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nodchip
reviewed
Dec 11, 2025
| class Solution: | ||
| def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]: | ||
| set1 = set(nums1) | ||
| set2= set(nums2) |
| if nums2[j] < nums1[i]: | ||
| j += 1 | ||
| continue | ||
| common = nums1[i] |
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132 行目に空行を入れているのであれば、ここにも空行を入れるのが自然だと思いました。
t9a-dev
reviewed
Dec 11, 2025
Comment on lines
+151
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+154
| nums1_set = set(nums1) | ||
| nums2_set = set(nums2) | ||
| # nums1_set.intersection(nums2_set)と同じ | ||
| return list(nums1_set & nums2_set) |
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nums*_set 変数に代入する理由があまりなさそうなので、そのままsetにしても良いのかなと思いました。& を利用するその場でsetにしていることが分かるというメリットも得られそうです。
Suggested change
| nums1_set = set(nums1) | |
| nums2_set = set(nums2) | |
| # nums1_set.intersection(nums2_set)と同じ | |
| return list(nums1_set & nums2_set) | |
| # return list(set(nums1).intersection(set(nums2))) | |
| return list(set(nums1) & (set(nums2))) |
Ryotaro25
reviewed
Dec 12, 2025
| seen.add(num) | ||
| intersection = set() | ||
| for num in nums2: | ||
| if num in seen and num not in intersection: |
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num not in intersectionの部分は不要かと思います。
Ryotaro25
reviewed
Dec 12, 2025
| nums2.sort() | ||
| i, j = 0, 0 | ||
| intersection = set() | ||
| last = None |
Ryotaro25
reviewed
Dec 12, 2025
| j += 1 | ||
| continue | ||
|
|
||
| common = nums1[i] |
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解く問題
Intersection Of Two Arrays
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