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

-1 - A wall or an obstacle. 0 - A gate. 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

Solution

public class Solution {
    public class Point {
        public int x;
        public int y;
        public Point(int x, int y) {
            this.x = x;
            this.y = y;
        }
    }

    public void wallsAndGates(int[][] rooms) {
        if(rooms.length == 0 || rooms[0].length == 0) return;
        int m = rooms.length, n = rooms[0].length;
        Queue<Point> queue = new LinkedList<Point>();
        for(int i = 0; i < m; i++) {
            for(int j = 0; j < n; j++) {
                if(rooms[i][j] == 0) {
                    queue.add(new Point(i, j));
                }
            }
        }

        int dirs[][] = new int[][]{{0,1},{0,-1},{1,0},{-1,0}};
        while(!queue.isEmpty()) {
            Point cur = queue.poll();
            int val = rooms[cur.x][cur.y];
            for(int[] dir: dirs) {
                if( cur.x + dir[0] >= 0 && cur.x + dir[0] < m &&
                    cur.y + dir[1] >= 0 && cur.y + dir[1] < n && 
                    rooms[cur.x + dir[0]][cur.y + dir[1]] > val + 1) {
                    rooms[cur.x + dir[0]][cur.y + dir[1]] = val + 1;
                    queue.offer(new Point(cur.x+dir[0],cur.y+dir[1]));
                }
            }
        }
    }
}