leecode/problems/211.add-and-search-word-data-structure-design.md

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2020-05-22 18:17:19 +08:00
## 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 .  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)