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. 【C#】图片处理(底片,黑白,锐化,柔化,浮雕,雾化)

    https://www.cnblogs.com/bomo/archive/2013/03/01/2939453.html --------------------------------------- ...

  2. 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 ...

  3. Codeforces 1054D Changing Array 贪心+异或和

    题意 给一个长度为\(n\)的位数为\(k\)的整数数列\(a\),一次操作可将任意\(a_i\)取反,问经过任意次操作后最多有多少个区间异或和不为\(0\) 分析 求出前缀异或和,区间异或和为\(0 ...

  4. Cow and Snacks

    ​ D. Cow and Snacks 参考:Codeforces 1209D. Cow and Snacks 思路:利用并查集,构建一个生成树,然后树的边数就是能够开心的客人的人数.用一个条件fin ...

  5. Java集合框架系列大纲

    ###Java集合框架之简述 Java集合框架之Collection Java集合框架之Iterator Java集合框架之HashSet Java集合框架之TreeSet Java集合框架之Link ...

  6. js基础( js嵌入方式、输出语句)

    s现在的作用 1.验证表单(以前的网速慢)  2.页面特效 (PC端的网页效果)  3.移动端 (移动 web 和app)  4.异步和服务器交互(ajax)  5.服务器端开发 (nodejs)   ...

  7. Java线程中的异常处理

    对于对线程,当主线程中有子线程运行出现异常时,主线程是不能捕获到该异常的,子线程会直接退出,不会记录任何日志. 解决: 1.子线程中try catch. 2.设置线程的未捕获异常处理器,Uncaugh ...

  8. css垂直居中布局总结

    简介 总结记录一下经常需要用到垂直居中布局,欢迎补充(空手套...O(∩_∩)O) 以下栗子如果未特别标注同一使用这样的html结构 <div class="container&quo ...

  9. Fastadmin 后台编辑,或者添加的时候,出现的问题

    1.情况如图:编辑的时候,这个关联id,默认查出来的是用户昵称,如果要显示用户名,该怎么修改,不要着急,听我慢慢道来 2.首先要找到 编辑页面,检查问题 3.完成

  10. react 的定义组件(了解)

    react 中定义组件的方法 1. 定义组件 React.createClass() (被淘汰了) 定义组件中的函数 methods 的中的 this 统统指向 组件 2. 函数定义组件 定义的组件时 ...