The Maze
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.

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.

- 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.
DFS
对于dfs,如果当前“决定”对后续有影响,可以使用第16行这种方法不断递归。
class Solution {
public boolean hasPath(int[][] maze, int[] start, int[] destination) {
int m = maze.length, n = maze[].length;
boolean[][] visited = new boolean[m][n];
return dfs(maze, visited, start, destination);
}
private boolean dfs(int[][] maze, boolean[][] visited, int[] start, int[] destination) {
int row = start[], col = start[];
if (row < || row >= maze.length || col < || col >= maze[].length || visited[row][col]) return false;
visited[row][col] = true;
if (row == destination[] && col == destination[]) return true;
int[] directions = { , , , -, };
for (int i = ; i < directions.length - ; i++) {
int[] newStart = roll(maze, start[], start[], directions[i], directions[i + ]);
if (dfs(maze, visited, newStart, destination)) return true;
}
return false;
}
private int[] roll(int[][] maze, int row, int col, int rowInc, int colInc) {
while (canRoll(maze, row + rowInc, col + colInc)) {
row += rowInc;
col += colInc;
}
return new int[]{row, col};
}
private boolean canRoll(int[][] maze, int row, int col) {
if (row >= maze.length || row < || col >= maze[].length || col < || maze[row][col] == ) return false;
return true;
}
}
BFS
class Solution {
public boolean hasPath(int[][] maze, int[] start, int[] destination) {
Deque<int[]> queue = new ArrayDeque<>();
boolean[][] visited = new boolean[maze.length][maze[].length];
queue.offer(start);
while (!queue.isEmpty()) {
int[] cur = queue.poll();
int row = cur[], col = cur[];
if (row == destination[] && col == destination[]) {
return true;
}
if (visited[row][col]) {
continue;
}
visited[row][col] = true;
int[] directions = { , , , -, };
for (int i = ; i < directions.length - ; i++) {
int[] newStart = roll(maze, row, col, directions[i], directions[i + ]);
queue.offer(newStart);
}
}
return false;
}
private int[] roll(int[][] maze, int row, int col, int rowInc, int colInc) {
while (canRoll(maze, row + rowInc, col + colInc)) {
row += rowInc;
col += colInc;
}
return new int[] { row, col };
}
private boolean canRoll(int[][] maze, int row, int col) {
if (row >= maze.length || row < || col >= maze[].length || col < || maze[row][col] == )
return false;
return true;
}
}
The Maze的更多相关文章
- Backtracking algorithm: rat in maze
Sept. 10, 2015 Study again the back tracking algorithm using recursive solution, rat in maze, a clas ...
- (期望)A Dangerous Maze(Light OJ 1027)
http://www.lightoj.com/volume_showproblem.php?problem=1027 You are in a maze; seeing n doors in fron ...
- 1204. Maze Traversal
1204. Maze Traversal A common problem in artificial intelligence is negotiation of a maze. A maze ...
- uva705--slash maze
/*这道题我原本是将斜线迷宫扩大为原来的两倍,但是在这种情况下对于在斜的方向上的搜索会变的较容易出错,所以参考了别人的思路后将迷宫扩展为原来的3倍,这样就变成一般的迷宫问题了*/ #include&q ...
- HDU 4048 Zhuge Liang's Stone Sentinel Maze
Zhuge Liang's Stone Sentinel Maze Time Limit: 10000/4000 MS (Java/Others) Memory Limit: 32768/327 ...
- Borg Maze(MST & bfs)
Borg Maze Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 9220 Accepted: 3087 Descrip ...
- poj 3026 bfs+prim Borg Maze
Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 9718 Accepted: 3263 Description The B ...
- HDU 4035:Maze(概率DP)
http://acm.split.hdu.edu.cn/showproblem.php?pid=4035 Maze Special Judge Problem Description When w ...
- POJ 3026 : Borg Maze(BFS + Prim)
http://poj.org/problem?id=3026 Borg Maze Time Limit: 1000MS Memory Limit: 65536K Total Submissions ...
- Borg Maze 分类: POJ 2015-07-27 15:28 5人阅读 评论(0) 收藏
Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 9971 Accepted: 3347 Description The B ...
随机推荐
- BZOJ 4814 Luogu P3699 [CQOI2017]小Q的草稿 (计算几何、扫描线、set)
题目链接 (BZOJ) http://lydsy.com/JudgeOnline/problem.php?id=4814 (Luogu) https://www.luogu.org/problem/P ...
- ettercap局域网DNS切换到恶意网址
ettercap -i eth0 -Tq -M arp:remote -P dns_spoof /// /// Dns欺骗--Ettercap工具进行Dns欺骗 转载至 https://blog.cs ...
- navicat for mysql安装
搜索一款navicat for mysql然后进行下载. 步骤阅读 2 当我们下载完成之后首先进行数据包的解压,同时可以运行navicat for mysql程序. 破解工具下载:https://pa ...
- 【软件工程】团队Git现场编程实战
组长博客链接 博客链接 组员职责分工 队员 职责分工 恩泽 进行任务的划分与安排,调用API,负责餐饮商铺及商圈信息的获取 金海 解析API返回的json数据,提取有关信息 君曦 部分算法编写 季城 ...
- bootstrap-table前端实现多条件时间段查询数据
实现思路:通过正则匹配到字段是否符合条件,时间段转换为时间戳比对. 这是大体的效果图: 页面的html代码 <div class="content-head mgb10"&g ...
- pure-ftpd搭建简单的Ubuntu FTP服务器
Linux下的ftpd很多,Ubuntu下常用vsftpd, proftpd和pure-ftpd,当初使用的就是proftpd. 不过前两者有个致命的问题就是内码转换,它们默认使用UTF-8编码,而W ...
- Https 协议解析
1 概述1.1 HTTPS 使用SSL协议,对HTTP协议通信过程中的明文数据加密,这就是HTTPS.网络分层结构如下: SSL协议负责数据加密,解密,位于HTPP协议和TCP协议之间. ...
- OGG-01201
OGG-01201 Table of Contents 1. OGG-01201 1.1. 案例1 1.2. 案例2 1 OGG-01201 这种错误,出现的场景之一是 direct load 加载数 ...
- java tfserving grpc 通信调用代码解析 【重点参考】
https://blog.csdn.net/shin627077/article/details/78592729/ [重点参考]
- pandas-赋值操作
1,pandas操作主要有对指定位置的赋值,如上一篇中的数据选择一样,根据loc,iloc,ix选择指定位置,直接赋值 2,插入,insert方法,插入行和列 3,添加 4,删除 drop方法 5,弹 ...