Trie树使用

leetcode题目：https://leetcode.com/problems/add-and-search-word-data-structure-design/

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

解：

struct MyTrie{//trie树的数据结构
    vector<MyTrie*> child;
    bool isWord;
    MyTrie():isWord(false),child(vector<MyTrie*>(26, NULL)){}
};

class WordDictionary {
public:

    MyTrie * root;
    unordered_set<string> mm;
    WordDictionary():root(new MyTrie()){}

    // Adds a word into the data structure.
    void addWord(string word) {
        MyTrie * cur=root;
        for(int i=0; i<word.size(); i++){
            int idx=word[i]-'a';
            if(cur->child[idx]==NULL) 
                cur->child[idx]=new MyTrie();
            cur=cur->child[idx];
        }
        cur->isWord=true;//在最后一个节点标识从根节点可以到这里形成一个单词
    }

    // 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 searchData(word.c_str(), root);
    }
    
    bool searchData(const char * c, MyTrie * cur){//单词的查找
        if(cur==NULL)   return false;
        if(*c=='\0') return cur->isWord;
        if(*c!='.')
            return searchData(c+1, cur->child[*c-'a']);
        else{
            for(int i=0; i<26; i++){
                if(searchData(c+1, cur->child[i]))
                    return true;
            }
        }
        return false;
    }
    
};

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