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.

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.

 

Approach #1: Trie. [C++]

class TrieNode {
public:
    TrieNode* children[26];
    bool word;
    TrieNode() {
        word = false;
        memset(children, NULL, sizeof(children));
    }
};

class WordDictionary {
public:
    /** Initialize your data structure here. */
    WordDictionary() {
        root = new TrieNode();
    }
    
    /** 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 query(word.c_str(), root);
    }
private:
    TrieNode* root;
    bool query(const char* word, TrieNode* node) {
        for (int i = 0; word[i] && node; ++i) {
            if (word[i] != '.') {
                node = node->children[word[i] - 'a'];
            } else {
                TrieNode* temp = node;
                for (int j = 0; j < 26; ++j) {
                    node = temp->children[j];
                    if (query(word+i+1, node)) {
                        return true;
                    }
                }
            }
        }
        return node && node->word;
    }
};

/**
 * Your WordDictionary object will be instantiated and called as such:
 * WordDictionary obj = new WordDictionary();
 * obj.addWord(word);
 * bool param_2 = obj.search(word);
 */

　　

Analysis:

这一题用到了一个叫做Trie的数据结构，虽然说以前学过但是当时自己也就是看一遍也没有真正的运用这种数据结构做过题。所以一开始看到这道题时完全没有思路，后来看了别人的题解。比较好奇的是如果将35行中的const关键字省略的话回编译出错，现在还不知道为什么会这样。虽然最近一直再看关于C++的书籍，但是看过书之后还是感觉对C++一点都不了解。