Create 122. Best Time to Buy and Sell Stock II.md#38
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oda
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Apr 22, 2025
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| if i == num_prices - 1: | ||
| break |
hroc135
reviewed
Apr 23, 2025
| if i == num_prices - 1: | ||
| break | ||
| profit -= prices[i] | ||
| # 頂点を見つける |
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こういうコメントのおかげもあって、一つの（外側の）ループで1連の株売買をしていることがすぐ読み取れますし、手でやるならこうなると思います。
一方、実務で書くなら前日からの値上がり分を累積していく方法の方がシンプルに書けるので、私はそちらを選択すると思いました。
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レビューありがとうございます！
実務なら私も累積和を選択すると思います。
| class Solution: | ||
| def maxProfit(self, prices: List[int]) -> int: | ||
| assert isinstance(prices, list) | ||
| if len(prices) <= 1: |
| result = 0 | ||
| holding = prices[0] | ||
| for i in range(1, len(prices) - 1): | ||
| print(holding) |
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print(f'i: {i}, holding: {holding}')
みたいにログを出すとデバッグに有用かもしれません。多くの情報をログに出すくらいならデバッガーを使った方がいいかもしれませんが。
hroc135
reviewed
Apr 23, 2025
| holding = prices[0] | ||
| for i in range(1, len(prices) - 1): | ||
| print(holding) | ||
| diff_next = prices[i+1] - prices[i] |
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ループ定義でiが1から始まっているので、iとi-1を比べるんだろうなと推測したのでここであれ？となりました。
動かないコードも貼っていただけると読み手にとっていい練習になるのでありがたいです！
nittoco
reviewed
Apr 25, 2025
Comment on lines
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| # 1. 谷 (Valley) を見つける | ||
| while i < num_prices - 1 and prices[i+1] <= prices[i]: | ||
| i += 1 | ||
| bottom = prices[i] | ||
| # 2. 山 (Peak) を見つける | ||
| while i < n - 1 and prices[i+1] >= prices[i]: | ||
| i += 1 | ||
| peak = prices[i] | ||
| total_profit += peak - bottom |
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この辺は個人的には関数化しますかね。
L96のnはどこから来ました？(num_pricesの間違い?)
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レビューありがとうございます。
nはタイポです。
確かに関数化するか迷いました。
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https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/description/