LeetCode-211 Add and Search Word - Data structure design

Posted 2022-01-22 heyn1

题目描述

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.

 

题目大意

实现两个操作：插入单词、查找单词（查找单词时的输入只能为 ‘a-z‘ 和 ‘.‘ ，其中 ‘.‘ 可以代表任何小写字母）。

 

示例

E

addWord("bad")
addWord("dad")
addWord("mad")
search("pad") -> false
search("bad") -> true
search(".ad") -> true
search("b..") -> true

 

解题思路

解题思路类似于LeetCode - 208 Implement Trie，设置一个node class用来记录保存树结点中的节点信息。

插入单词是建树的过程，查找单词是搜索树的过程。

 

复杂度分析

时间复杂度：O(N)

空间复杂度：O(N)

 

代码

class TrieNode 
public:
    //该节点代表的字母是否是某个单词的最后一个字母
    bool word;
    TrieNode* children[26];
    TrieNode() 
        word = false;
        memset(children, NULL, sizeof(children));
    
;

class WordDictionary 
public:
    /** Initialize your data structure here. */
    WordDictionary() 
        
    
    
    //建立一棵树
    /** Adds a word into the data structure. */
    void addWord(string word) 
        TrieNode* node = root;
        for (char c : word) 
            if (!node -> children[c - ‘a‘]) 
                node -> children[c - ‘a‘] = new TrieNode();
            
            node = node -> children[c - ‘a‘];
        
        node -> word = true;
    
    
    /** 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 search(word.c_str(), root);
    
private:
    TrieNode* root = new TrieNode();
    
    bool search(const char* word, TrieNode* node) 
        for (int i = 0; word[i] && node; i++) 
            //若搜索的单词该位置不是‘.’，则直接进行比较，否则对所有子结点进行遍历搜索
            if (word[i] != ‘.‘) 
                node = node -> children[word[i] - ‘a‘];
             else 
                TrieNode* tmp = node;
                for (int j = 0; j < 26; j++) 
                    node = tmp -> children[j];
                    if (search(word + i + 1, node)) 
                        return true;
                    
                
            
        
        return node && node -> word;
    
;