## 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。 ![](https://tva1.sinaimg.cn/large/006tNbRwly1gb5dmstsxxj30mz0gqmzh.jpg) 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): ```python 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: ```python 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: ```python 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: ```python 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 - [0208.implement-trie-prefix-tree](./208.implement-trie-prefix-tree.md) - [0212.word-search-ii](./212.word-search-ii.md) - [0472.concatenated-words](./problems/472.concatenated-words.md) - [0820.short-encoding-of-words](https://github.com/azl397985856/leetcode/blob/master/problems/820.short-encoding-of-words.md)