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oda
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Mar 14, 2026
| res = set() | ||
| for num in nums2: | ||
| if num_to_existance_in_nums1[num]: | ||
| res.add(num) |
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num_to_existance_in_nums1[num] = False
すれば res はいきなり list にできますね。
oda
reviewed
Mar 14, 2026
| class Solution: | ||
| def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]: | ||
| if len(nums1) > len(nums2): | ||
| return self.intersection(nums2, nums1) |
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私は、これは
nums1, nums2 = nums2, nums1
くらいにしますねえ。
https://discord.com/channels/1084280443945353267/1391962276122067044/1457227504270970952
(個人的な感覚として、スワップだけで再帰スタックを掘るのは少し重いです。)
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https://leetcode.com/problems/intersection-of-two-arrays/description/