JAVA第八周作业0423
package com.homework8; /*1、编写一个简单程序,要求数组长度为5,分别赋值10,20,30,40,50, 在控制台输出该数组的值。(知识点:数组定义和创建、一维数组初始化)[必做题]?*/ public class Exe8_1_1 { public static void main(String[] args) { int a[]={10,20,30,40,50}; //静态赋值 for (int i : a) { System.out.print(i +"\t"); } System.out.println();//动态赋值 for (int i = 0; i < a.length; i++) { a[i] =(i+1)*10; } for (int i : a) { System.out.print(i +"\t"); } } }
package com.homework8; /*2、将一个字符数组的值(neusofteducation)拷贝到另一个字符数组中。 (知识点:数组复制) [必做题]?*/ public class Exe8_1_2 { public static void main(String[] args) { char a[] ={'a','b','c','d','e','f','g'}; char b[] ={'A','B','C','D','E','F','G'}; System.arraycopy(a, 1, b, 1, 4); for (char c : b) { System.out.print(c +"\t"); } } }
package com.homework8; import java.util.Arrays; /*3、给定一个有9个整数(1,6,2,3,9,4,5,7,8)的数组,先排序, 然后输出排序后的数组的值。(知识点:Arrays.sort排序、冒泡排序)*/ public class Exe8_1_3 { public static void main(String[] args) { int a[] =new int[]{1,6,2,3,9,4,5,7,8}; Arrays.sort(a);//Arrays.sort排序 for (int i : a) { System.out.print(i +"\t"); } System.out.println();//冒泡排序 for (int i = 0; i < a.length; i++) { for (int j = i; j < a.length; j++) { if(a[i]>a[j]){ int x =a[i]; a[i] =a[j]; a[j] =x; } } } for (int i : a) { System.out.print(i +"\t"); } } }
package com.homework8; /*4、 输出一个double型二维数组(长度分别为5、4,值自己设定)的值。 (知识点:数组定义和创建、多维数组初始化、数组遍历)*/ public class Exe8_1_4 { public static void main(String[] args) { double a[][] =new double[5][4]; for (int i = 0; i < 5; i++) { for (int j = 0; j < 4; j++) { a[i][j] =i+j+2; } } for (int i = 0; i < 5; i++) { for (int j = 0; j < 4; j++) { System.out.print(a[i][j] +"\t"); } System.out.println(); } } }
package com.homework8; /*5、 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。(知识点:数组遍历、数组元素访问) [必做题]?*/ public class Exe8_1_5 { public static void main(String[] args) { int a[] =new int[]{18,25,7,36,13,2,89,63}; int max =0; int maxidx =0; for (int i = 0; i < a.length; i++) { if(a[i]>max) max =a[i]; maxidx =i; } System.out.println("max=" +max +"\nmaxidx="+maxidx); } }
package com.homework8; //6、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) public class Exe8_2_1 { public static void main(String[] args) { int a[] =new int[10]; for (int i = 0; i < a.length; i++) { a[i] =i; } for (int i : a) { System.out.print(i +" "); } System.out.println(); for (int i = 0; i < a.length; i++) { for (int j = i; j < a.length; j++) { int temp =a[i]; a[i] =a[j]; a[j] =temp; } } for (int i : a) { System.out.print(i +" "); } } }
package com.homework8; /*7. 将一个数组中的重复元素保留一个其他的清零。 (知识点:数组遍历、数组元素访问)*/ public class Exe8_2_2 { public static void main(String[] args) { int[] a ={1,2,2,3,4,4,5,5,6,6,6,7,7,8,8,4,1}; for (int i : a) { System.out.print(i +"\t"); } System.out.println(); for (int i = 0; i < a.length; i++) { for (int j = i+1; j < a.length; j++) { if (a[i] == a[j]) a[j] =0; } } for (int i : a) { System.out.print(i +"\t"); } } }
package com.homework8; /*8、给定一维数组{ -10,2,3,246,-100,0,5},计算出数组中的 平均值、最大值、最小值。(知识点:数组遍历、数组元素访问)*/ public class Exe8_2_3 { public static void main(String[] args) { int a[] =new int[]{-10,2,3,246,-100,0,5}; int sum = 0; int max =a[0]; int min =a[0]; for (int i = 0; i < a.length; i++) { if (max < a[i]) { max = a[i]; } if (min > a[i]) { min = a[i]; } sum +=a[i]; } System.out.println("最大值是" +max); System.out.println("最小值是" +min); System.out.println("平均值是:" +sum*1.0/a.length); } }
package com.homework8; //9、使用数组存放裴波那契数列的前20项 ,并输出 1 1 2 3 5 8 13 21 public class Exe8_2_4 { public static void main(String[] args) { int a[] =new int [20]; a[0] = 1; a[1] = 1; for (int i = 2; i < a.length; i++) { a[i] =a[i-1] +a[i-2]; } for (int i : a) { System.out.print(i +" "); } } }
package com.homework8; import java.util.Arrays; import java.util.Random; /*10、生成一个长度为10的随机整数数组(每个数都是0-100之间), 输出,排序后,再输出*/ public class Exe8_2_5 { public static void main(String[] args) { Random r =new Random();//生成随机数 int a[] =new int[10]; for (int i = 0; i < a.length; i++) { a[i] =r.nextInt(100); } Arrays.sort(a);//Arrays.sort排序及foreach遍历 for (int i : a) { System.out.print(i +"\t"); } System.out.println();//冒泡排序 for (int i = 0; i < a.length; i++) { for (int j = i; j < a.length; j++) { if (a[i] < a[j]) { int temp =a[i]; a[i] =a[j]; a[j] =temp; } } } for (int i : a) { System.out.print(i +"\t"); } } }