我正在尝试使用Java解决迷宫问题,代码仅使用两个方向(向右和向下)完全可以正常工作,但是,我想使其在所有方向上搜索/移动,从而引发错误(堆栈溢出) 。
以下代码:
public class Maze {
int n,m,startX,startY,endX,endY;
public Maze(int n,int m) {
this.n = n;
this.m = m;
}
public void findLocation(int[][]array)
{
for(int i=0; i < array.length;i++)
for(int j = 0 ; j <array[i].length;j++)
{if(array[i][j] == 3)
{
startX = i;
startY = j;
}
if(array[i][j] == 9)
{
endX = i;
endY = j;
}
}
}
public boolean isSafe(int[][]array,int x, int y)
{
return (x>=0 && x < n && y>=0 && y<m && array[x][y]!=0);
}
public boolean solution(int[][] array)
{
int[][]sol = new int[n][m];
findLocation(array);
if(solutionUtil(array,startX,startY,sol)==false)
{
System.out.println("no solution");
return false;
}
printsolution(sol);
return true;
}
private void printsolution(int[][] sol) {
// TODO Auto-generated method stub
for(int i=0; i < sol.length;i++)
{ for(int j = 0 ; j <sol[i].length;j++)
System.out.print(" " + sol[i][j] +
" ");
System.out.println();
}
}
private boolean solutionUtil(int[][] array, int x, int y, int[][] sol) {
// TODO Auto-generated method stub
if(x == endX && y == endY)
{
sol[x][y] = 1;
return true;
}
if(isSafe(array,x,y))
{
sol[x][y] = 1;
if(x!=0)
{
if(solutionUtil(array,x-1,y,sol)==true)
return true;
}
if(y!=0)
{
if(solutionUtil(array,x,y-1,sol)==true)
return true;
}
if(solutionUtil(array,x+1,y,sol)==true)
return true;
if(solutionUtil(array,x,y+1,sol)==true)
return true;
sol[x][y] =0;
return false;
}
return false;
}
public static void main(String[] args)
{
int[][] array = {{0, 0, 0, 0},
{1, 1, 3, 1},
{0, 1, 0, 0},
{1, 1, 1, 9}};
Maze run = new Maze(4,4);
run.solution(array);
}
}
起点是3,终点是9,1表示它可以移动,0表示它不能移动(障碍),我该如何避免错误,因此,可以遍历所有方向,我的代码有误吗?
答案 0 :(得分:1)
您无需检查之前是否未尝试过该步骤,因此您的代码可以进入重复步骤的无限循环。
将解决方案数组传递到function mysql_insert_array_serialised_test($conn, $table, $data, $addon = "mc_api") {
$random_code = substr(str_shuffle(MD5(microtime())), 0, 12);
$settings = serialize($data);
$uid = $_SESSION['uid'];
$sql_insert = 'INSERT INTO ' . $table . ' (uid, addon, sn, settings) VALUES (?, ?, ?, ?)';
if ($stmt = $conn->prepare($sql_insert)) {
$stmt->bind_param('ssss', $uid, $addon, $random_code, $settings);
$stmt->execute();
if (!$stmt->execute()) {
echo $stmt->error;
} else {
return array(
"mysqli_error" => false,
"sn" => $random_code
);
}
} else {
echo 'Could not prepare statement!';
}
}
并检查之前是否未访问过坐标:
isSafe