Create 33. Search in Rotated Sorted Array.md#43
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hroc135
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May 13, 2025
| break_point = search_break() | ||
| if target <= nums[-1]: | ||
| result = bisect_left(nums, target, lo=break_point, hi=len(nums)) | ||
| if result >= len(nums) or nums[result] != target: |
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result == len(nums) の場合、target > nums[-1] なので外側の if で弾かれており、不要な条件になると思います。
hroc135
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May 13, 2025
| if result >= len(nums) or nums[result] != target: | ||
| return -1 | ||
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| これの最初の条件いらなそう |
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compare関数の戻り値の最初の要素はnumsの末尾が必ず1になることを考慮すると、result == len(nums) になる場合は、compare(nums[i]) の結果が全て (0,0), (1,0) のいずれかになって、compare(target) の結果が (1,1) になる場合。でも、target <= nums[-1] の場合、nums[i] >= target となる i が必ず存在するので、result == len(nums) にはなり得ない、ということですね。
いらない理由を考えるのに結構頭を使ったのでコメントに書いた方が良さそうと思いました。
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レビューありがとうございます。
正確に書いていただいて非常に助かります。
３回書く中でコメントを書くと大変なので省略してしまっていますが、実務なら書いたほうが良いですね。
oda
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May 13, 2025
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| [TF,TF, TF] t= FTとなるのでresult=0となります | ||
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| key関数の命名がうまいものが思いつかないが、実務ならdocstringで説明を書くで十分な気がしている。 |
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https://leetcode.com/problems/search-in-rotated-sorted-array/