Here you can find the source of isEqualCollection(final Collection a, final Collection b)
public static boolean isEqualCollection(final Collection a, final Collection b)
//package com.java2s; //License from project: Apache License import java.util.*; public class Main { /**/*ww w . j av a2 s .com*/ * Returns <tt>true</tt> iff the given {@link Collection}s contain * exactly the same elements with exactly the same cardinality. * <p> * That is, iff the cardinality of <i>e</i> in <i>a</i> is * equal to the cardinality of <i>e</i> in <i>b</i>, * for each element <i>e</i> in <i>a</i> or <i>b</i>. */ public static boolean isEqualCollection(final Collection a, final Collection b) { if (a.size() != b.size()) { return false; } else { Map mapa = getCardinalityMap(a); Map mapb = getCardinalityMap(b); if (mapa.size() != mapb.size()) { return false; } else { Iterator it = mapa.keySet().iterator(); while (it.hasNext()) { Object obj = it.next(); if (getFreq(obj, mapa) != getFreq(obj, mapb)) { return false; } } return true; } } } /** * Returns a {@link Map} mapping each unique element in * the given {@link Collection} to an {@link Integer} * representing the number of occurances of that element * in the {@link Collection}. * An entry that maps to <tt>null</tt> indicates that the * element does not appear in the given {@link Collection}. */ public static Map getCardinalityMap(final Collection col) { HashMap count = new HashMap(); Iterator it = col.iterator(); while (it.hasNext()) { Object obj = it.next(); Integer c = (Integer) (count.get(obj)); if (null == c) { count.put(obj, new Integer(1)); } else { count.put(obj, new Integer(c.intValue() + 1)); } } return count; } private static final int getFreq(final Object obj, final Map freqMap) { try { return ((Integer) (freqMap.get(obj))).intValue(); } catch (NullPointerException e) { // ignored } catch (NoSuchElementException e) { // ignored } return 0; } }