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A fast and easy to use dense matrix linear algebra library written in Java.
There is a newer version: 0.25
/*
 * Copyright (c) 2009-2011, Peter Abeles. All Rights Reserved.
 *
 * This file is part of Efficient Java Matrix Library (EJML).
 *
 * EJML is free software: you can redistribute it and/or modify
 * it under the terms of the GNU Lesser General Public License as
 * published by the Free Software Foundation, either version 3
 * of the License, or (at your option) any later version.
 *
 * EJML is distributed in the hope that it will be useful,
 * but WITHOUT ANY WARRANTY; without even the implied warranty of
 * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
 * GNU Lesser General Public License for more details.
 *
 * You should have received a copy of the GNU Lesser General Public
 * License along with EJML.  If not, see .
 */

package org.ejml.alg.dense.misc;

import java.util.ArrayList;
import java.util.List;


/**
 * Generates a permutations of an integer set from 0 to N-1.  This can either be generated
 * all at once as a list of one at a time.
 *
 * @author Peter Abeles
 */
public class PermuteArray {

    // used by next
    private int level;

    private int data[];
    private int iter[];
    private int valk[];
    private int ret[];


    public PermuteArray( int N ) {
        level = 0;
        iter = new int[ N ];
        valk = new int[ N ];
        data = new int[ N ];
        ret = new int[ N ];
        for( int i = 0; i < data.length; i++ ) {
            data[i] = -1;
        }
    }

    /**
     * 
     * Returns signature of the permutation.  This is the sgn() operator and returns
     * -1 or 1 depending on it being odd or even.
     * 

     * sgn(σ) = ( − 1)^m

     * 

     * where m is the number of inversions.
     * 
     * 
     * NOTE: This implementation is painfully slow O(N!).  There is probably another algorithm out there
     * which is much faster.
     * 
     *
     * @return -1 or 1 for odd or even permutations.
     */
    public int sgn() {
        // Is there a way to compute the parity while performing the permutations
        // making this much less expensive
        int total = 0;

        for( int i = 0; i < ret.length; i++ ) {
            int val = ret[i];

            for( int j = i+1; j < ret.length; j++ ) {
                if( val > ret[j] ) {
                    total++;
                }
            }
        }

        if( total % 2 == 1 )
            return -1;
        return 1;
    }

    /**
     * Computes N factorial
     */
    public static int fact( int N ) {
        int ret = 1;

        while( N > 0 ) {
            ret *= N--;
        }

        return ret;
    }

    /**
     * Creates a list of all permutations for a set with N elements.
     *
     * @param N Number of elements in the list being permuted.
     * @return A list containing all the permutations.
     */
    public static List createList( int N )
    {
        int data[] = new int[ N ];
        for( int i = 0; i < data.length; i++ ) {
            data[i] = -1;
        }

        List ret = new ArrayList();

        createList(data,0,-1,ret);

        return ret;
    }


    /**
     * Internal function that uses recursion to create the list
     */
    private static void createList( int data[], int k , int level , List ret )
    {
        data[k] = level;

        if( level < data.length-1 ) {
            for( int i = 0; i < data.length; i++ ) {
                if( data[i] == -1 ) {
                    createList(data,i,level+1,ret);
                }
            }
        } else {
            int []copy = new int[data.length];
            System.arraycopy(data,0,copy,0,data.length);
            ret.add(copy);
        }
        data[k] = -1;
    }

    /**
     * Creates the next permutation in the sequence.
     *
     * @return An array containing the permutation.  The returned array is modified each time this function is called.
     */
    public int[] next()
    {
        boolean hasNewPerm = false;

        escape:while( level >= 0) {
//            boolean foundZero = false;
            for( int i = iter[level]; i < data.length; i = iter[level] ) {
                iter[level]++;

                if( data[i] == -1 ) {
                    level++;
                    data[i] = level-1;

                    if( level >= data.length ) {
                        // a new permutation has been created return the results.
                        hasNewPerm = true;
                        System.arraycopy(data,0,ret,0,ret.length);
                        level = level-1;
                        data[i] = -1;
                        break escape;
                    } else {
                        valk[level] = i;
                    }
                }
            }

            data[valk[level]] = -1;
            iter[level] = 0;
            level = level-1;

        }

        if( hasNewPerm )
            return ret;
        return null;
    }
}