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

最新推荐文章于 2019-10-31 15:24:59 发布

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#leetcode #Tire树 #python #DFS

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本文介绍了一种基于Trie树的数据结构设计，该结构能够支持单词的添加和搜索功能，其中搜索功能还支持包含字母a-z或'.'的正则表达式字符串。'.'可以代表任意一个字母。

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") -> false
search("bad") -> true
search(".ad") -> true
search("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