152. Maximum Product Subarray #73
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oda
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Jul 29, 2025
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| ## 他の解法 | ||
| 負の数が偶数個あると最大になりうる特性を活かして途中でswapすることで | ||
| 自分のstep1で行った3パターンから2パターンへ比較回数を減らすことができる(step2_2で実装) |
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step 2 が私は愚直で気に入りました。このあたりで解いている人たちがいます。
https://discord.com/channels/1084280443945353267/1196498607977799853/1358736384604766238
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ありがとうございます。
小田さんの解法はstep2_2に近いので理解できました。
野田さんの解法は他のコメント、解説を読んでギリギリ理解できたような感覚です。きちんと言語化できないですが負の数が奇数になる場合を両サイドから見ていくことで避けていると理解しました。
| int max_product_so_far = nums[0]; | ||
| int min_product_so_far = nums[0]; | ||
| for (int i = 1; i < nums.size(); i++) { | ||
| int temp_max_so_far = max(nums[i], max(max_product_so_far * nums[i], min_product_so_far * nums[i])); |
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Python のようにしたければ
tie(a, b) = make_pair(b, a);というのが一応ありますが、素直に両方の変数を new_max, new_min とでも一度置いてから更新のほうが読みやすいでしょう。
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| return max_product; | ||
| } | ||
| }; No newline at end of file |
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discord内で見つけた他の方のコメントです。
https://discord.com/channels/1084280443945353267/1196498607977799853/1240265127207632896
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問題へのリンク
https://leetcode.com/problems/maximum-product-subarray/description/
問題文(プレミアムの場合)
備考
次に解く問題の予告
15. 3Sum
フォルダ構成
LeetCodeの問題ごとにフォルダを作成します。
フォルダ内は、step1.cpp、step2.cpp、step2_2.cpp、step3.cppとmemo.mdとなります。
memo.md内に各ステップで感じたことを追記します。