Skip to content

Navigation Menu

Explore
By company size
By use case
By industry
View all solutions
Topics
- AI
- DevOps
- Security
- Software Development
- View all
Explore
- GitHub Sponsors
  Fund open source developers
- The ReadME Project
  GitHub community articles
Repositories
- Enterprise platform
  AI-powered developer platform
Available add-ons
Pricing

Search code, repositories, users, issues, pull requests...

Search

Clear

Search syntax tips

Provide feedback

We read every piece of feedback, and take your input very seriously.

Include my email address so I can be contacted

Saved searches

Use saved searches to filter your results more quickly

Name

Query

To see all available qualifiers, see our documentation.

You signed in with another tab or window. Reload to refresh your session. You signed out in another tab or window. Reload to refresh your session. You switched accounts on another tab or window. Reload to refresh your session.

Dismiss alert

SleepyBag / leetcode-python-in-one-line Public

Notifications You must be signed in to change notification settings
Fork 3
Star 29

Code
Issues
Pull requests
Actions
Projects
Security
Insights

Additional navigation options

Code
Issues
Pull requests
Actions
Projects
Security
Insights

Breadcrumbs

leetcode-python-in-one-line

/

58.length-of-last-word.py

Latest commit

History

3 lines (3 loc) · 127 Bytes

Breadcrumbs

leetcode-python-in-one-line

/

58.length-of-last-word.py

File metadata and controls

3 lines (3 loc) · 127 Bytes

1

2

3

class Solution:

def lengthOfLastWord(self, s: str) -> int:

return len(splited[-1]) if (splited := s.split()) else 0

Footer

© 2024 GitHub, Inc.

Footer navigation

Terms
Privacy
Security
Status
Docs
Contact

You can’t perform that action at this time.