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所以我正在編寫一個程序，生成並解決迷宮。除了這個小東西外，我已經做好了一切工作。迷宮中的標記將完成迷宮的第二步到最後一步，然後停下來，然後移動到不同的方向。我一直在爲此工作了大約2個小時，跟蹤我的代碼，似乎無法找到最後一步走向哪裏或爲何變得古怪。迷宮程序無法完成
迷宮的規則是X是牆壁，不能移動，O是可以移動的區域，.是起點。 -是我已經轉移到的路徑。標記可以以任何順序或主要方向（北，東北等）移動。
我有兩個矩陣，一個叫mat，向用戶顯示X和O和-'s。我還有另一個叫做visited的矩陣，其大小與mat相同，並且是跟蹤我可以或不可以移動到的座標的幕後矩陣。它爲牆壁存儲W，我已經訪問過的座標爲Y，我可以訪問的座標爲N。
我解決迷宮的方法是檢查可能的移動從北部開始，逆時針繞指南針移動，直到找到可能的移動。標記不能移動以發現它已經移動。
如果遇到有多個可能移動的分割路徑，我將該位置的座標放在兩個堆棧中，分別稱爲splitx，列爲splity，然後我可以稍後再回來。如果我遇到了每個方格都是牆壁或已經訪問過的死角，我會回溯到分割路徑堆棧中的最後一個座標，關閉剛剛移動到的路徑，並彈出堆棧的最高值。
我還有另一個堆棧，名爲visitStack，它存儲我爲北方的每個移動N，東北的NE，等等。這讓我回顧我的動作，並轉到我遇到的最後一條分割路徑。這裏是代碼和它下面的一些選擇輸出。如果您對代碼或輸出有任何疑問或者只是一般性的澄清，請離開。
import java.util.*;
public class MazeLab
{
public static void main(String args[])
{
Scanner input = new Scanner(System.in);
System.out.print("Enter random starting seed ===>> ");
int seed = input.nextInt();
Maze maze = new Maze(seed);
maze.displayMaze();
maze.solveMaze();
}
}
class Maze
{
private char mat[][]; // 2d character array that stores the maze display
private char visited[][]; // 2d character array that works behind to scenes to let me know where I can or can't move
private Stack<String> visitStack; // stack that stores every move I make as N, NE, E, SE, etc
private int nowR, nowC; // coordinates for current position in the matrix
private int startRow, startCol; // coordinates for the starting position
private boolean done = false; // not that important
private Stack<Integer> splitx = new Stack<Integer>(); // row coord for split paths
private Stack<Integer> splity = new Stack<Integer>(); // col coord for split paths
Random random = new Random();
public Maze(int seed)
// constructor which generates the random maze, random starting location
// and initializes Maze class values. If the random value equals 0 the maze
// store an 'X' otherwise it store an 'O' in the maze.
{
Random randomize = new Random(seed);
mat = new char[12][12];
visited = new char[12][12];
for (int r = 0; r < 12; r++)
for (int c = 0; c < 12; c++)
{
if (r == 0 || c == 0 || r == 11 || c == 11)
mat[r][c] = 'X';
else
{
int rndInt = randomize.nextInt(2);
if (rndInt == 0)
mat[r][c] = 'X';
else
mat[r][c] = 'O';
}
}
mat[0][0] = 'O';
startRow = randomize.nextInt(12);
startCol = 11;
nowR = startRow;
nowC = startCol;
mat[startRow][startCol] = '.';
visitStack = new Stack<String>();
for(int i = 0; i < mat.length; i++)
for(int k = 0; k < mat[i].length; k++)
if(mat[i][k] == 'X')
visited[i][k] = 'W';
else
visited[i][k] = 'N';
// Avoid going back to the starting point
visited[nowR][nowC] = 'Y';
}
void displayMaze()
// displays the current maze configuration
{
for (int r = 0; r < 12; r++)
{
for (int c = 0; c < 12; c++)
System.out.print(mat[r][c] + " ");
System.out.println();
}
System.out.println();
if(done == false)
pause();
}
public void solveMaze()
{
// for testing purposes, this is not the real method
for(int i = 0; i < 15; i++)
{
getMove();
displayMaze();
}
}
private int numMoves()
// This method determines if a position has a valid move or not
{
int moves = 0;
if(nowR != 0 && visited[nowR-1][nowC] == 'N')
moves++;
if(nowR != 0 && nowC != 11 && visited[nowR-1][nowC+1] == 'N')
moves++;
if(nowC != 11 && visited[nowR][nowC+1] == 'N')
moves++;
if(nowR != 11 && nowC != 11 && visited[nowR+1][nowC+1] == 'N')
moves++;
if(nowR != 11 && visited[nowR+1][nowC] == 'N')
moves++;
if(nowR != 11 && nowC != 0 && visited[nowR+1][nowC-1] == 'N')
moves++;
if(nowC != 0 && visited[nowR][nowC-1] == 'N')
moves++;
if(nowR != 0 && nowC != 0 && visited[nowR-1][nowC-1] == 'N')
moves++;
return moves;
}
private void getMove()
{
if(numMoves() > 1)
{
// checks for split paths
// north
if(nowR != 0 && visited[nowR-1][nowC] == 'N')
{
splitx.push(nowR);
splity.push(nowC);
}
// northwest
if(nowR != 0 && nowC != 0 && visited[nowR-1][nowC-1] == 'N')
{
splitx.push(nowR);
splity.push(nowC);
}
// west
if(nowC != 0 && visited[nowR][nowC-1] == 'N')
{
splitx.push(nowR);
splity.push(nowC);
}
// southwest
if(nowR != 11 && nowC != 0 && visited[nowR+1][nowC-1] == 'N')
{
splitx.push(nowR);
splity.push(nowC);
}
// south
if(nowR != 11 && visited[nowR+1][nowC] == 'N')
{
splitx.push(nowR);
splity.push(nowC);
}
// southeast
if(nowR != 11 && nowC != 11 && visited[nowR+1][nowC+1] == 'N')
{
splitx.push(nowR);
splity.push(nowC);
}
// east
if(nowC != 11 && visited[nowR][nowC+1] == 'N')
{
splitx.push(nowR);
splity.push(nowC);
}
// northeast
if(nowR != 0 && nowC != 11 && visited[nowR-1][nowC+1] == 'N')
{
splitx.push(nowR);
splity.push(nowC);
}
}
if(numMoves() > 0)
{
// moves the marker, oriented to the right
// north
if(nowR != 0 && visited[nowR-1][nowC] == 'N')
{
nowR--;
visited[nowR][nowC] = 'Y';
visitStack.push("N");
mat[nowR][nowC] = '-';
// System.out.println("PUSHING ON NORTH");
}
// northwest
else if(nowR != 0 && nowC != 0 && visited[nowR-1][nowC-1] == 'N')
{
nowR--;
nowC--;
visited[nowR][nowC] = 'Y';
visitStack.push("NW");
mat[nowR][nowC] = '-';
// System.out.println("PUSHING ON NORTHWEST");
}
// west
else if(nowC != 0 && visited[nowR][nowC-1] == 'N')
{
nowC--;
visited[nowR][nowC] = 'Y';
visitStack.push("W");
mat[nowR][nowC] = '-';
// System.out.println("PUSHING ON WEST");
}
// southwest
else if(nowR != 11 && nowC != 0 && visited[nowR+1][nowC-1] == 'N')
{
nowR++;
nowC--;
visited[nowR][nowC] = 'Y';
visitStack.push("SW");
mat[nowR][nowC] = '-';
// System.out.println("PUSHING ON SOUTHWEST");
}
// south
else if(nowR != 11 && visited[nowR+1][nowC] == 'N')
{
nowR++;
visited[nowR][nowC] = 'Y';
visitStack.push("S");
mat[nowR][nowC] = '-';
// System.out.println("PUSHING ON SOUTH");
}
// southeast
else if(nowR != 11 && nowC != 11 && visited[nowR+1][nowC+1] == 'N')
{
nowR++;
nowC++;
visited[nowR][nowC] = 'Y';
visitStack.push("SE");
mat[nowR][nowC] = '-';
// System.out.println("PUSHING ON SOUTHEAST");
}
// east
else if(nowC != 11 && visited[nowR][nowC+1] == 'N')
{
nowC++;
visited[nowR][nowC] = 'Y';
visitStack.push("E");
mat[nowR][nowC] = '-';
// System.out.println("PUSHING ON EAST");
}
// northeast
else if(nowR != 0 && nowC != 11 && visited[nowR-1][nowC+1] == 'N')
{
nowR--;
nowC++;
visited[nowR][nowC] = 'Y';
visitStack.push("NE");
mat[nowR][nowC] = '-';
// System.out.println("PUSHING ON NORTHEAST");
}
}
if(numMoves() == 0)
{
while(nowR != splitx.peek() && nowC != splity.peek())
{
switch(visitStack.pop())
{
// have to backtrack the opposite direction i previously went
case "N": visited[nowR][nowC] = 'N';
mat[nowR][nowC] = 'O';
nowR++;
break;
case "NE": visited[nowR][nowC] = 'N';
mat[nowR][nowC] = 'O';
nowR++;
nowC--;
break;
case "E": visited[nowR][nowC] = 'N';
mat[nowR][nowC] = 'O';
nowC--;
break;
case "SE": visited[nowR][nowC] = 'N';
mat[nowR][nowC] = 'O';
nowR--;
nowC--;
break;
case "S": visited[nowR][nowC] = 'N';
mat[nowR][nowC] = 'O';
nowR--;
break;
case "SW": visited[nowR][nowC] = 'N';
mat[nowR][nowC] = 'O';
nowR--;
nowC++;
break;
case "W": visited[nowR][nowC] = 'N';
mat[nowR][nowC] = 'O';
nowC++;
break;
case "NW": visited[nowR][nowC] = 'N';
mat[nowR][nowC] = 'O';
nowR++;
nowC++;
break;
default: System.out.println("SOMETHING WENT WRONG WITH BACKTRACKING");
}
}
// blocks off dead ends
// north
if(nowR != 0 && visited[nowR-1][nowC] == 'N')
visited[nowR-1][nowC] = 'Y';
// northwest
else if(nowR != 0 && nowC != 0 && visited[nowR-1][nowC-1] == 'N')
visited[nowR-1][nowC-1] = 'Y';
// west
else if(nowC != 0 && visited[nowR][nowC-1] == 'N')
visited[nowR][nowC-1] = 'Y';
// southwest
else if(nowR != 11 && nowC != 0 && visited[nowR+1][nowC-1] == 'N')
visited[nowR+1][nowC-1] = 'Y';
// south
else if(nowR != 11 && visited[nowR+1][nowC] == 'N')
visited[nowR+1][nowC] = 'Y';
// southeast
else if(nowR != 11 && nowC != 11 && visited[nowR+1][nowC+1] == 'N')
visited[nowR+1][nowC+1] = 'Y';
// east
else if(nowC != 11 && visited[nowR][nowC+1] == 'N')
visited[nowR][nowC+1] = 'Y';
// northeast
else if(nowR != 0 && nowC != 11 && visited[nowR-1][nowC+1] == 'N')
visited[nowR-1][nowC+1] = 'Y';
splitx.pop();
splity.pop();
}
}
private void pause()
{
Scanner input = new Scanner(System.in);
String dummy;
System.out.print("\nPress <Enter> to continue ===>> ");
dummy = input.nextLine();
}
}
這是標記移動通過它之前的迷宮。 
這是最後的迷宮。注意當標記應該向西北移動完成迷宮時，它會在下一個回合中停留在同一個地方，然後在回合之後向南移動。
您是否嘗試過調試它，單步執行代碼以查看發生了什麼？ – Andreas
是的，我追蹤了好幾遍，不明白爲什麼它不會在最後一步向西北方向移動。我試着在標記向西北方向移動時放置一個'System.out.println'語句，並在最後一步移動時打印它，但該點仍然是'O' –
@Andreas http：//i.imgur .com/4fkKYM7.png –