leetcode/1356: sort

[tony84727]

No description provided.

[gemini-code-assist]

Summary of Changes

Hello @tony84727, I'm Gemini Code Assist¹! I'm currently reviewing this pull request and will post my feedback shortly. In the meantime, here's a summary to help you and other reviewers quickly get up to speed!

This pull request introduces a solution to LeetCode problem 1356, 'Sort Integers by The Number of 1 Bits'. It includes the implementation of the sorting logic, comprehensive unit tests, and proper module integration within the existing project structure.

Highlights

• Implementation: Implemented a function sort_by_bits that sorts an array of integers based on the number of 1s in their binary representation, using Rust's sort_by method and count_ones.
• Testing: Added unit tests using the test_case crate to verify the correctness of the sorting algorithm with different input arrays.
• Module Integration: Integrated the new algorithm into the leetcode module by adding a new module algorithm_1356 and including it in the main leetcode module.

Activity

• A new module algorithm_1356 was added to the project.
• The sort_by_bits function was implemented to solve the LeetCode problem.
• Unit tests were added to ensure the correctness of the implemented function.
• The new module was integrated into the main leetcode module.

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[gemini-code-assist]

Code Review

The pull request adds a solution for LeetCode problem 1356. The implementation is correct and includes tests. I've provided one suggestion to improve performance and code clarity by using a more suitable sorting method from the standard library, which avoids redundant computations.

[gemini-code-assist]

The current implementation with sort_by re-calculates count_ones() for each element multiple times during the sorting process. For better performance and more idiomatic Rust, you can use sort_by_cached_key. This method computes the key for each element just once, which is more efficient when the key calculation is non-trivial.

        arr.sort_by_cached_key(|&v| (v.count_ones(), v));