Document algorithm for finding minimum in rotated array#14
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tokuhirat
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Sep 17, 2025
| #### step2 | ||
| https://github.com/KentaroJay/Leetcode/pull/18/files | ||
| https://github.com/Satorien/LeetCode/pull/42/files | ||
| - 半開区間じゃないコードを読み慣れていない。特に停止するかどうかが読み切れない。 |
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https://discord.com/channels/1084280443945353267/1196498607977799853/1269532028819476562
リンク先の考え方で整理すると色々なパターンでも対応できるようになると思います。
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Kaichi-Irie/leetcode-python#15 (comment)
この辺りも参考になると思います。
tokuhirat
reviewed
Sep 17, 2025
| - また、変数targetが必要なくなる | ||
| - この場合左が開いている半開区間で考えているから、最初から単調増加の場合も考えるならばleft = -1にする必要がある。 | ||
| - left_index -> left, right_index -> rightにしてもいいかもしれない。上から下に読むことを考えると、この変数はindexとしての振る舞いをしますというのが上の段階でわかるのは嬉しいかもしれないが、２行後にはその使い方をされているので。 | ||
| - indexといいつつ、left_index = -1とかくと違和感があるし、python の場合は最終項とも読めるので、ないほうがいいか。 |
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この違和感を解消するのであれば、左が閉じている区間で考えれば良いと思います。
tokuhirat
reviewed
Sep 17, 2025
| ```python | ||
| class Solution: | ||
| def findMin(self, nums: List[int]) -> int: | ||
| left, right = -1, len(nums) - 1 |
potrue
reviewed
Sep 18, 2025
| right_index = len(nums) - 1 | ||
| while right_index - left_index > 1: | ||
| mid_index = (left_index + right_index) // 2 | ||
| if nums[mid_index] > target: |
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>=のほうが自然だと思います。（この条件に関してTrueになる要素とFalseになる要素が並んでいる配列だとnumsを捉えることができるので）
実際には、mid_indexが0になる頃にはwhileが回っていないので動作はすると思うのですが。
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targetをnums[-1]にして、nums[mid_index] > targetとするとif nums[0] < nums[-1]の場合分けがいらなくなるかもしれません。
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This problem:
https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/
Next problem:
https://leetcode.com/problems/move-zeroes/