The Maze

Posted beiyeqingteng

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了The Maze相关的知识,希望对你有一定的参考价值。

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, 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.

技术图片

 
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.

技术图片

 
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.

DFS

 1 class Solution 
 2     public boolean hasPath(int[][] maze, int[] start, int[] destination) 
 3         int m = maze.length, n = maze[0].length;
 4         boolean[][] visited = new boolean[m][n];
 5         return dfs(maze, visited, start, destination);
 6     
 7     private boolean dfs(int[][] maze, boolean[][] visited, int[] start, int[] destination) 
 8         int row = start[0], col = start[1];
 9         if (row < 0 || row >= maze.length || col < 0 || col >= maze[0].length || visited[row][col]) return false;
10         visited[row][col] = true;
11         if (row == destination[0] && col == destination[1]) return true;
12         
13        int[] directions =  0, 1, 0, -1, 0 ;
14         for (int i = 0; i < directions.length - 1; i++) 
15             int[] newStart = roll(maze, start[0], start[1], directions[i], directions[i + 1]);
16             if (dfs(maze, visited, newStart, destination)) return true;
17         
18         return false;
19     
20     
21     private int[] roll(int[][] maze, int row, int col, int rowInc, int colInc) 
22         while (canRoll(maze, row + rowInc, col + colInc)) 
23             row += rowInc;
24             col += colInc;
25         
26         return new int[]row, col;
27     
28     
29     private boolean canRoll(int[][] maze, int row, int col) 
30         if (row >= maze.length || row < 0 || col >= maze[0].length || col < 0 || maze[row][col] == 1) return false;
31         return true;
32     
33 

BFS

 1 class Solution 
 2     public boolean hasPath(int[][] maze, int[] start, int[] destination) 
 3         Deque<int[]> queue = new ArrayDeque<>();
 4         boolean[][] visited = new boolean[maze.length][maze[0].length];
 5         queue.offer(start);
 6         while (!queue.isEmpty()) 
 7             int[] cur = queue.poll();
 8             int row = cur[0], col = cur[1];
 9             if (row == destination[0] && col == destination[1]) 
10                 return true;
11             
12             if (visited[row][col]) 
13                 continue;
14             
15             visited[row][col] = true;
16 
17             int[] directions =  0, 1, 0, -1, 0 ;
18             for (int i = 0; i < directions.length - 1; i++) 
19                 int[] newStart = roll(maze, row, col, directions[i], directions[i + 1]);
20                 queue.offer(newStart);
21             
22         
23         return false;
24     
25 
26     private int[] roll(int[][] maze, int row, int col, int rowInc, int colInc) 
27         while (canRoll(maze, row + rowInc, col + colInc)) 
28             row += rowInc;
29             col += colInc;
30         
31         return new int[]  row, col ;
32     
33 
34     private boolean canRoll(int[][] maze, int row, int col) 
35         if (row >= maze.length || row < 0 || col >= maze[0].length || col < 0 || maze[row][col] == 1)
36             return false;
37         return true;
38     
39 

 

以上是关于The Maze的主要内容,如果未能解决你的问题,请参考以下文章

The Maze

490. The Maze

The Maze II

505. The Maze II

[LC] 490. The Maze

505. The Maze II