@Yano
2019-09-20T10:45:51.000000Z
字数 3822
阅读 1484
Java
coding 笔记、点滴记录,以后的文章也会同步到公众号(Coding Insight)中,希望大家关注^_^
https://github.com/LjyYano/Thinking_in_Java_MindMapping
@Test
public void testLinkedHashMap() {
Map<String, Integer> map = new LinkedHashMap<String, Integer>() {
@Override
protected boolean removeEldestEntry(Map.Entry<String, Integer> eldest) {
return size() > 3;
}
};
map.put("1", 1);
map.put("2", 2);
map.put("3", 3);
map.put("4", 4);
for(Map.Entry<String, Integer> entry : map.entrySet()) {
System.out.println(entry.getKey() + ": " + entry.getValue());
}
}
输出
2: 2
3: 3
4: 4
迭代输出能够保持插入顺序。
LinkedHashMap继承自HashMap,内部额外维护了一个Entry的双向链表,用于记录访问和插入顺序。官方注释:
Hash table and linked list implementation of the Map interface, with predictable iteration order. This implementation differs from HashMap in that it maintains a doubly-linked list running through all of its entries. This linked list defines the iteration ordering, which is normally the order in which keys were inserted into the map (insertion-order). Note that insertion order is not affected if a key is re-inserted into the map.
需要注意:如果某个key已经存在,再次put不会改变插入的顺序。
LinkedHashMap.Entry只是比HashMap.Node多了两个指针而已,LinkedHashMap.Entry直接就是双向链表的元素了。
/**
* HashMap.Node subclass for normal LinkedHashMap entries.
*/
static class Entry<K,V> extends HashMap.Node<K,V> {
Entry<K,V> before, after;
Entry(int hash, K key, V value, Node<K,V> next) {
super(hash, key, value, next);
}
}
/**
* The head (eldest) of the doubly linked list.
*/
transient LinkedHashMap.Entry<K,V> head;
/**
* The tail (youngest) of the doubly linked list.
*/
transient LinkedHashMap.Entry<K,V> tail;
HashMap中定义的3个函数:
// Callbacks to allow LinkedHashMap post-actions
void afterNodeAccess(Node<K,V> p) { }
void afterNodeInsertion(boolean evict) { }
void afterNodeRemoval(Node<K,V> p) { }
LinkedHashMap继承于HashMap,重写了这3个函数。这3个函数分别为在访问节点、插入节点、删除节点时做一些事情。
在对LinkedHashMap进行put操作时,执行的是HashMap的put方法,多态调用LinkedHashMap的上述3个函数。
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;
}
public V get(Object key) {
Node<K,V> e;
if ((e = getNode(hash(key), key)) == null)
return null;
if (accessOrder)
afterNodeAccess(e);
return e.value;
}
可以看到其核心为这3个函数。
void afterNodeAccess(Node<K,V> e) { // move node to last
LinkedHashMap.Entry<K,V> last;
if (accessOrder && (last = tail) != e) {
LinkedHashMap.Entry<K,V> p =
(LinkedHashMap.Entry<K,V>)e, b = p.before, a = p.after;
p.after = null;
if (b == null)
head = a;
else
b.after = a;
if (a != null)
a.before = b;
else
last = b;
if (last == null)
head = p;
else {
p.before = last;
last.after = p;
}
tail = p;
++modCount;
}
}
void afterNodeInsertion(boolean evict) { // possibly remove eldest
LinkedHashMap.Entry<K,V> first;
if (evict && (first = head) != null && removeEldestEntry(first)) {
K key = first.key;
removeNode(hash(key), key, null, false, true);
}
}
void afterNodeRemoval(Node<K,V> e) { // unlink
LinkedHashMap.Entry<K,V> p =
(LinkedHashMap.Entry<K,V>)e, b = p.before, a = p.after;
p.before = p.after = null;
if (b == null)
head = a;
else
b.after = a;
if (a == null)
tail = b;
else
a.before = b;
}
时间复杂度应该是o(n),其中n为容量。因为put时在找到对应的value后,需要维护双向链表。
现在有一个疑问,LinkedHashMap存在有什么意义?既然要维护一个双向链表,就不可能做到HashMap o(1)的时间复杂度。LinkedHashMap和直接使用双向链表有什么区别?