java中的HashSet集合是怎么实现不重复的？

在做业务的时候经常会遇到需要对集合去重，最常用的就是HashSet。那HashSet是怎么实现的不重复存储的哪？各位看官往下看:

先看个最简单的构造方法

     * 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<>();
    }

很明显，HashSet底层是hashmap存储的。借大神的话

HashSet 就是HashMap的马甲       -----someone

很形象哈。

再看看add方法

// Dummy value to associate with an Object in the backing Map
 private transient HashMap<E,Object> map;
 private static final Object PRESENT = new Object();
/**
     * 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
     */
    public boolean add(E e) {
        return map.put(e, PRESENT)==null;
    }

add方法的参数（要存储的value）作为HashMap的key，PRESENT（Object PRESENT = new Object();）作为固定value。

重点看key（敲黑板）

HashMap中的put方法

    public V put(K key, V value) {
        return putVal(hash(key), key, value, false, true);
    }
   /**
     * Implements Map.put and related methods
     *
     * @param hash hash for key
     * @param key the key
     * @param value the value to put
     * @param onlyIfAbsent if true, don't change existing value
     * @param evict if false, the table is in creation mode.
     * @return previous value, or null if none
     */
    final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
                   boolean evict) {
        Node<K,V>[] tab; Node<K,V> p; int n, i;
        if ((tab = table) == null || (n = tab.length) == 0)
            n = (tab = resize()).length;
        if ((p = tab[i = (n - 1) & hash]) == null)
            tab[i] = newNode(hash, key, value, null);
        else {
            Node<K,V> e; K k;
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                e = p;
            else if (p instanceof TreeNode)
                e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
            else {
                for (int binCount = 0; ; ++binCount) {
                    if ((e = p.next) == null) {
                        p.next = newNode(hash, key, value, null);
                        if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
                            treeifyBin(tab, hash);
                        break;
                    }
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        break;
                    p = e;
                }
            }
            if (e != null) { // existing mapping for key
                V oldValue = e.value;
                if (!onlyIfAbsent || oldValue == null)
                    e.value = value;
                afterNodeAccess(e);
                return oldValue;
            }
        }
        ++modCount;
        if (++size > threshold)
            resize();
        afterNodeInsertion(evict);
        return null;
    }

这里边有两个看点：

• HashMap中key存储是hash后的值，对于String类型的相同值的hash值是一致的（其他接触类型类似，自定义对象类型需要重写hashcode方法与equel方法）。换句话说相同的值在hashMap中的存储位置是一样的。
• 基于上一点来看看怎么存储重复值的。如下代码对于hashMap中已经存在的key，key不变，新value覆盖就value。对于HashSet而言新旧value都是PRESENT对象，所以set在存储的时候就不会重复。

            if (e != null) { // existing mapping for key
              V oldValue = e.value;
                if (!onlyIfAbsent || oldValue == null)
                    e.value = value;
                afterNodeAccess(e);
                return oldValue;
            }

所以hashset中存储的值输出的顺序和存储的先后顺序不一致，而是按照值的hash顺序输出。

总结：

通过分析HashSet的实现原理，可以肯定的是它的去重效率是很高的，前提是去重对象需要有hashcode、equel方法的实现。除此外HashMap所拥有的大多数特性都适用于HashSet。