十年网站开发经验 + 多家企业客户 + 靠谱的建站团队
量身定制 + 运营维护+专业推广+无忧售后,网站问题一站解决
import java.util.Arrays;
public class Test {public static final int LEFT_UP = 0;//左上
public static final int UP_LEFT = 1;//上左
public static final int UP_RIGHT = 2;//上右
public static final int RIGHT_UP = 3;//右上
public static final int RIGHT_DOWN = 4;//右下
public static final int DOWN_RIGHT = 5;//下右
public static final int DOWN_LEFT = 6;//下左
public static final int LEFT_DOWN = 7;//左下
public static int currentNum = 1;//马的步数本来就位1,因为出生地算一个
public static int goalNum = 8;//寻找的目标数,如果goalNum为4,则目标数为16
public static void main(String[] args) {long start = System.currentTimeMillis();
for (int x = 0; x< goalNum; x++) {for (int y = 0; y< goalNum; y++) {int[] position = {x, y};//马儿的初始位置
int[][] map = new int[goalNum][goalNum];//地图
map[position[0]][position[1]] = 1;
for (int direction = LEFT_UP; direction<= LEFT_DOWN; direction++) {if (move(direction, position, map)) {//如果递归成功
long end = System.currentTimeMillis();
System.out.println(String.format("总耗时为%d秒", (end - start) / 1000));
return;
}
}
}
}
}
public static boolean judgePosition(int[] position, int[][] map) {//判断路况是否合法即每走完一步,落脚点必须在棋盘内,期盼的边界为 [0,0]-[3,3],且棋盘当前位置不能重新踩
return ((position[0] >= 0 && position[0]<= goalNum - 1) && (position[1] >= 0 && position[1]<= goalNum - 1)) && map[position[0]][position[1]] != 1;
}
//递归一次,currentNum就得+1
public static boolean move(int direction, int[] position, int[][] map) {//移动
if (currentNum == goalNum * goalNum) {//如果当前数量已经达到了递归的目标数,则返回true
printMap(position, map);
return true;
} else {currentNum++;//当前数量+1
int[] positionCopy = Arrays.copyOf(position, position.length);//一进入这个方法,就将position的值保存下来
int[][] mapCopy = new int[goalNum][goalNum];
for (int i = 0; i< goalNum; i++) {for (int j = 0; j< goalNum; j++) {mapCopy[i][j] = map[i][j];
}
}
modifyPosition(direction, position);//修改位置
if (!judgePosition(position, map)) {//如果修改的坐标不合法或者说原有地图此位置已经为1
currentNum--;//当前数量-1
restorePosition(direction, position);//将坐标进行还原
return false;
}
map[position[0]][position[1]] = 1;
for (int nextDirection = LEFT_UP; nextDirection<= LEFT_DOWN; nextDirection++) {if (move(nextDirection, position, map)) {//如果递归成功
printMap(positionCopy, mapCopy);
return true;
}
}
map[position[0]][position[1]] = 0;
restorePosition(direction, position);//将坐标进行还原
currentNum--;//当前数量-1
return false;
}
}
public static void modifyPosition(int direction, int[] position) {//修改位置
switch (direction) {case LEFT_UP:
position[0] -= 1;
position[1] -= 2;
break;
case UP_LEFT:
position[0] -= 2;
position[1] -= 1;
break;
case UP_RIGHT:
position[0] -= 2;
position[1] += 1;
break;
case RIGHT_UP:
position[0] -= 1;
position[1] += 2;
break;
case RIGHT_DOWN:
position[0] += 1;
position[1] += 2;
break;
case DOWN_RIGHT:
position[0] += 2;
position[1] += 1;
break;
case DOWN_LEFT:
position[0] += 2;
position[1] -= 1;
break;
case LEFT_DOWN:
position[0] += 1;
position[1] -= 2;
break;
}
}
public static void restorePosition(int direction, int[] position) {//还原位置
switch (direction) {case LEFT_UP:
position[0] += 1;
position[1] += 2;
break;
case UP_LEFT:
position[0] += 2;
position[1] += 1;
break;
case UP_RIGHT:
position[0] += 2;
position[1] -= 1;
break;
case RIGHT_UP:
position[0] += 1;
position[1] -= 2;
break;
case RIGHT_DOWN:
position[0] -= 1;
position[1] -= 2;
break;
case DOWN_RIGHT:
position[0] -= 2;
position[1] -= 1;
break;
case DOWN_LEFT:
position[0] -= 2;
position[1] += 1;
break;
case LEFT_DOWN:
position[0] -= 1;
position[1] += 2;
break;
}
}
public static void printMap(int[] position, int[][] map) {for (int i = 0; i< goalNum; i++) {for (int j = 0; j< goalNum; j++) {System.out.print(map[i][j] + " ");
}
System.out.println();
}
System.out.println(Arrays.toString(position));
System.out.println("=====================");
}
}
你是否还在寻找稳定的海外服务器提供商?创新互联www.cdcxhl.cn海外机房具备T级流量清洗系统配攻击溯源,准确流量调度确保服务器高可用性,企业级服务器适合批量采购,新人活动首月15元起,快前往官网查看详情吧
创新互联-专业网站定制、快速模板网站建设、高性价比沧州网站开发、企业建站全套包干低至880元,成熟完善的模板库,直接使用。一站式沧州网站制作公司更省心,省钱,快速模板网站建设找我们,业务覆盖沧州地区。费用合理售后完善,10多年实体公司更值得信赖。