[leetcode]211. Add and Search Word - Data structure design

제목 링크:https://leetcode.com/problems/add-and-search-word-data-structure-design/tabs/description
 
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

class TrieNode{
public:
    bool isKey;
    TrieNode* children[26];
    TrieNode():isKey(false){
        memset(children,NULL,sizeof(TrieNode*)*26);
    };
};

class WordDictionary {
public:
    /** Initialize your data structure here. */
    WordDictionary() {
        root=new TrieNode();
    }

    /** Adds a word into the data structure. */
    void addWord(string word) {
        TrieNode* run=root;
        for(char c:word)
        {
            if(!(run->children[c-'a']))
                run->children[c-'a']=new TrieNode();
            run=run->children[c-'a'];
        }
        run->isKey=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 query(word.c_str(),root);
    }
private:
    TrieNode* root;
    bool query(const char* word,TrieNode* node)
    {
        TrieNode* run=node;
        for(int i=0;word[i];i++)
        {
            if(run && word[i]!='.')
                run=run->children[word[i]-'a'];
            else if(run && word[i]=='.')
            {
                TrieNode* tmp=run;
                for(int j=0;j<26;j++)
                {
                    run=tmp->children[j];
                    if(query(word+i+1,run))
                        return true;
                }
            }
            else
                break;
        }
        return run && run->isKey;
    }
};