic directory with buildDict, and search methods.

For the method buildDict, you'll be given a list of non-repetitive words to build a dictionary.

For the method search, you'll be given a word, and judge whether if you modify exactly one character into another character in this word, the modified word is in the dictionary you just built.

Example 1:

Input: buildDict(["hello", "leetcode"]), Output: Null
Input: search("hello"), Output: False
Input: search("hhllo"), Output: True
Input: search("hell"), Output: False
Input: search("leetcoded"), Output: False

Note:

1. You may assume that all the inputs are consist of lowercase letters a-z
   .
2. For contest purpose, the test data is rather small by now. You could think about highly efficient algorithm after the contest.
3. Please remember to RESET your class variables declared in class MagicDictionary, as static/class variables are persisted across multiple test cases. Please see here for more details.

這道題讓我們設計一種神奇字典的資料結構，裡面有一些單詞，實現的功能是當我們搜尋一個單詞，只有存在和這個單詞只有一個位置上的字元不相同的才能返回true，否則就返回false，注意完全相同也是返回false，必須要有一個字元不同。博主首先想到了One Edit Distance那道題，只不過這道題的兩個單詞之間長度必須相等。所以只需檢測和要搜尋單詞長度一樣的單詞即可，所以我們用的資料結構就是根據單詞的長度來分，把長度相同相同的單詞放到一起，這樣就可以減少搜尋量。那麼對於和要搜尋單詞進行比較的單詞，由於已經保證了長度相等，我們直接進行逐個字元比較即可，用cnt表示不同字元的個數，初始化為0。如果當前遍歷到的字元相等，則continue；如果當前遍歷到的字元不相同，並且此時cnt已經為1了，則break，否則cnt就自增1。退出迴圈後，我們檢測是否所有字元都比較完了且cnt為1，是的話則返回true，否則就是跟下一個詞比較。如果所有詞都比較完了，則返回false，參見程式碼如下：

解法一：

class MagicDictionary {
public:
    /** Initialize your data structure here. */
    MagicDictionary() {}
    
    /** Build a dictionary through a list of words */
    void buildDict(vector<string> dict) {
        for (string word : dict) {
            m[word.size()].push_back(word);
        }
    }
    
    /** Returns if there is any word in the trie that equals to the given word after modifying exactly one character */
    bool search(string word) {
        for (string str : m[word.size()]) {
            int cnt = 0, i = 0;
            for (; i < word.size(); ++i) {
                if (word[i] == str[i]) continue;
                if (word[i] != str[i] && cnt == 1) break; 
                ++cnt;
            }
            if (i == word.size() && cnt == 1) return true;
        }
        return false;
    }

private:
    unordered_map<int, vector<string>> m;
};

下面這種解法實際上是用到了字首樹中的search的思路，但是我們又沒有整個用到prefix tree，博主感覺那樣寫法略複雜，其實我們只需要借鑑一下search方法就行了。我們首先將所有的單詞都放到一個集合中，然後在search函式中，我們遍歷要搜尋的單詞的每個字元，然後把每個字元都用a-z中的字元替換一下，形成一個新詞，當然遇到本身要跳過。然後在集合中看是否存在，存在的話就返回true。記得換完一圈字元後要換回去，不然就不滿足只改變一個字元的條件了，參見程式碼如下：

解法二：

class MagicDictionary {
public:
    /** Initialize your data structure here. */
    MagicDictionary() {}
    
    /** Build a dictionary through a list of words */
    void buildDict(vector<string> dict) {
        for (string word : dict) s.insert(word);
    }
    
    /** Returns if there is any word in the trie that equals to the given word after modifying exactly one character */
    bool search(string word) {
        for (int i = 0; i < word.size(); ++i) {
            char t = word[i];
            for (char c = 'a'; c <= 'z'; ++c) {
                if (c == t) continue;
                word[i] = c;
                if (s.count(word)) return true;
            }
            word[i] = t;
        }
        return false;
    }
    
private:
    unordered_set<string> s;
};

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