994. Rotting Oranges #49
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nodchip
reviewed
Jun 1, 2025
| def get_initial_rotten_oranges_num_and_coordinates(self, grid: List[List[int]]) -> Tuple[int, Set[Tuple[int, int]]]: | ||
| num_rotten = 0 | ||
| coordinates_rotten = set() | ||
| for y in range(len(grid)): |
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len(grid) と len(grid[0]) が複数回登場するため、それぞれ num_rows, num_columns に代入してあげると読みやすくなると思いました。
| return num_oranges | ||
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| def is_fresh(self, x: int, y: int, grid: List[List[int]], coordinates_rotten: Set[Tuple[int, int]]) -> bool: | ||
| if not 0 <= x < len(grid[0]) or not 0 <= y < len(grid): |
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if not (0 <= x < len(grid[0]) and 0 <= y < len(grid)): のほうがややシンプルだと思いました。
| FRESH = 1 | ||
| ROTTEN = 2 | ||
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| def get_initial_rotten_oranges_num_and_coordinates(self, grid: List[List[int]]) -> Tuple[int, Set[Tuple[int, int]]]: |
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protected method のため、先頭に _ を付けてあげるとよいと思いました。
| if not previously_rotten: | ||
| return -1 | ||
| for x, y in previously_rotten: | ||
| if self.is_fresh(x + 1, y, grid, coordinates_rotten): |
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deltas = [(0, 1), (0, -1), (1, 0), (-1, 0)]
...
for delta in deltas:
if (self.is_fresh(x + delta[0], y + delta[1], grid, coordinates_rotten)):といった感じにすると、コードの重複が減ると思いました。
oda
reviewed
Jun 1, 2025
| if any(cell == self.FRESH for row in rotting_grid for cell in row): | ||
| return -1 | ||
| return minites_plus_one - 1 | ||
| ``` |
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時間を進めるタイミングが難しいですね。素直に問題文の通りに表現すると
すべてのオレンジが腐っているかを確認して、その場合は return
時間を進めて、隣のオレンジを腐らせる。
状況が変わっていなかったら return -1
という無限ループでしょうか。
Owner
Author
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ありがとうございます。確かにこれが自然ですね
# 略
def orangesRotting(self, grid: List[List[Orange]]) -> int:
minutes = 0
rotten_oranges_so_far = set([(x, y) for y in range(len(grid)) for x in range(len(grid[0])) if grid[y][x] == Orange.ROTTEN])
num_oranges_all = sum(grid[y][x] != Orange.EMPTY for y in range(len(grid)) for x in range(len(grid[0])))
while True:
if len(rotten_oranges_so_far) == num_oranges_all:
return minutes
new_rotten_oranges = self._rot_next_oranges(grid, rotten_oranges_so_far)
minutes += 1
if len(new_rotten_oranges) == 0:
return -1
rotten_oranges_so_far |= new_rotten_orangesこれもアリかなと思ってました
増やせなかったときに、なのに全部腐らせることができなかったら-1
# 略
def orangesRotting(self, grid: List[List[Orange]]) -> int:
minutes = 0
rotten_oranges_so_far = set([(x, y) for y in range(len(grid)) for x in range(len(grid[0])) if grid[y][x] == Orange.ROTTEN])
num_oranges_all = sum(grid[y][x] != Orange.EMPTY for y in range(len(grid)) for x in range(len(grid[0])))
while True:
new_rotten_oranges = self._rot_next_oranges(grid, rotten_oranges_so_far)
if len(new_rotten_oranges) == 0:
if len(rotten_oranges_so_far) < num_oranges_all:
return -1
return minutes
rotten_oranges_so_far |= new_rotten_oranges
minutes += 1
hayashi-ay
reviewed
Jun 2, 2025
| - 意図は明確になるように感じるけど、長くなりがち？ | ||
| - minites_plus_oneは流石にわかりにくいかも。 | ||
| - while oranges_rottenのすぐ後でminites_plus_one += 1したいなあと思ってたらこうなってしまった。 | ||
| - あとで気づいたが、xとyが逆でした |
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個人的にはx,yで命名すると結構混乱する気がするんですよね。row, colとかのほうが個人的にはわかりやすいです。
Owner
Author
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ありがとうございます。rowとcolも自分は結構混乱するんですよね、、、(が、これは自分が覚えた方が良い気がしてきました)
| minites_plus_one = 0 | ||
| if all(cell == self.EMPTY for row in grid for cell in row): | ||
| return 0 | ||
| oranges_rotten = [(x, y) for y in range(len(grid[0])) for x in range(len(grid)) if grid[x][y] == self.ROTTEN] |
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URL: https://leetcode.com/problems/rotting-oranges/description/
You are given an m x n grid where each cell can have one of three values:
0 representing an empty cell,
1 representing a fresh orange, or
2 representing a rotten orange.
Every minute, any fresh orange that is 4-directionally adjacent to a rotten orange becomes rotten.
Return the minimum number of minutes that must elapse until no cell has a fresh orange. If this is impossible, return -1.
Example 1:
Input: grid = [[2,1,1],[1,1,0],[0,1,1]]
Output: 4
Example 2: