1. 题目

设计一个支持以下两种操作的数据结构：

void addWord(word)
bool search(word)
search(word) 可以搜索文字或正则表达式字符串，字符串只包含字母 . 或 a-z 。 . 可以表示任何一个字母。

示例:
addWord("bad")
addWord("dad")
addWord("mad")
search("pad") -> false
search("bad") -> true
search(".ad") -> true
search("b..") -> true
说明:
你可以假设所有单词都是由小写字母 a-z 组成的。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/add-and-search-word-data-structure-design
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

2. Trie解题

• 构建Trie树
• 回溯查找，遇见.在所有的子树里查找，没有遇见.在，当前相等的情况下，再继续在所有的子树中递归查找

class TrieNode
{
public:char ch;TrieNode *next[26];bool isEnd;TrieNode(char c = '/'):ch(c),isEnd(false) {memset(next, 0, sizeof(TrieNode*)*26);}
};
class Trie
{
public:TrieNode *root;Trie(){root = new TrieNode();}~Trie(){destroy(root);}void destroy(TrieNode *root){if(root == NULL)return;for(int i = 0; i < 26; i++)destroy(root->next[i]);delete root;}void insert(string str){TrieNode *cur = root;for(char s:str){if(cur->next[s-'a'] == NULL)cur->next[s-'a'] = new TrieNode(s);cur = cur->next[s-'a'];}cur->isEnd = true;}
};
class WordDictionary {Trie tree;
public:/** Initialize your data structure here. */WordDictionary() {}/** Adds a word into the data structure. */void addWord(string word) {tree.insert(word);}/** 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) {TrieNode *cur = tree.root;bool found = false;for(int i = 0; i < 26; ++i){find(word,cur->next[i],0,found);}return found;}void find(string &word, TrieNode *root, int idx, bool &found){if(found || !root)return;if(idx == word.size()-1){if(root->isEnd)if(word[idx] == '.' || word[idx] == root->ch)found = true;return;}if((word[idx] != '.'&&root->ch == word[idx])|| word[idx] == '.'){	for(int i = 0; i < 26; ++i){find(word,root->next[i],idx+1,found);}}}
};