java扫雷小游戏---仅控制台

淡淡的烟草味﹌ 2024-03-25 17:40 68阅读 0赞

由八个方向，控制计数，

我们通过，二维数组的坐标变化来模拟八个方向的变化

![ba87b9d1582029b1070a1a0f79a39537.png][]

下面处理越界的时候，如果超边界肯定就不执行就ok

随机雷阵就是注释掉的那一块，然后随机需要把n也改为控制台输入

下面的n变为n-1

在自定义函数的时候添加输入数组

package 小玩意;
    
    import java.util.Random;
    import java.util.Scanner;
    
    import javax.swing.text.StyledEditorKit.ForegroundAction;
    
    public class 扫雷 {
        static int[] inx = {0,-1,-1,-1,0,1,1,1};
        static int[] iny = {1,1,0,-1,-1,-1,0,1};
        static int count = 0;
        static int n = 6;    
        public static int[][] arr= {
                {0,1,0,0,0,1,0},
                {1,0,1,1,0,0,0},
                {0,1,0,0,0,0,0},
                {0,1,0,0,1,0,0},
                {0,0,0,1,0,0,0},
                {1,0,0,0,0,0,1},
                {0,0,0,0,0,1,0},
        };
    
        public static void main(String[] args) {
    
            //        Random ra = new Random();
            //        for (int i = 0; i < n; i++) {
            //            for (int j = 0; j < n; j++) {
            //                int num =  0;
            //                int temp = ra.nextInt(10);
            //                if ((temp==4||temp==8)&&num<=10) {
            //                    num++;
            //                    arr[i][j]==ra.nextInt(1);
            //                }
            //            }
            //        }
            // TODO Auto-generated method stub
            Scanner sc = new Scanner(System.in);;
            while (true) {
                int x = sc.nextInt();int y = sc.nextInt();
                if (arr[x][y]==1) {
                    System.out.println("游戏结束");
                    break;
                }else {
                    game(x,y);                            
                }
                System.out.println(count);        
            }
        }
        private static void game(int x, int y) {
            // TODO Auto-generated method stub
            count = 0;
            for (int k = 0; k < 8; k++) {
                int xx = x + inx[k];
                int yy = y + iny[k];
                if (xx >= 0 && yy >= 0&&xx<=n&&yy<=n) {
                    if (arr[xx][yy] == 1) {
                        count = count + 1;
                    }
                }
            }
        }
    }

[ba87b9d1582029b1070a1a0f79a39537.png]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/25/473e25282d3340b19869167ac8ae8be3.png