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:
  1. There is only one ball and one destination in the maze.
  2. Both the ball and the destination exist on an empty space, and they will not be at the same position initially.
  3. 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.
  4. 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

  1. Backtracking algorithm: rat in maze

    Sept. 10, 2015 Study again the back tracking algorithm using recursive solution, rat in maze, a clas ...

  2. (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 ...

  3. 1204. Maze Traversal

    1204.   Maze Traversal A common problem in artificial intelligence is negotiation of a maze. A maze ...

  4. 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 ...

  5. 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 ...

  6. Borg Maze(MST & bfs)

    Borg Maze Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9220   Accepted: 3087 Descrip ...

  7. poj 3026 bfs+prim Borg Maze

    Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9718   Accepted: 3263 Description The B ...

  8. HDU 4035:Maze(概率DP)

    http://acm.split.hdu.edu.cn/showproblem.php?pid=4035 Maze Special Judge Problem Description   When w ...

  9. POJ 3026 : Borg Maze(BFS + Prim)

    http://poj.org/problem?id=3026 Borg Maze Time Limit: 1000MS   Memory Limit: 65536K Total Submissions ...

  10. 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

  1. docker stack /swarm 替代 docker-compose 进行部署

    之前一直用docker-compose开发了几个单例的service, 今天开始压力测试, 结果发现postgres的CPU负载很重, 就想设置cpus 结果发现docker-compose V3之后 ...

  2. hdu 5821 Ball 思维题

    Ball Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submis ...

  3. K8S中Service

    Service 的概念Kubernetes  Service  定义了这样一种抽象:一个  Pod  的逻辑分组,一种可以访问它们的策略 —— 通常称为微服务. 这一组  Pod  能够被  Serv ...

  4. 顺序表应用4-2:元素位置互换之逆置算法(数据改进)(SDUT 3663)

    Problem Description 一个长度为len(1<=len<=1000000)的顺序表,数据元素的类型为整型,将该表分成两半,前一半有m个元素,后一半有len-m个元素(1&l ...

  5. linux命令---vi编辑器快速定位行数

    linux命令—vi编辑器快速定位行数.删除当前行.和删除当前行后面的全部内容 1.vi 编辑器如何快速定位到第N行 命令方式下 :n http://bbs.chinaunix.net/thread- ...

  6. hive序列化和反序列化serde

    一.简介 SerDe是Serializer/Deserializer的缩写.SerDe允许Hive读取表中的数据,并将其以任何自定义格式写回HDFS. 任何人都可以为自己的数据格式编写自己的SerDe ...

  7. EBS AP 创建会计科目失败

    路径:应付款系统管理员/发票/发票 问题描述: 在AP发票界面录入一笔发票,验证发票,撤销发票,然后在AP发票界面的“活动”中创建会计科目,发现带出的并发请求“创建会计科目”正常跑完,但是通过“报表/ ...

  8. 通过ID获取元素

    网页由标签将信息组织起来,而标签的id属性值是唯一的,就像是每人有一个身份证号一样,只要通过身份证号就可以找到相对应的人.那么在网页中,我们通过id先找到标签,然后进行操作. 语法: document ...

  9. android下载网络图片,设置宽高,等比缩放

    使用Picasso组件去下载图片会发现图片宽高会变形不受等比缩放控制,即使设置了图片的 scaleType,可能是对Picasso的api没有用对, Picasso.with(this.activit ...

  10. LC 963. Minimum Area Rectangle II

    Given a set of points in the xy-plane, determine the minimum area of any rectangle formed from these ...