Create 0050-powx-n.md #46
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potrue
reviewed
Oct 30, 2025
| result = 1 | ||
| bit = 1 | ||
| power = x | ||
| abs_n = abs(n) |
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absで回さずに、if n < 0: return 1 / myPow(x, -n)を最初の方に持ってきてここにくる段階ではn > 0の時を考えれば良いようにした方が見やすくなる気がしました。
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ありがとうございます。
確かにそうですね。その手があったんですね。
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関数を一段階深くするのもいいんですが、ちょっと重い(といっても数ナノ秒、どちらかというとスタックが深くなるのでデバッグが面倒になるほうが気にしているかもな)ので条件分岐で済むならそちらも一つでしょう。
potrue
reviewed
Oct 30, 2025
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| result = 1 | ||
| bit = 1 | ||
| power = x |
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xを何乗かしたものであるというのを明らかにしておくために、x_poweredみたいな名前にしてもいいと思いました。
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この問題：https://leetcode.com/problems/powx-n/
次の問題：https://leetcode.com/problems/k-th-symbol-in-grammar/