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oda
reviewed
Sep 27, 2025
| 後から文字列のスライスのコストを考慮してなかったことに気づいた。 | ||
| 文字列のスライスコストを入れるとN * W * (L + N)なのでL * N^2になる。そうなるとRubyでギリギリかもしれないが、LeetCode上では想定以上に速い気がする。 | ||
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| Rubyのスライスメソッドを見たところ、ASCII文字列のスライスはO(1)であるが、マルチバイト文字列のスライスはO(N)らしい。 |
tokuhirat
reviewed
Sep 28, 2025
| # @return {Boolean} | ||
| def word_break(str, word_dict) | ||
| unreachable_suffixes = Set.new | ||
| is_reachable = ->(str){ |
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Ruby でどの程度スタイルガイドが参照されているのかわかっていませんが、スタイルガイドでは複数行の場合は lambda を使うのが良いとされていました。
https://github.com/rubocop/ruby-style-guide?tab=readme-ov-file#lambda-multi-line
tokuhirat
reviewed
Sep 28, 2025
| word_hashes.add(hash) | ||
| end | ||
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| reachable = Array.bnew(str_size, false) |
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おっしゃる通りです。
コピーする時に誤って入ったのかもしれないです。
tokuhirat
reviewed
Sep 28, 2025
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| hash = 0 | ||
| (i...str_size).each do |j| | ||
| hash = (hash * base + byte_of.call(str.getbyte(j))) % mod |
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この処理はL53にもあるため、関数にしたら良さそうと思いました。
next_rolling_hash = lambda do |hash, c, base, mod|
(hash * base + byte_of.call(c)) % mod
endc の型は文字列でも良いかもしれません。
https://discord.com/channels/1084280443945353267/1201211204547383386/1235165575592939521
| false | ||
| } | ||
| is_reachable.call(str) | ||
| end |
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自分は再帰で練習しませんでしたが、変数名含めて読みやすかったです。
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解いた問題
139. Word Break
使用言語
Ruby
次に解く問題
322. Coin Change