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
- 【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 --------------------------------------- ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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) ...
- 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 ...
- 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 ...
- 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
- 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 ...