1、定义长度为5的整形数组,输入他们的值,用冒泡排序后输出。
package nn; public class nmz { public static void main(String[] args) { // TODO Auto-generated method stub int[] a = { 8, 6, 9, 2, 5 }; for (int i = 0; i < a.length - 1; i++) { for (int j = 0; j < a.length - 1; j++) if (a[j + 1] > a[j]) { int x = a[j + 1]; a[j + 1] = a[j]; a[j] = x; } } for (int i = 0; i < a.length; i++) { System.out.println(a[i] + ""); } } }
2、定义数组{34、22、35、67、45、66、12、33},输入一个数a,查找在数组中是否存在,如果存在,输出下标,不存在输出“not found”。
package nn; import java.util.Scanner; public class nmz { public static void main(String[] args) { // TODO Auto-generated method stub int[] a = { 34, 22, 35, 67, 45, 66, 12, 33 }; Scanner input = new Scanner(System.in); System.out.println("请输入一个数:"); int b = input.nextInt(); int xiabiao = -1; for (int i = 0; i < a.length; i++) { if (a[i] == b) { xiabiao = i; System.out.println(i); break; } } if (xiabiao == -1) System.out.println("not found"); } }
3、以矩阵的形式输出一个double型二维数组(长度分别为5、4,值自己设定)的值。
package nn; public class nmz { public static void main(String[] args) { // TODO Auto-generated method stub double[][] a= {{1,2,3,4,5},{6,7,8,9,10}, {11,12,13,14,15},{16,17,18,19,20}}; for (int i = 0; i < a.length; i++) { for (int j = 0; j < 5; j++) { System.out.print(a[i][j]+" "); } System.out.print(" "); } } }
4、定义一个二维数组(长度分别为3、4,值自己设定),求该二维数组的最大值。
package nn; public class nmz { public static void main(String[] args) { // TODO Auto-generated method stub int[][]a={{1,2,3,4},{5,6,7,8},{66,98,77,88}}; int max=0; for (int i = 0; i < 3; i++) { for (int j = 0; j < 4; j++) { if(a[i][j]>max){ max=a[i][j]; } } } System.out.println(max); } }