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Given an array nums of distinct integers, return all the possible permutations. You can return the answer in any order.
oda
reviewed
Jun 29, 2024
| class Solution: | ||
| def permute(self, nums: List[int]) -> List[List[int]]: | ||
| def helper(index): | ||
| if(index == len(nums)): |
oda
reviewed
Jun 29, 2024
Comment on lines
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| for index in range(i+1, len(next_permute_nums)): | ||
| if next_permute_nums[index] < next_permute_nums[i] or next_permute_nums[index] > bigger_min_value: | ||
| continue | ||
| bigger_min_value = next_permute_nums[index] | ||
| bigger_min_index = index |
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この for 意味的に 上の for の外にいて欲しいですよね。たぶん、それを表しているのが、この後の必ずする return です。
関数にするほか、
for A:
if B:
break
C
else:
return None
Dという方法などがあるでしょう。Discord を関数化かなんかで調べると出てくると思います。
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https://discord.com/channels/1084280443945353267/1221030192609493053/1225674901445283860
これですね。読んだことあるのに今回考慮に出なかったの反省です。
後でその方法で書いてみます。ありがとうございます
oda
reviewed
Jun 29, 2024
| class Solution: | ||
| def permute(self, nums: List[int]) -> List[List[int]]: | ||
| def make_next_permutation(permutation): | ||
| def find_decreasing_index_from_last(permutation): |
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これはここまでは関数化しない方が好みということでしょうか。
インデントが深くなりすぎると見にくいからでしょうか。
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私は、三重にせず、二重にします。
↑これ意味勘違いしてました。
find_decreasing_index_from_lastを外に出す(make_next_permutationと同じ階層にする)ってことですね。
Mike0121
reviewed
Jun 30, 2024
| def permute(self, nums: List[int]) -> List[List[int]]: | ||
| def helper(permutation): | ||
| if len(permutation) == len(nums): | ||
| all_permutations.append(permutation.copy()) |
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確かにその選択肢、言われればわかるもののあんまり頭になかったです。ありがとうございます
Mike0121
reviewed
Jun 30, 2024
| helper(permutation) | ||
| permutation.pop() | ||
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問題文: https://leetcode.com/problems/permutations/description/
Given an array nums of distinct integers, return all the possible permutations. You can return the answer in any order.