LeetCode *** 211. Add and Search Word - Data structure design（字典树）

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 lettersa-z or .. A . means it can represent any one letter.

For example:

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

Note:
You may assume that all words are consist of lowercase letters a-z.

分析：

代码：

 struct TrieNode { public: TrieNode *next[26]; bool isStr = false; TrieNode() { memset(next, 0, sizeof(next)); } }; class WordDictionary { public: TrieNode *root = new TrieNode(); // Adds a word into the data structure. void addWord(string word) { int len = word.length(); TrieNode *tmp = root; for (int i = 0; i<len; ++i) { if (tmp->next[word[i] - 'a'] == NULL)tmp->next[word[i] - 'a'] = new TrieNode(); tmp = tmp->next[word[i] - 'a']; } tmp->isStr = 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) { if (word.size() == 0)return true; TrieNode *tmp = root; if (word[0] == '.') { bool res = false; for (int i = 0; i<26; ++i) if (tmp->next[i] != NULL)res |= (search(word.substr(1), tmp->next[i])); return res; } if (tmp->next[word[0] - 'a'] == NULL)return false; return search(word.substr(1), tmp->next[word[0] - 'a']); } bool search(string word, TrieNode *tmp) { if (word.size() == 0 && tmp->isStr)return true; if (word.size() == 0 || (word.size() && tmp == NULL))return false; if (word[0] == '.') { bool res = false; for (int i = 0; i<26; ++i) if (tmp->next[i] != NULL)res |= (search(word.substr(1), tmp->next[i])); return res; } if (tmp->next[word[0] - 'a'] == NULL)return false; return search(word.substr(1), tmp->next[word[0] - 'a']); } };