LeetCode----Add and Search Word - Data structure design

Add and Search Word - Data structure design

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") -> falsesearch("bad") -> truesearch(".ad") -> truesearch("b..") -> true

Note:
You may assume that all words are consist of lowercase letters a-z.

click to show hint.

You should be familiar with how a Trie works. If not, please work on this problem: Implement Trie (Prefix Tree) first.

分析：

同上一题，实现Trie树。增加了“.”的支持。使用递归完成。

代码：

class TrieNode(object):    def __init__(self):        """        Initialize your data structure here.        """        self.tnlst = {}        self.terminal = False    def __str__(self):        return str(self.tnlst.items())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        """        r = self.root        for w in word:            if w not in r.tnlst:                r.tnlst[w] = TrieNode()            r = r.tnlst[w]        r.terminal = 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.searchLeaf(self.root, word)    def searchLeaf(self, curNode, word):        if not word:            if not curNode.terminal:                return False            return True        ch = word[0]        if ch == '.':            for i in curNode.tnlst.keys():                if self.searchLeaf(curNode.tnlst[i], word[1:]):                    return True            return False        if ch in curNode.tnlst:            return self.searchLeaf(curNode.tnlst[ch], word[1:])        else:            return False