我試圖用遞歸方法來解決由Java中的1和0的int數組表示的迷宮。 wasHere []檢查該方法是否已經解析了該特定座標集,correctPath []記錄了該迷宮的回答路徑;它們都被初始化爲每個索引都是假的。StackOverflow錯誤,由於Java中的遞歸
import java.awt.Point;
class mazeSolver{
int[][] maze;
int startX, startY;
int endX, endY;
int width, height;
boolean[][] wasHere, correctPath;
mazeSolver(int[][] m, int sX,int sY,int nX,int nY){
maze = m;
width = maze[0].length;
height = maze.length;
startX = sX;
startY = sY;
endX = nX;
endY = nY;
System.out.println("Height: " + height + "\t Width: " + width);
correctPath = new boolean[height][width];
wasHere = new boolean[height][width];
for (int row = 0; row < maze.length; row++)
for (int col = 0; col < maze[row].length; col++){
wasHere[row][col] = false;
correctPath[row][col] = false;
}
solveMaze(startX,startY);
}
public boolean solveMaze(int x, int y){
boolean solvable = recursiveSolve(x,y);
return solvable;
}
private boolean recursiveSolve(int x, int y){ //1s are walls, 0s are free space
if (x == endX && y == endY)
return true;
if (wasHere[y][x] || maze[y][x] == 1)
return false;
wasHere[y][x] = true;
if (y < height-1){
if (solveMaze(x,y+1))
return true;
}
if (y > 0){
if (solveMaze(x,y-1))
return true;
}
if (x > 0){
if (solveMaze(x-1,y))
return true;
}
if (x < width-1){
if (solveMaze(x+1,y))
return true;
}
return false;
}
public int[][] getSolvedMaze(){
for(int y = 0; y < height; y++)
for (int x = 0; x< width; x++)
if(correctPath[y][x])
maze[y][x] = 2;
return maze;
}
}
我一直堅持這個錯誤的過去幾個小時,任何幫助將不勝感激。
粘貼您的完整代碼。 – vinayc
有可能沒有錯誤。你的迷宮有多大?堆棧是有限的,所以每個堆棧都有一個足夠大的迷宮來溢出它。 –
@BarryFruitman我的測試迷宮是300x400。我最好喜歡處理那些是2000x2000,但也是 – potatochemist