153. Find Minimum in Rotated Sorted Array#40
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hemispherium wants to merge 4 commits into
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liquo-rice
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May 8, 2026
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| ### step1 | |||
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| とりあえずmin_elementでいいんじゃないかと思ったら通ってしまった。調べたところO(log n)の計算量になっていそう。（https://cpprefjp.github.io/reference/algorithm/min_element.html） | |||
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ドキュメントには、以下のように記載がありますが…
計算量
max((last - first) - 1, 0) 回の比較を行う
そもそも任意の配列で最小値をO(logN)で見つけるのは可能ですかね？
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失礼しました。そこに書いてある通りO(n)ですね。ソート済みなどの条件がなければ素朴に考えると全ての要素を見ないといけないと思うのでO(n)は掛かってしまいそうな気がしますね。
liquo-rice
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May 9, 2026
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| ### step1 | |||
5ky7
reviewed
May 29, 2026
| int right = nums.size() - 1; | ||
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| while (left < right) { | ||
| int mid = left + (right -left) / 2; |
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演算子周りの空白の使い方は統一することをお勧めします
Suggested change
| int mid = left + (right -left) / 2; | |
| int mid = left + (right - left) / 2; |
| int left = 0; | ||
| int right = nums.size() - 1; | ||
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| while (left < right) { |
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今回はnumsの中身に重複がないのでこれでも成立しますが，2分探索における区間の意味を確認しておくと良いかもしれません（典型コメント集の二分探索の項に色々まとまってます）
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この問題：https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/description/
次に解く問題：https://leetcode.com/problems/search-in-rotated-sorted-array/description/