public V put(K key, V value) {
        return putVal(hash(key), key, value, false, true);
    }

    // jdk8中hash方法较为简单，因为即便是在有冲突的情况下，红黑树的查询效率依然很高
    static final int hash(Object key) {
        int h;
        return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
    }

    final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
                   boolean evict) {
        Node<K,V>[] tab; Node<K,V> p; int n, i;
        // 如果table为空
        if ((tab = table) == null || (n = tab.length) == 0)
            n = (tab = resize()).length;
        // 根据hash值进行索引
        if ((p = tab[i = (n - 1) & hash]) == null)
            tab[i] = newNode(hash, key, value, null);
        else {
            Node<K,V> e; K k;
            // 判断链表头结点或者树的根节点是否为我们需要的值
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                e = p;
            // 如果p节点为树节点，执行红黑树的插入逻辑
            else if (p instanceof TreeNode)
                e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
            // 如果p节点为链表节点，执行链表的插入操作（尾插法）
            else {
                for (int binCount = 0; ; ++binCount) {
                    if ((e = p.next) == null) {
                        p.next = newNode(hash, key, value, null);
                        // 如果链表的节点数目大于等于8个，则转化为红黑树
                        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;
                }
            }
            // 找到了key对应的节点Node，根据onlyIfAbsent决定是否更新
            if (e != null) { // existing mapping for key
                V oldValue = e.value;
                if (!onlyIfAbsent || oldValue == null)
                    e.value = value;
                afterNodeAccess(e);
                return oldValue;
            }
        }
        // 未找到key，则表示节点为新加入的
        ++modCount;
        if (++size > threshold)
            // 扩容方法
            resize();
        afterNodeInsertion(evict);
        return null;
    }

public V get(Object key) {
        Node<K,V> e;
        return (e = getNode(hash(key), key)) == null ? null : e.value;
    }

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) {
                // 红黑树查询
                if (first instanceof TreeNode)
                    return ((TreeNode<K,V>)first).getTreeNode(hash, key);
                // 链表查询
                do {
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        return e;
                } while ((e = e.next) != null);
            }
        }
        return null;
    }

final Node<K,V> removeNode(int hash, Object key, Object value,
 boolean matchValue, boolean movable) {
 Node<K,V>[] tab; Node<K,V> p; int n, index;

 if ((tab = table) != null && (n = tab.length) > 0 &&
 (p = tab[index = (n - 1) & hash]) != null) {
 Node<K,V> node = null, e; K k; V v;
 // 先查询看看是够有该节点
 // 1.头结点查询
 if (p.hash == hash &&
 ((k = p.key) == key || (key != null && key.equals(k))))
 node = p;
 else if ((e = p.next) != null) {
 // 2.树节点查询
 if (p instanceof TreeNode)
 node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
 else {
 // 3.链表节点查询
 do {
 if (e.hash == hash &&
 ((k = e.key) == key ||
 (key != null && key.equals(k)))) {
 node = e;
 break;
 }
 p = e;
 } while ((e = e.next) != null);
 }
 }
 // 查询到了之后开始执行删除操作
 if (node != null && (!matchValue || (v = node.value) == value ||
 (value != null && value.equals(v)))) {
 // 1.删除树节点
 if (node instanceof TreeNode)
 ((TreeNode<K,V>)node).removeTreeNode(this, tab, movable);
 // 2.删除链表节点（如果节点为链表的头结点）
 else if (node == p)
 tab[index] = node.next;
 // 3.删除链表非头节点
 else
 p.next = node.next;
 ++modCount;
 --size;
 afterNodeRemoval(node);
 return node;
 }
 }
 return null;
 }