HashSet类是如何实现添加元素保证不重复的

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参考：!HashSet类是如何实现添加元素保证不重复的---哈希码的原理

总结

向HashSet中add()元素时，判断元素是否存在的依据，不仅要比较hash值，同时还要结合equles方法比较。
HashSet中的add()方法会使用HashMap的add()方法。

源码：

类中map和PARENT的定义：

private transient HashMap<E,Object> map;
// Dummy value to associate with an Object in the backing Map
//用来匹配Map中后面的对象的一个虚拟值
private static final Object PRESENT = new Object();

HashSet 类中的add()方法：

//boolean java.util.HashSet.add(E e)
public boolean add(E e) {
        return map.put(e, PRESENT)==null;
    }

map.put(e, PRESENT)是来自HashMap中的put方法

//java.util.HashMap.put(K key, V value)
public V put(K key, V value) {
        return putVal(hash(key), key, value, false, true);
    }

hash()函数

//java.util.HashMap.put(K key, V value)
static final int hash(Object key) {
        int h;
        return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
    }

//java.util.HashMap.put(K key, V value)
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;
    }