Leetcode: Walls ang Gates

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 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.


 public class Solution {
    public static final int[] d = {0, 1, 0, -1, 0};

    public void wallsAndGates(int[][] rooms) {
        if (rooms.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) bfs(rooms, i, j);
    }

    private void bfs(int[][] rooms, int i, int j) {
        int m = rooms.length, n = rooms[0].length;
        Deque queue = new ArrayDeque<>();
        queue.offer(i * n + j); // Put gate in the queue
        while (!queue.isEmpty()) {
            int x = queue.poll();
            i = x / n; j = x % n;
            for (int k = 0; k < 4; ++k) {
                int p = i + d[k], q = j + d[k + 1];
                if (0 <= p && p < m && 0 <= q && q < n && rooms[p][q] > rooms[i][j] + 1) {
                    rooms[p][q] = rooms[i][j] + 1;
                    queue.offer(p * n + q);
                }
            }
        }
    }
}



 public class Solution {
    private static int[] d = {0, 1, 0, -1, 0};

    public void wallsAndGates(int[][] rooms) {
        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);
    }

    public void dfs(int[][] rooms, int i, int j) {
        for (int k = 0; k < 4; ++k) {
            int p = i + d[k], q = j + d[k + 1];
            if (0<= p && p < rooms.length && 0<= q && q < rooms[0].length &&
                rooms[p][q] > rooms[i][j] + 1) {
                rooms[p][q] = rooms[i][j] + 1;
                dfs(rooms, p, q);
            }
        }
    }
}