注意 ， 本文基于JDK 1.8

HashMap#get()

    /** * Returns the value to which the specified key is mapped, * or {@code null} if this map contains no mapping for the key. * * <p>More formally, if this map contains a mapping from a key * {@code k} to a value {@code v} such that {@code (key==null ? k==null : * key.equals(k))}, then this method returns {@code v}; otherwise * it returns {@code null}. (There can be at most one such mapping.) * * <p>A return value of {@code null} does not <i>necessarily</i> * indicate that the map contains no mapping for the key; it's also * possible that the map explicitly maps the key to {@code null}. * The {@link #containsKey containsKey} operation may be used to * distinguish these two cases. * * @see #put(Object, Object) */
    public V get(Object key) {
        Node<K,V> e;
        return (e = getNode(hash(key), key)) == null ? null : e.value;
    }
    /** * Implements Map.get and related methods * * @param hash hash for key * @param key the key * @return the node, or null if none */
    final Node<K,V> getNode(int hash, Object key) {
        Node<K,V>[] tab; Node<K,V> first, e; int n; K k;
        if ((tab = table) != null && (n = tab.length) > 0 &&
            (first = tab[(n - 1) & hash]) != null) {
            if (first.hash == hash && // always check first node
                ((k = first.key) == key || (key != null && key.equals(k))))
                return first;
            if ((e = first.next) != null) {
                if (first instanceof TreeNode)
                    return ((TreeNode<K,V>)first).getTreeNode(hash, key);
                do {
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        return e;
                } while ((e = e.next) != null);
            }
        }
        return null;
    }

上面注释中说的比较重要一点就是：如果返回值是null，并不一定是没有这种KV映射，也有可能是该key映射的值value是null，即key-null的映射。也就是说，使用get方法并不能判断这个key是否存在，只能通过containsKey方法来实现。
由此可见get方法调用的是getNode方法，返回一个Node。getNode方法接受两个参数hash值和key值。
首先判断first Node，在判断的时候，先看hash值是否相等，再看地址是否相等，再看equals的返回值。这个写的挺好。
然后再遍历，判断first是不是树节点，是的话，在树中查找；否则，遍历链表。
关于红黑树的操作，有时间另写博文讨论。

HashMap#containsKey()

    /** * Returns <tt>true</tt> if this map contains a mapping for the * specified key. * * @param key The key whose presence in this map is to be tested * @return <tt>true</tt> if this map contains a mapping for the specified * key. */
    public boolean containsKey(Object key) {
        return getNode(hash(key), key) != null;
    }