Ascending and Descending Order number in java

Question

I am doing an ascending and descending order number in java and here is my code:

System.out.print("Enter How Many Inputs: "); int num1 = Integer.parseInt(in.readLine()); int arr[] = new int[num1]; for (int i = 0; i<num1; i++) { System.out.print("Enter Value #" + (i + 1) + ":"); arr[i] =Integer.parseInt(in.readLine()); } System.out.print("Numbers in Ascending Order:" ); for(int i = 0; i < arr.length; i++) { Arrays.sort(arr); System.out.print( " " +arr[i]); } System.out.println(" "); System.out.print("Numbers in Descending Order: " ); 

The code currently generates the following:

 Enter How Many Inputs: 5 Enter Value #1:3 Enter Value #2:5 Enter Value #3:6 Enter Value #4:11 Enter Value #5:2 Numbers in Ascending Order: 2 3 5 6 11 Numbers in Descending Order: 

So the call to Arrays.sort(arr) seems to work - but I'm looking for a similarly simple way to provide a downward sort, and can't find it in the documentation. Any ideas?

Answer 1

Three possible solutions come to my mind:

1. Change the order:

 //convert the arr to list first Collections.reverse(listWithNumbers); System.out.print("Numbers in Descending Order: " + listWithNumbers); 

2. Go back and print it:

 Arrays.sort(arr); System.out.print("Numbers in Descending Order: " ); for(int i = arr.length - 1; i >= 0; i--){ System.out.print( " " +arr[i]); } 

3. Sort it with the "oposite" comparator:

 Arrays.sort(arr, new Comparator<Integer>(){ int compare(Integer i1, Integer i2) { return i2 - i1; } }); // or Collections.reverseOrder(), could be used instead System.out.print("Numbers in Descending Order: " ); for(int i = 0; i < arr.length; i++){ System.out.print( " " +arr[i]); } 

Answer 2

 public static void main(String[] args) { Scanner input =new Scanner(System.in); System.out.print("enter how many:"); int num =input.nextInt(); int[] arr= new int [num]; for(int b=0;b<arr.length;b++){ System.out.print("enter no." + (b+1) +"="); arr[b]=input.nextInt(); } for (int i=0; i<arr.length;i++) { for (int k=i;k<arr.length;k++) { if(arr[i] > arr[k]) { int temp=arr[k]; arr[k]=arr[i]; arr[i]=temp; } } } System.out.println("******************\n output\t accending order"); for (int i : arr){ System.out.println(i); } } } 

Answer 3

you can make two functions for the upstream and another for lowering the following two functions: after converting the array to a list

 public List<Integer> sortDescending(List<Integer> arr){ Comparator<Integer> c = Collections.reverseOrder(); Collections.sort(arr,c); return arr; } 

next function

 public List<Integer> sortAscending(List<Integer> arr){ Collections.sort(arr); return arr; } 

Answer 4

 package pack2; import java.util.Scanner; public class group { public static void main(String[] args) { // TODO Auto-generated method stub Scanner data= new Scanner(System.in); int value[]= new int[5]; int temp=0,i=0,j=0; System.out.println("Enter 5 element of array"); for(i=0;i<5;i++) value[i]=data.nextInt(); for(i=0;i<5;i++) { for(j=i;j<5;j++) { if(value[i]>value[j]) { temp=value[i]; value[i]=value[j]; value[j]=temp; } } } System.out.println("Increasing Order:"); for(i=0;i<5;i++) System.out.println(""+value[i]); } 

Answer 5

 int arr[] = { 12, 13, 54, 16, 25, 8, 78 }; for (int i = 0; i < arr.length; i++) { Arrays.sort(arr); System.out.println(arr[i]); } 

Answer 6

Sort the array in the same way as before, but print the elements in reverse order using a loop that counts, not counts.

Also, move the sort out of the loop - you are currently sorting the array again and again, when you only need to sort it once.

  Arrays.sort(arr); for(int i = 0; i < arr.length; i++){ //Arrays.sort(arr); // not here System.out.print( " " +arr[i]); } for(int i = arr.length-1; i >= 0; i--){ //Arrays.sort(arr); // not here System.out.print( " " +arr[i]); } 

Answer 7

Just sort the array in ascending order and print it back.

 Arrays.sort(arr); for(int i = arr.length-1; i >= 0 ; i--) { //print arr[i] } 

Answer 8

First you can sort the array and then skip it twice, once in both directions:

 Arrays.sort(arr); System.out.print("Numbers in Ascending Order:" ); for(int i = 0; i < arr.length; i++){ System.out.print( " " + arr[i]); } System.out.print("Numbers in Descending Order: " ); for(int i = arr.length - 1; i >= 0; i--){ System.out.print( " " + arr[i]); } 

Answer 9

 Arrays.sort(arr, Collections.reverseOrder()); for(int i = 0; i < arr.length; i++){ System.out.print( " " +arr[i]); } 

And move Arrays.sort() from this loop. You are sorting the same array at each iteration.

Answer 10

You can take an upstream array and print it in reverse order, so replace the second statement:

 for(int i = arr.length - 1; i >= 0; i--) { ... } 

If you have Apache commons-lang in the classpath, it has an ArrayUtils.reverse (int []) method that you can use.

By the way, you probably don't want to sort it in every for loop.

Answer 11

Why are you using array and worried about the first question about the number of numbers you need?

Prefer ArrayList associated with the corresponding comparator:

 List numbers = new Arraylist(); //add read numbers (int (with autoboxing if jdk>=5) or Integer directly) into it //Initialize the associated comparator reversing order. (since Integer implements Comparable) Comparator comparator = Collections.reverseOrder(); //Sort the list Collections.sort(numbers,comparator); 

Answer 12

I did it this way (I'm new to java (also in programming))

 import java.util.Scanner; 

public class SortingNumbers {

 public static void main(String[] args) { Scanner scan1=new Scanner(System.in); System.out.print("How many numbers you want to sort: "); int a=scan1.nextInt(); int i,j,k=0; // i and j is used in various loops. int num[]=new int[a]; int great[]= new int[a]; //This array elements will be used to store "the number of being greater." Scanner scan2=new Scanner(System.in); System.out.println("Enter the numbers: "); for(i=0;i<a;i++) num[i] = scan2.nextInt(); for (i=0;i<a;i++) { for(j=0;j<a;j++) { if(num[i]>num[j]) //first time when executes this line, i=0 and j=0 and then i=0;j=1 and so on. each time it finishes second for loop the value of num[i] changes. k++;} great[i]=k++; //At the end of each for loop (second one) k++ contains the total of how many times a number is greater than the others. k=0;} // And then, again k is forced to 0, so that it can collect (the total of how many times a number is greater) for another number. System.out.print("Ascending Order: "); for(i=0;i<a;i++) for(j=0;j<a;j++) if(great[j]==i) System.out.print(num[j]+","); //there is a fixed value for each great[j] that is, from 0 upto number of elements(input numbers). System.out.print("Discending Order: "); for(i=0;i<=a;i++) for(j=0;j<a;j++) if(great[j]==ai) System.out.print(+num[j]+","); } 

}