求助！想把如下c++编写的骑士周游问题转成java代码，求论坛大神帮助

[wayaoliang2864]

代码如下。因为本次作业要求加一个图形化界面，但是本人暂时没有学会c++图形化界面的编写，想转用java，但是转换代码这里遇到了困难！主要是重载部分和优先级队列部分不会转化！求帮助！

#include <iostream>
#include <stdlib.h>
#include <iomanip>
#include <queue>

using namespace std;

typedef struct
{
	int x;
	int y;
} Step;

Step step[8] = { { -2, -1 },{ -1, -2 },{ 1, -2 },{ 2, -1 },{ 2, 1 },{ 1, 2 },{ -1, 2 },{ -2,1 } };

typedef struct NextPos
{
	int nextPosSteps; //表示下一位置有多少种走法；走法少的优先考虑
	int nextPosDirection; //下一位置相对于当前位置的方位
	int nextPosToMidLength; //表示当前位置距中间点距离；距离中间点远的优先考虑

							//
	bool operator < (const NextPos &a) const
	{
		return nextPosSteps > a.nextPosSteps && nextPosToMidLength < a.nextPosToMidLength;
	}

};

int board[8][8];
//int N; //棋盘大小
int MIDPOS = 3; // 中点

	   //检测这个位置是否可以走
bool check(int x, int y)
{
	if (x >= 0 && x < 8 && y >= 0 && y < 8 && board[x][y] == 0)
		return true;
	else return false;
}
//下一位置有多少种走法
int nextPosHasSteps(int x, int y)
{
	int steps = 0;
	for (int i = 0; i < 8; ++i)
	{
		if (check(x + step[i].x, y + step[i].y))
			steps++;
	}
	return steps;
}
//判断是否回到起点
bool returnStart(int x, int y)
{
	//校验最后是否可以回到起点，也就是棋盘的中间位置
	//int midpos = 3;
	//midx = N / 2 - 1;
	//midy = midx;
	for (int i = 0; i < 8; ++i)
		if (x + step[i].x == MIDPOS && y + step[i].y == MIDPOS)
			return true;
	return false;
}

//输出结果
void outputResult(int xstart, int ystart)
{
	int num = 64;
	int k = num - board[xstart][ystart];
	for (int i = 1; i <= 8; ++i)
	{
		cout << endl << endl;
		for (int j = 1; j <= 8; ++j)
		{
			board[i][j] = (board[i][j] + k) % num + 1;
			cout << setw(5) << board[i][j];
		}
	}
	cout << endl << endl;
}

//某一位置距离棋盘中心的距离
int posToMidLength(int x, int y)
{
	//int midx = N / 2 - 1;
	//int midy = midx;;
	return (abs(x - MIDPOS) + abs(y - MIDPOS));
}

void BackTrace(int t, int x, int y, int xstart, int ystart)
{
	//找到结果
	if (t == 64 && returnStart(x, y)) //遍历了棋盘的所以位置，并且最后可以回到起点，形成回路
	{
		outputResult(xstart, ystart);
		exit(1);
	}
	else
	{
		priority_queue<NextPos> nextPosQueue;
		for (int i = 0; i < 8; ++i)
		{
			if (check(x + step[i].x, y + step[i].y))
			{
				NextPos aNextPos;
				aNextPos.nextPosSteps = nextPosHasSteps(x + step[i].x, y + step[i].y);
				aNextPos.nextPosDirection = i;
				aNextPos.nextPosToMidLength = posToMidLength(x + step[i].x, y + step[i].y);
				nextPosQueue.push(aNextPos);
			}
		}

		while (nextPosQueue.size())
		{
			int d = nextPosQueue.top().nextPosDirection;
			nextPosQueue.pop();

			x += step[d].x;
			y += step[d].y;
			board[x][y] = t + 1;
			BackTrace(t + 1, x, y, xstart, ystart);
			//回溯
			board[x][y] = 0;
			x -= step[d].x;
			y -= step[d].y;
		}
	}
}


void horseRun(int xstart, int ystart)
{
	//初始化棋盘
	for (int i = 0; i < 8; i++)
		for (int j = 0; j < 8; j++)
			board[i][j] = 0;
	//int midx = N / 2 - 1;
	//int midy = midx;
	board[MIDPOS][MIDPOS] = 1; //从棋盘的中间的位置开始马周游
	BackTrace(1, MIDPOS, MIDPOS, xstart, ystart);
}

int main()
{
	//马周游起始位置
	int x, y;

	//cout << "请输入棋盘大小N:";
	//cin >> N;

	cout << "请输入马周游起始位置:";
	cin >> x >> y;

	horseRun(x, y); //执行马周游
	return 0;
}