JDK 1.8 HashMap 源码分析

Java 面试 About 6,030 words

数据结构

数组 + 单链表 + 红黑树

哈希冲突解决方法

尾插方式

if ((p = tab[i = (n - 1) & hash]) == null)
    tab[i] = newNode(hash, key, value, null);

Node<K,V> newNode(int hash, K key, V value, Node<K,V> next) {
    return new Node<>(hash, key, value, next);
}

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;
    }
}

null key

空键插入到数组索引0的位置上。

static final int hash(Object key) {
    int h;
    return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}

转换红黑树

条件:链表长度大于等于8且数组长度大于等于64

static final int MIN_TREEIFY_CAPACITY = 64;

static final int TREEIFY_THRESHOLD = 8;

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;
}

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) {
        // ...
    }
}

扩容

数组扩容为原始数组的2倍:·newCap = oldCap << 1

阈值扩容为原始阈值的2倍: newThr = oldThr << 1

如果数组索引位置上的元素没有后继节点,则直接使用e.hash & (newCap - 1)取模(&2n次幂等价于取模)获取到索引位置,新索引位置为什么可以直接赋值呢?因为扩容后的节点再hash只会出现在原索引位置,或索引位置+先前数组容量大小的值的索引。

对于形成链表的索引位置上,对(e.hash & oldCap) == 0(容量都是2n次幂,所以二进制位只有一位是1,其余位都是0),则还是在原来索引位置newTab[j] = loHead。反之,在原来索引的基础上再加上先前容量大小的值newTab[j + oldCap] = hiHead;

loHeadhiHead形成链表后分别赋值给原始索引原始索引+原始容量位置处。

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
    }
    // ...
    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;
}

移除节点

当红黑树高度不够高时会退化为链表。移除节点时只有高度不够高时退化为链表,没有使用到退化阈值6

public V remove(Object key) {
    Node<K,V> e;
    return (e = removeNode(hash(key), key, null, false, true)) == null ?
        null : e.value;
}

final Node<K,V> removeNode(int hash, Object key, Object value, boolean matchValue, boolean movable) {
    Node<K,V>[] tab; Node<K,V> p; int n, index;
    if ((tab = table) != null && (n = tab.length) > 0 &&
        (p = tab[index = (n - 1) & hash]) != null) {
        Node<K,V> node = null, e; K k; V v;
        if (p.hash == hash &&
            ((k = p.key) == key || (key != null && key.equals(k))))
            node = p;
        else if ((e = p.next) != null) {
            if (p instanceof TreeNode)
                node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
            else {
                do {
                    if (e.hash == hash &&
                        ((k = e.key) == key ||
                         (key != null && key.equals(k)))) {
                        node = e;
                        break;
                    }
                    p = e;
                } while ((e = e.next) != null);
            }
        }
        if (node != null && (!matchValue || (v = node.value) == value || (value != null && value.equals(v)))) {
            if (node instanceof TreeNode)
                ((TreeNode<K,V>)node).removeTreeNode(this, tab, movable);
            else if (node == p)
                tab[index] = node.next;
            else
                p.next = node.next;
            ++modCount;
            --size;
            afterNodeRemoval(node);
            return node;
        }
    }
    return null;
}
Views: 1,847 · Posted: 2021-10-20

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