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.

click to show hint.

字典樹+dfs時間復雜度 查詢的時候復雜度O(26^n) n為‘.’的個數

class WordDictionary {
public
 
:
    /** Initialize your data structure here. */
    WordDictionary() {
        
    }
    class node {
      public:
        node* next[26];
        int end;
        node() {
            for (int i = 0; i < 26; ++i) {
                next[i] = nullptr;
            }
            end = 0
 
;
        }
    };
    
    class Trie {
      public:
        node* root;
        Trie() {
            root = new node();
        }
        void add(string s) {
            if (s.size() == 0) return ;
            node* p = root;
            for (int i = 0; i < s.size(); ++i) {
                int x = s[i] - ‘a‘;
                if (p->next[x] == nullptr) {
                    p->next[x] = new node();
                    p = p->next[x];
                } else {
                    p = p->next[x];
                }
            }
            p->end = 1;
        }
        bool search(string& s, int pos, node *p) {
            if (s.size() == pos && p->end == 1) {
                return true;
            }
            if (s[pos] == ‘.‘) {
                for (int j = 0; j < 26; ++j) {
                    if (p->next[j] != nullptr) {
                        if (search(s, pos + 1, p->next[j])) return true;
                    }
                }
            } else {
                int x = (int)(s[pos] - ‘a‘);
                if (x >= 0 && x < 26 && p->next[x] != nullptr && search(s, pos + 1, p->next[x])) return true;
            }
            return false;  
        }
    };
    
    /** Adds a word into the data structure. */
    void addWord(string word) {
        ac.add(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) {
        if (ac.search(word, 0, ac.root)) return true;
        return false;
    }
    private:
        Trie ac;
};

/**
 * Your WordDictionary object will be instantiated and called as such:
 * WordDictionary obj = new WordDictionary();
 * obj.addWord(word);
 * bool param_2 = obj.search(word);
 */

leetcode 211. Add and Search Word - Data structure design