There is aballin a maze with empty spaces and walls. The ball can go through empty spaces by rollingup,down,leftorright, but it won't stop rolling until hitting a wall. When the ball stops, it could choose the next direction.

Given the ball'sstart position, thedestinationand themaze, determine whether the ball could stop at the destination.

The maze is represented by a binary 2D array. 1 means the wall and 0 means the empty space. You may assume that the borders of the maze are all walls. The start and destination coordinates are represented by row and column indexes.

Example 1

Input 1: a maze represented by a 2D array

0 0 1 0 0
0 0 0 0 0
0 0 0 1 0
1 1 0 1 1
0 0 0 0 0

Input 2: start coordinate (rowStart, colStart) = (0, 4)
Input 3: destination coordinate (rowDest, colDest) = (4, 4)

Output: true
Explanation: One possible way is : left -> down -> left -> down -> right -> down -> right.

Example 2

Input 1: a maze represented by a 2D array

0 0 1 0 0
0 0 0 0 0
0 0 0 1 0
1 1 0 1 1
0 0 0 0 0

Input 2: start coordinate (rowStart, colStart) = (0, 4)
Input 3: destination coordinate (rowDest, colDest) = (3, 2)

Output: false
Explanation: There is no way for the ball to stop at the destination.

Note:

1. There is only one ball and one destination in the maze.
2. Both the ball and the destination exist on an empty space, and they will not be at the same position initially.
3. The given maze does not contain border (like the red rectangle in the example pictures), but you could assume the border of the maze are all walls.
4. The maze contains at least 2 empty spaces, and both the width and height of the maze won't exceed 100.

tag: BFS

class Solution {
    public boolean hasPath(int[][] maze, int[] start, int[] destination) {
        if (maze == null || maze.length == 0) return false;

        Set<Integer> seen = new HashSet<>();
        Queue<int[]> q = new LinkedList<>();
        int[][] dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
        int n = maze.length, m = maze[0].length;

        seen.add(start[0] * m + start[1]);
        q.offer(start);

        while (!q.isEmpty()){
            int[] cur = q.poll();
            if (cur[0] == destination[0] && cur[1] == destination[1]) return true;

            for (int i = 0; i < 4; i++){
                int xx = cur[0] + dirs[i][0];
                int yy = cur[1] + dirs[i][1];

                if (xx < 0 || xx >= n || yy < 0 || yy >= m || maze[xx][yy] == 1) continue;

                while ( maze[xx][yy] != 1 ){
                    xx += dirs[i][0];
                    yy += dirs[i][1];

                    if (xx < 0 || xx >= n || yy < 0 || yy >= m) break;
                }

                xx -= dirs[i][0];
                yy -= dirs[i][1];

                int[] temp = {xx, yy};

                if (!seen.contains(xx * m + yy)){
                    seen.add(xx * m + yy);
                    q.offer(temp);

                    //System.out.println("adding: " + " x: " + temp[0] + " y: " + temp[1]);
                }
            }
        }

        return false;
    }
}