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hashmap原理解析

hashmap原理解析

作者: nich | 来源:发表于2022-11-22 18:06 被阅读0次

    HashMap<K,V> extends AbstractMap<K,V>
    implements Map<K,V>, Cloneable, Serializable

    1.重要属性

    threshold:是 HashMap 所能容纳的最⼤的键值对的个数,threshold = capacity *loadFactor,也就是说 capacity 数组⼀定的情况下,负载因⼦越⼤,所能容纳的键值对个数越多,超出 threshold 这个数⽬就重新 resize(扩容),扩容后的 HashMap的容量是之前的2倍。size 是 HashMap 中实际存在的键值对的数量,注意和 Node[]table 的长度 capacity 、容纳最⼤键值对数量 threshold 的区别
    loadFactor:负载因子

    2.构造方法

    2.1

    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);
        }
     //任何一个int 数字,都能找到离他最近的 2 的幂次方数字(且比他大
     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;
        }
    

    2.2

    public HashMap(int initialCapacity) {
            this(initialCapacity, DEFAULT_LOAD_FACTOR);
        }
    
        /**
         * Constructs an empty <tt>HashMap</tt> with the default initial capacity
         * (16) and the default load factor (0.75).
         */
        public HashMap() {
            this.loadFactor = DEFAULT_LOAD_FACTOR; // all other fields defaulted
        }
    

    2.3

    太烦了可以先不用看
     public HashMap(Map<? extends K, ? extends V> m) {
            this.loadFactor = DEFAULT_LOAD_FACTOR;
            putMapEntries(m, false);
        }
    

    3.常用方法

    3.1put(K key, V value)

    final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
                       boolean evict) {
            Node<K,V>[] tab; Node<K,V> p; int n, I;
         //步骤1初始化进来
            if ((tab = table) == null || (n = tab.length) == 0)
                //初始化
                n = (tab = resize()).length;
            if ((p = tab[i = (n - 1) & hash]) == null)
                tab[i] = newNode(hash, key, value, null)
            else {
                步骤1
                Node<K,V> e; K k;
               当前hash的key值和hash值一摸一样
                if (p.hash == hash &&
                    ((k = p.key) == key || (key != null && key.equals(k))))
                    e = p;
                else if (p instanceof TreeNode)
               //如果是二叉树
                    e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
                else {
               //如果是链表
                    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;
        }
    
    
    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;
                }
                //步骤4 newCap==oldCap*2 ,newThr = oldThr*2
                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
                //初始化newCap =16和newThr = 0.75*16
                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;
    //步骤3重新分散分布
            if (oldTab != null) {
                for (int j = 0; j < oldCap; ++j) {
                    Node<K,V> e;
                    if ((e = oldTab[j]) != null) {
                        oldTab[j] = null;
                        //如果node后面没有链表或者数组,直接存在新数组的指定位置
                        if (e.next == null)
                            newTab[e.hash & (newCap - 1)] = e;
                        else if (e instanceof TreeNode)
                            //在新的Node数组中,将该红黑树进行拆分,(如果拆分后的子树过小(子树的节点小于等于6个),则取消树化,即将其转为链表结构);
                            ((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
                        else { // preserve order
                           //链表处理
                            Node<K,V> loHead = null, loTail = null;
                            Node<K,V> hiHead = null, hiTail = null;
                            Node<K,V> next;
                            do {
                                next = e.next;
    //如果e.hash&oldCap进行与运算,算出的结果是为0,即说明该Node节点所对应的数组下标不需要改变
                                if ((e.hash & oldCap) == 0) {
                                    if (loTail == null)
    //如果loTail为null,说明该链表没有头节点
                                        loHead = e;
                                    else
    //尾插
                                        loTail.next = e;
                                    loTail = e;
                                }
                                else {
    //如果e.hash&oldCap进行与运算,算出的结果不为0,则更新该Node节点所对应的数组下标
                                    if (hiTail == null)
                                        hiHead = e;
                                    else
                                        hiTail.next = e;
                                    hiTail = e;
                                }
                            } while ((e = next) != null);
    //该Node节点所对应的数组下标不需要改变,直接把数组下标对应的节点指向新Node数组下标位置链表的头节点
                            if (loTail != null) {
                                loTail.next = null;
                                newTab[j] = loHead;
                            }
                        //该Node节点所对应的数组下标需要改变,重新计算出所对应的数组下标值,然后指向新Node数组下标位置链表的头节点
    
                            if (hiTail != null) {
                                hiTail.next = null;
                                newTab[j + oldCap] = hiHead;
                            }
                        }
                    }
                }
            }
            return newTab;
        }
    

    1.初始化进来tab和p都是空,执行n = (tab = resize()).length,
    然后执行 newCap = 16 ,newThr = 16*0.75=12来着,
    然后执行Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];
    初始化长度为16的table,然后执行tab[i] = newNode(hash, key, value, null);


    image.png

    2.当再有数据进来的时候,如果没有位置没有占用走tab[i] = newNode(hash, key, value, null);如果占用走putval下面一个else就是步骤1,如果是首节点走首节点,树走树,链表走链表
    如果链表长度>=7走treeifyBin(tab, hash)

    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);
            如果tab的长度<64,走resize()走到步骤3重新分散分布,不然就转化成二叉树
    最后执行if (++size > threshold) resize();当putval次数大于阀值,走resize()
    就会执行resize的()步骤4扩容,然后数据发散分布
    

    3.2get(Object key)

    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) {
                if (first.hash == hash && // always check first node
                    ((k = first.key) == key || (key != null && key.equals(k))))
                    return first;
                if ((e = first.next) != null) {
                    if (first instanceof TreeNode)
                        return ((TreeNode<K,V>)first).getTreeNode(hash, key);
                    do {
                        if (e.hash == hash &&
                            ((k = e.key) == key || (key != null && key.equals(k))))
                            return e;
                    } while ((e = e.next) != null);
                }
            }
            return null;
        }
    先查询头节点是不是,不是的话再往下查分节点的东西
    

    扰动函数
    static final int hash(Object key) {
    int h;
    return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
    }
    减少hash冲突
    最后hash&(n-1)

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