Leetcode 211. Add and Search Word - Data structure design

最新推荐文章于 2024-04-12 12:36:48 发布

xinqrs01

最新推荐文章于 2024-04-12 12:36:48 发布

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文章标签： trie 搜索string 正则表达式通配符

本文链接：https://blog.youkuaiyun.com/xinqrs01/article/details/54961848

本文介绍了一种使用Trie数据结构和深度优先搜索来实现支持正则表达式搜索的功能。具体实现了addWord和search两个核心操作，其中search能够处理包含‘.’字符的正则表达式。

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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);
 */