This is the instruction in one of the exercises in our Java class. Before anything else, I would like to say that I 'do my homework' and I'm not just being lazy asking someone on Stack Overflow to answer this for me. This specific item has been my problem out of all the other exercises because I've been struggling to find the 'perfect algorithm' for this.
Write JAVA program that will input 10 integer values and display either in ascending or descending order. Note: Arrays.sort() is not allowed.
This is the code I have come up with, it works but it has one obvious flaw. If I enter the same value twice or more, for example:
5, 5, 5, 4, 6, 7, 3, 2, 8, 10
Only one of the three 5s entered would be counted and included in the output. The output I get (for the ascending order) is:
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]+" ");
}
}
}
}
You can find so many different sorting algorithms in internet, but if you want to fix your own solution you can do following changes in your code:
Instead of:
orderedNums[greater]=tenNums[indexL];
you need to do this:
while (orderedNums[greater] == tenNums[indexL]) {
greater++;
}
orderedNums[greater] = tenNums[indexL];
This code basically checks if that particular index is occupied by a similar number, then it will try to find next free index.
Note: Since the default value in your sorted array elements is 0, you need to make sure 0 is not in your list. otherwise you need
to initiate your sorted array with an especial number that you sure is
not in your list e.g: Integer.MAX_VALUE
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]);
}
}
}
Bubble sort can be used here:
//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;
}
I would recommend looking at Selection sort or Insertion sort if you aren't too worried about performance. Maybe that will give you some ideas.
here is the Sorting Simple Example try it
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;
}
}
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]);
}
This will surely 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");
}
int[] arr = {111, 111, 110, 101, 101, 102, 115, 112};
/* for ascending order */
System.out.println(Arrays.toString(getSortedArray(arr)));
/*for descending order */
System.out.println(Arrays.toString(getSortedArray(arr)));
private int[] getSortedArray(int[] k){
int localIndex =0;
for(int l=1;l<k.length;l++){
if(l>1){
localIndex = l;
while(true){
k = swapelement(k,l);
if(l-- == 1)
break;
}
l = localIndex;
}else
k = swapelement(k,l);
}
return k;
}
private int[] swapelement(int[] ar,int in){
int temp =0;
if(ar[in]<ar[in-1]){
temp = ar[in];
ar[in]=ar[in-1];
ar[in-1] = temp;
}
return ar;
}
private int[] getDescOrder(int[] byt){
int s =-1;
for(int i = byt.length-1;i>=0;--i){
int k = i-1;
while(k >= 0){
if(byt[i]>byt[k]){
s = byt[k];
byt[k] = byt[i];
byt[i] = s;
}
k--;
}
}
return byt;
}
output:-
ascending order:-
101, 101, 102, 110, 111, 111, 112, 115
descending order:-
115, 112, 111, 111, 110, 102, 101, 101
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]
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 output is:
Values before sorting:
12
9
-4
-1
3
10
34
12
11
Values after sorting:
-4
-1
3
9
10
11
12
12
34
Simple sorting algorithm Bubble sort:
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;
}
}
}
}
Array Sorting without using built in functions in java ......just make new File unsing 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]);
}
}
}
Source: Stackoverflow.com