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potrue
reviewed
Feb 19, 2026
| for (int r = 0; r < num_row; r++) { | ||
| for (int c = 0; c < num_col; c++) { | ||
| if (obstacleGrid[r][c] == kObstacle) { | ||
| unique_paths[r][c] = 0; |
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動きますね。自分は入れておいた方がわかりやすいかなと思って入れました。
oda
reviewed
Mar 14, 2026
| 例外処理。そもそもベクトルが blank の場合への対処と、ある行すべてに障害物があった場合に打ち切り処理を入れる方法。 | ||
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| - [参照](https://github.com/irohafternoon/LeetCode/pull/37/files)。 | ||
| - 後者のケースってあんまり思いつかない。この処理を入れることによる計算の効率化の幅と、可読性の低下幅と、どっちが大きいだろう？都度判定を入れることになるので計算の効率化もそこまでされないのではと思う。 |
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「打ち切り処理を入れるか」といったことは結局は、そのプログラムがどのようなものを扱うかの分布との兼ね合いです。
たとえば、ライブラリーのソート関数は、意外とすでにソート済みが入力に来ることが多い、と判断されて、すでにソート済みの場合は何もしない処理が入っていたりすることがあります。しかし、これは分布を試して確認しないことにはなんとも言えない話です。
なので、とりあえず、シンプルに作って、必要性との兼ね合いで後から入れることが割と好まれるのですね。
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問題
https://leetcode.com/problems/unique-paths-ii/
次に解く問題
198. House Robber