173 lines
4.8 KiB
Markdown
173 lines
4.8 KiB
Markdown
## 题目地址（211. 添加与搜索单词 - 数据结构设计）
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https://leetcode-cn.com/problems/add-and-search-word-data-structure-design/description/
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## 题目描述
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```
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设计一个支持以下两种操作的数据结构：
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void addWord(word)
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bool search(word)
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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。
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示例:
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addWord("bad")
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addWord("dad")
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addWord("mad")
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search("pad") -> false
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search("bad") -> true
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search(".ad") -> true
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search("b..") -> true
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说明:
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你可以假设所有单词都是由小写字母 a-z 组成的。
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```
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## 思路
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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。
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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。
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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。
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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。
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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)：
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```python
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def search(self, word):
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"""
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Returns if the word is in the trie.
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:type word: str
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:rtype: bool
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"""
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curr = self.Trie
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for i, w in enumerate(word):
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if w == '.':
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wizards = []
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for k in curr.keys():
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if k == '#':
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continue
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wizards.append(self.search(word[:i] + k + word[i + 1:]))
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return any(wizards)
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if w not in curr:
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return False
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curr = curr[w]
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return "#" in curr
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```
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标准的前缀树搜索我也贴一下代码，大家可以对比一下：
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```python
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def search(self, word):
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"""
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Returns if the word is in the trie.
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:type word: str
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:rtype: bool
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"""
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curr = self.Trie
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for w in word:
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if w not in curr:
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return False
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curr = curr[w]
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return "#" in curr
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```
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## 关键点
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- 前缀树（也叫字典树），英文名 Trie（读作 tree 或者 try）
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## 代码
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- 语言支持：Python3
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Python3 Code：
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关于 Trie 的代码:
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```python
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class Trie:
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def __init__(self):
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"""
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Initialize your data structure here.
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"""
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self.Trie = {}
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def insert(self, word):
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"""
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Inserts a word into the trie.
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:type word: str
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:rtype: void
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"""
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curr = self.Trie
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for w in word:
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if w not in curr:
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curr[w] = {}
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curr = curr[w]
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curr['#'] = 1
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def search(self, word):
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"""
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Returns if the word is in the trie.
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:type word: str
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:rtype: bool
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"""
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curr = self.Trie
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for i, w in enumerate(word):
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if w == '.':
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wizards = []
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for k in curr.keys():
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if k == '#':
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continue
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wizards.append(self.search(word[:i] + k + word[i + 1:]))
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return any(wizards)
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if w not in curr:
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return False
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curr = curr[w]
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return "#" in curr
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```
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主逻辑代码：
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```python
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class WordDictionary:
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def __init__(self):
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"""
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Initialize your data structure here.
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"""
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self.trie = Trie()
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def addWord(self, word: str) -> None:
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"""
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Adds a word into the data structure.
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"""
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self.trie.insert(word)
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def search(self, word: str) -> bool:
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"""
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Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter.
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"""
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return self.trie.search(word)
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# Your WordDictionary object will be instantiated and called as such:
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# obj = WordDictionary()
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# obj.addWord(word)
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# param_2 = obj.search(word)
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```
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## 相关题目
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- [0208.implement-trie-prefix-tree](./208.implement-trie-prefix-tree.md)
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- [0212.word-search-ii](./212.word-search-ii.md)
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- [0472.concatenated-words](./problems/472.concatenated-words.md)
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- [0820.short-encoding-of-words](https://github.com/azl397985856/leetcode/blob/master/problems/820.short-encoding-of-words.md)
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