- html - 出于某种原因,IE8 对我的 Sass 文件中继承的 html5 CSS 不友好?
- JMeter 在响应断言中使用 span 标签的问题
- html - 在 :hover and :active? 上具有不同效果的 CSS 动画
- html - 相对于居中的 html 内容固定的 CSS 重复背景?
我现在已经让它停止无限重复,但它只是一遍又一遍地尝试相同的错误路径。有谁知道有什么方法让它尝试不同的路径吗?
数字键:0 已开放1是一堵墙2 是路径的一部分3是迷宫的尽头
public class Maze{
public static void main(String[] args){
int[][] maze = {{1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1},
{0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1},
{1,0,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,1,1,0,1,0,1},
{1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,1,0,1},
{1,0,1,0,1,1,1,0,1,0,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,0,1,1,1,1,1,0,1,1,1,0,1,0,1,0,1,1,1,0,1,1,1,0,1},
{1,0,1,0,1,0,1,0,1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1},
{1,0,1,1,1,0,1,0,1,1,1,0,1,0,1,0,1,0,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,0,1},
{1,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,1,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,1},
{1,1,1,0,1,1,1,1,1,1,1,1,1,0,1,0,1,0,1,0,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1},
{1,0,1,0,1,0,0,0,0,0,0,0,1,0,1,0,1,0,1,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,1},
{1,0,1,0,1,0,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,0,1,1,1},
{1,0,1,0,1,0,0,0,1,0,1,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1},
{1,0,1,0,1,1,1,0,1,0,1,0,1,0,1,1,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1},
{1,0,1,0,1,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1},
{1,0,1,0,1,0,1,0,1,0,1,1,1,0,1,0,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,0,1,1,1,1,1,0,1},
{1,0,0,0,1,0,1,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,1},
{1,0,1,1,1,0,1,0,1,1,1,1,1,1,1,0,1,0,1,1,1,1,1,0,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,0,1,1,1,1,1,0,1,1,1,0,1},
{1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,1,0,1},
{1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,0,1,0,1,0,1,1,1,1,1,1,1,0,1,0,1,1,1,1,1,0,1,0,1,0,1,0,1,1,1,1,1,0,1,0,1},
{1,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,1},
{1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,0,1,1,1,0,1,0,1,1,1,0,1,1,1,1,1,0,1,0,1,1,1,1,1,1,1,0,1,0,1,0,1,0,1},
{1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,1,0,1,0,1,0,1},
{1,1,1,0,1,0,1,1,1,0,1,1,1,0,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,0,1,0,1,1,1,0,1,0,1,0,1,0,1,0,1,1,1,0,1},
{1,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,1},
{1,0,1,1,1,0,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,1,1,0,1,1,1,0,1,0,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,0,1,0,1},
{1,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,1},
{1,0,1,1,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,0,1,0,1,0,1,0,1,1,1,0,1,0,1,0,1,1,1,1,1,0,1,0,1,1,1,0,1,0,1,0,1},
{1,0,0,0,0,0,0,0,1,0,1,0,1,0,1,0,0,0,1,0,1,0,1,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,1,0,1,0,0,0,1,0,1,0,1},
{1,0,1,1,1,1,1,0,1,0,1,0,1,0,1,0,1,1,1,0,1,0,1,0,1,0,1,1,1,1,1,0,1,0,1,0,1,1,1,0,1,0,1,0,1,1,1,1,1,0,1},
{1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,1},
{1,1,1,1,1,0,1,1,1,1,1,1,1,0,1,0,1,0,1,1,1,1,1,1,1,1,1,0,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,1,1,1,1,0,1,0,1},
{1,0,0,0,1,0,1,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,1},
{1,0,1,1,1,0,1,1,1,0,1,0,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,1,0,1,0,1,0,1,0,1},
{1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,1},
{1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1}};
boolean[][] posCheck = new boolean[maze.length][maze[0].length];
int r = 0;
int c = 0;
for(int row = 0; row < maze.length; row++){
for(int col = 0; col < maze[row].length; col++){
if(maze[row][col]==0){
r = row;
c = col;
}
}
}
maze[r][c] = 3;
mazeSolver(1, 0, maze, posCheck);
}
public static boolean mazeSolver(int r, int c, int[][]maze, boolean[][] posCheck){
posCheck[r][c] = true;
maze[r][c] = 2;
if(maze[r][c] == 3){
print(maze);
return true;
}
if((c+1 < maze.length) && maze[r][c+1]==0 && !posCheck[r][c+1] && (mazeSolver(r, c + 1, maze, posCheck))){
maze[r][c] = 2;
return true;
}
if((r-1 >= 0) && maze[r-1][c]==0 && !posCheck[r-1][c] && (mazeSolver(r - 1, c, maze, posCheck))){
maze[r][c] = 2;
return true;
}
if((c-1 >= 0) && maze[r][c-1]==0 && !posCheck[r][c-1] && (mazeSolver(r, c - 1, maze, posCheck))){
maze[r][c] = 2;
return true;
}
if((r+1 < maze.length) && maze[r+1][c]==0 && !posCheck[r+1][c] && (mazeSolver(r + 1, c, maze, posCheck))){
maze[r][c] = 2;
return true;
}
print(maze);
return false;
}
public static void print(int[][] maze){
for(int row = 0; row<maze.length; row++){
for(int col = 0; col<maze[row].length; col++)
System.out.print(maze[row][col]);
System.out.println();
}
}
}
最佳答案
我已将调试输出添加到您的 mazeSolver
函数中的每个分支。正如您所看到的,没有分支被调用,因为第三个 if block 中的递归调用永远不会完成评估。因此,它是无限递归,这就是 StackOverflowError
的原因。它还表明它仅在两个不同的值之间来回跳动,这可能是一个问题。
您需要确定您的 mazeSolver 实际上在做什么 - 为什么无限递归永远不会展开,并更改它以满足其他条件之一。
换句话说,您需要修复“递归终止符”,并找出为什么除了 [1, 1] 和 [0, 1] 之外没有其他值被传递。
public static boolean mazeSolver(int r, int c, int[][] maze){
System.out.println("mazeSolver(" + r + ", " + c + ", maze)");
if(maze[r][c] == 3) {
System.out.println(" Equals 3 -- return true");
return true;
} else if((c+1 < maze.length) && maze[r][c+1]==0 && (mazeSolver(r, c + 1, maze))) {
maze[r][c] = 2;
System.out.println(" Set to 2 (a) -- return true");
return true;
} else if((r-1 >= 0) && maze[r-1][c]==0 && (mazeSolver(r - 1, c, maze))) {
maze[r][c] = 2;
System.out.println(" Set to 2 (b) -- return true");
return true;
}
System.out.println(" Between first and second if-block");
if((c-1 >= 0 && maze[r][c-1]==0) && (mazeSolver(r, c - 1, maze))) {
maze[r][c] = 2;
System.out.println(" Set to 2 (c) -- return true");
return true;
}
System.out.println(" Between second and third if-block");
if((r+1 < maze.length) && maze[r+1][c]==0 && (mazeSolver(r + 1, c, maze))){
maze[r][c] = 2;
System.out.println(" Set to 2 (d) -- return true");
return true;
}
System.out.println(" return false");
return false;
}
}
输出:
[R:\jeffy\programming\sandbox\xbnjava]java Maze
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
...等等...
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
mazeSolver(1, 0, maze)
mazeSolver(1, 1, maze)
Between first and second if-block
Exception in thread "main" java.lang.StackOverflowError
at sun.nio.cs.SingleByte.withResult(Unknown Source)
at sun.nio.cs.SingleByte.access$000(Unknown Source)
at sun.nio.cs.SingleByte$Encoder.encodeArrayLoop(Unknown Source)
at sun.nio.cs.SingleByte$Encoder.encodeLoop(Unknown Source)
at java.nio.charset.CharsetEncoder.encode(Unknown Source)
at sun.nio.cs.StreamEncoder.implWrite(Unknown Source)
at sun.nio.cs.StreamEncoder.write(Unknown Source)
at java.io.OutputStreamWriter.write(Unknown Source)
at java.io.BufferedWriter.flushBuffer(Unknown Source)
at java.io.PrintStream.write(Unknown Source)
at java.io.PrintStream.print(Unknown Source)
at java.io.PrintStream.println(Unknown Source)
at Maze.mazeSolver(Maze.java:64)
at Maze.mazeSolver(Maze.java:82)
at Maze.mazeSolver(Maze.java:69)
at Maze.mazeSolver(Maze.java:82)
at Maze.mazeSolver(Maze.java:69)
at Maze.mazeSolver(Maze.java:82)
at Maze.mazeSolver(Maze.java:69)
at Maze.mazeSolver(Maze.java:82)
at Maze.mazeSolver(Maze.java:69)
at Maze.mazeSolver(Maze.java:82)
at Maze.mazeSolver(Maze.java:69)
at Maze.mazeSolver(Maze.java:82)
at Maze.mazeSolver(Maze.java:69)
at Maze.mazeSolver(Maze.java:82)
...等等...
关于java - Java 中的迷宫求解器问题,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/22236578/
我正在使用混合效应模型,并且由于我的方法的特殊性我需要解决下面模型的积分,然后制作图表获得的估计值。 换句话说,我需要求解下面的积分: 其中,di^2 是我模型中的 Var3,dh 是混合效应模型对应
我有一个方程组,我想用数值方法求解它。给定起始种子,我想得到一个接近的解决方案。让我解释。 我有一个常量向量,X,值: X <- (c(1,-2,3,4)) 和一个向量 W 的权重: W <- (c(
假设我有以下方程组: a * b = 5 sqrt(a * b^2) = 10 如何求解 R 中 a 和 b 的这些方程? 我想这个问题可以说是一个优化问题,具有以下功能......? fn <- f
我在 R 中有一个简单的通量模型。它归结为两个微分方程,对模型中的两个状态变量进行建模,我们将它们称为 A和 B .它们被计算为四个分量通量的简单差分方程 flux1-flux4 , 5 个参数 p1
R有什么办法吗?求解给定单变量函数的反函数?动机是我以后告诉R使用值向量作为反函数的输入,以便它可以吐出反函数值。 例如,我有函数 y(x) = x^2 ,逆是 y = sqrt(x) .有没有办法R
我在字符串中有以下方程 y = 18774x + 82795 求解x我会这样做:- x = (y-82795) / 18774 我知道y的值 但是方程一直在变化,并且始终采用字符串格式 是否可以简单地
如果我用 diophantine(2*x+3*y-5*z-77) 我收到了这个结果。 {(t_0, -9*t_0 - 5*t_1 + 154, -5*t_0 - 3*t_1 + 77)} 到目前为止还
我正在尝试求解仅限于正解的 ODE,即: dx/dt=f(x) x>=0。 在 MATLAB 中这很容易实现。 R 是否有任何变通方法或包来将解决方案空间限制为仅正值? 这对我来说非常重要,不幸的是没
下面的 ANTLR 文法中的 'expr' 规则显然是相互左递归的。作为一个 ANTLR 新手,我很难解决这个问题。我已经阅读了 ANTLR 引用书中的“解决非 LL(*) 冲突”,但我仍然没有看到解
我有一个关于在 R 中求解函数的可能性的非常基本的问题,但知道答案确实有助于更好地理解 R。 我有以下等式: 0=-100/(1+r)+(100-50)/(1+r)^2+(100-50)/(1+r)^
我正在编写使用递归回溯来解决 8 个皇后问题的代码(将 n 个国际象棋皇后放在 n × n 的棋盘上,这样皇后就不会互相攻击)。 我的任务是创建两个方法:编写一个公共(public)solveQuee
我不知道在以下情况下如何进行,因为最后一个方程没有所有 4 个变量。所以使用了等式下面的代码,但这是错误的......有谁知道如何进行? 方程: 3a + 4b - 5c + d = 10 2a +
假设我们有这个递归关系,它出现在 AVL 树的分析中: F1 = 1 F2 = 2 Fn = Fn - 1 + Fn - 2 + 1(其中 n ≥ 3) 你将如何解决这个递归以获得 F(n) 的封闭形
在Maple中,有谁知道是否存在一个函数来求解变量?例如,我正在尝试求解 r 的 solve4r=(M-x^y)*(r^(-1)) mod (p-1)。所以我知道 M、x、y 和 p 的值,但不知道
我也问过这个here在声音设计论坛上,但问题是沉重的计算机科学/数学,所以它实际上可能属于这个论坛: 因此,通过读取文件中的二进制文件,我能够成功地找到关于 WAV 文件的所有信息,除了 big si
我有以下问题: 设 a 和 b 为 boolean 变量。是否可以设置 a 和 b 的值以使以下表达式的计算结果为 false? b or (((not a) or (not a)) or (a or
我需要用 C 求解这个超越方程: x = 2.0 - 0.5sen(x) 我试过这个: double x, newx, delta; x = 2.0 - 0.5; newx = sin(x); del
我在 Windows 上使用 OpenCV 3.1。 一段代码: RNG rng; // random number generator cv::Mat rVec = (cv::Mat_(3, 1)
我正在尝试求解一个包含 3 个变量和数量可变的方程的方程组。 基本上,系统的长度在 5 到 12 个方程之间,无论有多少个方程,我都试图求解 3 个变量。 看起来像这样: (x-A)**2 + (y-
我正在尝试为有限差分法设计一种算法,但我有点困惑。所讨论的 ODE 是 y''-5y'+10y = 10x,其中 y(0)=0 且 y(1)=100。所以我需要一种方法来以某种方式获得将从关系中乘以“
我是一名优秀的程序员,十分优秀!