Notice
Recent Posts
Recent Comments
일 | 월 | 화 | 수 | 목 | 금 | 토 |
---|---|---|---|---|---|---|
1 | 2 | 3 | 4 | 5 | 6 | 7 |
8 | 9 | 10 | 11 | 12 | 13 | 14 |
15 | 16 | 17 | 18 | 19 | 20 | 21 |
22 | 23 | 24 | 25 | 26 | 27 | 28 |
29 | 30 |
Tags
- 우분투
- leetcode
- dacon
- ubuntu
- 더현대서울 맛집
- 편스토랑 우승상품
- programmers
- hackerrank
- 백준
- Real or Not? NLP with Disaster Tweets
- Git
- 프로그래머스
- Baekjoon
- PYTHON
- 코로나19
- gs25
- 자연어처리
- AI 경진대회
- 캐치카페
- github
- SW Expert Academy
- 데이콘
- 금융문자분석경진대회
- Kaggle
- 프로그래머스 파이썬
- 파이썬
- 맥북
- Docker
- 편스토랑
- ChatGPT
Archives
- Today
- Total
솜씨좋은장씨
[leetCode] 717. 1-bit and 2-bit Characters (Python) 본문
728x90
반응형
We have two special characters. The first character can be represented by one bit 0. The second character can be represented by two bits (10 or 11).
Now given a string represented by several bits. Return whether the last character must be a one-bit character or not. The given string will always end with a zero.
Example 1:
Input:
bits = [1, 0, 0]
Output: True
Explanation:
The only way to decode it is two-bit character and one-bit character. So the last character is one-bit character.
Example 2:
Input:
bits = [1, 1, 1, 0]
Output: False
Explanation:
The only way to decode it is two-bit character and two-bit character. So the last character is NOT one-bit character.
Note:
- 1 <= len(bits) <= 1000.
- bits[i] is always 0 or 1.
Solution
class Solution:
def isOneBitCharacter(self, bits: List[int]) -> bool:
i = 0
while i < len(bits)-1:
if bits[i] == 0:
i+=1
else:
i+=2
return i == len(bits)-1
'Programming > 코딩 1일 1문제' 카테고리의 다른 글
[leetCode] 1592. Rearrange Spaces Between Words (Python) (0) | 2021.01.19 |
---|---|
[leetCode] 876. Middle of the Linked List (Python) (0) | 2021.01.18 |
[leetCode] 908. Smallest Range I (Python) (0) | 2021.01.16 |
[leetCode] 520. Detect Capital (Python) (0) | 2021.01.15 |
[leetCode] 237. Delete Node in a Linked List (Python) (0) | 2021.01.14 |
Comments