33. Search in Rotated Sorted Array #41
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oda
reviewed
Feb 26, 2025
| #### 2c | ||
| - 2bと同じロジックを構造体を使って実装 | ||
| - numsの尻より大きいか -> targetとの大小比較 の順に調べる | ||
| - https://github.com/Yoshiki-Iwasa/Arai60/pull/36/files#r1712955053 |
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見直したら cmp いらなくて
def priority(x):
return (x <= nums[-1], target <= x)これでよかったです。ただ、Go で書くと少し面倒ですね。
|
良いと思いました。 |
TORUS0818
reviewed
Mar 19, 2025
| tail := nums[len(nums)-1] | ||
| isTargetLargerThanTail := target > tail | ||
| for left < right { | ||
| middle := left + (right-left)/2 |
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ここで
if nums[middle] == target {
return middle
}のようにするのはどうでしょうか。
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ありがとうございます。こちらでの指摘も含めて違う書き方ができますね。今回の問題ではnumsに重複要素が含まれないことが保証されていますが、そうでない場合はtargetに一致する要素のうちランダムなものが返ることに注意ですね。
| } else { | ||
| right = middle | ||
| } | ||
| continue |
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ここは好みかと思うのですが、私はif-elseにした方が好きです。
ちなみに上でアーリーリターンしておけばright = middle - 1とできると思います。
oda
reviewed
Mar 27, 2025
| としてnumsを[false,...,false,true,...,true]のような配列とみなした時、 | ||
| 一番左のtrueの位置を探す問題と捉える | ||
| - nums=[6,0,2,4], target=2 -> [false,false,true,true] | ||
| - nums=[6,0,2,4], target=1 -> [false,false,false,false] |
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おっしゃる通りですね。
T/F を target 以上の nums の要素のうち最小のものとそれより右のもの、と定義した方がいいですかね。
nums=[6,0,2,4], target=2 -> [false, false, true, true]
nums=[6,0,2,4], target=1 -> [false, false, true, true]
nums=[6,0,2,4], target=5 -> [true, true, true, true]
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https://leetcode.com/problems/search-in-rotated-sorted-array/description/