Java: sort an integer array without using Array.sort ()

Question

This is an instruction in one of the exercises in our Java class. First of all, I would like to say that I am “doing my homework”, and I'm not just lazy, asking someone from Qaru to answer this question for me. This particular element was my problem from all the other exercises, because I struggled to find the “perfect algorithm” for this.

Write a JAVA program that will enter 10 integer values ​​and display them in ascending or descending order. Note. Arrays .sort () are not allowed.

This is the code I came up with, it works, but it has one obvious flaw. If I enter the same value two or more, for example:

5, 5, 5, 4, 6, 7, 3, 2, 8, 10

Only one of the three entered 5s will be counted and included in the output. The output I get (for upstream):

2 3 4 5 0 0 6 7 8 10.

import java.util.Scanner; public class Exer3AscDesc { public static void main(String args[]) { Scanner scan = new Scanner(System.in); int tenNums[]=new int[10], orderedNums[]=new int[10]; int greater; String choice; //get input System.out.println("Enter 10 integers : "); for (int i=0;i<tenNums.length;i++) { System.out.print(i+1+"=> "); tenNums[i] = scan.nextInt(); } System.out.println(); //imperfect number ordering algorithm for(int indexL=0;indexL<tenNums.length;indexL++) { greater=0; for(int indexR=0;indexR<tenNums.length;indexR++) { if(tenNums[indexL]>tenNums[indexR]) { greater++; } } orderedNums[greater]=tenNums[indexL]; } //ask if ascending or descending System.out.print("Display order :\nA - Ascending\nD - Descending\nEnter your choice : "); choice = scan.next(); //output the numbers based on choice if(choice.equalsIgnoreCase("a")) { for(greater=0;greater<orderedNums.length;greater++) { System.out.print(orderedNums[greater]+" "); } } else if(choice.equalsIgnoreCase("d")) { for(greater=9;greater>-1;greater--) { System.out.print(orderedNums[greater]+" "); } } } } 

Answer 1

You can find so many different sorting algorithms on the Internet, but if you want to fix your decision, you can make the following changes to your code:

Instead:

  orderedNums[greater]=tenNums[indexL]; 

you need to do this:

 while (orderedNums[greater] == tenNums[indexL]) { greater++; } orderedNums[greater] = tenNums[indexL]; 

This code basically checks to see if this particular index is occupied by a similar number, then tries to find the next free index.

Note. Since the default value in your sorted array elements is 0, you need to make sure that 0 is not in your list. otherwise, you need to run a sorted array with a special number that you are not sure about your list, for example: Integer.MAX_VALUE

Answer 2

Simple sorting algorithm Sorting bubbles :

 public static void main(String[] args) { int[] arr = new int[] { 6, 8, 7, 4, 312, 78, 54, 9, 12, 100, 89, 74 }; for (int i = 0; i < arr.length; i++) { for (int j = i + 1; j < arr.length; j++) { int tmp = 0; if (arr[i] > arr[j]) { tmp = arr[i]; arr[i] = arr[j]; arr[j] = tmp; } } } } 

Answer 3

Here is one simple solution.

 public static void main(String[] args) { //Without using Arrays.sort function int i; int nos[] = {12,9,-4,-1,3,10,34,12,11}; System.out.print("Values before sorting: \n"); for(i = 0; i < nos.length; i++) System.out.println( nos[i]+" "); sort(nos, nos.length); System.out.print("Values after sorting: \n"); for(i = 0; i <nos.length; i++){ System.out.println(nos[i]+" "); } } private static void sort(int nos[], int n) { for (int i = 1; i < n; i++){ int j = i; int B = nos[i]; while ((j > 0) && (nos[j-1] > B)){ nos[j] = nos[j-1]; j--; } nos[j] = B; } } 

And the result:

Values ​​before sorting:

12
nine
-4
-one
3
10
34
12
eleven

Values ​​after sorting:

-4
-one
3
nine
10
eleven
12
12
34

Answer 4

I would recommend looking at Sort Sort or Insert Sort if you are not too worried about performance. Maybe this will give you some ideas.

Answer 5

A simple way:

 int a[]={6,2,5,1}; System.out.println(Arrays.toString(a)); int temp; for(int i=0;i<a.length-1;i++){ for(int j=0;j<a.length-1;j++){ if(a[j] > a[j+1]){ // use < for Descending order temp = a[j+1]; a[j+1] = a[j]; a[j]=temp; } } } System.out.println(Arrays.toString(a)); Output: [6, 2, 5, 1] [1, 2, 5, 6] 

Answer 6

Here you can use the bubble sort:

  //Time complexity: O(n^2) public static int[] bubbleSort(final int[] arr) { if (arr == null || arr.length <= 1) { return arr; } for (int i = 0; i < arr.length; i++) { for (int j = 1; j < arr.length - i; j++) { if (arr[j - 1] > arr[j]) { arr[j] = arr[j] + arr[j - 1]; arr[j - 1] = arr[j] - arr[j - 1]; arr[j] = arr[j] - arr[j - 1]; } } } return arr; } 

Answer 7

Sorting an array without using the built-in functions in java ...... just create a new file by canceling this name → (ArraySorting.java) ..... Run the project and enjoy it !!!!!

  import java.io.*; import java.util.Arrays; import java.util.Scanner; public class ArraySorting { public static void main(String args[]) { int temp=0; Scanner user_input=new Scanner(System.in); System.out.println("enter Size of Array..."); int Size=user_input.nextInt(); int[] a=new int[Size]; System.out.println("Enter element Of an Array..."); for(int j=0;j<Size;j++) { a[j]=user_input.nextInt(); } for(int index=0;index<a.length;index++) { for(int j=index+1;j<a.length;j++) { if(a[index] > a[j] ) { temp = a[index]; a[index] = a[j]; a[j] = temp; } } } System.out.print("Output is:- "); for(int i=0;i<a.length;i++) { System.out.println(a[i]); } } } 

Answer 8

 int x[] = { 10, 30, 15, 69, 52, 89, 5 }; int max, temp = 0, index = 0; for (int i = 0; i < x.length; i++) { int counter = 0; max = x[i]; for (int j = i + 1; j < x.length; j++) { if (x[j] > max) { max = x[j]; index = j; counter++; } } if (counter > 0) { temp = x[index]; x[index] = x[i]; x[i] = temp; } } for (int i = 0; i < x.length; i++) { System.out.println(x[i]); } 

Answer 9

 class Sort { public static void main(String[] args) { System.out.println("Enter the range"); java.util.Scanner sc=new java.util.Scanner(System.in); int n=sc.nextInt(); int arr[]=new int[n]; System.out.println("Enter the array values"); for(int i=0;i<=n-1;i++) { arr[i]=sc.nextInt(); } System.out.println("Before sorting array values are"); for(int i=0;i<=n-1;i++) { System.out.println(arr[i]); } System.out.println(); for(int pass=1;pass<=n;pass++) { for(int i=0;i<=n-1;i++) { if(i==n-1) { break; } int temp; if(arr[i]>arr[i+1]) { temp=arr[i]; arr[i]=arr[i+1]; arr[i+1]=temp; } } } System.out.println("After sorting array values are"); for(int i=0;i<=n-1;i++) { System.out.println(arr[i]); } } } 

Answer 10

Here is a sorting example. A simple example.

 public class SortingSimpleExample { public static void main(String[] args) { int[] a={10,20,1,5,4,20,6,4,2,5,4,6,8,-5,-1}; a=sort(a); for(int i:a) System.out.println(i); } public static int[] sort(int[] a){ for(int i=0;i<a.length;i++){ for(int j=0;j<a.length;j++){ int temp=0; if(a[i]<a[j]){ temp=a[j]; a[j]=a[i]; a[i]=temp; } } } return a; } } 

Answer 11

This will help you.

 int n[] = {4,6,9,1,7}; for(int i=n.length;i>=0;i--){ for(int j=0;j<n.length-1;j++){ if(n[j] > n[j+1]){ swapNumbers(j,j+1,n); } } } printNumbers(n); } private static void swapNumbers(int i, int j, int[] array) { int temp; temp = array[i]; array[i] = array[j]; array[j] = temp; } private static void printNumbers(int[] input) { for (int i = 0; i < input.length; i++) { System.out.print(input[i] + ", "); } System.out.println("\n"); }