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社区首页 >专栏 >[Java 8 HashMap 详解系列] 2.HashMap 中 Key 的 index 是怎样计算的?

[Java 8 HashMap 详解系列] 2.HashMap 中 Key 的 index 是怎样计算的?

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发布2020-03-24 16:45:43
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发布2020-03-24 16:45:43
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2.HashMap 中 Key 的 index 是怎样计算的?

HashMap中的 table 是怎样确定数组索引位置的?

对于HashMap内的所有实现来说,首先第一步是定位对键值对所在数组的索引下标位置,这是后续所有操作的基础.

如下代码是展示索引下标获取的基本逻辑:

代码语言:javascript
复制
    /* ---------------- Static utilities -------------- */

    /**
     * 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);
    }

然后, 数组元素的下标算法是:

代码语言:javascript
复制
    public static final int index(int hash, int n) {
        return (n - 1) & hash;
    }

代码拆解:

代码语言:javascript
复制
    public static final int hash(Object key) {
        if (key == null) {
            return 0;
        }

        int h = key.hashCode();
        int rh = h >>> 16;
        out.println("===================== hash(" + key + ") ===========================");
        out.println("key                  \t\t" + key);
        out.println("h=key.hashCode()     \t\t" + toBinary(h) + "\t\t" + h);
        out.println("rh=h>>>16            \t\t" + toBinary(rh) + "\t\t" + rh);

        return h ^ rh;
    }

    public static final int index(int hash, int n) {
        out.println("hash=h ^ rh          \t\t" + toBinary(hash) + "\t\t" + hash);
        out.println("n    =               \t\t" + toBinary(n) + "\t\t" + (n));
        out.println("n - 1=               \t\t" + toBinary(n - 1) + "\t\t" + (n - 1));
        int index = (n - 1) & hash;
        out.println("index=(n - 1) & hash \t\t" + toBinary(index) + "\t\t" + index);
        out.println();
        return index;
    }

运行测试数据:

代码语言:javascript
复制
===================== hash(a) ===========================
key                         a
h=key.hashCode()            00000000000000000000000001100001        97
rh=h>>>16                   00000000000000000000000000000000        0
hash(a)=97
hash=h ^ rh                 00000000000000000000000001100001        97
n    =                      00000000000000000000000000000001        1
n - 1=                      00000000000000000000000000000000        0
index=(n - 1) & hash        00000000000000000000000000000000        0

===================== hash(b) ===========================
key                         b
h=key.hashCode()            00000000000000000000000001100010        98
rh=h>>>16                   00000000000000000000000000000000        0
hash(b)=98
hash=h ^ rh                 00000000000000000000000001100010        98
n    =                      00000000000000000000000000000010        2
n - 1=                      00000000000000000000000000000001        1
index=(n - 1) & hash        00000000000000000000000000000000        0

===================== hash(abc) ===========================
key                         abc
h=key.hashCode()            00000000000000010111100001100010        96354
rh=h>>>16                   00000000000000000000000000000001        1
hash(abc)=96355
hash=h ^ rh                 00000000000000010111100001100011        96355
n    =                      00000000000000000000000000000100        4
n - 1=                      00000000000000000000000000000011        3
index=(n - 1) & hash        00000000000000000000000000000011        3

===================== hash(abcde) ===========================
key                         abcde
h=key.hashCode()            00000101100001001111010001100011        92599395
rh=h>>>16                   00000000000000000000010110000100        1412
hash(abcde)=92598759
hash=h ^ rh                 00000101100001001111000111100111        92598759
n    =                      00000000000000000000000000001000        8
n - 1=                      00000000000000000000000000000111        7
index=(n - 1) & hash        00000000000000000000000000000111        7

===================== hash(abcdefg) ===========================
key                         abcdefg
h=key.hashCode()            10111000000110010111010001100100        -1206291356
rh=h>>>16                   00000000000000001011100000011001        47129
hash(abcdefg)=-1206268803
hash=h ^ rh                 10111000000110011100110001111101        -1206268803
n    =                      00000000000000000000000000001000        8
n - 1=                      00000000000000000000000000000111        7
index=(n - 1) & hash        00000000000000000000000000000101        5

算法图解:

源码分析:

代码语言:javascript
复制
    /**
     * 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) { // index = (n - 1) & hash
            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;
    }

HashMap 中的 tableSizeFor() 方法详解

输入: int cap; 返回: n = tableSizeFor(cap) 其中, n % 2 ==0, and cap <=n.

代码语言:javascript
复制
    /**
     * Returns a power of two size for the given target capacity.
     */
    static final int tableSizeFor(int cap) {
        // 先 cap 减 1
        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; // 再加 1
    }

我们来运行测试一下:

代码语言:javascript
复制
===============tableSizeFor(2)==================
cap = 2
n = cap - 1         00000000000000000000000000000001
n = n | (n >>> 1)   00000000000000000000000000000001
n = n | (n >>> 2)   00000000000000000000000000000001
n = n | (n >>> 4)   00000000000000000000000000000001
n = n | (n >>> 8)   00000000000000000000000000000001
n = n | (n >>> 16)  00000000000000000000000000000001
tableSizeFor = 2
2
===============tableSizeFor(5)==================
cap = 5
n = cap - 1         00000000000000000000000000000100
n = n | (n >>> 1)   00000000000000000000000000000110
n = n | (n >>> 2)   00000000000000000000000000000111
n = n | (n >>> 4)   00000000000000000000000000000111
n = n | (n >>> 8)   00000000000000000000000000000111
n = n | (n >>> 16)  00000000000000000000000000000111
tableSizeFor = 8
8
===============tableSizeFor(10)==================
cap = 10
n = cap - 1         00000000000000000000000000001001
n = n | (n >>> 1)   00000000000000000000000000001101
n = n | (n >>> 2)   00000000000000000000000000001111
n = n | (n >>> 4)   00000000000000000000000000001111
n = n | (n >>> 8)   00000000000000000000000000001111
n = n | (n >>> 16)  00000000000000000000000000001111
tableSizeFor = 16
16
===============tableSizeFor(25)==================
cap = 25
n = cap - 1         00000000000000000000000000011000
n = n | (n >>> 1)   00000000000000000000000000011100
n = n | (n >>> 2)   00000000000000000000000000011111
n = n | (n >>> 4)   00000000000000000000000000011111
n = n | (n >>> 8)   00000000000000000000000000011111
n = n | (n >>> 16)  00000000000000000000000000011111
tableSizeFor = 32
32
===============tableSizeFor(58)==================
cap = 58
n = cap - 1         00000000000000000000000000111001
n = n | (n >>> 1)   00000000000000000000000000111101
n = n | (n >>> 2)   00000000000000000000000000111111
n = n | (n >>> 4)   00000000000000000000000000111111
n = n | (n >>> 8)   00000000000000000000000000111111
n = n | (n >>> 16)  00000000000000000000000000111111
tableSizeFor = 64
64
===============tableSizeFor(100)==================
cap = 100
n = cap - 1         00000000000000000000000001100011
n = n | (n >>> 1)   00000000000000000000000001110011
n = n | (n >>> 2)   00000000000000000000000001111111
n = n | (n >>> 4)   00000000000000000000000001111111
n = n | (n >>> 8)   00000000000000000000000001111111
n = n | (n >>> 16)  00000000000000000000000001111111
tableSizeFor = 128
128
===============tableSizeFor(896)==================
cap = 896
n = cap - 1         00000000000000000000001101111111
n = n | (n >>> 1)   00000000000000000000001111111111
n = n | (n >>> 2)   00000000000000000000001111111111
n = n | (n >>> 4)   00000000000000000000001111111111
n = n | (n >>> 8)   00000000000000000000001111111111
n = n | (n >>> 16)  00000000000000000000001111111111
tableSizeFor = 1024
1024
===============tableSizeFor(1073741000)==================
cap = 1073741000
n = cap - 1         00111111111111111111110011000111
n = n | (n >>> 1)   00111111111111111111111011100111
n = n | (n >>> 2)   00111111111111111111111111111111
n = n | (n >>> 4)   00111111111111111111111111111111
n = n | (n >>> 8)   00111111111111111111111111111111
n = n | (n >>> 16)  00111111111111111111111111111111
tableSizeFor = 1073741824
1073741824

参考资料

https://blog.csdn.net/fan2012huan/article/details/51097331 https://www.jianshu.com/p/cbe3f22793be https://www.jianshu.com/p/dbbecc36200d

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  • 2.HashMap 中 Key 的 index 是怎样计算的?
    • HashMap中的 table 是怎样确定数组索引位置的?
      • HashMap 中的 tableSizeFor() 方法详解
      • 参考资料
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