• 周六. 8月 20th, 2022

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java集合浅谈——Map之HashMap

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11月 28, 2021

Map概览

Map提供的是对象和对象的关联。

  • HashMap

HashMap是以哈希表来实现的,查找对象时通过哈希函数计算其位置。

  • LinkedHashMap

LinkedHashMap继承自HashMap,其定义如下:

public class LinkedHashMap<K,V>
    extends HashMap<K,V>
    implements Map<K,V>
{
  • WeakHashMap

WeakHashMap是一种改进的HashMap。

public class WeakHashMap<K,V>
    extends AbstractMap<K,V>
    implements Map<K,V> {
  • TreeMap

TreeMap实现了SortedMap接口,其内部是以红黑树来实现的。

  • Hashtable

Hashtable是以哈希表来实现的,解决冲突的方式与HashMap一样,也是采用了散列链表的形式,不过性能比HashMap要低。

源码解读

HashMap无参构造函数的定义如下:

    /**
     * Constructs an empty {@code HashMap} with the default initial capacity
     * (16) and the default load factor (0.75).
     */
    public HashMap() {
        this.loadFactor = DEFAULT_LOAD_FACTOR; // all other fields defaulted
    }

HashMap是一个数组和链表的结合体(在数据结构称“链表散列”),其中数组的定义如下:

    /**
     * The table, initialized on first use, and resized as
     * necessary. When allocated, length is always a power of two.
     * (We also tolerate length zero in some operations to allow
     * bootstrapping mechanics that are currently not needed.)
     */
    transient Node<K,V>[] table;
    /**
     * Basic hash bin node, used for most entries.  (See below for
     * TreeNode subclass, and in LinkedHashMap for its Entry subclass.)
     */
    static class Node<K,V> implements Map.Entry<K,V> {
        final int hash;
        final K key;
        V value;
        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 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;
        }
    }

put()

当往HashMap中添加元素的时候,先根据key的hash值得到这个元素在数组中的位置(即下标),然后把这个元素放到对应的位置中。

如果这个元素所在的位置上已经存放有其他元素,那么在同一个位置上的元素将以链表的形式存放,新加入的元素放在链头,之前的元素放在链尾,如下图所示:

对应源代码如下:

    /**
     * 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 {@code key}, or
     *         {@code null} if there was no mapping for {@code key}.
     *         (A {@code null} return can also indicate that the map
     *         previously associated {@code null} with {@code key}.)
     */
    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()

方法的示意图如下:

对应源代码如下:

    /**
     * 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;
    }

Note

(1)如果使用自定义对象作为Key,需要确保其实现了hashCode()和equals()方法。

(2)HashMap中hash数组的默认大小是16;默认负载因子是0.75。

(3)HashMap根据需要可能会对元素重新哈希,元素的顺序会被打散,因此不同时间迭代同一个HashMap的顺序可能会不同。

(4)针对null元素,HashMap在存储Entry对象时检查键是否为空,如果键为空,则始终将其映射到桶0。

    /**
     * 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 key) {
        int h;
        return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
    }

(5)当get()方法返回null时,既可以表示 HashMap中没有该键,也可以表示该键所对应的值为null,所以不能由get()方法来判断是否存在某个键,而应该用containsKey()方法来判断

(6)java.util.concurrent.ConcurrentHashMap是HashMap的线程安全版

变更

HashMap implementation changes in Java 8

在 Hash 函数将对象均匀分布在桶中的理想情况下,HashMap为 get() 和 put() 方法提供了恒定的时间性能 O(1)。

但是,如果 hashCode() 存在大量冲突,则性能可能会恶化。

在散列冲突的情况下,对象被存储为链表中的一个结点,并且使用 equals() 方法来比较Key。

在链表中找到正确Key的比较是一个线性操作,在最坏的情况下,时间复杂度变为 O(n)。

Java 8 为了解决这个问题,在达到某个阈值后,使用平衡树。

这意味着 HashMap 初始将 Entry 对象存储在链表中,但当链表中的元素数目大于某个阈值后,将链表变为平衡树,这使最坏情况下的性能为 O(log n)。

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