转载请注明出处:http://blog.csdn.net/crazy1235/article/details/75579654
与JDK1.7中HashMap的实现相比,JDK1.8做了如下改动:
hash()函数算法修改
table数组的类型,由Entry改成了Node
HashMap存储数据的结构由数组+链表,进化成了 数组+链表+(RBT)红黑树
即使加载因子和Hash算法设计的再合理,也免不了出现拉链过长的情况,所以JDK1.8中,当拉链超过一定长度(8)时,会将链表转成一颗红黑树!
(图片转自网络)
重点关注与1.7中实现不同的地方!
构造函数
public HashMap(int initialCapacity, float loadFactor) {
if (initialCapacity < 0)
throw new IllegalArgumentException("Illegal initial capacity: " +
initialCapacity);
if (initialCapacity > MAXIMUM_CAPACITY)
initialCapacity = MAXIMUM_CAPACITY;
if (loadFactor <= 0 || Float.isNaN(loadFactor))
throw new IllegalArgumentException("Illegal load factor: " +
loadFactor);
this.loadFactor = loadFactor;
this.threshold = tableSizeFor(initialCapacity); // 初始化阈值
}
/**
* 根据容量计算阈值(临界值)
*/
static final int tableSizeFor(int cap) {
int n = cap - 1;
n |= n >>> 1;
n |= n >>> 2;
n |= n >>> 4;
n |= n >>> 8;
n |= n >>> 16;
return (n < 0) ? 1 : (n >= MAXIMUM_CAPACITY) ? MAXIMUM_CAPACITY : n + 1;
}
Node
// 与1.7中 Entry的内容大同小异,只是换了个名称而已!
static class Node<K,V> implements Map.Entry<K,V> {
final int hash;
final K key;
V value;
Node<K,V> next; // 指向下一个节点
Node(int hash, K key, V value, Node<K,V> next) {
this.hash = hash;
this.key = key;
this.value = value;
this.next = next;
}
public final K getKey() { return key; }
public final V getValue() { return value; }
public final String toString() { return key + "=" + value; }
public final int hashCode() {
return Objects.hashCode(key) ^ Objects.hashCode(value);
}
public final V setValue(V newValue) {
V oldValue = value;
value = newValue;
return oldValue;
}
public final boolean equals(Object o) {
if (o == this)
return true;
if (o instanceof Map.Entry) {
Map.Entry<?,?> e = (Map.Entry<?,?>)o;
if (Objects.equals(key, e.getKey()) &&
Objects.equals(value, e.getValue()))
return true;
}
return false;
}
}
hash
相比JDK1.7中的hash()函数,1.8做了简化!
static final int hash(Object key) {
int h;
return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}
如果key是null,则返回0
如果key不是null,则得到key的hashCode值,右移16位之后,做异或运算!(高位参与运算)
Hash算法计算得到的结果越分散,发生碰撞的几率就越小,map的存取效率就会越高!
(图片转自网路)
put
// 这个常量的意思就是,当一个bucket是一个链表,链表个数大于等于8时,就要树状化,也就是要从链表结构变成红黑树结构
static final int TREEIFY_THRESHOLD = 8;
先来看看put()的实现
// 如果已经存在key对应的节点,则覆盖value值
public V put(K key, V value) {
return putVal(hash(key), key, value, false, true);
}
// 如果已经存在key对应的节点,不覆盖value值
@Override
public V putIfAbsent(K key, V value) {
return putVal(hash(key), key, value, true, true);
}
重点来看 putVal() 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) // 如果map为空时,调用resize()进行初始化!
n = (tab = resize()).length;
if ((p = tab[i = (n - 1) & hash]) == null) // 如果没有在数组中找到对应的节点,则直接插入一个Node (未发生碰撞)
tab[i] = newNode(hash, key, value, null);
else { // 找到了(n - 1) & hash 对应下标的数组(tab)中的节点 ,也就是发生了碰撞
Node<K,V> e; K k;
// 1. hash值一样,key值一样,则找到目标Node
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
// 2. 数组中找到的这个节点p是TreeNode类型,则需要插入到RBT里面一个节点
else if (p instanceof TreeNode)
e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
else {
// 3. 不是TreeNode类型,则表示是一个链表,这里就类似与jdk1.7中的操作
for (int binCount = 0; ; ++binCount) { // 遍历链表
if ((e = p.next) == null) {
// 4. 此时查找当前链表的次数已经超过7个,则需要链表RBT化!
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)))) // 5. 找到链表中对应的节点
break;
p = e;
}
}
// 如果e不为空,则表示在HashMap中找到了对应的节点
if (e != null) { // existing mapping for key
V oldValue = e.value;
// 当onlyIfAbsent=false 或者key对应的旧value为空时,用新的value替换旧value
if (!onlyIfAbsent || oldValue == null)
e.value = value;
afterNodeAccess(e);
return oldValue;
}
}
++modCount; // 操作次数+1
if (++size > threshold) // hashmap节点个数+1,并判断是否超过阈值,如果超过则重建结构!
resize();
afterNodeInsertion(evict);
return null;
}
下面主要关注是三个函数:
putTreeVal(this, tab, hash, key, value);
treeifyBin(tab, hash);
treeify();
resize();
putTreeVal() 函数的目的就是往RBT中插入一个节点,但是牵涉到平衡化的方法,所以相对来讲难一些!
treeifyBin
final void treeifyBin(Node<K,V>[] tab, int hash) {
int n, index; Node<K,V> e;
if (tab == null || (n = tab.length) < MIN_TREEIFY_CAPACITY)
resize();
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);
}
}
// 将Node对象转成TreeNode对象
TreeNode<K,V> replacementTreeNode(Node<K,V> p, Node<K,V> next) {
return new TreeNode<>(p.hash, p.key, p.value, next);
}
treeify 主要涉及二叉树的创建,就是一个一个add节点,涉及到平衡的算法!
先来说下一红黑树的定义:
任何一个节点都是有颜色的,黑色或者红色
根节点是黑色的
父子结点之间不能出现两个连续的红色节点
任何一个节点向下遍历到其子孙的叶子节点,所经历的黑色结点个数必须相等
空节点被认为是黑色的
final void treeify(Node<K,V>[] tab) {
TreeNode<K,V> root = null;
for (TreeNode<K,V> x = this, next; x != null; x = next) {
next = (TreeNode<K,V>)x.next;
x.left = x.right = null;
if (root == null) {
x.parent = null;
x.red = false; // 根节点是黑色的
root = x;
}
else {
K k = x.key;
int h = x.hash;
Class<?> kc = null;
// 遍历root,插入到RBTree中
for (TreeNode<K,V> p = root;;) {
int dir, ph;
K pk = p.key;
if ((ph = p.hash) > h)
dir = -1;
else if (ph < h)
dir = 1;
else if ((kc == null &&
(kc = comparableClassFor(k)) == null) ||
(dir = compareComparables(kc, k, pk)) == 0)
dir = tieBreakOrder(k, pk);
TreeNode<K,V> xp = p;
if ((p = (dir <= 0) ? p.left : p.right) == null) {
x.parent = xp;
if (dir <= 0)
xp.left = x;
else
xp.right = x;
// 修正红黑树
root = balanceInsertion(root, x);
break;
}
}
}
}
moveRootToFront(tab, root);
}
插入结点,只有插入节点的父节点是空色的,才需要考虑变换结点的顺序进行平衡。
总结下来,有三种情况需要反转
- 叔叔节点也是红色
- 叔叔节点为空,且祖父结点、父节点和新节点在一条斜线上
- 叔叔节点为空,且祖父结点、父节点和新节点不在一条斜线上
static <K,V> TreeNode<K,V> balanceInsertion(TreeNode<K,V> root,
TreeNode<K,V> x) {
// 新节点置为红色
x.red = true;
for (TreeNode<K,V> xp, xpp, xppl, xppr;;) {
// x的父节点是空时,x即为根节点,将x变为黑色
if ((xp = x.parent) == null) {
x.red = false;
return x;
}
//此时不需要平衡,直接返回结合!
else if (!xp.red || (xpp = xp.parent) == null)
return root;
// 如果x的父节点是x的祖父节点的左节点
if (xp == (xppl = xpp.left)) {
// 如果xp的兄弟节点是红色,此时需要反转
if ((xppr = xpp.right) != null && xppr.red) {
xppr.red = false;
xp.red = false;
xpp.red = true;
x = xpp;
}
else {
如果x是父节点的右节点,则此时需要先左旋后右旋
if (x == xp.right) {
root = rotateLeft(root, x = xp); // 左旋
xpp = (xp = x.parent) == null ? null : xp.parent;
}
if (xp != null) {
xp.red = false;
if (xpp != null) {
xpp.red = true;
root = rotateRight(root, xpp); // 右旋
}
}
}
}
// 这里else分支是上面if分支的镜像,对称操作即可
else {
if (xppl != null && xppl.red) {
xppl.red = false;
xp.red = false;
xpp.red = true;
x = xpp;
}
else {
if (x == xp.left) {
root = rotateRight(root, x = xp);
xpp = (xp = x.parent) == null ? null : xp.parent;
}
if (xp != null) {
xp.red = false;
if (xpp != null) {
xpp.red = true;
root = rotateLeft(root, xpp);
}
}
}
}
}
}
get
// 根据key值找到value
public V get(Object key) {
Node<K,V> e;
return (e = getNode(hash(key), key)) == null ? null : e.value;
}
/**
* Implements Map.get and related methods
*
* @param hash hash for key
* @param key the key
* @return the node, or null if none
*/
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) {
// always check first node
if (first.hash == hash &&
((k = first.key) == key || (key != null && key.equals(k))))
return first;
if ((e = first.next) != null) {
// 如果是红黑树结构则去查询红黑树
if (first instanceof TreeNode)
return ((TreeNode<K,V>)first).getTreeNode(hash, key);
// 如果不是则遍历链表
do {
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
return e;
} while ((e = e.next) != null);
}
}
return null;
}
此时主要关注在RBT中查找一个节点的函数原理
/**
* Calls find for root node.
*/
final TreeNode<K,V> getTreeNode(int h, Object k) {
return ((parent != null) ? root() : this).find(h, k, null);
}
// 实际上就是一个遍历红黑树查找节点的函数
final TreeNode<K,V> find(int h, Object k, Class<?> kc) {
TreeNode<K,V> p = this;
do {
int ph, dir; K pk;
TreeNode<K,V> pl = p.left, pr = p.right, q;
if ((ph = p.hash) > h) // h小于p(根节点)的hash值,则h代表的节点在p的左子树上
p = pl;
else if (ph < h) // h大于ph值时,h代表的节点在p的右子树上
p = pr;
// 如果hash中相等,且key值相等,则命中节点p。
else if ((pk = p.key) == k || (k != null && k.equals(pk)))
return p;
else if (pl == null)
p = pr;
else if (pr == null)
p = pl;
else if ((kc != null ||
(kc = comparableClassFor(k)) != null) &&
(dir = compareComparables(kc, k, pk)) != 0)
p = (dir < 0) ? pl : pr;
else if ((q = pr.find(h, k, kc)) != null)
return q;
else
p = pl;
} while (p != null);
return null;
}
- 发生了碰撞的意思就是遇到了在同一个bucket上的节点
resize
扩容
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) {
// 如果超过了最大容量之则直接返回旧的数组(也就是不扩容李处了)
if (oldCap >= MAXIMUM_CAPACITY) {
threshold = Integer.MAX_VALUE;
return oldTab;
}
// 没有超过最大值,左移一位(扩大1倍)
else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
oldCap >= DEFAULT_INITIAL_CAPACITY)
// 将阈值也扩大一倍
newThr = oldThr << 1; // double threshold
}
else if (oldThr > 0) // initial capacity was placed in threshold
newCap = oldThr;
else { // 这个分支只有在调用空构造函数时,初始化走这个
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);
}
threshold = newThr;
@SuppressWarnings({"rawtypes","unchecked"})
// 创建新的数组
Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];
table = newTab;
if (oldTab != null) { //扩容时走这个if
for (int j = 0; j < oldCap; ++j) { // 遍历旧数组
Node<K,V> e;
if ((e = oldTab[j]) != null) {
oldTab[j] = null; //释放旧数组中占据的资源
if (e.next == null)
newTab[e.hash & (newCap - 1)] = e; //hash运算并取模,放到新数组中
else if (e instanceof TreeNode)
((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
else { // preserve order
Node<K,V> loHead = null, loTail = null;
Node<K,V> hiHead = null, hiTail = null;
Node<K,V> next;
do {
next = e.next;
// 原索引
if ((e.hash & oldCap) == 0) {
if (loTail == null)
loHead = e;
else
loTail.next = e;
loTail = e;
}
// 原索引+oldCap
else {
if (hiTail == null)
hiHead = e;
else
hiTail.next = e;
hiTail = e;
}
} while ((e = next) != null);
// 原索引放到bucket里
if (loTail != null) {
loTail.next = null;
newTab[j] = loHead;
}
// 原索引+oldCap放到bucket里
if (hiTail != null) {
hiTail.next = null;
newTab[j + oldCap] = hiHead;
}
}
}
}
}
return newTab;
}
参考
http://www.cnblogs.com/dennyzhangdd/p/6745282.html
https://tech.meituan.com/java-hashmap.html
http://www.jianshu.com/p/e694f1e868ec
关于红黑树的具体实现操作可以参考美团的这篇文章:
https://zhuanlan.zhihu.com/p/24367771