Create best-time-to-buy-and-sell-stock-ii.md#22
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naoto-iwase
reviewed
Jan 4, 2026
| return total_profit | ||
| ``` | ||
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| いや、貪欲でできる理由がよくわかっていなかった。最適解を $\sum_{k=1}^{n-1} \max(0, p_k - p_{k-1})$ と分解できることが本質だった。 |
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なので、ifによる条件分岐を外し、単にtotal_profit += max(0, prices[today] - prices[today - 1])としてもいいですね。
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| next_profit_without_stock = max( | ||
| profit_without_stock, # そのまま買わない | ||
| profit_with_stock + prices[today] # 売る | ||
| ) | ||
| next_profit_with_stock = max( | ||
| profit_with_stock, # 売らずに持ち続ける | ||
| profit_without_stock - prices[today] # 新たに買う | ||
| ) | ||
| profit_without_stock, profit_with_stock = ( | ||
| next_profit_without_stock, next_profit_with_stock | ||
| ) |
| ``` | ||
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| ## step3 | ||
| 最初の step1 がもっとも素直だと思ったので、これを採用する |
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売買に回数制限がある場合などはDPの方法が有用になったりします。
Arai 60には含まれていないですが、よければ以下もどうぞ。
https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/
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ありがとうございます、いったん完走したらやってみたいです！
| いや、貪欲でできる理由がよくわかっていなかった。最適解を $\sum_{k=1}^{n-1} \max(0, p_k - p_{k-1})$ と分解できることが本質だった。 | ||
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| ## step2 | ||
| ほかの回答は長くてなかなか読む気にならない… |
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こういう意見もあります。個人的にも同意できる考え方です。
ついでなのでここに書いておくが、一般的にコードは書くよりも読むほうがずっと大切で重要である。読んで理解するためには書くよりもどうしてもある面では高い能力が必要になる。自分で書くときにはどう書くかは制限できるが、他人がどう書くかは読んで理解するまでは分からないからである。
https://nuc.hatenadiary.org/entry/2025/11/29#二分探索を読めるか
読む方が大事
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ありがとうございます、少しずつ読めるようになりたいです
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関数型っぽく書いてみました。
import itertools
class Solution1:
def maxProfit(self, prices: list[int]) -> int:
return sum(
price_today - price_yesterday
for price_yesterday, price_today
in itertools.pairwise(prices)
if price_today > price_yesterday
)
class Solution2:
def maxProfit(self, prices: List[int]) -> int:
return sum(map(lambda p: max(0, p[1] - p[0]), itertools.pairwise(prices)))
tokuhirat
reviewed
Jan 5, 2026
| if prices[today] > prices[today - 1]: | ||
| total_profit += prices[today] - prices[today - 1] | ||
| return total_profit | ||
| ``` |
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https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/