Java递归迷宫
Java递归迷宫
提问于 2015-04-17 21:26:01
在这个任务中，我应该用Java创建一个迷宫解算器。我决定应用的算法工作方式如下:这是一种递归方法，每次找到路径时都会再次调用自己。如果它运行到死胡同，它将调用第二个递归方法"goBack"，该方法一直返回，直到找到一条新路径。墙壁为0，路径为1s，步行路径为2s，两次行走的路径为3s。这个想法很简单，但我就是不能让它起作用。ArrayOutOfBounds异常一直在出现。有人对此有什么想法吗？
public class Project5v2 {
static String mazecsv = "/Users/amorimph/Documents/COMP 182/Project 5/mazeinput.csv";
static File solvedMaze = new File("/Users/amorimph/Documents/COMP 182/Project 5/solvedMaze.txt");
static int[][] maze = new int[50][50];
static int trigger = 0;
static int mazeWidth;
static int mazeHeight;
public static void main(String[] args) {
readCSV(mazecsv);
start(maze);
mazeToString(maze);
}
public static void readCSV(String csvfile) {
BufferedReader br = null;
String line = "";
String csvSplitBy = ",";
int x = 1;
int y = 0;
try {
br = new BufferedReader(new FileReader(csvfile));
br.readLine();
while ((line = br.readLine()) != null) {
String[] info = line.split(csvSplitBy);
for (x = 1; x < info.length; x++) {
maze[y][x] = Integer.parseInt(info[x]);
}
mazeWidth = info.length;
y++;
mazeHeight = y;
}
} catch (FileNotFoundException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
} finally {
if (br != null) {
try {
br.close();
} catch (IOException e) {
e.printStackTrace();
}
}
}
}
public static void start(int[][] maze) {
int i = 0;
while(maze[0][i] != 1) {
i++;
}
System.out.println(i);
move(maze,i,1);
}
public static void move(int[][] maze, int x, int y) {
for (int i = 0; i < 4; i++) {
switch(i) {
case 0: if(maze[x][y-1] == 1) {
maze[x][y] = 2;
maze[x][y-1] = 2;
move(maze, x, y-1);
break;
}
case 1: if(maze[x-1][y] == 1) {
maze[x][y] = 2;
maze[x-1][y] = 2;
move(maze, x-1, y);
break;
}
case 2: if(maze[x+1][y] == 1) {
maze[x][y] = 2;
maze[x+1][y] = 2;
move(maze, x+1, y);
break;
}
case 3: if(maze[x][y+1] == 1) {
maze[x][y] = 2;
maze[x][y+1] = 2;
move(maze, x, y+1);
break;
}
//case 4:
// maze[x][y] = 2;
// goBack(maze, y, x);
// break;
}
}
}
public static void goBack(int[][] maze, int x, int y) {
for (int i = 0; i < 7; i++) {
switch(i) {
case 0: if(maze[x][y-1] == 1) {
maze[x][y] = 2;
maze[x][y-1] = 2;
move(maze, x, y-1);
break;
}
case 1: if(maze[x-1][y] == 1) {
maze[x][y] = 2;
maze[x-1][y] = 2;
move(maze, x-1, y);
break;
}
case 2: if(maze[x+1][y] == 1) {
maze[x][y] = 2;
maze[x+1][y] = 2;
move(maze, x+1, y);
break;
}
case 3: if(maze[x][y+1] == 1) {
maze[x][y] = 2;
maze[x][y+1] = 2;
move(maze, x, y+1);
break;
}
case 4: if(maze[x][y+1] == 2) {
maze[x][y] = 3;
goBack(maze, x, y+1);
break;
}
case 5: if(maze[x+1][y] == 2) {
maze[x][y] = 3;
goBack(maze, x+1, y);
break;
}
case 6: if(maze[x-1][y] == 2) {
maze[x][y] = 3;
goBack(maze, x-1, y);
break;
}
case 7: if(maze[x][y-1] == 2) {
maze[x][y] = 3;
goBack(maze, x, y-1);
break;
}
}
}
}
public static void FWriter(String content, File file) {
try {
if (!file.exists()) {
file.createNewFile();
}
FileWriter fw = new FileWriter(file.getAbsoluteFile(), true);
BufferedWriter bw = new BufferedWriter(fw);
bw.write(content);
bw.close();
}
catch (IOException e) {
e.printStackTrace();
}
}
public static void mazeToString(int[][] maze) {
for(int i = 0; i < mazeHeight; i++) {
for(int j = 0; j < mazeWidth; j++) {
FWriter(Integer.toString(maze[i][j]), solvedMaze);
}
FWriter("\n", solvedMaze);
}
}
}回答 1
Stack Overflow用户
回答已采纳
发布于 2015-04-17 21:32:56
你的迷宫是50*50。
看看你的移动方法，我看不出有什么东西能阻止你从迷宫中掉下来(也就是说，移动到一个指数大于49或低于0)。
更普遍的情况是，如果您想使用递归，并且不想最终陷入麻烦，则必须检查完成后的情况。您的move方法将永远不会停止对自身的调用，因此请考虑移动方法应该在哪一点终止。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/29709605
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