LinkedHashMap 源码分析，底层竟这么简单！

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;
}

// possibly remove eldest
void afterNodeInsertion(boolean evict) {
  LinkedHashMap.Entry<K,V> first;
  if (evict && (first = head) != null && removeEldestEntry(first)) {
    K key = first.key;
    removeNode(hash(key), key, null, false, true);
  }
}

void afterNodeAccess(Node<K,V> e) {
  LinkedHashMap.Entry<K,V> last;
  //需要改变顺序 并且 当前节点不是最后一个
  if (accessOrder && (last = tail) != e) {
    // b 当前节点之前的节点
    // a 当前节点之后的节点
    LinkedHashMap.Entry<K,V> p =
      (LinkedHashMap.Entry<K,V>)e, b = p.before, a = p.after;
    //需要将p节点置为最后一个节点，所以置 p节点的 after 为 null
    p.after = null;
    B->P->A ===> B->P->E
    //如果没有前一个节点，所以 后一个节点置为 头节点
    if (b == null)
      head = a;
    else
      //否则 将 b.after 置为 a
      b.after = a;
    // B->P->A ===> B->A
    if (a != null)
      a.before = b;
    else
      // B->P->NULL ===> B->A
      last = b;
    //如果 last 为 null，将 p 置为头结点
    if (last == null)
      head = p;
    else {
      //B -> P -> NULL
      p.before = last;
      last.after = p;
    }
    //最后将tail置为 p 节点
    tail = p;
    ++modCount;
  }
}