面试一、jdk1.8之hashMap

 

1、hashmap

  1.1、线程不安全

  1.2、数据结构

    jdk1.7版本为Table数组+ Entry链表、jdk1.8版本为Table数组+ Node链表/红黑树

  1.3、重要变量

    DEFAULT_INITIAL_CAPACITY(初始化默认长度):Table数组的初始化长度: 1 << 42^4=16

    MAXIMUM_CAPACITY(最大长度):2^30

    DEFAULT_LOAD_FACTOR(默认负载因子):默认值为0.75,当map的size>table数组长度*负载因子时,table数组会扩容为原来的2倍

    TREEIFY_THRESHOLD(链表转红黑树阀值):默认为8,表示table一个node下的数量超过8个时,链表结构会转换为红黑树

    UNTREEIFY_THRESHOLD(红黑树转链表阀值):默认为6,当table数组自动扩容时,如果table一个node下的数量小于6个,红黑树结构会转换为链表

    MIN_TREEIFY_CAPACITY(最小树化阀值):默认为64,当map的size>64时才会进行红黑树转换,防止map前期扩容频繁和树化冲突

  1.4、实现原理(jdk1.8)

    采用table数组+链表,HashMap采?Node数组来存储key-value对,每?个键值对组成了?个Node实体,

    Node类实际上是?个单向的链表结构,它具有Next指针,可以连接下?个Node实体。 只是在JDK1.8中,链表?度?于8的时候,链表会转成红?树!

    面试一、jdk1.8之hashMap

  1.5、基本方法

    put:对key取hash(),和当前table数组大小-1相与(hash(key) & n - 1),作为坐标。

      1、如果当前位置为空,则new Node,将Node插入第一个

      2、如果不为空,判断key是否和当前Node链表第一个Node的key相等,如果是则替换value

      3、如果不是,判断是否为红黑树,是红黑树直接插入

      4、如果是链表,则遍历查询key是否在当前链表中,如果是则替换value,

      5、如果不是,则new Node,并插入到最后。此过程如果一个链表的数量大于TREEIFY_THRESHOLD会触发树化,当table数组小于MIN_TREEIFY_CAPACITY时用扩容代替树化

      6、当map.size>threshold时,会触发扩容。threshold见resize()的计算newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY,即table数组大小*扩容因子

/**
     * Returns <tt>true</tt> if this map contains a mapping for the
     * specified key.
     *
     * @param   key   The key whose presence in this map is to be tested
     * @return <tt>true</tt> if this map contains a mapping for the specified
     * key.
     */
    public boolean containsKey(Object key) {
        return getNode(hash(key), key) != null;
    }

    /**
     * Associates the specified value with the specified key in this map.
     * If the map previously contained a mapping for the key, the old
     * value is replaced.
     *
     * @param key key with which the specified value is to be associated
     * @param value value to be associated with the specified key
     * @return the previous value associated with <tt>key</tt>, or
     *         <tt>null</tt> if there was no mapping for <tt>key</tt>.
     *         (A <tt>null</tt> return can also indicate that the map
     *         previously associated <tt>null</tt> with <tt>key</tt>.)
     */
    public V put(K key, V value) {
        return putVal(hash(key), key, value, false, true);
    }

    /**
     * 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;
        if ((tab = table) == null || (n = tab.length) == 0)
            n = (tab = resize()).length;
        if ((p = tab[i = (n - 1) & hash]) == null)
            tab[i] = newNode(hash, key, value, null);
        else {
            Node<K,V> e; K k;
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                e = p;
            else if (p instanceof TreeNode)
                e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
            else {
                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;
    }

    get:对key取hash(),和当前table数组大小-1相与(hash(key) & n - 1),作为坐标取出值Node,如果Node是树结构则调用TreeNode查询,否则遍历Node整个链表取出key相等的值。

/**
     * Returns the value to which the specified key is mapped,
     * or {@code null} if this map contains no mapping for the key.
     *
     * <p>More formally, if this map contains a mapping from a key
     * {@code k} to a value {@code v} such that {@code (key==null ? k==null :
     * key.equals(k))}, then this method returns {@code v}; otherwise
     * it returns {@code null}.  (There can be at most one such mapping.)
     *
     * <p>A return value of {@code null} does not <i>necessarily</i>
     * indicate that the map contains no mapping for the key; it‘s also
     * possible that the map explicitly maps the key to {@code null}.
     * The {@link #containsKey containsKey} operation may be used to
     * distinguish these two cases.
     *
     * @see #put(Object, Object)
     */
    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) {
            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;
    }
/**
         * Finds the node starting at root p with the given hash and key.
         * The kc argument caches comparableClassFor(key) upon first use
         * comparing keys.
         */
        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)
                    p = pl;
                else if (ph < h)
                    p = pr;
                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;
        }

        /**
         * Calls find for root node.
         */
        final TreeNode<K,V> getTreeNode(int h, Object k) {
            return ((parent != null) ? root() : this).find(h, k, null);
        }  

    resize:当map.size大于table数组长度*负载因子时会触发自动扩容,table数组长度会扩容为原先的2倍。

      扩容后遍历原先table数组,如果Node是红黑树则调用split方法分裂原先树,如果是链表则计算新位置。

      计算:hash(key)&oldCap == 0判断是否在原位置(参考图例),如果等于1则说明需要在新扩展的列表中,新坐标为原先坐标+oldCap

      loTail和loHead表示原链表

      hiTail和hiHead表示扩容出来的链表

/**
     * 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;
    }

  面试一、jdk1.8之hashMap

    treeifyBin:将Node链表替换为红黑树,只有在table数组长度大于MIN_TREEIFY_CAPACITY会进行替换,否则进行扩容(防止在前期table数组大小较小时,频繁扩容和树化冲突)

    /**
     * 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;
        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);
        }
    }  

    remove:

 

面试一、jdk1.8之hashMap

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