Design a data structure that supports the following two operations:

void addWord(word)
bool search(word)

search(word) can search a literal word or a regular expression string containing only letters a-z or .. A . means it can represent any one letter.

For example:

addWord("bad")
addWord("dad")
addWord("mad")
search("pad") -> false
search("bad") -> true
search(".ad") -> true
search("b..") -> true

Note:
You may assume that all words are consist of lowercase letters a-z.

LeetCode出新題的速度越來越快了，有點跟不上節奏的感覺了。這道題如果做過之前的那道 Implement Trie (Prefix Tree) 實現字典樹(字首樹)的話就沒有太大的難度了，還是要用到字典樹的結構，唯一不同的地方就是search的函式需要重新寫一下，因為這道題裡面'.'可以代替任意字元，所以一旦有了'.'，就需要查詢所有的子樹，只要有一個返回true，整個search函式就返回true，典型的DFS的問題，其他部分跟上一道實現字典樹沒有太大區別，程式碼如下：

class WordDictionary {
public:
    struct TrieNode {
    public:
        TrieNode *child[26];
        bool isWord;
        TrieNode() : isWord(false) {
            for (auto &a : child) a = NULL;
        }
    };
    
    WordDictionary() {
        root = new TrieNode();
    }
    
    
 
// Adds a word into the data structure.
    void addWord(string word) {
        TrieNode *p = root;
        for (auto &a : word) {
            int i = a - 'a';
            if (!p->child[i]) p->child[i] = new TrieNode();
            p = p->child[i];
        }
        p->isWord = true;
    }

    // Returns if the word is in the data structure. A word could
    // contain the dot character '.' to represent any one letter.
    bool search(string word) {
        return searchWord(word, root, 0);
    }
    
    bool searchWord(string &word, TrieNode *p, int i) {
        if (i == word.size()) return p->isWord;
        if (word[i] == '.') {
            for (auto &a : p->child) {
                if (a && searchWord(word, a, i + 1)) return true;
            }
            return false;
        } else {
            return p->child[word[i] - 'a'] && searchWord(word, p->child[word[i] - 'a'], i + 1);
        }
    }
    
private:
    TrieNode *root;
};

// Your WordDictionary object will be instantiated and called as such:
// WordDictionary wordDictionary;
// wordDictionary.addWord("word");
// wordDictionary.search("pattern");

討論：這道題有個很好的Follow up，就是當搜尋的單詞中存在星號怎麼搞，星號的定義和Wildcard Matching中一樣，可以代表任意的字串，包括空字串，請參見評論區1樓。

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