Walls and 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 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

DFS

    public void wallsAndGates(int[][] rooms) {
        if(rooms.length == 0) return ;

        boolean [][] visited = new boolean[rooms.length][rooms[0].length];

        for(int i = 0 ; i < rooms.length; i++)
            for(int j = 0 ; j < rooms[0].length; j ++)
                if(rooms[i][j] == 0)
                    dfs(rooms, visited, i, j, 0);
    }

    private void dfs(int [][] rooms, boolean visited[][], int i, int j, int dist){

        if(i < 0 || j < 0 || i >= rooms.length || j >= rooms[0].length || visited[i][j] || rooms[i][j] == -1) return ;

        //when meet a root value is smaller than current dist, jump out of DFS. 
        //Because no matter how to move, the current dist will always greater than room value since both of them should plus one to the next room.
        if(rooms[i][j] < dist) return ;

        visited[i][j] = true;

        rooms[i][j] = dist;

        dfs(rooms,visited, i - 1, j, dist + 1);
        dfs(rooms,visited, i + 1, j, dist + 1);
        dfs(rooms,visited, i, j - 1, dist + 1);
        dfs(rooms,visited, i, j + 1, dist + 1);

        visited[i][j] = false;

        return;
    }