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tokuhirat
reviewed
Oct 5, 2025
| # @param {Integer[]} nums | ||
| # @return {Integer[][]} | ||
| def subsets(nums) | ||
| subsets = [] |
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関数名と同じ変数名は避けたいと思いました。冪集合なので power_set もしくは単に results はいかがでしょうか。
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コメントありがとうございます。power_setが良いですね。
potrue
reviewed
Oct 5, 2025
| end | ||
| subsets_helper.call(0, []) | ||
| subsets | ||
| end |
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読みやすいです。
bit全探索という考え方があり、それを使っても解けるので見てみると良いと思います。
整数をビットの列として解釈して、i番目のビットが1だったらi番目の要素が含まれる、という風に解釈する感じです。
def subsets(nums)
subsets = []
n = nums.size
# 0 から 2^n - 1 までの全パターンを試す
(0...(1 << n)).each do |bit|
subset = []
n.times do |i|
# i番目のビットが立っていれば、nums[i]を含める
subset << nums[i] if (bit >> i) & 1 == 1
end
subsets << subset
end
subsets
end
potrue
reviewed
Oct 5, 2025
| # @param {Integer[]} nums | ||
| # @return {Integer[][]} | ||
| def subsets(nums) | ||
| nums.inject([[]]) { |subsets, num| subsets.concat(subsets.map { |subset| subset + [num] }) } |
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reduceっぽい関数を使ってこのように書く書き方は思いついていませんでした。勉強になります 👀
コンパクトでいいですね。
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解いた問題
78. Subsets
使用言語
Ruby
次に解く問題
https://leetcode.com/problems/combination-sum/description/