题目：

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.

click to show hint.

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

设计一个数据结构支持addWord和search操作。search操作支持正则表达式‘.‘查找。

思路：

采用trie树，参考：http://blog.csdn.net/u012243115/article/details/47252213。

查找过程中，如果遇到字符为‘.’ ，则对子节点进行深度优先遍历。

代码：

class Trie
{
public:
    bool isWord;
    Trie *next[26];
    Trie()
    {
        isWord = false;
        memset(next , 0 , sizeof(next));
    }
};

class WordDictionary {
public:
    WordDictionary()
    {
        root = new Trie();
    }
    // Adds a word into the data structure.
    void addWord(string word)
    {
        Trie *p = root;
        int len = word.size();
        for(int i = 0 ; i < len ; i++)
        {
            if(p->next[word[i] - 'a'] == NULL)
                p->next[word[i] - 'a'] = new Trie();
            p = p->next[word[i] - 'a'];
        }
        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 DFS(word , root);
    }
    bool DFS(string word , Trie *root)
    {
        Trie *p = root;
        int len = word.size();
        bool result = false;
        for(int i = 0 ; i < len && p ; i++)
        {
            if(word[i] != '.')
                p = p->next[word[i] - 'a'];
            else
            {
                Trie *tmp = p;  //临时把p保存下来，下面对p的next结点使用深搜
                for(int j = 0 ; j < 26 ; j++)
                {
                    p = tmp->next[j];
                    if(p)
                        result = DFS(word.substr(i+1) , p); //对i后面的子串递归调用
                    if(result)
                        return result;
                }
            }
        }
        return p && p->isWord;
    }
private:
    Trie *root;
};

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

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