505. The Maze II

最新推荐文章于 2024-07-16 21:50:31 发布

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最新推荐文章于 2024-07-16 21:50:31 发布

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本文介绍了一种求解球从起点到终点穿越迷宫的最短路径算法。该算法考虑了球滚动直到撞墙才会停止的特点，并通过广度优先搜索来确定最短路径。文章提供了具体的例子和实现代码。

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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, 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:

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.

类似于 490. The Maze，需要加一些限制条件避免不必要的遍历：当len>length[destination[0]][destination[1]]时，继续下一轮循环；当len<length[x][y]时，更新length[x][y]为len，同时offer进当前点。代码如下：

public class Solution {
    class Point {
        int x;
        int y;
        int 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;
        int n = maze[0].length;
        int[][] length = new int[m][n];
        for (int i = 0; i < m; i ++) {
            Arrays.fill(length[i], Integer.MAX_VALUE);
        }
        Queue<Point> queue = new LinkedList<Point>();
        queue.offer(new Point(start[0], start[1], 0));
        int[][] dirs = new int[][]{{0,1},{0,-1},{1,0},{-1,0}};
        while (!queue.isEmpty()) {
            Point p = queue.poll();
            for (int[] dir: dirs) {
                int xx = p.x, yy = p.y, ll = p.l;
                while (xx >= 0 && yy >= 0 && xx < m && yy < n && maze[xx][yy] == 0) {
                    xx += dir[0];
                    yy += dir[1];
                    ll ++;
                }
                xx -= dir[0];
                yy -= dir[1];
                ll --;
                if (ll > length[destination[0]][destination[1]]) {
                    continue;
                }
                if (ll < length[xx][yy]) {
                    length[xx][yy] = ll;
                    queue.offer(new Point(xx, yy, ll));
                }
            }
        }
        return length[destination[0]][destination[1]] == Integer.MAX_VALUE? -1: length[destination[0]][destination[1]];
    }
}