HashMap源码分析

HashMap

JDK1.7 和1.8中关于对HashMap的实现,有了一些变化,其中很重要的一个变化,就是在解决Hash冲突的时候,存储数据结构有所调整。

1.7版本:

主要实现方式: 通过数组+ 链表的方式实现。当hash冲突的时候,使用链表来解决冲突。但是当hash不均匀的时候,可能会导致数据倾斜到某个数组槽位。那么对集合的更新、查找操作最后转变为线性查找,失去了hash查找的特性。

//使用数组式的链表,如果key的hash值一样,则通过List结构来解决冲突,当hash不均匀,可能会导致最后的数据变为线性查找List,性能无法保证
transient Entry<K,V>[] table;

    static class Entry<K,V> implements Map.Entry<K,V> {
        final K key;
        V value;
        Entry<K,V> next;
        int hash;
        /**其他方法**/
    }

    public V put(K key, V value) {
        if (key == null)
            return putForNullKey(value);
        int hash = hash(key);
        int i = indexFor(hash, table.length);
        //当该数组的hash槽位有数据时,则通过链表的方式追加到链表的结尾
        for (Entry<K,V> e = table[i]; e != null; e = e.next) {
            Object k;
            if (e.hash == hash && ((k = e.key) == key || key.equals(k))) {
                V oldValue = e.value;
                e.value = value;
                e.recordAccess(this);
                return oldValue;
            }
        }

        modCount++;
        addEntry(hash, key, value, i);
        return null;
    }

    void addEntry(int hash, K key, V value, int bucketIndex) {
        if ((size >= threshold) && (null != table[bucketIndex])) {
            resize(2 * table.length);
            hash = (null != key) ? hash(key) : 0;
            bucketIndex = indexFor(hash, table.length);
        }

        createEntry(hash, key, value, bucketIndex);
    }

    void createEntry(int hash, K key, V value, int bucketIndex) {
        Entry<K,V> e = table[bucketIndex];
        table[bucketIndex] = new Entry<>(hash, key, value, e);
        size++;
    }

1.8 版本

在1.8的版本中,同样是通过数组+链表的方式存储结构。但是1.7的Entry 被命名为Node,并且 当Node容量到达8的时候,会将Node转换为TreeNode(红黑树结构),查找效率大大提高

    /**
     * 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;
        /**其他方法**/
     }
     
    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;
        //不存在,直接新建并赋值到数组对应槽位
        if ((p = tab[i = (n - 1) & hash]) == null)
            tab[i] = newNode(hash, key, value, null);
        else {
            Node<K,V> e; K k;
            //如果已经有该key值,则直接返回该Node
            if (p.hash == hash &&
                    ((k = p.key) == key || (key != null && key.equals(k))))
                e = p;
            //如果该Node 是TreeNode,则直接放入到TreeNode结构中
            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);
                        //如果该槽位的值大于等于7的时候,需要转换成TreeNode数据结构来存储;TREEIFY_THRESHOLD等于8
                        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;
    }
     
    /**
     * 将Node数组中,对应hash槽位的Node转换为TreeNode数据结构
     * 
     * 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);
        }
    }     
     
posted @ 2019-01-06 20:13  IT小工~  阅读(462)  评论(0编辑  收藏  举报