Design Add and Search Words Data Structure

Description

Design a data structure that supports adding new words and finding if a string matches any previously added string.

Implement the WordDictionary class:

• WordDictionary() Initializes the object.
• void addWord(word) Adds word to the data structure, it can be matched later.
• bool search(word) Returns true if there is any string in the data structure that matches word or false otherwise. word may contain dots '.' where dots can be matched with any letter.

Example:

Input [“WordDictionary”,“addWord”,“addWord”,“addWord”,“search”,“search”,“search”,“search”] [[],[“bad”],[“dad”],[“mad”],[“pad”],[“bad”],[“.ad”],[“b..“]] Output [null,null,null,null,false,true,true,true]

Explanation WordDictionary wordDictionary = new WordDictionary(); wordDictionary.addWord(“bad”); wordDictionary.addWord(“dad”); wordDictionary.addWord(“mad”); wordDictionary.search(“pad”); // return False wordDictionary.search(“bad”); // return True wordDictionary.search(“.ad”); // return True wordDictionary.search(“b..”); // return True

Constraints:

• 1 <= word.length <= 25
• word in addWord consists of lowercase English letters.
• word in search consist of '.' or lowercase English letters.
• There will be at most 3 dots in word for search queries.
• At most 104 calls will be made to addWord and search.

Code

使用在 Implement Trie (Prefix Tree) 中學到的 trie，唯一不一樣的是要如何處理遇到 . 的情形，這裡採用 recursion 。

class WordDictionary {
    vector<WordDictionary*> children;
    bool isEndofWord;
public:
    WordDictionary(): isEndofWord(false) {
        children = vector<WordDictionary*>(26, nullptr);
    }
    
    void addWord(string word) {
        WordDictionary* cur = this;
        for(auto c: word) {
            c -= 'a';
            if(cur->children[c] == nullptr) 
                cur->children[c] = new WordDictionary();
            cur = cur->children[c];
        }
        cur->isEndofWord = true;
    }
    
    bool search(string word) {
        WordDictionary* cur = this;
        for(int i = 0; i < word.length(); i++) {
            char c = word[i];
            if(word[i] == '.') {
                string subStr = word.substr(i + 1);
                for(auto ch: cur->children) {
                    if(ch && ch->search(subStr)) 
                        return true;
                }
                return false;
            } else {
	            c -= 'a';
	            if(cur->children[c] == nullptr) return false;
	            cur = cur->children[c];
            }
            
        } 
        return cur->isEndofWord;
    }
};
 
/**
 * Your WordDictionary object will be instantiated and called as such:
 * WordDictionary* obj = new WordDictionary();
 * obj->addWord(word);
 * bool param_2 = obj->search(word);
 */

Source

• Design Add and Search Words Data Structure - LeetCode