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);
*/