Walls and Gates(Medium)

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

private void dfs(int[][] rooms, int i, int j, int distance) {
    if (i < 0 || i >= rooms.length || j < 0 ||  j >= rooms[0].length) {
        return ;
    }
    if (rooms[i][j] < distance) {
        return ;
    } else {
        rooms[i][j] = distance;
        dfs(rooms, i + 1, j, distance + 1);
        dfs(rooms, i - 1, j, distance + 1);
        dfs(rooms, i, j + 1, distance + 1);
        dfs(rooms, i, j - 1, distance + 1);
    }
}
public void wallsAndGates(int[][] rooms) {
    if (rooms == null || rooms.length == 0 || rooms[0].length == 0) {
        return ;
    }
    for (int i = 0; i < rooms.length; ++i) {
        for (int j = 0; j < rooms[0].length; ++j) {
            if (rooms[i][j] == 0) {
                dfs(rooms, i, j, 0);
            }
        }
    }
}

References:

1 [LeetCode]Walls and Gates