994. Rotting Oranges

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.

class Solution {
    public int orangesRotting(int[][] grid) {
        if(grid == null || grid.length == 0) return 0;
        int rows = grid.length;
        int cols = grid[0].length;
        Queue<int[]> queue = new LinkedList<>();
        int count_fresh = 0;
        //Put the position of all rotten oranges in queue
        //count the number of fresh oranges
        for(int i = 0 ; i < rows ; i++) {
            for(int j = 0 ; j < cols ; j++) {
                if(grid[i][j] == 2) {
                    queue.offer(new int[]{i , j});
                }
                else if(grid[i][j] == 1) {
                    count_fresh++;
                }
            }
        }
        //if count of fresh oranges is zero --> return 0 
        if(count_fresh == 0) return 0;
        int count = 0;
        int[][] dirs = {{1,0},{-1,0},{0,1},{0,-1}};
        //bfs starting from initially rotten oranges
        while(!queue.isEmpty()) {
            ++count;
            int size = queue.size();
            for(int i = 0 ; i < size ; i++) {
                int[] point = queue.poll();
                for(int dir[] : dirs) {
                    int x = point[0] + dir[0];
                    int y = point[1] + dir[1];
                    //if x or y is out of bound
                    //or the orange at (x , y) is already rotten
                    //or the cell at (x , y) is empty
                        //we do nothing
                    if(x < 0 || y < 0 || x >= rows || y >= cols || grid[x][y] == 0 || grid[x][y] == 2) continue;
                    //mark the orange at (x , y) as rotten
                    grid[x][y] = 2;
                    //put the new rotten orange at (x , y) in queue
                    queue.offer(new int[]{x , y});
                    //decrease the count of fresh oranges by 1
                    count_fresh--;
                }
            }
        }
        return count_fresh == 0 ? count-1 : -1;
    }
}

典型bfs

class Solution {
    public int orangesRotting(int[][] grid) {
        if(grid == null || grid.length == 0) return 0;
        int[][] dirs = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
        int m = grid.length, n = grid[0].length;
        Queue<int[]> q = new LinkedList();
        int fresh = 0;
        
        for(int i = 0; i < m; i++) {
            for(int j = 0; j < n; j++) {
                if(grid[i][j] == 2) q.offer(new int[]{i, j});
                else if(grid[i][j] == 1) fresh++;
            }
        }
        if(fresh == 0) return 0;
        
        int res = 0;
        System.out.print(q.size());
        while(!q.isEmpty()) {
            res++;
            int si = q.size();
            for(int i = 0; i < si; i++) {
                int[] cur = q.poll();
                for(int[] dir : dirs) {
                    int newi = cur[0] + dir[0];
                    int newj = cur[1] + dir[1];
                    if(newi >= 0 && newi < m && newj >= 0 && newj < n && grid[newi][newj] == 1) {
                        grid[newi][newj] = 2;
                        q.offer(new int[]{newi, newj});
                        fresh--;
                    }
                }
            }
        }
        return fresh == 0 ? res - 1 : -1;
    }
}

每一步扩展1，注意res要减1，因为只要多算了一次（如果只有rotten的也会进循环）