Leetcode 第211題：Add and Search Word - Data--添加與搜索單詞（C++、Python）

題目地址：Add and Search Word - Data structure design

---

題目簡介：

設計一個支持以下兩種操作的數據結構：     

void addWord(word)
bool search(word)

可以搜索文字或正則表達式字符串，字符串只包含字母 . 或 a-z 。其中'.' 可以表示任何一個字母。

示例:     

addWord("bad")
addWord("dad")
addWord("mad")
search("pad") -> false
search("bad") -> true
search(".ad") -> true
search("b..") -> true

說明: 你可以假設所有單詞都是由小寫字母 a-z 組成的。

---

題目解析：

C++提供的String類型便能夠很好完成這個任務，這裏需要自己實現。結合前面剛學到的Implement Trie (Prefix Tree），這裏只需要把search函數定義好即可。怎麼判斷是否出現過呢？最主要的是'.'的處理方法，因爲'.'可以代表任何一個字符，所以出現時將其代表爲任意字符的情況便是遍歷所有的子節點，只要出現一個滿足條件的即可。於是進行修改Implement Trie (Prefix Tree）代碼如下：

C++：

class TrieNode {
public:
    vector<TrieNode*> child;
    bool isWord;
    TrieNode() : isWord(false), child(26, nullptr) {
    }
    ~TrieNode() {
        for (auto& c : child)
            delete c;
    }
};
class WordDictionary {
public:
    /** Initialize your data structure here. */
    WordDictionary() {
        root = new TrieNode();
    }
    
    /** Adds a word into the data structure. */
    void addWord(string word) {
        TrieNode* p = root;
        for (char a : word) {
            int i = a - 'a';
            if (!p -> child[i])
                p -> child[i] = new TrieNode();
            p = p -> child[i];
        }
        p -> isWord = 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) {
        return searchWord(word, root, 0);
    }
    bool searchWord(string &word, TrieNode *p, int i) {
        if (i == word.size()) 
            return p -> isWord;
        if (word[i] == '.') 
        {
            for (auto &a : p->child) 
            {
                if (a && searchWord(word, a, i + 1)) 
                    return true;
            }
            return false;
        } 
        else 
        {
            return p -> child[word[i] - 'a'] && searchWord(word, p -> child[word[i] - 'a'], i + 1);
        }
    }
private:
    TrieNode* root;
};

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

Python:

class Node(object):
    def __init__(self):
        self.children = collections.defaultdict(Node)
        self.isword = False
class WordDictionary(object):

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.root = Node()

    def addWord(self, word):
        """
        Adds a word into the data structure.
        :type word: str
        :rtype: None
        """
        current = self.root
        for w in word:
            current = current.children[w]
        current.isword = True

    def search(self, word):
        """
        Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter.
        :type word: str
        :rtype: bool
        """
        def helper(word, root, i):
            if i == len(word):
                return root.isword
            curr_ch = word[i]
            if curr_ch == '.':
                for _, next_node in root.children.items():
                    if helper(word, next_node, i + 1):
                        return True
                return False
            if curr_ch in root.children:
                return helper(word, root.children[curr_ch], i + 1)
            return False
            
        return helper(word, self.root, 0)
        
        


# Your WordDictionary object will be instantiated and called as such:
# obj = WordDictionary()
# obj.addWord(word)
# param_2 = obj.search(word)