Solved: 63. Unique Paths II #34
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oda
reviewed
Apr 21, 2025
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| ### 感想 | ||
| - h>0, w>0 に加算を分けるとかなりスッキリ変形できて良い | ||
| - 地味なところだが、 意図がない限りは height -> width の順で操作することが統一されてる方が適切だと思った |
| class Solution: | ||
| def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int: | ||
| OBSTACLE = 1 | ||
| if obstacleGrid[0][0] == OBSTACLE: |
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[0][0]にアクセスする前に一応チェックしてもいいかもしれませんね。問題文に制約があるにせよ。
Owner
Author
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レビューありがとうございます。
空のGridの考慮は完全に抜けてました。
obstacleGrid is None の場合は弾いておくべきですね。
fuga-98
reviewed
Apr 21, 2025
| - https://github.com/sakupan102/arai60-practice/pull/35/files | ||
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| ### 感想 | ||
| - h>0, w>0 に加算を分けるとかなりスッキリ変形できて良い |
fuga-98
reviewed
Apr 21, 2025
| sum_paths[0][0] = 1 | ||
| for row in range(num_rows): | ||
| for col in range(num_cols): | ||
| if obstacleGrid[row][col] == OBSTACLE: |
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十分わかりやすいですが、ループ内が長いのでメソッド化したらより見やすくなりそつです。
Owner
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レビューありがとうございます。
おっしゃる通りですね。
自分はネストを減らす意識を持つべきかもしれないです。
| return 0 | ||
| num_rows = len(obstacleGrid) | ||
| num_cols = len(obstacleGrid[0]) | ||
| sum_paths = [[0] * num_cols for _ in range(num_rows)] |
nodchip
reviewed
Apr 28, 2025
| height = len(obstacleGrid) | ||
| sum_paths = [[0] * width for _ in range(height)] | ||
| sum_paths[0][0] = 1 | ||
| for h in range(1, height): |
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自分は (width, height) に対しては (x, y) という変数を使うことが多いです。チームの平均的な書き方に合わせることをお勧めいたします。
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| width = len(obstacleGrid[0]) | ||
| height = len(obstacleGrid) | ||
| sum_paths = [[0] * width for _ in range(height)] |
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sum_paths を二次元ではなく一次元のリストにしても解くことができると思います。挑戦してみてもよいと思います。
Owner
Author
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レビューありがとうございます。
ある地点の結果が上と左の結果の和であることを利用して、
ある列の結果を、次の列で再利用する形ですね。
class Solution:
def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
OBSTACLE = 1
if obstacleGrid is None:
return -1
if obstacleGrid[0][0] == OBSTACLE:
return 0
height = len(obstacleGrid)
width = len(obstacleGrid[0])
num_paths = [0] * width
num_paths[0] = 1
for y in range(height):
for x in range(width):
if obstacleGrid[y][x] == OBSTACLE:
num_paths[x] = 0
continue
if x > 0:
num_paths[x] += num_paths[x - 1]
return num_paths[width - 1]
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63. Unique Paths II