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

Given the ball's start position, the destination and the maze, find the shortest distance for the ball to stop at the destination. The distance is defined by the number of empty spaces traveled by the ball from the start position (excluded) to the destination (included). If the ball cannot stop at the destination, return -1.

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.

基本框架同refer 490 The Maze I : //www.greatytc.com/p/eb61a466c556

Solution：BFS + dijkstra

思路：
Time Complexity: O(mn) Space Complexity: O(mn)

Solution Code：

public class Solution {
    class Point {
        int x, y;
        int l; // minimal length (current)
        public Point(int _x, int _y, int _l) {
            x = _x;
            y = _y;
            l = _l;
        }
    }
    
    public int shortestDistance(int[][] maze, int[] start, int[] destination) {
        int m = maze.length, n = maze[0].length;
        int[][] length = new int[m][n]; // record length (Dijkstra idea)
        
        for (int i = 0; i < m * n; i++) length[i / n][i % n]=Integer.MAX_VALUE;
        int[][] dir = new int[][] {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
        
        PriorityQueue<Point> list = new PriorityQueue<>((o1, o2) -> o1.l - o2.l); // using priority queue
        
        
        list.offer(new Point(start[0], start[1], 0));
        while (!list.isEmpty()) {
            Point p = list.poll();
            if (length[p.x][p.y] <= p.l) continue; // if we have already found a route shorter
            length[p.x][p.y] = p.l;
            for (int i = 0; i < 4; i++) {
                int xx = p.x, yy = p.y, l = p.l;
                while (xx >= 0 && xx < m && yy >= 0 && yy < n && maze[xx][yy] == 0) {
                    xx += dir[i][0];
                    yy += dir[i][1];
                    l++;
                }
                xx -= dir[i][0];
                yy -= dir[i][1];
                l--;
                list.offer(new Point(xx, yy, l));
            }
        }
        return length[destination[0]][destination[1]] == Integer.MAX_VALUE ? -1 : length[destination[0]][destination[1]];
    }
}