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| ### ① | ||
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| 解法を覚えていた。lisが何の略かコメントで書くくらいしても良いと思った。 |
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個人的にはほしいかなあと思います(phase3ぐらいでいいと思います)
| if not nums: | ||
| return 0 | ||
| lis_lengths = [1] * len(nums) | ||
| for i in reversed(range(len(nums))): |
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ちょっと細かい感じになっちゃいますが
for i in range(1, len(nums)):
for j in range(i):
if nums[j] >= nums[i]:
continue
lis_lengths[i] = max(lis_lengths[i], lis_lengths[j] + 1)というふうにも変形できますね。こうするとreversedの文の新しくメモリ領域を作る操作と要素を逆転させる処理の文が節約できます。(全体がO(n^2)なので少しのチューニングにしかならないかもですが..)
oda
reviewed
Jul 4, 2024
| lis = [] # longest increasing subsequence | ||
| for num in nums: | ||
| i = bisect_left(lis, num) | ||
| if i == len(lis): |
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私は、
if i < len(lis):
lis[i] = num
continue
lis.append(num)にします。趣味の範囲です。
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https://leetcode.com/problems/longest-increasing-subsequence/description/