Finding Permutations of an Array

Question

I was hoping I could get a little help on finding a way to cycle through an array and print all of the permutations of its elements. I have a short example of what I am trying to do. If someone could fill in the blank, I could understand how to use it in future programs.
The goal of the code below is to write a line to line, comma delimited listing of the elements of an int array 1 through 5. The order of the output does not matter to me, I'd just like to try and get all 120 permutations listed into a new file passed as an arg. I've tried thinking of how to switch the elements around, but I don't know the most efficient way of doing so. Any help would be appreciated.
// PermutationTest.java: Create a comma delimited text file of permutations of an array
import java.io.*;
public class PermutationTest {
/** Main method: args[0] is the output file **/
public static void main(String[] args) {
// Declare print stream
PrintWriter pw = null;
// Check usage
if (args.length != 1) {
System.out.println("Usage: java PermutationTest file");
System.exit(0);
File tempFile = new File(args[0]);
if (tempFile.exists()) {
System.out.println("The file " + args[0] +
" already exists, delete it, rerun the program");
System.exit(0);
// int array
int[] permArray = {1, 2, 3, 4, 5};
// Write data
try {
// Create data output stream for tempFile
pw = new PrintWriter(new FileOutputStream(tempFile), true);
* loop/sort of some kind to generate permutation *
// After finding a permutation, print a comma delimited output and repeat until all 120 permutations were written.
pw.println(permArray[0] + "," + permArray[1] + "," + permArray[2] + "," + permArray[3] + "," + permArray[4]);
catch (IOException ex) {
System.out.println(ex.getMessage());
finally {
// Close files
if (pw != null) pw.close();
}

Answer 1

Given a current permutation of a not necessarily unique sequence
p[0] p[1] p[2] ... p[n-1], the next permutation can be found as follows:
1) find a largest i such that p[ i ] < p[ i+1 ]; if no such i can be found, stop;
2) find a largest j > i such that p[ i ] < p[ j ]; such a j can always be found;
3) swap elements p[ i ] and p[ j ]
4) reverse the tail p[ i+1 ]... p[ n-1 ].
Repeat the steps above until step 1) fails.
kind regards,
Jos