update: 添加问题“1458.两个子序列的最大点积”的代码和题解 #1311
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1458: AC.cpp (#1310) + word(en) + fix(pic) cpp - AC,84.23%,29.53% py - AC,97.99%,39.85% go - AC,76.92%,38.46% java - AC,65.24%,32.38% rust - AC,100.00%,60.00% Signed-off-by: LetMeFly666 <814114971@qq.com>
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LGTM
昨晚忘记测试api-hook-listener了
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waiting for copilot and test for hook |
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Pull request overview
This PR adds a solution for LeetCode problem 1458 "两个子序列的最大点积" (Max Dot Product of Two Subsequences) in five programming languages (C++, Python, Java, Go, and Rust). The solution uses dynamic programming to find the maximum dot product between two subsequences.
Key changes:
- Implemented the max dot product solution in C++, Python, Java, Go, and Rust
- Updated README.md with the new problem entry and links to solutions
- Added three new English vocabulary words to the learning notes
Reviewed changes
Copilot reviewed 12 out of 12 changed files in this pull request and generated 3 comments.
Show a summary per file
| File | Description |
|---|---|
| toSay.md | Removed outdated announcement about Cloudflare proxy changes |
| Solutions/Other-English-LearningNotes-SomeWords.md | Added three new English vocabulary words (defy, pasture, tread) |
| README.md | Added entry for problem 1458 with links to solutions on multiple platforms |
| Codes/lib.rs | Updated include statement to reference the new Rust solution file |
| Codes/1458-max-dot-product-of-two-subsequences.rs | New Rust implementation using DP with O(n*m) time complexity |
| Codes/1458-max-dot-product-of-two-subsequences.py | New Python implementation using DP with enumerate |
| Codes/1458-max-dot-product-of-two-subsequences.java | New Java implementation using DP with Arrays.fill for initialization |
| Codes/1458-max-dot-product-of-two-subsequences.go | New Go implementation with custom max helper functions |
| Codes/1458-max-dot-product-of-two-subsequences.cpp | New C++ implementation using space-optimized DP approach |
| .commitmsg | Commit message with performance metrics for all language implementations |
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| } | ||
| for (int i = 1; i <= n; i++) { | ||
| for (int j = 1; j <= m; j++) { | ||
| dp[i][j] = Math.max(nums1[i-1] * nums2[j-1], Math.max(dp[i-1][j], Math.max(dp[i][j - 1], dp[i-1][j-1] + nums1[i-1] * nums2[j-1]))); |
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Inconsistent spacing around the subtraction operator. The expression uses dp[i][j - 1] with spaces but nums1[i-1] and nums2[j-1] without spaces. For better consistency and readability, consider using uniform spacing: either spaces around all minus operators or no spaces for all.
| dp[i][j] = Math.max(nums1[i-1] * nums2[j-1], Math.max(dp[i-1][j], Math.max(dp[i][j - 1], dp[i-1][j-1] + nums1[i-1] * nums2[j-1]))); | |
| dp[i][j] = Math.max(nums1[i - 1] * nums2[j - 1], Math.max(dp[i - 1][j], Math.max(dp[i][j - 1], dp[i - 1][j - 1] + nums1[i - 1] * nums2[j - 1]))); |
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不如把dp[i][j - 1]换成·、dp[i][j-1]
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Inconsistent spacing around the subtraction operator. The expression uses dp[i][j - 1] with spaces but nums1[i-1] and nums2[j-1] without spaces. For better consistency and readability, consider using uniform spacing: either spaces around all minus operators or no spaces for all.减法算符周围的间距不一致。表达式使用 dp[i][j - 1]，带有空格，但 nums1[i-1] 和 nums2[j-1]，不含空格。为了提高一致性和可读性，建议使用均匀间距：要么在所有负算符周围空格，要么全都不加空格。
@copilot 把这次java的dp[i][j - 1]换成dp[i][j-1]，pr中不要包含有多余文件
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@LetMeFly666 I've opened a new pull request, #1312, to work on those changes. Once the pull request is ready, I'll request review from you. |
bye! 先合了哈哈 下次merge到新火车上 |
…rd(en) + fix(pic) 420c93d
By newSolution.py using GH on MacOS | close: #1310