122. Best Time To Sell and Buy Stock Ⅱ #44
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https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/description/ You are given an integer array prices where prices[i] is the price of a given stock on the ith day. On each day, you may decide to buy and/or sell the stock. You can only hold at most one share of the stock at any time. However, you can buy it then immediately sell it on the same day. Find and return the maximum profit you can achieve.
| is_having = False | ||
| prev_price = None | ||
| for price in prices: | ||
| if prev_price is None: |
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if prev_price is None:を使うのはループの1回目だけだと思いますので、prev_priceをprice[0]で初期化するのもアリかと思いました。prices空の場合の制御は必要になりますがループ内のコードは短くなると思います。
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ループの方をprice[1]からスタートする感じでしょうか。
確かにその方がシンプルですね
| return profit | ||
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| - テストケースを考える(実は今まであんまりやってなかった😅) |
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| - 関数を定める方法、DPでやる方法などがある | ||
| - 関数を定める方法で書いてみる | ||
| - find_next_peakとfind_next_bottomはよく似て関数だが、ここを無理やり共通化しようとするのは流石にややこしい？ |
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C++ には adjacent_find というのがありますね。
共通化しなくてもいいように思いますが。
| - [6, 5, 4, 5, 6, 5, 4, 5, 6,,,,,] | ||
| - [1, 2, 3, 4, 5, 6,,,,] | ||
| - [100, 99, 98, 97, ,,,,,,] | ||
| - くらいかな。うーん。 |
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自分は以前、理事の川中さんが「テストケースはシステマティックに生成すると良い」とおっしゃっていたのを参考にテストケースを考えています
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ありがとうございます。
そうおっしゃってましたよね〜、あとは全ての行を通るケース、空とかですかねえ
| - peakとbottomを交互にループするのは、最初にbottomをやった方が綺麗そう？気づかなかった。 | ||
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| ```python | ||
| python |
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そうですね、ありがとうございます(よくあるのでlinterとか導入したほうがいい気もしてきました)
| for i in range(len(prices) - 1): | ||
| diff = prices[i + 1] - prices[i] | ||
| if diff > 0: | ||
| profit += diff |
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ifを消して
profit += max(0, diff) でもよさそうです。
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確かに。ありがとうございます。
簡潔な一方、やや読み手に推理が必要な気もしますが、これぐらいなら許容範囲内ですかね
https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/description/
You are given an integer array prices where prices[i] is the price of a given stock on the ith day.
On each day, you may decide to buy and/or sell the stock. You can only hold at most one share of the stock at any time. However, you can buy it then immediately sell it on the same day.
Find and return the maximum profit you can achieve.