505. The Maze II

Description

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.

image

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.

image

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.

Solution

Dijkstra, time O(mn * log(mn)), space O(mn)

在"490. The Maze"的BFS算法基础上，稍微扩展下，就能想到这道题就是Dijkstra算法的应用。注意需要去掉visited（因为一个节点会被放进队列多次），同时需要用int[][] distance去保存所有位置的shortest distance。

image.png

class Solution {
    public static final int[][] DIRECTIONS = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
    
    public int shortestDistance(int[][] maze, int[] start, int[] destination) {
        int m = maze.length;
        int n = maze[0].length;
        int[][] distances = new int[m][n];
        for (int[] dist : distances) {
            Arrays.fill(dist, Integer.MAX_VALUE);
        }
        
        PriorityQueue<int[]> queue
            = new PriorityQueue<>((a, b) -> a[2] - b[2]);
        queue.offer(new int[] {start[0], start[1], 0});
        
        while (!queue.isEmpty()) {
            int[] arr = queue.poll();
            if (arr[0] == destination[0] && arr[1] == destination[1]) {
                return arr[2];
            }
            
            if (distances[arr[0]][arr[1]] <= arr[2]) {  // ignore longer distance
                continue;
            }
            
            distances[arr[0]][arr[1]] = arr[2];     // update distance
            
            for (int[] direction : DIRECTIONS) {
                int i = arr[0];
                int j = arr[1];
                int dist = arr[2];
                
                while (isEmpty(maze, i + direction[0], j + direction[1])) {
                    i += direction[0];
                    j += direction[1];
                    ++dist;
                }
                
                queue.offer(new int[] {i, j, dist});
            }
        }
        
        return -1;
    }
    
    public boolean isEmpty(int[][] maze, int i, int j) {
        return isValid(maze, i, j) && maze[i][j] != 1;
    }
    
    public boolean isValid(int[][] maze, int i, int j) {
        return i >= 0 && i < maze.length && j >= 0 && j < maze[0].length;
    }
}

也可以省略掉distance[][]，队列中第一次取出某节点p时，意味着此时p的最短路径已经找到，在maze中标记为-1即可。

class Solution {
    public static final int[][] DIRECTIONS = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
    
    public int shortestDistance(int[][] maze, int[] start, int[] destination) {
        PriorityQueue<int[]> queue = new PriorityQueue<>((a, b) -> a[2] - b[2]);
        queue.offer(new int[] {start[0], start[1], 0});
        
        while (!queue.isEmpty()) {
            int[] arr = queue.poll();
            if (arr[0] == destination[0] && arr[1] == destination[1]) {
                return arr[2];
            }
           
            if (maze[arr[0]][arr[1]] == -1) {   // min dis already found
                continue;
            }
            
            maze[arr[0]][arr[1]] = -1;  // first time visit, arr[2] is minDis
            
            for (int[] d : DIRECTIONS) {
                int x = arr[0];
                int y = arr[1];
                int k = 0;
                
                while (isEmpty(maze, x + d[0], y + d[1])) {
                    x += d[0];
                    y += d[1];
                    ++k;
                }
                
                queue.offer(new int[] {x, y, arr[2] + k});
            }
        }
        
        return -1;
    }
    
    private int getIndex(int i, int j, int cols) {
        return i * cols + j;
    }
    
    public boolean isEmpty(int[][] maze, int i, int j) {
        return isValid(maze, i, j) && maze[i][j] != 1;
    }
    
    public boolean isValid(int[][] maze, int i, int j) {
        return i >= 0 && i < maze.length && j >= 0 && j < maze[0].length;
    }
}