490. The Maze

最新推荐文章于 2025-03-17 11:27:53 发布

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最新推荐文章于 2025-03-17 11:27:53 发布

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本文介绍了一种使用广度优先遍历(BFS)算法解决迷宫中球从起点到终点寻径的问题。通过构建二进制2D数组来表示迷宫，其中1表示墙壁，0表示空地，并给出球的起始位置和目的地坐标，利用BFS算法找到一条从起点到终点的有效路径。

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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.

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.

Note:

There is only one ball and one destination in the maze.
Both the ball and the destination exist on an empty space, and they will not be at the same position initially.
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.
The maze contains at least 2 empty spaces, and both the width and height of the maze won't exceed 100.

广度优先遍历去寻找路径。路径很像一个数，根节点是start点，子节点是能够到达的点，进行层序遍历，搜索到destination时返回true，否则返回false。代码如下：

public class Solution {
    class Point {
        int x;
        int y;
        public Point(int _x, int _y) { x = _x; y = _y;}
    }
    public boolean hasPath(int[][] maze, int[] start, int[] destination) {
        int m = maze.length;
        int n = maze[0].length;
        boolean[][] visited = new boolean[m][n];
        Queue<Point> queue = new LinkedList<Point>();
        queue.offer(new Point(start[0], start[1]));
        int[][] dirs = new int[][]{{1,0},{-1,0},{0,1},{0,-1}};
        while (!queue.isEmpty()) {
            Point p = queue.poll();
            for (int[] dir: dirs) {
                int xx = p.x, yy = p.y;
                while (xx >= 0 && yy >= 0 && xx < m && yy < n && maze[xx][yy] == 0) {
                    xx += dir[0];
                    yy += dir[1];
                }
                xx -= dir[0];
                yy -= dir[1];
                if (visited[xx][yy]) {
                    continue;
                }
                visited[xx][yy] = true;
                if (xx == destination[0] && yy == destination[1]) {
                    return true;
                }
                queue.offer(new Point(xx, yy));
            }
        }
        return false;
    }
}