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;if (p.hash == hash &&((k = p.key) == key || (key != null && key.equals(k)))) //删除节点在数组上node = p;else if ((e = p.next) != null) {if (p instanceof TreeNode) //删除节点在树上node = ((TreeNode<K,V>)p).getTreeNode(hash, key);else { // 删除节点在链表上do {if (e.hash == hash &&((k = e.key) == key ||(key != null && key.equals(k)))) {node = e;break;}p = e;} while ((e = e.next) != null);}}//matchValue的作用是指现在是否需要值匹配。因为可能没有传入value,所以判断一下if (node != null && (!matchValue || (v = node.value) == value ||(value != null && value.equals(v)))) { //if (node instanceof TreeNode) // 红黑树执行删除元素操作((TreeNode<K,V>)node).removeTreeNode(this, tab, movable);else if (node == p) // 如果要删除的node正好是首个元素tab[index] = node.next;elsep.next = node.next; //执行链表元素的删除操作++modCount;--size;afterNodeRemoval(node); //空方法 留给子类拓展 比如LinkedHashMapreturn node;}}return null;}
