499. The Maze III

There is a ball in a maze with empty spaces and walls. The ball can go through empty spaces by rolling up (u), down (d), left (l) or right (r), but it won't stop rolling until hitting a wall. When the ball stops, it could choose the next direction. There is also a hole in this maze. The ball will drop into the hole if it rolls on to the hole.

Given the ball position, the hole position and the maze, find out how the ball could drop into the hole by moving the shortest distance. The distance is defined by the number of empty spaces traveled by the ball from the start position (excluded) to the hole (included). Output the moving directions by using 'u', 'd', 'l' and 'r'. Since there could be several different shortest ways, you should output the lexicographically smallest way. If the ball cannot reach the hole, output "impossible".

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 ball and the hole coordinates are represented by row and column indexes.

Solution：

思路： 和505 The Maze II 思路相似，加一个route record
Time Complexity: O(N) Space Complexity: O(N)

Solution Code：

public class Solution {
    class Point implements Comparable {
        int x,y,l;
        String s; // record route
        public Point(int _x, int _y) {x=_x;y=_y;l=Integer.MAX_VALUE;s="";}
        public Point(int _x, int _y, int _l,String _s) {x=_x;y=_y;l=_l;s=_s;}
        public int compareTo(Point p) {return l==p.l ? s.compareTo(p.s):l-p.l;}
    }
    public String findShortestWay(int[][] maze, int[] ball, int[] hole) {
        int m=maze.length, n=maze[0].length;
        Point[][] points=new Point[m][n];
        for (int i=0;i list=new PriorityQueue<>(); // using priority queue
        list.offer(new Point(ball[0], ball[1], 0, ""));
        while (!list.isEmpty()) {
            Point p=list.poll();
            if (points[p.x][p.y].compareTo(p)<=0) continue; // if we have already found a route shorter
            points[p.x][p.y]=p;
            for (int i=0;i<4;i++) {
                int xx=p.x, yy=p.y, l=p.l;
                while (xx>=0 && xx=0 && yy