LeetCode 994. Rotting Oranges

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Words count in article: 279 | Reading time ≈ 1

题目

In a given grid, each cell can have one of three values:

  • the value 0 representing an empty cell;
  • the value 1 representing a fresh orange;
  • the value 2 representing a rotten orange.

Every minute, any fresh orange that is adjacent (4-directionally) 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 instead.

Example 1:

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Input: [[2,1,1],[1,1,0],[0,1,1]]
Output: 4

Example 2:

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Input: [[2,1,1],[0,1,1],[1,0,1]]
Output: -1
Explanation:  The orange in the bottom left corner (row 2, column 0) is never rotten, because rotting only happens 4-directionally.

Example 3:

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Input: [[0,2]]
Output: 0
Explanation:  Since there are already no fresh oranges at minute 0, the answer is just 0.

Note:

  1. 1 <= grid.length <= 10
  2. 1 <= grid[0].length <= 10
  3. grid[i][j] is only 0, 1, or 2.

思路

BFS.

代码

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class Solution(object):
    def orangesRotting(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        row, col = len(grid), len(grid[0])
        rotting = {(i, j) for i in range(row) for j in range(col) if grid[i][j] == 2}
        fresh = {(i, j) for i in range(row) for j in range(col) if grid[i][j] == 1}
        timer = 0
        while fresh:
            if not rotting: 
                return -1
            rotting = {(i+di, j+dj) for i, j in rotting for di, dj in [(0, 1), (1, 0), (0, -1), (-1, 0)] if (i+di, j+dj) in fresh}
            fresh -= rotting
            timer += 1
        return timer