【Java-集合】HashMap-Hash冲突解决

た入场券 2022-02-28 07:10 298阅读 0赞

# 【Java-集合】HashMap-Hash冲突解决 #

2017年10月22日 20:49:58 [binke\_mxb][binke_mxb] 阅读数：5371

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);
            }
        }

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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");

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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;
    }

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## 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.**

![HashMap中的实际数据结构][HashMap]

(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);  
    }

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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;
    }

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[binke_mxb]: https://me.csdn.net/u012712901
[HashMap]: /images/20220228/91909ca4ae684115ac9474c04a70bdf4.png
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