[Leetcode] 211. Add and Search Word

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") -> falsesearch("bad") -> truesearch(".ad") -> truesearch("b..") -> true

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

思路：

这道题目的正解就是实现Trie这种数据结构（请参考我的博文《[Leetcode] 208. Implement Trie (Prefix Tree) 解题报告》）。唯一不同的地方在于本题目中出现了通配符‘.’，也就是说'.'可以通配a-z中的任意一个字符。因此在search中需要采用DFS进行穷举：一旦发现了通配符，则尝试将其替换成每一个可能的字符，一旦匹配成功就返回true；如果所有的匹配都失败，则返回false。

代码：

class TrieNode {public:    bool isComplete;    TrieNode* ch[26];    TrieNode() {        isComplete = false;        for(int i = 0; i < 26; ++i) {            ch[i] = NULL;        }    }};class Trie {public:    Trie() {        root = new TrieNode();    }        void insert(string word) {        TrieNode* p = root;        for(auto c : word) {            if(p->ch[c - 'a'] == NULL) {                p->ch[c - 'a'] = new TrieNode();            }            p = p->ch[c - 'a'];        }        p->isComplete = true;    }        bool search(string word) {        return dfs(root, word, 0);    }        bool dfs(TrieNode* p, string& word, int startIndex) {        if(p == NULL) {            return false;        }        if(startIndex == word.length()) {            return p->isComplete;        }        char c = word[startIndex];        if(c == '.') {            for(int i = 0; i < 26; ++i) {                if(p->ch[i] != NULL && dfs(p->ch[i], word, startIndex + 1)) {                    return true;                }            }            return false;        }        else {            return dfs(p->ch[c-'a'], word, startIndex + 1);        }    }private:    TrieNode* root;};class WordDictionary {public:    /** Initialize your data structure here. */    WordDictionary() {            }        /** Adds a word into the data structure. */    void addWord(string word) {        trie.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) {        return trie.search(word);    }private:    Trie trie;};/** * Your WordDictionary object will be instantiated and called as such: * WordDictionary obj = new WordDictionary(); * obj.addWord(word); * bool param_2 = obj.search(word); */