63 unique paths ii - #33
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nodchip
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Aug 15, 2026
| def uniquePathsWithObstacles(self, obstacleGrid: list[list[int]]) -> int: | ||
| num_rows = len(obstacleGrid) | ||
| num_columns = len(obstacleGrid[0]) | ||
| num_paths = [[0 for _ in range(num_columns)] for _ in range(num_rows)] |
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[[0] * num_columns for _ in range(num_rows)]のほうがシンプルだと思います。
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一応、このあたりのオブジェクトの理屈が分かっているかは確認しておいてください。
https://docs.google.com/document/d/11HV35ADPo9QxJOpJQ24FcZvtvioli770WWdZZDaLOfg/edit?tab=t.0#heading=h.wrv8idqm5j2b
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お二人ともありがとうございます。確認しておきます。
| def uniquePathsWithObstacles(self, obstacleGrid: list[list[int]]) -> int: | ||
| num_rows = len(obstacleGrid) | ||
| num_columns = len(obstacleGrid[0]) | ||
| num_paths = [[0 for _ in range(num_columns)] for _ in range(num_rows)] |
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空間計算量 O(num_columns) で解く方法もあります。興味があればぜひ考えてみて下さい。
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ありがとうございます。こうやれば良さそうというのが頭にあるのでそれを実装してみます。
MA-yo-TA
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Aug 21, 2026
| アイディア: | ||
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| num_paths は各行を持つ必要はない。 | ||
| 各マスへの行き方は、そのマスの上のマスへの行き方と左のマスへの行き方の合計であるが、ある行を計算し終わったとき、次の行の各マスにおいて、上のマスからの寄与は今保持している行そのもの。左からの分だけ計算すればいい。 |
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考えてみれば当たり前で、 step3 のやり方は
- 最初に 0 を詰める
- 上の行を
+=で足す
なので、そもそもコピーしていただけだったと考えると無駄なデータを保存していた。
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この問題：https://leetcode.com/problems/unique-paths-ii/
次の問題：https://leetcode.com/problems/house-robber/