回溯法解决迷宫问题

import java.awt.Point;import java.util.ArrayList;/*回溯法走迷宫 * 定义一个二维数组：int maze[5][5] = {0, 1, 0, 1, 0,0, 1, 0, 1, 0,0, 0, 0, 0, 0,0, 1, 1, 1, 0,0, 0, 0, 1, 0,};它表示一个迷宫，其中的1表示墙壁，0表示可以走的路，只能横着走或竖着走，不能斜着走，要求编程序找出从左上角到右下角的最短路线。 输入:一个5 × 5的二维数组，表示一个迷宫。数据保证有唯一解。 输出:左上角到右下角的最短路径，格式如样例所示。样例输入:0 1 0 1 00 1 0 1 00 0 0 0 00 1 1 1 00 0 0 1 0样例输出:(0, 0)(1, 0)(2, 0)(2, 1)(2, 2)(2, 3)(2, 4)(3, 4)(4, 4) */public class Maze { private int[][] maze; private int maxLine= 5; private int maxRow= 5; private ArrayList<Point> list=null;  public static void main(String[] args) { new Maze().initialise();} private void initialise(){ maze=new int[][] {{0, 1, 0, 1, 0},{0, 1, 0, 1, 0},{0, 0, 0, 0, 0},{0, 1, 1, 1, 0},{0, 0, 0, 1, 0},};list=new ArrayList<Point>();check(0, 0);  } private void check(int i,int j){ if(i==maxRow-1&&j==maxLine-1) {print();return; } //向上走 if(canMove(i, j, i-1, j)){ Point p=new Point(i, j); maze[i][j]=-1; list.add(p); check(i-1, j); maze[i][j]=0; list.remove(p); } //向下走 if(canMove(i, j, i+1, j)){ Point p=new Point(i, j); maze[i][j]=-1; list.add(p); check(i+1, j); maze[i][j]=0; list.remove(p); } //向左走 if(canMove(i, j, i, j-1)){ Point p=new Point(i, j); maze[i][j]=-1; list.add(p); check(i, j-1); maze[i][j]=0; list.remove(p); } //向右走 if(canMove(i, j, i, j+1)){ Point p=new Point(i, j); maze[i][j]=-1; list.add(p); check(i, j+1); maze[i][j]=0; list.remove(p); } }  private boolean canMove(int i,int j,int targetI,int targetJ){ if(targetI<0||targetJ<0||targetI>maxRow-1||targetJ>maxLine-1) return false; if(maze[targetI][targetJ]==1) return false; if(maze[targetI][targetJ]==-1) return false; return true; }  private void print(){ for (Point point : list) System.out.println("("+(int)point.getX()+", "+(int)point.getY()+")"); System.out.println("(4, 4)"); }}