Longest increasing subsequence #28
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nodchip
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Jun 10, 2024
| class Solution: | ||
| def lengthOfLIS(self, nums: List[int]) -> int: | ||
| lis_so_far = [1] * len(nums) # lis: longest increasing subsequence | ||
| for i in range(len(nums)-1): |
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| lis_so_far = [1] * len(nums) # lis: longest increasing subsequence | ||
| for i in range(len(nums)-1): | ||
| for j in range(i+1, len(nums)): | ||
| if nums[i] < nums[j] and lis_so_far[j] < lis_so_far[i] + 1: |
| class Solution: | ||
| def lengthOfLIS(self, nums: List[int]) -> int: | ||
| MAX_NUM = 10 ** 4 | ||
| increasing_subsequence = [MAX_NUM + 1] * len(nums) |
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個人的には長さ 0 から始めて、どこにも挿入できない場合は末尾に追加する、といったロジックのほうが好みです。ですが、好みの問題だと思います。
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私は、10 ** 4 が(問題文の制約からくる)マジックナンバーであることが気になりますね。1でも大きくなったら動かなくなるわけですよねえ。
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確かにそうですね, 訂正いたします。ところで, 逆にマジックナンバーをあえて使う場面などはあるのでしょうか...?
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いや、読み手が分かるように書くので、マジックナンバーになったらコメントを書いてください。
有名なマジックナンバーとしては 0x5f3759df があります。
https://en.wikipedia.org/wiki/Fast_inverse_square_root
| for j in range(i): | ||
| if nums[j] < nums[i]: | ||
| lis_so_far[i] = max(lis_so_far[i], lis_so_far[j] + 1) | ||
| max_lis = max(max_lis, lis_so_far[i]) |
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step2 のように、最後に 1 回 max() 関数を使ったほうがシンプルで読みやすい思います。
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問題
300. Longest Increasing Subsequence