0994. Rotting Oranges (M)

Rotting Oranges (M)

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:

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

Example 2:

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:

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.

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题意

给定一个矩阵包含若干个新鲜橘子和烂橘子。每一分钟每个烂橘子相邻四个橘子都会腐烂，问经过几分钟只剩下烂橘子。

思路

先遍历一遍矩阵，将所有烂橘子的位置加入队列，并统计新鲜橘子的个数。每一分钟的过程相当于将当前队列中位置全部出栈，判断相邻4个位置是否有新鲜橘子，有则将其变为烂橘子并将其位置加入队列以便下一分钟的处理。最后判断还有没有新鲜橘子。

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代码实现

Java

class Solution {
    public int orangesRotting(int[][] grid) {
        int count = 0;
        int minutes = 0;
        int[] xPlus = { 0, 1, 0, -1 };
        int[] yPlus = { 1, 0, -1, 0 };
        int m = grid.length, n = grid[0].length;
        Queue q = new LinkedList<>();

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 2) {
                    q.offer(new Coor(i, j));
                } else if (grid[i][j] == 1) {
                    count++;
                }
            }
        }

        while (!q.isEmpty()) {
            int size = q.size();

            for (int j = 0; j < size; j++) {
                Coor cur = q.poll();
                for (int i = 0; i < 4; i++) {
                    int nextX = cur.x + xPlus[i];
                    int nextY = cur.y + yPlus[i];
                    if (isValid(m, n, nextX, nextY) && grid[nextX][nextY] == 1) {
                        grid[nextX][nextY] = 2;
                        count--;
                        q.offer(new Coor(nextX, nextY));
                    }
                }
            }

            if (!q.isEmpty()) minutes++;
        }

        return count == 0 ? minutes : -1;
    }

    private boolean isValid(int m, int n, int i, int j) {
        return i >= 0 && i < m && j >= 0 && j < n;
    }
}

class Coor {
    int x;
    int y;

    public Coor(int x, int y) {
        this.x = x;
        this.y = y;
    }
}