211. Add and Search Word - Data structure design

最新推荐文章于 2019-08-05 10:40:12 发布

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最新推荐文章于 2019-08-05 10:40:12 发布

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本文链接：https://blog.youkuaiyun.com/fengyuhe1001/article/details/73730616

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本文介绍了一种支持添加单词和搜索单词的数据结构设计，其中搜索功能支持正则表达式的查询方式，包括任意单个字符的匹配。文章通过Trie树实现这一功能，并提供了完整的C++代码实现。

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211. Add and Search Word - Data structure design

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.

代码如下：

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

class WordDictionary {
public:
    WordDictionary() {
        root = new TrieNode();
    }

    // Adds a word into the data structure.
    void addWord(string word) {
        TrieNode* run = root;
        for (char c : word) {
            if (!(run -> children[c - 'a'])) 
                run -> children[c - 'a'] = new TrieNode();
            run = run -> children[c - 'a'];
        }
        run -> isKey = 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) {
        TrieNode* run = node;
        for (int i = 0; word[i]; i++) {
            if (run && word[i] != '.')
                run = run -> children[word[i] - 'a'];
            else if (run && word[i] == '.') { 
                TrieNode* tmp = run;
                for (int j = 0; j < 26; j++) {
                    run = tmp -> children[j];
                    if (query(word + i + 1, run))
                        return true;
                }
            }
            else break;
        }
        return run && run -> isKey; 
    }
};