HashSet结合源码分析

一、HashSet

1 /**
2  * This class implements the <tt>Set</tt> interface, backed by a hash table
3  * (actually a <tt>HashMap</tt> instance).  It makes no guarantees as to the
4  * iteration order of the set; in particular, it does not guarantee that the
5  * order will remain constant over time.  This class permits the <tt>null</tt>
6  * element.
7  */

　　HashSet是实现的set的接口，其底层主要是由HashMap来实现的，不保证存入的顺序。详细的HashMap详解http://www.cnblogs.com/houstao/p/8271362.html

 1  // Dummy value to associate with an Object in the backing Map
 2 //new了一个常量用量存value值
 3     private static final Object PRESENT = new Object();
 4 
 5     /**
 6      * Constructs a new, empty set; the backing <tt>HashMap</tt> instance has
 7      * default initial capacity (16) and load factor (0.75).
 8     *默认的构造器底层是new了一个HashMap，默认的数组的容量是16，加载因子为0.75
 9     *
10      */
11     public HashSet() {
12         map = new HashMap<>();
13     }

1  /**
2      * Returns the number of elements in this set (its cardinality).
3      *Hashset的长度就是map的长度
4      * @return the number of elements in this set (its cardinality)
5      */
6     public int size() {
7         return map.size();
8     }

   /**
     * Returns <tt>true</tt> if this set contains no elements.
     *判断是否为空，底层也是调用的map的isEmpty()方法
     * @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&nbsp;?&nbsp;e==null&nbsp;:&nbsp;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
    *调用的map底层的containsKey()的方法，map中的调用的getEntry，获    
     *取key-value键值对的方法，如果返回值为空话，则不包含该元素，否则包       
     *含
     */
    public boolean contains(Object o) {
        return map.containsKey(o);
    }

    /**
     * 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&nbsp;?&nbsp;e2==null&nbsp;:&nbsp;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
     *add方法是调用hashMap中的put方法，从这里你可以发现，set中add的 *元素相当于map中的key，value是由Object的对象常量存的。如果添加的元**素在map中存在，通过hash(key)计算在数组中存放位置，计算该元素的*hash是都与存在的元素相同，再通过equals判断，并且返回旧的value，故在*set中是不等于null的，因此添加不上，所以说，set中的元素是不重复的，也**相当于map中的key。如果在map中不存在该元素，则返回null，故set添***加。
     *
     *
     */
    public boolean add(E e) {
        return map.put(e, PRESENT)==null;
    }

    /**
     * Removes the specified element from this set if it is present.
     * More formally, removes an element <tt>e</tt> such that
     * <tt>(o==null&nbsp;?&nbsp;e==null&nbsp;:&nbsp;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;
    }

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

二、总结：

　　　　1、Hashset的中的元素就是HashMap中的key，因此，元素是唯一的不重复的，同时也是无序的。

　　　　2、Hashset的最底层实现是通过HashMap来实现的。

如有错误请多多指点。

转载于:https://www.cnblogs.com/houstao/p/8299328.html