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良さそうです！ |
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| // sum(nums[0:j]) - k = sum(nums[0:i]) となる i (0 <= i < j) が x 個存在すれば | ||
| // k と等しい部分配列 sum(nums[i:j]) も x 個存在する | ||
| count += cumulativeSumFrequency.getOrDefault(cumulativeSum - k, 0); |
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感想: cumulativeSumCountでもいいのかなと。countに向けて加算してますし
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確かに、全然 ~Count でも意味通りますね
nittoco
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Aug 8, 2024
| - 累積和の出現回数を格納する map はそのまま cumulativeSumFrequency, 累積和を格納する整数値は cumulativeSum とした | ||
| - 最近 Go のコードをよく読んでいたので冗長に感じる気持ちがないでもないが、まあ Java なのでこれにした | ||
| - あとはなるべく素直に実装したつもり | ||
| - 61-63 行は `count += cumulativeSumFrequency.getOrDefault(cumulativeSum - k, 0);` とも書けるが、これだとこの関数の処理の詳細に加えて getOrDefault() の挙動の両方をちゃんと把握していないと内容の理解が出来ないので、若干認知負荷が高い気がしたので今回はこのようにしてみた |
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ありがとうございます！結局次のステップで getOrDefault() を使い一行で書くように修正したのですが、認知負荷高まらないとのことで安心しました。
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みました！良さそうです！ |
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拝見しました。 |
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https://leetcode.com/problems/subarray-sum-equals-k/description/