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5ky7
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Mar 4, 2026
5ky7
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全体的に読みやすかったです！
動的計画でやる場合「一方通行の更新は1次元配列のみでできる」ということを見落としていたので良い気づきになりました．
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| final rows = obstacleGrid.length; | ||
| final cols = obstacleGrid[0].length; |
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今回のコードの長さであれば問題ないと思いますが，個人的には中身を明確に表すnum_rows, num_colsの方が好みです．
mamo3gr
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Mar 5, 2026
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| for (var row = 0; row < rows; row++) { | ||
| for (var col = 0; col < cols; col++) { |
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行数 rows と、インデックスである row が字面の上で似ているので、インデックスを r にするとか、行数を num_rows にするなどして区別しやすいようにすると良いと思いました。
mamo3gr
reviewed
Mar 5, 2026
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| for (var row = 0; row < rows; row++) { | ||
| for (var col = 0; col < cols; col++) { | ||
| if (obstacleGrid[row][col] == obstacle || (row == 0 && col == 0)) { |
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結果としては同じ (continue) になるものの、状況が違うので、if文を分けたほうが分かりやすいと感じました。
mamo3gr
reviewed
Mar 5, 2026
| uniquePathsCount[0][0] = 1; | ||
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| int uniquePaths(int row, int col) { | ||
| if (row < 0 || row >= rows || col < 0 || col >= cols) { |
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Dartはchained comparisonを採用していなさそうですね(cf)．
not (0 <= row && row < rows)のように「『正しいケース』ではない」という自然言語の説明に近いコードの方が意図がわかりやすいという点は同感です．
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相変わらず読みやすかったです！ |
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解いた問題：https://leetcode.com/problems/unique-paths-ii/description/
次に解く問題：https://leetcode.com/problems/house-robber/description/