本文主要是介绍LeetCode *** 211. Add and Search Word - Data structure design（字典树），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
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.
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']);}};这篇关于LeetCode *** 211. Add and Search Word - Data structure design（字典树）的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
