[LC] 490. The Maze
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.
class Solution {
public boolean hasPath(int[][] maze, int[] start, int[] destination) {
int row = maze.length;
int col = maze[0].length;
Queue<Cell> queue = new LinkedList<>();
boolean[][] visited = new boolean[row][col];
queue.offer(new Cell(start[0], start[1]));
visited[start[0]][start[1]] = true;
int[][] directions = new int[][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
while (!queue.isEmpty()) {
Cell cur = queue.poll();
int curX = cur.x;
int curY = cur.y;
for (int[] direction: directions) {
int newX = curX;
int newY = curY;
while (newX >= 0 && newX < row && newY >= 0 && newY < col && maze[newX][newY] == 0) {
newX += direction[0];
newY += direction[1];
}
newX -= direction[0];
newY -= direction[1];
if (visited[newX][newY]) {
continue;
}
if (newX == destination[0] && newY ==destination[1]) {
return true;
}
queue.offer(new Cell(newX, newY));
visited[newX][newY] = true;
}
}
return false;
}
} class Cell {
int x;
int y;
public Cell(int x, int y) {
this.x = x;
this.y = y;
}
}
[LC] 490. The Maze的更多相关文章
- [LeetCode] 490. The Maze 迷宫
There is a ball in a maze with empty spaces and walls. The ball can go through empty spaces by rolli ...
- LeetCode 490. The Maze
原题链接在这里:https://leetcode.com/problems/the-maze/ 题目: There is a ball in a maze with empty spaces and ...
- LC 499. The Maze III 【lock,hard】
There is a ball in a maze with empty spaces and walls. The ball can go through empty spaces by rolli ...
- [LC] 505. The Maze II
There is a ball in a maze with empty spaces and walls. The ball can go through empty spaces by rolli ...
- 490. The Maze
原题链接:https://leetcode.com/articles/the-maze/ 这道题目是需要冲会员才能使用的,然而我个穷逼现在还是失业状态根本冲不起...以后如果把免费题目都刷完了的话,再 ...
- 【LeetCode】490. The Maze 解题报告 (C++)
作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 BFS 日期 题目地址:https://leetcod ...
- [LeetCode] 490. The Maze_Medium tag: BFS/DFS
There is a ball in a maze with empty spaces and walls. The ball can go through empty spaces by rolli ...
- [LeetCode] 499. The Maze III 迷宫 III
There is a ball in a maze with empty spaces and walls. The ball can go through empty spaces by rolli ...
- [LeetCode] 505. The Maze II 迷宫 II
There is a ball in a maze with empty spaces and walls. The ball can go through empty spaces by rolli ...
随机推荐
- c#连接sql server数据库字符串
第一种方式 Data Source=数据库地址;Initial Catalog=数据库名称;User Id=数据库登录名;Password=数据库密码;[Integrated Security=SSP ...
- 可能对Flutter应用程序开发有用的代码/库/专有技术列表
当我开始使用Flutter实施该应用程序时,我开始担心“如何最好地编写?”以及“如何使其更好地放置?”. 在这种情况下,您将需要参考GitHub上发布的代码和应用程序. 因此,我收集了似乎对Flu ...
- 结点选择(树形DP)
Description 有一棵 n 个节点的树,树上每个节点都有一个正整数权值.如果一个点被选择了,那么在树上和它相邻的点都不能被选择.求选出的点的权值和最大是多少? Input 接下来的一行包含 n ...
- canvas实现粒子星空连线
<!DOCTYPE html><html> <head> <meta charset="UTF-8"> <title>离 ...
- Spring使用Rabbitmq (简单使用)
1.pom.xml jar包引用 <dependencies> <dependency> <groupId>org.springframework</grou ...
- 专业程序设计part2
05tue 乘以1.0使得int*int!=0 today:缩放 和计算机图形学关联 已知:currentdataset ask for:两个方向的缩放比例.保存路径.重采样方法(necessary) ...
- nginx中rewrite flag
rewrite 正则表达式 新URI [flag]; [flag] 选项用于调控重写的行为,它的取值可能是: last:重写完成后,会停止继续处理当前区块所有属于ngx_http_rewrite ...
- 2019年java后端年终总结(六年开发经验),送给正在努力的你
长大之后,时间总是过得飞快,转眼之间,今年已经只剩下1天了.小时候总感觉遥不可及.只在科幻小说里面出现的2020年,已经开始进入蓄力期了. 这篇文章主要和大家聊一聊分析2019年java技术的更新给大 ...
- 1016D.Vasya And The Matrix#矩阵存在
题目出处:http://codeforces.com/contest/1016/problem/D #include<iostream> #define ll long long int ...
- windows之anaconda导入torch失败和pip install命令执行read time out
昨天用jupyter导入torch还好好的呢,今天用就不行了,先是ImportError: DLL load failed: 找不到指定的模块.再是No such comm target regist ...