The Maze II

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.

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: 12
         Explanation: One shortest way is : left -> down -> left -> down -> right -> down -> right.
                      The total distance is 1 + 1 + 3 + 1 + 2 + 2 + 2 = 12.

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: -1
         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.

Tips:

和maze1很相似，不同的是mazeI只需要判断是否能到，mazeII要计算最短距离。此时只需要把代表visited数组变成distance数组，保存到这点的最小距离即可，当需要更新最小距离时，再往下dfs。

bfs的做法可以直接用heap来找到到每个点的最小距离。

Code:

dfs:

public class Solution {
    private static final int[][] directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};    
    public int shortestDistance(int[][] maze, int[] start, int[] destination) {
        int[][] distance = new int[maze.length][maze[0].length];
        for (int i = 0; i < maze.length; i++) {
            for (int j = 0; j < maze[0].length; j++) {
                distance[i][j] = Integer.MAX_VALUE;
            }
        }
        distance[start[0]][start[1]] = 0;
        dfs(maze, start, destination, distance);
        return distance[destination[0]][destination[1]] == Integer.MAX_VALUE ? -1 : distance[destination[0]][destination[1]];
    }

    private void dfs(int[][] maze, int[] start, int[] destination, int[][] distance) {
        if (start[0] == destination[0] && start[1] == destination[1]) return;
        for (int[] d : directions) {
            int row = start[0], col = start[1];
            while (row >= 0 && row < maze.length && col >= 0 && col < maze[0].length && maze[row][col] == 0) {
                row += d[0];
                col += d[1];
            }
            row -= d[0];
            col -= d[1];
            int[] newStart = {row, col};
            int dis = distance[start[0]][start[1]] + Math.abs(start[0] - row) + Math.abs(start[1] - col);
            if (distance[row][col] > dis) {
                distance[row][col] = dis;
                dfs(maze, newStart, destination, distance);
            }           
        }
    } 
}

bfs:

public class Solution {
    class Point {
        int x,y,l;
        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
        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]];
    }
}