leetcode Add and Search Word - Data structure design

最新推荐文章于 2020-04-25 11:21:31 发布

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本文介绍了一个支持添加单词和模糊搜索的单词数据结构的设计与实现。使用字典树(trie)来存储单词，并针对包含通配符'.'的搜索提供递归解决方案。

题目

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

分析

关于字典树（trie tree）的概念和分析见：http://blog.youkuaiyun.com/u010902721/article/details/45749447
主要挑战性在于’.’的匹配。遇到一个’.’就得匹配当前next[]数组里面所有的有效分支。用个递归函数，递归匹配吧。

代码

class TrieNode{
public:
    int flag;
    TrieNode* next[26];

    TrieNode(){
        flag = 0;
        for(int i = 0; i < 26; i++){
            next[i] = NULL;
    }
}
};
class WordDictionary {
private:
    TrieNode *root;
public:
    WordDictionary(){
        root = new TrieNode();
    }
    // Adds a word into the data structure.
    void addWord(string word) {
        TrieNode* p = root;
        int len = word.length();
        if(len <= 0)
            return;
        for(int i = 0; i < len; i++){
            int d = word.at(i) - 'a';
            if((p->next)[d] == NULL){
                p->next[d] = new TrieNode();
                p->next[d]->flag = 0;
            }
            p = p->next[d];
        }
        p->flag = 1;
    }

    // Returns if the word is in the data structure. A word could
    // contain the dot character '.' to represent any one letter.
    bool searchHelp(string word, struct TrieNode *root){
        struct TrieNode *p = root;
        int len = word.length();
        if(len <= 0){//为了匹配字符串中字符'.'是最后一个字符的情形。
            if(p->flag == 1)
                return true;
            else
                return false;
        }
        if(root == NULL)
            return false;
        for(int i = 0; i < len; i++){
            if(word.at(i) == '.'){
                bool result = false;
                for(int j = 0; j < 26; j++){
                    if(p->next[j] != NULL){
                        result = result || searchHelp(word.substr(i+1), p->next[j]);
                    }
                }
                return result;
            }
            else
            {
                int d = word.at(i) - 'a';
                if(p->next[d] == NULL)
                    return false;
                p = p->next[d];
            }
        }
        if(p->flag == 1)
                return true;
            else
                return false;
    }
    bool search(string word) {
        int len = word.length();
        if(len <= 0)
            return true;
        return searchHelp(word, root);
    }
};
// Your WordDictionary object will be instantiated and called as such:
// WordDictionary wordDictionary;
// wordDictionary.addWord("word");
// wordDictionary.search("pattern");