publicVput(K key, Vvalue){// 对key为null的处理if(key ==null)returnputForNullKey(value);// 计算hashinthash = hash(key);// 得到在table中的indexinti = indexFor(hash, table.length);// 遍历table[index],是否key已经存在,存在则替换,并返回旧值for(Entry<K,V> e = table[i]; e !=null...
LinkedHashMap没有提供remove方法,所以调用的是HashMap的remove方法,实现如下: 代码语言:javascript 代码运行次数:0 运行 AI代码解释 public V remove(Object key) { Entry<K,V> e = removeEntryForKey(key); return (e == null ? null : e.value); } final Entry<K,V> removeEntryForKey(Object key)...
void addEntry(int hash, K key, V value, int bucketIndex) { // 调用父类的addEntry,增加一个Entry到HashMap中 super.addEntry(hash, key, value, bucketIndex); // removeEldestEntry方法默认返回false,不用考虑 Entry<K,V> eldest = header.after; if (removeEldestEntry(eldest)) { removeEntryForKey...
Node<K,V> e = next;// 下一个元素要访问节点if(modCount != expectedModCount)thrownewConcurrentModificationException();if(e ==null)thrownewNoSuchElementException();// 首先迭代自身所在链表if((next = (current = e).next) ==null&& (t = table) !=null) {// 如果链表访问结束,遍历哈希桶数组...
if(removeEldestEntry(eldest)) {removeEntryForKey(eldest.key); }else{//扩容到原来的2倍if(size>= threshold)resize(2* table.length); } }voidcreateEntry(inthash, Kkey, V value,intbucketIndex) {// 向哈希表中插入Entry,这点与HashMap中相同//创建新的Entry并将其链入到数组对应桶的链表的头结点...
removeEntryForKey(eldest.key); } else { //扩容到原来的2倍 if (size >= threshold) resize(2 * table.length); } } void createEntry(int hash, K key, V value, int bucketIndex) { // 向哈希表中插入Entry,这点与HashMap中相同 //创建新的Entry并将其链入到数组对应桶的链表的头结点处, ...
null : e.value; } final Node<K,V> removeNode(int hash, Object key, Object value, boo...
(except through the iterator's ownremoveoperation, or through thesetValueoperation on a map entry returned by the iterator) the results of the iteration are undefined. The set supports element removal, which removes the corresponding mapping from the map, via theIterator.remove,Set.remove,...
Notice howthe order of elements in the key set is transformed as we perform access operations on the map. Simply put, any access operation on the map results in an order such that the element that was accessed would appear last if an iteration were to be carried out right away. ...
* Returns <tt>true</tt> if this map should remove its eldest entry. * This method is invoked by <tt>put</tt> and <tt>putAll</tt> after * inserting a new entry into the map. It provides the implementor * with the opportunity to remove the eldest entry each time a new one ...