LeetCode 211. Add and Search Word - Data structure design

原题

设计一个包含下面两个操作的数据结构：addWord(word), search(word)
addWord(word)会在数据结构中添加一个单词。而search(word)则支持普通的单词查询或是只包含. 和a-z的简易正则表达式的查询。一个 . 可以代表一个任何的字母。

样例

addWord("bad")
addWord("dad")
addWord("mad")
search("pad")  // return false
search("bad")  // return true
search(".ad")  // return true
search("b..")  // return true

解题思路

• 本题跟Implement Trie十分相似，同样需要使用字典树
• 由于本题里中'.'可以代替任意字符，所以当某一个字符是'.'，就需要查找所有的子树，只要有一个最终能够存在，就返回True
• 一个字母一个字母的递归

完整代码

class TrieNode(object):
    
    def __init__(self):
        
        self.child = {}
        self.isword = False

class WordDictionary(object):

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

    def addWord(self, word):
        """
        Adds a word into the data structure.
        :type word: str
        :rtype: void
        """
        node = self.root
        for i in word:
            if not node.child.get(i):
                node.child[i] = TrieNode()
            node = node.child[i]
        node.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
        """
        return self.find(self.root, word)
        
    def find(self, node, word):
        if word == "":
            return node.isword
        if word[0] == '.':
            for key in node.child:
                if self.find(node.child.get(key), word[1:]):
                    return True
        else:
            if not node.child.get(word[0]):
                return False
            return self.find(node.child.get(word[0]),word[1:])
        return False    #防止44行里面的节点后面没有key值或是返回false

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