LC: 490. The Maze
https://leetcode.com/problems/the-maze/
490. The Maze
There is a ball in a maze with empty spaces (represented as 0) and walls (represented as 1). The ball can go through the 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 m x n maze, the ball's start position and the destination, where start = [startrow, startcol] and destination = [destinationrow, destinationcol], return true if the ball can stop at the destination, otherwise return false.
You may assume that the borders of the maze are all walls (see examples).
Example 1:
Input: maze = [[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]], start = [0,4], destination = [4,4]
Output: true
Explanation: One possible way is : left -> down -> left -> down -> right -> down -> right.Example 2:
Input: maze = [[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]], start = [0,4], destination = [3,2]
Output: false
Explanation: There is no way for the ball to stop at the destination. Notice that you can pass through the destination but you cannot stop there.Example 3:
Input: maze = [[0,0,0,0,0],[1,1,0,0,1],[0,0,0,0,0],[0,1,0,0,1],[0,1,0,0,0]], start = [4,3], destination = [0,1]
Output: falseConstraints:
m == maze.lengthn == maze[i].length1 <= m, n <= 100maze[i][j]is0or1.start.length == 2destination.length == 20 <= startrow, destinationrow <= m0 <= startcol, destinationcol <= nBoth the ball and the destination exist in an empty space, and they will not be in the same position initially.
The maze contains at least 2 empty spaces.
The Essence:
Anstatt einzelne Gitterfelder als Knoten eines Graphs zu betrachten, soll der Problemlöser nur die stabilen Felder, wo der Ball nach dem Rutschen halten kann, für Knoten halten. Das Labyrinth als ein Graph muss also nur durch die stabilen Felder durchgelaufen werden.
Details:
Hierbei kann sowohl Tiefensuche als auch Breitensuche verwendet werden. Während des Durchlaufs soll die instabilen Felder in Richtung der Bewegung übersprungen und nicht verarbeitet werden, was eigentlich das Rutschen des Balls simuliert. Man kann dafür einfach While-Schleifen benutzen.
Solution(s): The implementation can be found here:
Default Code:
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