Question

/**
 * Returns the value of the mapping with the specified key.
 *
 * @param key
 *      the key.
 * @return the value of the mapping with the specified key, or {@code null}
 *     if no mapping for the specified key is found.
 */
@Override public V get(Object key) {
  /*
   * This method is overridden to eliminate the need for a polymorphic
   * invocation in superclass at the expense of code duplication.
   */
  if (key == null) {
    HashMapEntry<K, V> e = entryForNullKey;
    if (e == null)
      return null;
    if (accessOrder)
      makeTail((LinkedEntry<K, V>) e);
    return e.value;
  }

  int hash = Collections.secondaryHash(key);
  HashMapEntry<K, V>[] tab = table;
  for (HashMapEntry<K, V> e = tab[hash & (tab.length - 1)];
     e != null; e = e.next) {
    K eKey = e.key;
    if (eKey == key || (e.hash == hash && key.equals(eKey))) {
      if (accessOrder)
        makeTail((LinkedEntry<K, V>) e);
      return e.value;
    }
  }
  return null;
}

其实仔细看 if (accessOrder) 的逻辑即可，如果 accessOrder=true 那么每次 get 都会执行 N 次 makeTail(LinkedEntry<K, V> e) 。

接下来看看：