---------[ HahsMap ]-------------
JDK 1.8
1、K、V 可以为 null，非线程安全
2、实现原理
- 数组(Node[] table) + 链表(Node) + 红黑树(TreeNode)
- HashMap使用 Node[] table 数组存放 Node<K, V>节点；
- 初始化HashMap的时候，table容量默认为16 （1 << 4）；
- Key hash值相同的节点，存入table相同位置，以链表结构存储（Node就是链表结构），新值存放在链表末尾；
- 当某个节点的链表长度超过8时，会转为红黑树TreeNode（若table长度未超过最小阀值（MIN_TREEIFY_CAPACITY = 64）时，优先扩容）；
读取时：
首先计算所在数组的位置 index = hash(K) & (n - 1)，然后再遍历链表或红黑树，取K相同得节点
3、何时触发扩容？扩容时，是否需要重新计算hash？
static final int DEFAULT_INITIAL_CAPACITY = 1 << 4; // aka 16
static final float DEFAULT_LOAD_FACTOR = 0.75f; // 负载因子
/**
* The smallest table capacity for which bins may be treeified.
* (Otherwise the table is resized if too many nodes in a bin.)
* Should be at least 4 * TREEIFY_THRESHOLD to avoid conflicts between resizing and treeification thresholds.
*/
static final int MIN_TREEIFY_CAPACITY = 64;
if (++size > threshold) resize();
// 双倍扩容 oldCap << 1
if (oldCap > 0) {
if (oldCap >= MAXIMUM_CAPACITY) {
threshold = Integer.MAX_VALUE;
return oldTab;
} else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY && oldCap >= DEFAULT_INITIAL_CAPACITY) {
newThr = oldThr << 1;
}
}
扩容时，不会重新计算key的hash值，但是会重新计算该节点在数组中的位置，并放入新的 Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];
4、什么时候由链表转为红黑树？
链表长度超过8，并且数组table容量超过64，若未超过64，则会触发数组扩容，直到容量大于64，才会转为红黑树
5、HashMap容量为什么是2的幂次方？
put(K key, V value) 操作时，存放的位置是按照key的hash值 与 （table的容量-1） 按位与（&）计算出来的
// index 一定小于 table的容量（2^n）
index = (2^n - 1) & hash(K)
/**
* e.hash & 2^n == 0，迁移至新数组时，index = j
* e.hash & 2^n != 0，迁移至新数组时，index = j + oldCap
* 二进制 & 按位与 运算结果
* 111 1000 1111 111 1000 1111
* 1111 1111 1111 111 111 111
* -------------- --------------
* 0111 1000 1111 111 0000 0111
*/
6、key的hash计算方法
(key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16)
7、源码解读
/**
* Implements Map.put and related methods
*
* @param hash hash for key
* @param key the key
* @param value the value to put
* @param onlyIfAbsent if true, don't change existing value
* @param evict if false, the table is in creation mode.
* @return previous value, or null if none
*/
final V putVal(int hash, K key, V value, boolean onlyIfAbsent, boolean evict) {
Node<K,V>[] tab; Node<K,V> p; int n, i;
// tab为空
if ((tab = table) == null || (n = tab.length) == 0)
n = (tab = resize()).length;
// i位置的节点为空，则添加一个新节点p
if ((p = tab[i = (n - 1) & hash]) == null)
tab[i] = newNode(hash, key, value, null);
else {
Node<K,V> e; K k;
// key的hash相同，key也相同，则节点替换
if (p.hash == hash && ((k = p.key) == key || (key != null && key.equals(k))))
e = p;
// p为TreeNode，红黑树
else if (p instanceof TreeNode)
e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
else {
// 链表
for (int binCount = 0; ; ++binCount) {
// 循环找到链表的末节点，添加新节点
if ((e = p.next) == null) {
p.next = newNode(hash, key, value, null);
// 该链表节点数超过阀值，将链表转为红黑树
if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
treeifyBin(tab, hash);
break;
}
// 判断新节点hash、key是否与当前遍历的节点一致，作替换
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;
}
扩容
/**
* 初始化 或 扩容
* Initializes or doubles table size.
* If null, allocates in accord with initial capacity target held in field threshold.
* Otherwise, because we are using power-of-two expansion,
* the elements from each bin must either stay at same index,
* or move with a power of two offset in the new table.
*
* @return the table
*/
final Node<K,V>[] resize() {
Node<K,V>[] oldTab = table;
int oldCap = (oldTab == null) ? 0 : oldTab.length;
int oldThr = threshold;
int newCap, newThr = 0;
if (oldCap > 0) {
if (oldCap >= MAXIMUM_CAPACITY) {
threshold = Integer.MAX_VALUE;
return oldTab;
}
// 双倍扩容 double threshold
else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY && oldCap >= DEFAULT_INITIAL_CAPACITY)
newThr = oldThr << 1;
}
else if (oldThr > 0) // initial capacity was placed in threshold
newCap = oldThr;
else { // zero initial threshold signifies using defaults
newCap = DEFAULT_INITIAL_CAPACITY;
newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);
}
if (newThr == 0) {
float ft = (float)newCap * loadFactor;
newThr = (newCap < MAXIMUM_CAPACITY && ft < (float)MAXIMUM_CAPACITY ? (int)ft : Integer.MAX_VALUE);
}
// 扩容后，新建数组，并赋值给table，迁移节点 至新数组
threshold = newThr;
@SuppressWarnings({"rawtypes","unchecked"})
Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];
table = newTab;
if (oldTab != null) {
// 遍历旧数组oldTab
for (int j = 0; j < oldCap; ++j) {
Node<K,V> e;
if ((e = oldTab[j]) != null) {
oldTab[j] = null;
// 当前节点链表只有一个元素，则重新计算在数组中的位置，赋值
if (e.next == null)
newTab[e.hash & (newCap - 1)] = e;
// 红黑树
else if (e instanceof TreeNode)
((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
else { // preserve order
// 链表反转，并重新计算该节点在数组中的位置
Node<K,V> loHead = null, loTail = null;
Node<K,V> hiHead = null, hiTail = null;
Node<K,V> next;
do {
next = e.next;
/**
* e.hash & 2^n == 0，迁移至新数组时，index = j
* e.hash & 2^n != 0，迁移至新数组时，index = j + oldCap
* 二进制 & 按位与 运算结果
* 111 1000 1111 111 1000 1111
* 1111 1111 1111 111 111 111
* -------------- --------------
* 0111 1000 1111 111 0000 0111
*/
if ((e.hash & oldCap) == 0) {
if (loTail == null)
loHead = e;
else
loTail.next = e;
loTail = e;
}
else {
if (hiTail == null)
hiHead = e;
else
hiTail.next = e;
hiTail = e;
}
} while ((e = next) != null);
if (loTail != null) {
loTail.next = null;
newTab[j] = loHead;
}
if (hiTail != null) {
hiTail.next = null;
newTab[j + oldCap] = hiHead;
}
}
}
}
}
return newTab;
}
链表转化为红黑树
/**
* 树化桶 - 将链表转为红黑树
* Replaces all linked nodes in bin at index for given hash
* unless table is too small, in which case resizes instead.
*/
final void treeifyBin(Node<K,V>[] tab, int hash) {
int n, index; Node<K,V> e;
// 如果table为空、或者tab.length < 树化的阀值，则扩容
if (tab == null || (n = tab.length) < MIN_TREEIFY_CAPACITY)
resize();
// index 为根据该hash值计算出来的table数组中存放的位置
else if ((e = tab[index = (n - 1) & hash]) != null) {
TreeNode<K,V> hd = null, tl = null;
do {
TreeNode<K,V> p = replacementTreeNode(e, null);
if (tl == null)
hd = p;
else {
p.prev = tl;
tl.next = p;
}
tl = p;
} while ((e = e.next) != null);
if ((tab[index] = hd) != null)
hd.treeify(tab);
}
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