LeetCode题解211.add-and-search-word-data-structure-design

题目地址(211. 添加与搜索单词 - 数据结构设计)

https://leetcode-cn.com/problems/add-and-search-word-data-structure-design/description/

题目描述

设计一个支持以下两种操作的数据结构:

void addWord(word)
bool search(word)
search(word) 可以搜索文字或正则表达式字符串,字符串只包含字母 . 或 a-z 。 . 可以表示任何一个字母。

示例:

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

你可以假设所有单词都是由小写字母 a-z 组成的。

思路

我们首先不考虑字符"."的情况。这种情况比较简单,我们 addWord 直接添加到数组尾部,search 则线性查找即可。

接下来我们考虑特殊字符“.”,其实也不难,只不过 search 的时候,判断如果是“.”, 我们认为匹配到了,继续往后匹配即可。

上面的代码复杂度会比较高,我们考虑优化。如果你熟悉前缀树的话,应该注意到这可以使用前缀树来进行优化。前缀树优化之后每次查找复杂度是$O(h)$, 其中 h 是前缀树深度,也就是最长的字符串长度。

关于前缀树,LeetCode 有很多题目。有的是直接考察,让你实现一个前缀树,有的是间接考察,比如本题。前缀树代码见下方,大家之后可以直接当成前缀树的解题模板使用。

由于我们这道题需要考虑特殊字符".",因此我们需要对标准前缀树做一点改造,insert 不做改变,我们只需要改变 search 即可,代码(Python 3):

def search(self, word):
        """
        Returns if the word is in the trie.
        :type word: str
        :rtype: bool
        """
        curr = self.Trie
        for i, w in enumerate(word):
            if w == '.':
                wizards = []
                for k in curr.keys():
                    if k == '#':
                        continue
                    wizards.append(self.search(word[:i] + k + word[i + 1:]))
                return any(wizards)
            if w not in curr:
                return False
            curr = curr[w]
        return "#" in curr

标准的前缀树搜索我也贴一下代码,大家可以对比一下:

def search(self, word):
        """
        Returns if the word is in the trie.
        :type word: str
        :rtype: bool
        """
        curr = self.Trie
        for w in word:
            if w not in curr:
                return False
            curr = curr[w]
        return "#" in curr

关键点

  • 前缀树(也叫字典树),英文名 Trie(读作 tree 或者 try)

代码

  • 语言支持:Python3

Python3 Code:

关于 Trie 的代码:

class Trie:

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.Trie = {}

    def insert(self, word):
        """
        Inserts a word into the trie.
        :type word: str
        :rtype: void
        """
        curr = self.Trie
        for w in word:
            if w not in curr:
                curr[w] = {}
            curr = curr[w]
        curr['#'] = 1

    def search(self, word):
        """
        Returns if the word is in the trie.
        :type word: str
        :rtype: bool
        """
        curr = self.Trie
        for i, w in enumerate(word):
            if w == '.':
                wizards = []
                for k in curr.keys():
                    if k == '#':
                        continue
                    wizards.append(self.search(word[:i] + k + word[i + 1:]))
                return any(wizards)
            if w not in curr:
                return False
            curr = curr[w]
        return "#" in curr

主逻辑代码:

class WordDictionary:

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.trie = Trie()

    def addWord(self, word: str) -> None:
        """
        Adds a word into the data structure.
        """
        self.trie.insert(word)

    def search(self, word: str) -> bool:
        """
        Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter.
        """
        return self.trie.search(word)


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

相关题目

  • 208.implement-trie-prefix-tree
  • 212.word-search-ii
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