mirror of
https://github.com/prdlk/leetcode.git
synced 2026-11-02 21:22:43 +00:00
66 lines
1.7 KiB
Plaintext
66 lines
1.7 KiB
Plaintext
---
|
|
title: '211. Design Add and Search Words Data Structure'
|
|
description: Design a data structure that supports adding new words and finding if a string matches any previously added string
|
|
sidebar:
|
|
label: 'Design Add and Search Words Data Structure'
|
|
badge: 'Medium'
|
|
---
|
|
|
|
<Badge variant="accent">Trie</Badge>
|
|
|
|
### Constraints:
|
|
|
|
- `1 <= word.length <= 25`
|
|
- `word` in addWord consists of lowercase English letters.
|
|
- `word` in search consist of '.' or lowercase English letters.
|
|
- There will be at most 2 dots in `word` for search queries.
|
|
- At most 10^4 calls will be made to addWord and search.
|
|
|
|
## Solution
|
|
|
|
```py
|
|
class TrieNode:
|
|
def __init__(self):
|
|
self.children = {}
|
|
self.endOfWord = False
|
|
|
|
|
|
class WordDictionary:
|
|
def __init__(self):
|
|
self.root = TrieNode()
|
|
|
|
def addWord(self, word: str) -> None:
|
|
cur = self.root
|
|
for c in word:
|
|
if c not in cur.children:
|
|
cur.children[c] = TrieNode()
|
|
cur = cur.children[c]
|
|
cur.endOfWord = True
|
|
|
|
# Leverage Recursion
|
|
def search(self, word: str) -> bool:
|
|
def dfs(j, root):
|
|
cur = root
|
|
for i in range(j, len(word)):
|
|
c = word[i]
|
|
|
|
if c == ".":
|
|
for child in cur.children.values():
|
|
if dfs(i + 1, child):
|
|
return True
|
|
return False
|
|
else:
|
|
if c not in cur.children:
|
|
return False
|
|
cur = cur.children[c]
|
|
return cur.endOfWord
|
|
|
|
return dfs(0, self.root)
|
|
|
|
|
|
# Your WordDictionary object will be instantiated and called as such:
|
|
# obj = WordDictionary()
|
|
# obj.addWord(word)
|
|
# param_2 = obj.search(word)
|
|
```
|