HashSet代码分析

 HashSet （jdk 1.7）的继承关系如下：

HashSet是使用HashMap实现的一个没有重复元素的集合。HashSet用法如下：

HashSet<String> hashSet = new HashSet<String>();  
hashSet.add("java001");  
hashSet.add("java01");  
hashSet.add("java011");  
hashSet.add("java002");  
hashSet.add("java004");  

从HashSet的add()方法可以看出，只有一个参数，并没有【键-值对】。

其实是HashSet只使用了HashMap的key，value统一是一个固定的Object，因此保证没有重复元素的方法，也是使用的HashMap的key来保证的。

既然HashSet是使用HashMap来实现的，那么HashSet必然会有一个HashMap的对象，即：

    private transient HashMap<E,Object> map;

    // Dummy value to associate with an Object in the backing Map
    private static final Object PRESENT = new Object();

    /**
     * Constructs a new, empty set; the backing <tt>HashMap</tt> instance has
     * default initial capacity (16) and load factor (0.75).
     */
    public HashSet() {
        map = new HashMap<>();
    }

其中，PRESENT = new Object()是用来作为HashMap的value使用的。可以通过其add()方法看到

    /**
     * Adds the specified element to this set if it is not already present.
     * More formally, adds the specified element <tt>e</tt> to this set if
     * this set contains no element <tt>e2</tt> such that
     * <tt>(e==null ? e2==null : e.equals(e2))</tt>.
     * If this set already contains the element, the call leaves the set
     * unchanged and returns <tt>false</tt>.
     *
     * @param e element to be added to this set
     * @return <tt>true</tt> if this set did not already contain the specified
     * element
     */
    public boolean add(E e) {
        return map.put(e, PRESENT)==null;
    }

在add()方法，添加的元素作为key，PRESENT作为value。

remove()方法也是调用的HashMap的remove()方法

    /**
     * Removes the specified element from this set if it is present.
     * More formally, removes an element <tt>e</tt> such that
     * <tt>(o==null ? e==null : o.equals(e))</tt>,
     * if this set contains such an element.  Returns <tt>true</tt> if
     * this set contained the element (or equivalently, if this set
     * changed as a result of the call).  (This set will not contain the
     * element once the call returns.)
     *
     * @param o object to be removed from this set, if present
     * @return <tt>true</tt> if the set contained the specified element
     */
    public boolean remove(Object o) {
        return map.remove(o)==PRESENT;
    }

　　

同样，size() / isEmpty() / contains() / clear() 都是调用的HashMap相应的方法。

    /**
     * Returns the number of elements in this set (its cardinality).
     *
     * @return the number of elements in this set (its cardinality)
     */
    public int size() {
        return map.size();
    }

    /**
     * Returns <tt>true</tt> if this set contains no elements.
     *
     * @return <tt>true</tt> if this set contains no elements
     */
    public boolean isEmpty() {
        return map.isEmpty();
    }
    /**
     * Returns <tt>true</tt> if this set contains the specified element.
     * More formally, returns <tt>true</tt> if and only if this set
     * contains an element <tt>e</tt> such that
     * <tt>(o==null ? e==null : o.equals(e))</tt>.
     *
     * @param o element whose presence in this set is to be tested
     * @return <tt>true</tt> if this set contains the specified element
     */
    public boolean contains(Object o) {
        return map.containsKey(o);
    }

    /**
     * Removes all of the elements from this set.
     * The set will be empty after this call returns.
     */
    public void clear() {
        map.clear();
    }

　　

另外，HashSet中有个更明显的方法来说明【HashSet只使用了HashMap的key】，即iterator()方法

    /**
     * Returns an iterator over the elements in this set.  The elements
     * are returned in no particular order.
     *
     * @return an Iterator over the elements in this set
     * @see ConcurrentModificationException
     */
    public Iterator<E> iterator() {
        return map.keySet().iterator();
    }