Sorting Array In Javaneed source code
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12 Replies - 14150 Views - Last Post: 21 November 2008 - 01:03 PM
#1
Sorting Array In Java
Posted 15 January 2005 - 03:04 AM
1, a program that reads frm user 12 integer numbers,store them in an array and then extracts fro this array the largest and the smallest number within it then display these numbers to the console.
2.modify this program so that it uses a vector class from the java.util.*class
Replies To: Sorting Array In Java
#2
Re: Sorting Array In Java
Posted 15 January 2005 - 08:34 AM
Also, do you have any code? Unfortunatley, we can't help you unless you provide us with what you have so far or atleast the pseudo code.
#3
Re: Sorting Array In Java
Posted 30 October 2008 - 11:02 PM
skyhawk133, on 15 Jan, 2005 - 08:34 AM, said:
Also, do you have any code? Unfortunatley, we can't help you unless you provide us with what you have so far or atleast the pseudo code.
Hi , Could you help me by adding an insert sort method to my code that generates an array of 100 random integers and if possible can I know the time it took to sort?
MANY THANKS
Geetanjali
import java.util.Random;
public class RandomArrayToSort {
public static void main(String[] args) {
new RandomArrayToSort();
}
public void create()
{
Random r = new Random();
// generate a uniformly distributed int random numbers
int[] integers = new int[100];
for (int i = 0; i < integers.length; i++) {
integers[i] = r.nextInt();
}
for (int i : integers) {
System.out.print(i + ", "+"\n");
}
System.out.println("");
}
}
** Edit **
#4
Re: Sorting Array In Java
Posted 30 October 2008 - 11:44 PM
Please help
Thanks
Geetanjali Dewan
import java.util.Random;
public class RandomArrayToSort {
public static void main(String[] args) {
new RandomArrayToSort();
}
public void create()
{
Random r = new Random();
// generate a uniformly distributed int random numbers
int[] integers = new int[100];
for (int i = 0; i < integers.length; i++) {
integers[i] = r.nextInt();
}
for (int i : integers) {
System.out.print(i +"\n");
}
System.out.println("");
int last = integers.length;
for ( int i = 0 ; i < last ; i++)
for (int j = i+1 ; j < last +1; j++)
if (integers[i] < integers[j])
{
int temp = integers[i];
integers[i] = integers[j];
integers[j] = temp;
}
System.out.println("Records Sorted");
}
}
** Edit **
#5
Re: Sorting Array In Java
Posted 31 October 2008 - 08:44 AM
for(int i=0; i<integers.length-1; i++) {
for(int j=0; j<integers.length-1-i; j++) {
if(integers[j] > integers[j+1]) {
int temp = integers[j];
integers[j] = integers[j+1];
integers[j+1] = temp;
}
}
}
#6
Re: Sorting Array In Java
Posted 31 October 2008 - 09:35 AM
Im kinda desperate pleeease.Thanks again.
Geetanjali
BigAnt, on 31 Oct, 2008 - 08:44 AM, said:
for(int i=0; i<integers.length-1; i++) {
for(int j=0; j<integers.length-1-i; j++) {
if(integers[j] > integers[j+1]) {
int temp = integers[j];
integers[j] = integers[j+1];
integers[j+1] = temp;
}
}
}
#7
Re: Sorting Array In Java
Posted 31 October 2008 - 09:47 AM
-394590862
-1211847198
-624061471
962627250
-2060369180
941358999
-341520185
1927901582
-1045353644
126383582
The output of the array after the sort is:
-2060369180
-1211847198
-1045353644
-624061471
-394590862
-341520185
126383582
941358999
962627250
1927901582
Records Sorted
which is sorted correctly.
#8
Re: Sorting Array In Java
Posted 01 November 2008 - 12:19 AM
Am I doing something wrong in the bit where I generate an array of random numbers.
I am submitting my complete code for you to see .Sorry for being so slow in understanding.
Thanks in advance
Geetanjali
My code is as below:
import java.util.Random;
public class RandomArrayToSort {
public static void main(String[] args) {
new RandomArrayToSort();
}
public void create()
{
Random r = new Random();
// generate a uniformly distributed int random numbers
int[] integers = new int[1000];
for (int i = 0; i < integers.length; i++) {
integers[i] = r.nextInt();
}
for (int i : integers) {
System.out.print(i +"\n");
}
System.out.println("");
int last = integers.length;
for(int i=0; i<integers.length-1; i++) {
for(int j=0; j<integers.length-1-i; j++) {
if(integers[j] > integers[j+1]) {
int temp = integers[j];
integers[j] = integers[j+1];
integers[j+1] = temp;
}
}
}
System.out.println("Records Sorted");
}
}
Please help.
BigAnt, on 31 Oct, 2008 - 09:47 AM, said:
-394590862
-1211847198
-624061471
962627250
-2060369180
941358999
-341520185
1927901582
-1045353644
126383582
The output of the array after the sort is:
-2060369180
-1211847198
-1045353644
-624061471
-394590862
-341520185
126383582
941358999
962627250
1927901582
Records Sorted
which is sorted correctly.
#9
Re: Sorting Array In Java
Posted 01 November 2008 - 07:49 AM
import java.util.Random;
public class RandomArrayToSort {
public static void main(String[] args) {
RandomArrayToSort theArray = new RandomArrayToSort();
theArray.create();
}
public void create(){
Random r = new Random();
// generate a uniformly distributed int random numbers
int[] integers = new int[1000];
for (int i = 0; i < integers.length; i++) {
integers[i] = r.nextInt();
}
for (int i : integers) {
System.out.println(i);//System.out.print(i +"\n"); Just use the println() method as it does this for you
}
System.out.println(); //System.out.println(""); If you just want to print a new line the argument isn't needed
//int last = integers.length; //Not used, using array's length directly in loops
for(int i=0; i<integers.length-1; i++) {
for(int j=0; j<integers.length-1-i; j++) {
if(integers[j] > integers[j+1]) {
int temp = integers[j];
integers[j] = integers[j+1];
integers[j+1] = temp;
}
}
}
System.out.println("Records Sorted");
System.out.println("\nSorted array:");
for(int i: integers){
System.out.println(i);
}
}
}
#10
Re: Sorting Array In Java
Posted 02 November 2008 - 06:56 AM
I get it now.
Much appreciated
Geetanjali
BigAnt, on 1 Nov, 2008 - 07:49 AM, said:
import java.util.Random;
public class RandomArrayToSort {
public static void main(String[] args) {
RandomArrayToSort theArray = new RandomArrayToSort();
theArray.create();
}
public void create(){
Random r = new Random();
// generate a uniformly distributed int random numbers
int[] integers = new int[1000];
for (int i = 0; i < integers.length; i++) {
integers[i] = r.nextInt();
}
for (int i : integers) {
System.out.println(i);//System.out.print(i +"\n"); Just use the println() method as it does this for you
}
System.out.println(); //System.out.println(""); If you just want to print a new line the argument isn't needed
//int last = integers.length; //Not used, using array's length directly in loops
for(int i=0; i<integers.length-1; i++) {
for(int j=0; j<integers.length-1-i; j++) {
if(integers[j] > integers[j+1]) {
int temp = integers[j];
integers[j] = integers[j+1];
integers[j+1] = temp;
}
}
}
System.out.println("Records Sorted");
System.out.println("\nSorted array:");
for(int i: integers){
System.out.println(i);
}
}
}
#11
Re: Sorting Array In Java
Posted 20 November 2008 - 09:26 PM
Its me again.The bubble sort code seemed to work perfectly on my computer all this while.
Now when I try to run it on another computer I get an error in the for loop line
for (int i : integers)
it expects me to write a ; as is written in for loops.What do I do and how come it works just linke this on my laptop?
Secondly , is it possible to know the time taken by a sort code to process when I make my array size large , say 10,0000..Is there a method I should write or should I go to control panel performance and observe ?
Pl help.
Im sure I will get the best solution to both my problems on this amazing forum.
Thanks in advance
Geetanjali
geetanjalidewan, on 2 Nov, 2008 - 05:56 AM, said:
I get it now.
Much appreciated
Geetanjali
BigAnt, on 1 Nov, 2008 - 07:49 AM, said:
import java.util.Random;
public class RandomArrayToSort {
public static void main(String[] args) {
RandomArrayToSort theArray = new RandomArrayToSort();
theArray.create();
}
public void create(){
Random r = new Random();
// generate a uniformly distributed int random numbers
int[] integers = new int[1000];
for (int i = 0; i < integers.length; i++) {
integers[i] = r.nextInt();
}
for (int i : integers) {
System.out.println(i);//System.out.print(i +"\n"); Just use the println() method as it does this for you
}
System.out.println(); //System.out.println(""); If you just want to print a new line the argument isn't needed
//int last = integers.length; //Not used, using array's length directly in loops
for(int i=0; i<integers.length-1; i++) {
for(int j=0; j<integers.length-1-i; j++) {
if(integers[j] > integers[j+1]) {
int temp = integers[j];
integers[j] = integers[j+1];
integers[j+1] = temp;
}
}
}
System.out.println("Records Sorted");
System.out.println("\nSorted array:");
for(int i: integers){
System.out.println(i);
}
}
}
#12
Re: Sorting Array In Java
Posted 21 November 2008 - 01:46 AM
Quote
for (int i : integers)
it expects me to write a ; as is written in for loops.What do I do and how come it works just linke this on my laptop?
You are working with a different version of Java installed on each.
for (int i : integers)
will only work for Java >= 1.5. Use normal for loops if you want it to work on either
#13
Re: Sorting Array In Java
Posted 21 November 2008 - 01:03 PM
Quote
You could do this by calling System.currentTimeMillis() before and after the method was called and then subtracting them to get the execution time.
**Also you may want to check out a benchmarking framework such as JBench.
This post has been edited by BigAnt: 21 November 2008 - 01:07 PM
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