286. Walls and Gates

You are given a m x n 2D grid initialized with these three possible values.

1. -1 - A wall or an obstacle.
2. 0 - A gate.
3. INF - Infinity means an empty room. We use the value 231 - 1 = 2147483647 to represent INF as you may assume that the distance to a gate is less than 2147483647.

Fill each empty room with the distance to its nearest gate. If it is impossible to reach a gate, it should be filled with INF.

For example, given the 2D grid:

INF	-1	0	INF
INF	INF	INF	-1
INF	-1	INF	-1
0	-1	INF	INF

 

After running your function, the 2D grid should be:

  3  -1   0   1
  2   2   1  -1
  1  -1   2  -1
  0  -1   3   4

Similar: 

• 130. Surrounded Regions
• 200. Number of Islands
• 317. Shortest Distance from All Buildings

public class Solution {
    public void wallsAndGates(int[][] rooms) {
        if (rooms.length == 0 || rooms[0].length==0)
            return;
        Queue<int[]> queue = new LinkedList<int[]>();
        int m = rooms.length;
        int n = rooms[0].length;
        int[] shift = {1, 0, -1, 0, 1};
        for (int i=0; i<m; i++)
            for (int j=0; j<n; j++) {
                if (rooms[i][j] == 0) {
                    queue.offer(new int[] {i,j});
                }
            }
        int level = 1; 
        while (!queue.isEmpty()) {
            int size = queue.size();
            while (size > 0) {
                int[] pt = queue.poll();
                // rooms[pt[0]][pt[1]] = level; 
                for (int k = 0; k < 4; k++) {
                    int r = pt[0] + shift[k];
                    int c = pt[1] + shift[k+1];
                    if (r>=0 && r<m && c>=0 && c<n && rooms[r][c] == Integer.MAX_VALUE) {
                        rooms[r][c] = level; // set here in case point [1][1] would put [1][0] in the queue Twice
                        queue.offer(new int[] {r,c});
                    }
                }
                size--;
            }
            level++;
        }
        return;
    }
}