Java堆栈实现迷宫求解

package stack;

忘记上课了,正好好在寝室看到一个迷宫求解的问题,没想到真正实现还有点复杂,下面是实现源码,不过这并非最短路径解法
  转载请注明出处,谢谢合作！
public class Maze {
  static int[][] dedale={    //用二维数组表示地图
        {0,0,0,0,0,0,0,0,0,0},
        {0,1,1,0,1,1,1,0,1,0},
        {0,1,1,0,1,1,1,0,1,0},
        {0,1,1,1,1,0,0,1,1,0},
        {0,1,0,0,0,1,1,1,1,0},
        {0,1,1,1,0,1,1,0,1,0},
        {0,1,0,1,1,1,0,1,1,0},
        {0,1,0,0,0,0,0,1,0,0},
        {0,0,1,1,1,1,1,1,1,0},
        {0,0,0,0,0,0,0,0,0,0}};
 

  public static void main(String[] args) {
    Stack<Position> stack=new Stack<Position>();
    Position start=new Position(1, 1);
    Position current=start;  //设置当前位置为起始位置
 
    do{
      if(dedale[current.x][current.y]==1&&validate(current)){  //当前位置可通（注：当前位置是标记过的位置时,路线肯定不对）
    dedale[current.x][current.y]=9;  //已经过的地方做个标记
    stack.push(current);
    if(current.x==8&&current.y==8) break;  //到达终点
   
    current=nextPosition(current);  //将当前指向下一个位置
      }else{
    if(stack.size()!=0){
      current=stack.peek();  //取出当前栈顶元素
      if(current.di>=4){    //此时此位置四个方向的路均不通时
    dedale[current.x][current.y]=4;  //留下不能通过的标记
    stack.pop();      //pop出栈顶元素
    current=stack.peek();    //获取此时栈顶元素
    continue;
        }
      if(current.di<4) current=nextPosition(current);
    }
       }  //end else
     }while(stack.size()>0);

    printPath(dedale);
   }
 
  public static void printPath(int[][] a){
for(int i=0;i<a.length;i++){
  for(int j=0;j<a[i].length;j++)
System.out.print(a[i][j]+",");
  System.out.println();
}
   }
 
  public static Position nextPosition(Position p){  //获取下一个Position
if(p.di==1){         //东
  p.di++;
  Position pos=new Position(p.x,p.y+1);
  if(!validate(pos)) return nextPosition(p);
  return pos;
}else if(p.di==2){   //南
  p.di++;
  Position pos=new Position(p.x+1,p.y);
  if(!validate(pos)) return nextPosition(p);
  return pos;  
}else if(p.di==3){   //西
  p.di++;
  Position pos=new Position(p.x,p.y-1);
  if(!validate(pos)) return nextPosition(p);
  return pos;  
}else if(p.di==4){   //北
        p.di++;
      Position pos=new Position(p.x-1,p.y);
  return pos;  
}else return null;
 
   }
 
  public static boolean validate(Position p){
    if(dedale[p.x][p.y]==9||dedale[p.x][p.y]==4){   //排除已走过的部分
      return false;
     }
    return true;
   }

}

class Position{ //表示点实体
  int x;
  int y;
  int di=1; //方向(direct)
 
  public Position(){}
 
  public Position(int x,int y){
this.x=x;
this.y=y;
   }
 
}


最终结果：
0,0,0,0,0,0,0,0,0,0,
0,9,9,0,4,4,4,0,1,0,
0,1,9,0,4,4,4,0,1,0,
0,9,9,4,4,0,0,1,1,0,
0,9,0,0,0,1,9,9,9,0,
0,9,9,9,0,9,9,0,9,0,
0,1,0,9,9,9,0,9,9,0,
0,1,0,0,0,0,0,9,0,0,
0,0,1,1,1,1,1,9,9,0,
0,0,0,0,0,0,0,0,0,0,