Java集合类之基于拉链法的哈希表HashMap

发表于 | 更新于 | 分类于 | 浏览 次
字数统计: 2.6k

HashMap底层的数据结构为数组+单链表+红黑树，它最多只允许一个记录的键为null，但可以有多条记录的值为null。

存储结构

数组

1
transient Node<K,V>[] table;

单链表

1
2
3
4
5
6
7
8
9
10
11
12
13
14
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;
    }
    ...
}

红黑树

1
2
3
4
5
6
7
8
9
10
11
12
// LinkedHashMap.Entry<K,V>又继承了HashMap.Node<K,V>
static final class TreeNode<K,V> extends LinkedHashMap.Entry<K,V> {
    TreeNode<K,V> parent;  // red-black tree links
    TreeNode<K,V> left;
    TreeNode<K,V> right;
    TreeNode<K,V> prev;    // needed to unlink next upon deletion
    boolean red;
    TreeNode(int hash, K key, V val, Node<K,V> next) {
        super(hash, key, val, next);
    }
    ...
}

静态常量

threshold = capacity × load factor

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
// 默认情况下，表的初始化容量
static final int DEFAULT_INITIAL_CAPACITY = 1 << 4; // aka 16

static final int MAXIMUM_CAPACITY = 1 << 30;
// 默认的负载因子
static final float DEFAULT_LOAD_FACTOR = 0.75f;

// 当链表中的结点个数=该阈值时，考虑将链表转换为红黑树
static final int TREEIFY_THRESHOLD = 8;

// 当红黑树中的结点个数<=该阈值时，红黑树被还原为链表
static final int UNTREEIFY_THRESHOLD = 6;

// 当表的容量>=该阈值时，才将链表转换为红黑树；否则，仅执行resize操作
static final int MIN_TREEIFY_CAPACITY = 64;

构造方法

1
2
3
4
5
6
7
8
9
10
11
12
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);
}

tableSizeFor方法用于计算满足initialCapacity <= x的最小x。其中，x是2的整数次幂。

1
2
3
4
5
6
7
8
9
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;
}

也就是说，不管给定的初始容量是多少，最终HashMap的容量都将是2的整数次幂。

put操作

1
2
3
public V put(K key, V value) {
    return putVal(hash(key), key, value, false, true);
}

hash方法的目的在于将给定的key转换成一个整数。如果key为null（最多只允许有一条记录的key为null），那么将其放入第0个桶中。否则，让键哈希码h的低16位与高16位做异或运算，使得HashMap中记录分布更加均匀。

1
2
3
4
static final int hash(Object key) {
    int h;
    return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}

得到hash值以后，通过除留余数法，即对表的长度n进行取模运算，来确定桶下标。

已知当n为2的整数次幂时，hash % n等价于hash & (n - 1)，因此，可以使用位运算来优化取模运算。

这也是为什么HashMap的容量必须为2的整数次幂的原因。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
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 {
            // 对单链表采用尾插法
            for (int binCount = 0; ; ++binCount) {
                if ((e = p.next) == null) {
                    p.next = newNode(hash, key, value, null);
                    // 若桶中的结点个数>=树化阈值，则考虑将单链表转变为红黑树
                    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;
}

扩容操作

扩容后的大小为原来容量的2倍。

设原来的容量为n，则扩容以后新的容量为2n。

假设hash % n = j，当对表中元素再次定址时，有以下两种情况：

1.如果hash < n 或者 hash = n × 2a + j (1 <= a < n/2)，即hash & n = 0，那么hash % (2n) = j。

此时，之前位于第j个桶中的元素，扩容以后依然放在第j个桶中。

2.否则，hash % (2n) = n + j。

此时，原来放在第j个桶中的元素，扩容以后将放在第n + j个桶中。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
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;
    // 扩容
    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;
}

按键取值

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
public V get(Object key) {
    Node<K,V> e;
    return (e = getNode(hash(key), key)) == null ? null : e.value;
}
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) {
        // 1.若为桶中的第一个结点
        if (first.hash == hash && // always check first node
            ((k = first.key) == key || (key != null && key.equals(k))))
            return first;
        if ((e = first.next) != null) {
            // 2.若桶中为红黑树
            if (first instanceof TreeNode)
                return ((TreeNode<K,V>)first).getTreeNode(hash, key);
            // 3.若桶中为单链表
            do {
                if (e.hash == hash &&
                    ((k = e.key) == key || (key != null && key.equals(k))))
                    return e;
            } while ((e = e.next) != null);
        }
    }
    return null;
}

remove操作

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
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;
        // 1.若待删除结点是桶中的第一个结点
        if (p.hash == hash &&
            ((k = p.key) == key || (key != null && key.equals(k))))
            node = p;
        else if ((e = p.next) != null) {
            // 2.若桶中为红黑树 
            if (p instanceof TreeNode)
                node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
            else {
                // 3.若桶中为单链表
                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)))) {
            // 在红黑树中删除
            if (node instanceof TreeNode)
                ((TreeNode<K,V>)node).removeTreeNode(this, tab, movable);
            else if (node == p) // 桶中的第一个结点
                tab[index] = node.next;
            else // 在单链表中删除
                p.next = node.next;
            ++modCount;
            --size;
            afterNodeRemoval(node);
            return node;
        }
    }
    return null;
}

红黑树与单链表的相互转变

单链表=>红黑树

当执行插入操作时，若桶中的结点个数>=树化阈值（8），并且表的容量>=最小树化容量（64），才会将单链表转变为红黑树。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
final void treeifyBin(Node<K,V>[] tab, int hash) {
    int n, index; Node<K,V> e;
    // 1.若散列表尚未初始化，或者表的容量小于最小树化容量，则执行初始化（扩容）
    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;
        // 2.将链表中的结点转变为树结点
        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);
        // 3.执行树化操作
        if ((tab[index] = hd) != null)
            hd.treeify(tab);
    }
}
TreeNode<K,V> replacementTreeNode(Node<K,V> p, Node<K,V> next) {
    return new TreeNode<>(p.hash, p.key, p.value, next);
}

如果只有桶中的结点个数>=树化阈值（此时，将调用treeifyBin方法）时，可能是因为表容量过小，导致散列不均匀，过多的结点集中在部分桶中。为了防止这种情况的发生，HashMap将尝试扩大容量。

红黑树=>单链表

1.当执行扩容操作时，若桶中的结点个数<=不树化阈值（6），则将红黑树还原为单链表。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
// TreeNode的方法，在resize()方法中被调用
final void split(HashMap<K,V> map, Node<K,V>[] tab, int index, int bit) {
    TreeNode<K,V> b = this;
    // Relink into lo and hi lists, preserving order
    TreeNode<K,V> loHead = null, loTail = null;
    TreeNode<K,V> hiHead = null, hiTail = null;
    int lc = 0, hc = 0;
    for (TreeNode<K,V> e = b, next; e != null; e = next) {
        next = (TreeNode<K,V>)e.next;
        e.next = null;
        if ((e.hash & bit) == 0) {
            if ((e.prev = loTail) == null)
                loHead = e;
            else
                loTail.next = e;
            loTail = e;
            ++lc;
        }
        else {
            if ((e.prev = hiTail) == null)
                hiHead = e;
            else
                hiTail.next = e;
            hiTail = e;
            ++hc;
        }
    }

    if (loHead != null) {
        // 还原为单链表
        if (lc <= UNTREEIFY_THRESHOLD)
            tab[index] = loHead.untreeify(map);
        else {
            tab[index] = loHead;
            if (hiHead != null) // (else is already treeified)
                loHead.treeify(tab);
        }
    }
    if (hiHead != null) {
        // 还原为单链表
        if (hc <= UNTREEIFY_THRESHOLD)
            tab[index + bit] = hiHead.untreeify(map);
        else {
            tab[index + bit] = hiHead;
            if (loHead != null)
                hiHead.treeify(tab);
        }
    }
}

2.当执行删除操作时，若桶中结点太少，则将红黑树还原为单链表。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
// TreeNode的方法，在remove方法中被调用
final void removeTreeNode(HashMap<K,V> map, Node<K,V>[] tab,
                          boolean movable) {
    int n;
    if (tab == null || (n = tab.length) == 0)
        return;
    int index = (n - 1) & hash;
    TreeNode<K,V> first = (TreeNode<K,V>)tab[index], root = first, rl;
    TreeNode<K,V> succ = (TreeNode<K,V>)next, pred = prev;
    if (pred == null)
        tab[index] = first = succ;
    else
        pred.next = succ;
    if (succ != null)
        succ.prev = pred;
    if (first == null)
        return;
    if (root.parent != null)
        root = root.root();
    // 若桶中结点太少，则还原为单链表
    if (root == null
        || (movable
            && (root.right == null
                || (rl = root.left) == null
                || rl.left == null))) {
        tab[index] = first.untreeify(map);  // too small
        return;
    }
    ...
}

----------本文结束感谢您的阅读----------
坚持原创技术分享，您的支持将鼓励我继续创作！