java.util.HashMap

Posted 2020-10-06 zno2

1. HashMap 如何实现？
2. put 做了什么？
3. get 做了什么？
4. 初始化容量，满载率，扩展？
5. hash？
6. 类似的结构及相似处，不同点？

 

load factor 满载率

capacity 容量

threshold 临界值

 

 

 

import java.util.Map;

public class Test {

    public static void main(String[] args) {
        Map<Key, String> map = new HashMap<Key, String>();
        map.put(new Key("A"), "1");
        map.put(new Key("B"), "2");
        map.put(new Key("C"), "3");
        System.out.println(map);
    }
}

class Key {
    private String name;

    Key(String name) {
        this.name = name;
    }

    @Override
    public int hashCode() {
        return 1;
    }

    @Override
    public String toString() {
        return "Key [name=" + name + "]";
    }
}

 

 

import java.util.Map;

public class Test {

    public static void main(String[] args) {
        Map<Key, Integer> map = new HashMap<Key, Integer>();
        for (int i = 0; i < 11; i++) {
            map.put(new Key(System.nanoTime()), i);
        }
        System.out.println(map);
    }
}

class Key {
    private long name;

    Key(long name) {
        this.name = name;
    }

    @Override
    public int hashCode() {
        return 1;
    }

    @Override
    public String toString() {
        return "Key [name=" + name + "]";
    }
}

 

 

    static final int TREEIFY_THRESHOLD = 8;
    static final int UNTREEIFY_THRESHOLD = 6;
    static final int MIN_TREEIFY_CAPACITY = 64;

控制treeify的临界值是8 ，当bin中链大于8时，则尝试treeify

• 1）如果此时表容量不足64，则会扩表。因此添加第9个元素时，由16->32 ，添加第10个元素时，由32->64
• 2）添加第11个元素时，此时转为了TreeNode

 

 

 

 

 

 ==============================================

 

put

    final V putVal(int hash, K key, V value, boolean onlyIfAbsent, boolean evict) {
        Node<K, V>[] tab;
        Node<K, V> p;
        int n, i;
        // 1 table 为空时重置大小
        if ((tab = table) == null || (n = tab.length) == 0)
            n = (tab = resize()).length;
        // 2  落到空bin 直接添加节点
        if ((p = tab[i = (n - 1) & hash]) == null)
            tab[i] = newNode(hash, key, value, null);
        // 3  落到非空bin ,判断 a , b ,c,
        else {
            Node<K, V> e;
            K k;
            // a) header与入参key相同，替换旧值
            if (p.hash == hash && ((k = p.key) == key || (key != null && key.equals(k))))
                e = p;
            // b) header是红黑树去添加树节点
            else if (p instanceof TreeNode)
                e = ((TreeNode<K, V>) p).putTreeVal(this, tab, hash, key, value);
            // c) header是链
            else {
                // 遍历链，逐项检查 ，判断 I ,II
                for (int binCount = 0;; ++binCount) {
                    // I) 末尾追加节点并检查是否需要转为红黑树
                    if ((e = p.next) == null) {
                        p.next = newNode(hash, key, value, null);
                        if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
                            treeifyBin(tab, hash);
                        break;
                    }
                    // II) key相同则替换旧值 返回
                    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;
        // 表中元素个数大于临界值时，扩展表X2
        if (++size > threshold)
            resize();
        afterNodeInsertion(evict);
        return null;
    }

 

get

    final Node<K, V> getNode(int hash, Object key) {
        Node<K, V>[] tab;
        Node<K, V> first, e;
        int n;
        K k;
        // table 存在，header节点存在
        if ((tab = table) != null && (n = tab.length) > 0 && (first = tab[(n - 1) & hash]) != null) {
            // 与 header 节点key相同，返回header 节点
            if (first.hash == hash && ((k = first.key) == key || (key != null && key.equals(k))))
                return first;
            // 如果存在header的子节点
            if ((e = first.next) != null) {
                // 如果 header 是红黑树去取树节点
                if (first instanceof TreeNode)
                    return ((TreeNode<K, V>) first).getTreeNode(hash, key);
                // 如果 header 是链
                do {
                    // 遍历节点，找到相同key返回
                    if (e.hash == hash && ((k = e.key) == key || (key != null && key.equals(k))))
                        return e;
                } while ((e = e.next) != null);
            }
        }
        // 未查到相同key
        return null;
    }

 

treeify

    final void treeifyBin(Node<K, V>[] tab, int hash) {
        int n, index;
        Node<K, V> e;
        // 如果table达不到最小treeify 容量(64)则扩容X2
        if (tab == null || (n = tab.length) < MIN_TREEIFY_CAPACITY)
            resize();
        // header 节点存在时
        else if ((e = tab[index = (n - 1) & hash]) != null) {
            TreeNode<K, V> hd = null, tl = null;
            // 遍历链将 Node -> TreeNode
            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)
                // 绘制红黑树 
                hd.treeify(tab);
        }
    }

 

resize

    /**
     * Initializes or doubles table size.  If null, allocates in
     * accord with initial capacity target held in field threshold.
     * Otherwise, because we are using power-of-two expansion, the
     * elements from each bin must either stay at same index, or move
     * with a power of two offset in the new table.
     *
     * @return the table
     */
    final Node<K, V>[] resize() {
         ...
        return newTab;
    }

 

hash

    /**
     * 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);
    }

 

关于编写方法时使用的缩略词猜想

• mc = modCount
• tab = table
• bin = table[i]
• tl = tree left
• hd = header
• p = present
• i = index
• prev = previous
• next = next
• k = key
• val = value
• v = value
• n = length
• e = entry
• simple -> simplify
• tree -> treeify
• do -> undo
• untreeify -> untreeify