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mamo3gr
reviewed
Mar 19, 2026
| - 長さが2の場合 | ||
| - 0番目の金額と1番目の金額の大きい方 | ||
| - 長さが3の場合 | ||
| - 0番目の金額と3番目の金額の和と1番目の金額の大きい方 |
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Suggested change
| - 0番目の金額と3番目の金額の和と1番目の金額の大きい方 | |
| - 0番目の金額と2番目の金額の和と1番目の金額の大きい方 |
mamo3gr
reviewed
Mar 19, 2026
| if len(nums) == 1: | ||
| return nums[0] | ||
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| max_robbed_amounts = [0] * len(nums) |
mamo3gr
reviewed
Mar 19, 2026
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| ## Step2 | ||
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| 配列を利用する方が読みやすいと思った。 |
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この感覚の言語化はいいと思いました。ちなみに私もこれに同意です（漸化式に対応してすんなり入ってくる）。メモリ使用量が許すならこちらを書きたいです。
mamo3gr
reviewed
Mar 19, 2026
| max_robbed_two_houses_ago = 0 # 2つ前の家までで得られた最大金額 | ||
| max_robbed_one_house_ago = 0 # 1つ前の家までで得られた最大金額 | ||
| for n in nums: | ||
| amount = max(max_robbed_one_house_ago, max_robbed_two_houses_ago + n) |
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個人的には robbed_last, skipped_last が好みです。直前をスキップしている（盗んでいない）のだから、n が盗めるでしょう、という理屈が分かりやすいからです。
Suggested change
| amount = max(max_robbed_one_house_ago, max_robbed_two_houses_ago + n) | |
| amount = max(robbed_last, skipped_last + n) |
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ありがとうございます。
何の値かと、処理の文脈がわかりやすいと思いました。
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今回の問題
House Robber - LeetCode
使用言語
Python