[Leetcode]211. Add and Search Word - Data structure design @python

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”) -> false
search(“bad”) -> true
search(“.ad”) -> true
search(“b..”) -> true
Note:
You may assume that all words are consist of lowercase letters a-z.

题目要求

要求建立一种数据结构，方便单词的模式匹配。模式包含字母和字符’.’，’.’可以表示任意的字符。

解题思路

此题参考了书影博客的解题思路。
采用了字典树作为此题的数据结构。

代码

class TrieNode(object):    def __init__(self):        self.childs = {}        self.isWord = Falseclass 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 letter in word:            child = node.childs.get(letter)            if child == None:                child = TrieNode()                node.childs[letter] = child            node = child        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 _,child in node.childs.items():                if self.find(child,word[1:]):                    return True        elif word[0] in node.childs:            return self.find(node.childs.get(word[0]),word[1:])        return False