.1.编写一个简单程序,要求数组长度为5,静态赋值10,20,30,40,50,在控制台输出该数组的值。
package test1; public class finish { public static void main(String[] args) { // TODO Auto-generated method stub int[] arr={10,20,30,40,50}; for(int i=0;i<5;i++){ System.out.println(arr[i]); } } }
2.编写一个简单程序,要求数组长度为5,动态赋值10,20,30,40,50,在控制台输出该数组的值。
package test1; public class finish { public static void main(String[] args) { // TODO Auto-generated method stub int [] a = new int[5]; a[0] = 10; a[1] = 20; a[2] = 30; a[3] = 40; a[4] = 50; for(int i = 0; i < a.length; i++) { System.out.println(a[i]); } } }
3.编写一个简单程序,定义整型数组,里面的元素是{23,45,22,33,56},求数组元素的和、平均值
package test1; public class finish { public static void main(String[] args) { // TODO Auto-generated method stub int [] a = {23,45,22,33,56}; double sum = 0; for(int i = 0; i < a.length; i++) { sum = sum + a[i]; } System.out.println("数组元素的和是:" + sum); System.out.println("数组元素的平均值是:" + sum / a.length); } }
4.在一个有8个整数(18,25,7,36,13,2,89,63)的数组中找出其中最大的数及其下标
package test1; public class finish { public static void main(String[] args) { // TODO Auto-generated method stub int [] arr = {18,25,7,36,13,2,89,63}; int max = arr[0]; int a=0; for(int i = 1; i < arr.length; i++) { if(arr[i] > max){ max = arr[i]; a=i; } } System.out.println("数组中最大值是:"+ max); System.out.println("下标为:" + a); } }
5. 将一个数组中的元素逆序存放(知识点:数组遍历、数组元素访问)
package test1; public class finish { public static void main(String[] args) { // TODO Auto-generated method stub int [] a = {23,56,7,94,15,2,35,73}; for (int i = 0; i < a.length / 2; i++) { int t = a[i]; a[i] = a[a.length - 1 - i]; a[a.length - 1 - i] =t; } for (int i = 0; i < a.length; i++) { System.out.print(a[i] + " "); } } }
6.有一个已经排好序的数组。现输入一个数,要求按原来的规律将它插入数组中。
package test1; public class finish { public static void main(String[] args) { // TODO Auto-generated method stub int[] m = {14,23,56,81}; int[] n = new int[5]; int a = 32; for(int i =0; i< m.length; i++){ if(m[i] > a){ n[i] = a; for(i = i + 1; i < n.length; i++ ){ n[i] = m[i-1]; } } else{ n[i] = m[i]; } } for(int i = 0; i<5; i++){ System.out.println(n[i]); } } }