HashMap-Hash冲突解决

转载自：https://blog.csdn.net/u012712901/article/details/78313130

用途为个人学习，如有冲突请提醒！

背景：我们常用HashMap作为我们Java开发时的K-V数据存储结构(如id-person，这个ID对应这个人)。我们知道他们的数据结构么，它的Hash值是什么意义。Hash冲突是怎么解决的。我们带着这2个问题将HashMap做个整体剖析。（其实还有一个问题是，它怎么进行动态扩容的）

一、HashMap的数据结构是什么。

下面是HashMap中的源码。其实HashMap的本质是Node数组；K-V结构对应的基础数据结构就是以下源码

  
  
transient HashMap.Node<K, V>[] table;

static class Node<K,V> implements Map.Entry<K,V> {
//Hash值
final int hash;
//Key值
final K key;
//Value值
V value;
//当前Node对应的下个Node(用于解决Hash冲突；稍后讲解)
Node<K,V> next;

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 int hashCode() {
return Objects.hashCode(key) ^ Objects.hashCode(value);
}
}
  
  

根据上面代码我们知悉，HashMap中基本的数据结构是Node。

  
  
Map persons = new HashMap< String, String>();
//put方法会新建一个Node
//hash=k.hashCode() ^ k >>> 16 ; hash是数组所在位置
//K="1",V="jack";next=null;(无Hash冲突情况)
persons.put( "1", "jack");
persons.put( "2", "john");
  
  

上述的Node数据结构中的hash值的意义是将Node均匀的放置在数组[]中。获得hash值后使用在table[(n - 1) & hash] 位置置值。

  
  
/**
* Computes key.hashCode() and spreads (XORs) higher bits of hash
* to lower. Because the table uses power-of-two masking, sets of
* hashes that vary only in bits above the current mask will
* always collide. (Among known examples are sets of Float keys
* holding consecutive whole numbers in small tables.) So we
* apply a transform that spreads the impact of higher bits
* downward. There is a tradeoff between speed, utility, and
* quality of bit-spreading. Because many common sets of hashes
* are already reasonably distributed (so don't benefit from
* spreading), and because we use trees to handle large sets of
* collisions in bins, we just XOR some shifted bits in the
* cheapest possible way to reduce systematic lossage, as well as
* to incorporate impact of the highest bits that would otherwise
* never be used in index calculations because of table bounds.
*/
static final int hash(Object var0) {
int var1;
return var0 == null? 0:(var1 = var0.hashCode()) ^ var1 >>> 16;
}
  
  

二、HashMap怎么处理hash冲突

上述说道，Node中的hash值是用hash算法得到的，目的是均匀放置，那如果put的过于频繁，会造成不同的key-value计算出了同一个hash，这个时候hash冲突了，那么这2个Node都放置到了数组的同一个位置。HashMap是怎么处理这个问题的呢？

解决方案：HashMap的数据结构是：数组Node[]与链表Node中有next Node.

(1)如果上述的 persons.put(“1”,”jack”);persons.put(“2”,”john”); 同时计算到的hash值都为123，那么jack先放在第一列的第一个位置Node-jack，persons.put(“2”,”john”);执行时会将Node-jack的next(Node) = Node(john)，Jack的下个节点将指向Node(john)。

(2)那么取的时候呢，persons.get(“2”)，这个时候取得的hash值是123，即table[123]，这时table[123]其实是Node-jack，Key值不相等，取Node-jack的next下个Node，即Node-John，这时Key值相等了，然后返回对应的person.

三、HashMap怎么进行动态扩容

JDK1.7的逻辑相对简单，JDK1.8使用了红黑树TreeMap相对复杂，现在用1.7的进行讲解。本质上其实一样。

  
  
void resize(int newCapacity) { //传入新的容量
Entry[] oldTable = table; //引用扩容前的Entry数组
int oldCapacity = oldTable.length;
if (oldCapacity == MAXIMUM_CAPACITY) { //扩容前的数组大小如果已经达到最大(2^30)了
threshold = Integer.MAX_VALUE; //修改阈值为int的最大值(2^31-1)，这样以后就不会扩容了
return;
}

Entry[] newTable = new Entry[newCapacity]; //初始化一个新的Entry数组
transfer(newTable); //！！将数据转移到新的Entry数组里
table = newTable; //HashMap的table属性引用新的Entry数组
threshold = ( int) (newCapacity * loadFactor); //修改阈值
}

void transfer(Entry[] newTable) {
Entry[] src = table; //src引用了旧的Entry数组
int newCapacity = newTable.length;
for ( int j = 0; j < src.length; j++) { //遍历旧的Entry数组
Entry<K, V> e = src[j]; //取得旧Entry数组的每个元素
if (e != null) {
src[j] = null; //释放旧Entry数组的对象引用（for循环后，旧的Entry数组不再引用任何对象）
do {
Entry<K, V> next = e.next;
int i = indexFor(e.hash, newCapacity); //！！重新计算每个元素在数组中的位置
e.next = newTable[i]; //标记[1]
newTable[i] = e; //将元素放在数组上
e = next; //访问下一个Entry链上的元素
} while (e != null);
}
}
}

static int indexFor(int h, int length) {
return h & (length - 1);
}
  
  

扩容的方式是新建一个newTab，是oldTab的2倍。遍历oldTab，将oldTab赋值进对应位置的newTab。与ArrayList中的扩容逻辑基本一致，只不过ArrayList是当前容量+(当前容量>>1)。

JDK1.8实现resize()方式

  
  
/*
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;
}
else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
oldCap >= DEFAULT_INITIAL_CAPACITY)
newThr = oldThr << 1; // double threshold
}
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);
}
threshold = newThr;
@SuppressWarnings({ "rawtypes", "unchecked"})
Node<K,V>[] newTab = (Node<K,V>[]) new Node[newCap];
table = newTab;
if (oldTab != null) {
for ( int j = 0; j < oldCap; ++j) {
Node<K,V> e;
if ((e = oldTab[j]) != null) {
oldTab[j] = null;
if (e.next == null)
newTab[e.hash & (newCap - 1)] = e;
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;
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;
}
  
  

转载自：https://blog.csdn.net/u012712901/article/details/78313130