Leetcode 211. 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.

s思路：
1. 搜索题，而且有regular expression,正则表达式的匹配搜索。用”.”代表一个字母的情况如何在代码或数学上模拟呢？
2. 难道是遇到”.”就遍历所有child指针不为NULL的下家吗？试一试先！

//方法1：用trie, 用dfs尝试26种情况，如果遇到"."class WordDictionary {private:    struct node{        node* child[26];        bool isWord;        node(){            for(int i=0;i<26;i++)                child[i]=NULL;            isWord=false;           }    };    node* root;public:    /** Initialize your data structure here. */    WordDictionary() {        root=new node();//新建一个node，让指针指向，不能搞没有实体的指针呀！    }    /** Adds a word into the data structure. */    void addWord(string word) {        //        node* cur=root;        for(char c:word){            int idx=c-'a';            if(!cur->child[idx])                    cur->child[idx]=new node();            cur=cur->child[idx];        }        cur->isWord=true;    }    /** Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter. */    bool helper(string word,node* cur,int i){        if(!cur) return false;        if(i==word.size()) return cur->isWord;          if(word[i]=='.'){            for(int j=0;j<26;j++)                if(cur->child[j]&&helper(word,cur->child[j],i+1)) return true;            return false;        }else{            return helper(word,cur->child[word[i]-'a'],i+1);                }    }    bool search(string word) {        //        node* cur=root;        return helper(word,cur,0);    }};/** * Your WordDictionary object will be instantiated and called as such: * WordDictionary obj = new WordDictionary(); * obj.addWord(word); * bool param_2 = obj.search(word); */