994. Rotting Oranges[Medium]

994. Rotting Oranges

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.

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 10
• grid[i][j] is 0, 1, or 2.

Example

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

思路

先遍历一边二维数组，有两个目的

• 统计新鲜橘子数量，到最后如果还有剩余的新鲜橘子，那就说明没有全部腐烂，直接返回-1
• 把已经腐烂的橘子放进队列中，这就是首批腐烂橘子，对应着 res -〉0

接下来就是依次取出队列中腐烂橘子，做广度遍历，只要在边界范围内并且是新鲜橘子的，就将其腐烂，然后也入队列，每一批次队列的橘子取完，那这回合的腐烂就结束了，res++

题解

    Integer freshCount = 0;

    public int orangesRotting(int[][] grid) {
        int res;
        Queue<int[]> queue = new LinkedList<>();
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == 1)
                    freshCount++;
                if (grid[i][j] == 2)
                    queue.add(new int[]{i, j});
            }
        }
        res = bfs(grid, queue);
        return freshCount == 0 ? res : -1;
    }

    public int bfs(int[][] grid, Queue<int[]> queue) {
        int res = 0;
        int[][] dire = new int[][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        while (!queue.isEmpty() && freshCount != 0) {
            int size = queue.size();
            for (int j = 0; j < size; j++) {
                int[] curPos = queue.poll();
                for (int i = 0; i < dire.length; i++) {
                    int curRow = curPos[0] + dire[i][0];
                    int curCol = curPos[1] + dire[i][1];
                    if (curRow < 0 || curRow >= grid.length || curCol < 0 || curCol >= grid[0].length
                            || grid[curRow][curCol] != 1)
                        continue;
                    queue.add(new int[]{curRow, curCol});
                    grid[curRow][curCol] = 2;
                    freshCount--;
                }
            }
            res++;
        }
        return res;
    }