题目链接：Add and Search Word - Data structure design

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.

You should be familiar with how a Trie works. If not, please work on this problem: Implement Trie (Prefix Tree) first.

这道题的要求是设计一种支持添加单词和查找单词的数据结构。其中查找单词时包含小写字母a-z和’.’（表示任意字母）。

存储和查找单词，可以考虑Hash表或者Trie树以提升查询效率。而由于查询时可以用’.’表示任意单个字母，因此采用Trie实现。和之前的Implement Trie (Prefix Tree)基本差不多吧。

其中插入单词，和Implement Trie (Prefix Tree)完全相同，同样是根据单词中的字母决定接下来的子树，如果不存在就建立新节点。而在查找的时候，需要对’.’进行特殊处理：需要依次遍历下面的26个子树（这里采用递归方式）。

时间复杂度：O(???)

空间复杂度：O(???)

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class TrieNode
{
public:
    TrieNode *next[26];
    bool is_word;
    
    // Initialize your data structure here.
    TrieNode(bool b = false)
    {
        memset(next, 0, sizeof(next));
        is_word = b;
    }
};

class WordDictionary
{
    TrieNode *root;
public:
    WordDictionary()
    {
        root = new TrieNode();
    }

    // Adds a word into the data structure.
    void addWord(string word)
    {
        TrieNode *p = root;
        for(int i = 0; i < word.size(); ++ i)
        {
            if(p -> next[word[i] - 'a'] == NULL)
                p -> next[word[i] - 'a'] = new TrieNode();
            p = p -> next[word[i] - 'a'];
        }
        p -> is_word = 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 search(root, word, 0);
    }
private:
    bool search(TrieNode *r, string &word, int b)
    {
        if(b == word.size())
            return r != NULL && r -> is_word;
        
        TrieNode *p = r;
        for(int i = b; i < word.size() && p != NULL; ++ i)
        {
            if(word[i] != '.')
                p = p -> next[word[i] - 'a'];
            else
            {
                for(int j = 0; j < 26; ++ j)
                    if(p -> next[j] != NULL && search(p -> next[j], word, i + 1))
                        return true;
                return false;
            }
        }
        return p != NULL && p -> is_word;
    }
};