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 zhang {
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 zhang {
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 zhang {
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 zhang {
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 zhang {
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]);
}
}
}