[Java Collections Framework] 浅析java 集合框架(五) : Map,HashMap,LinkedHashMap

Map

这部分可以参见官方文档的介绍

The Map Interface

view

有意思的是这个 视图集 View,

• keySet — the Set of keys contained in the Map.
• values — The Collection of values contained in the Map. This Collection is not a Set, because multiple keys can map to the same value.
• entrySet — the Set of key-value pairs contained in the Map. The Map interface provides a small nested interface called Map.Entry, the type of the elements in this Set.

注意 values 并不是一个 Set 而是一个Collection

multimap

Map里并没有实现 multimap，不过它提供了一种很简单的实现方式，即把key 与一个 list对应。

Map.entry

JDK 1.8 更新了几个提供比较器的方法，用它完全可以实现 类似c++中的 pair;

        public static <K extends Comparable<? super K>, V> Comparator<Map.Entry<K,V>> comparingByKey() {
            return (Comparator<Map.Entry<K, V>> & Serializable)
                (c1, c2) -> c1.getKey().compareTo(c2.getKey());
        }

        public static <K, V extends Comparable<? super V>> Comparator<Map.Entry<K,V>> comparingByValue() {
            return (Comparator<Map.Entry<K, V>> & Serializable)
                (c1, c2) -> c1.getValue().compareTo(c2.getValue());
        }

        public static <K, V> Comparator<Map.Entry<K, V>> comparingByKey(Comparator<? super K> cmp) {
            Objects.requireNonNull(cmp);
            return (Comparator<Map.Entry<K, V>> & Serializable)
                (c1, c2) -> cmp.compare(c1.getKey(), c2.getKey());
        }
        public static <K, V> Comparator<Map.Entry<K, V>> comparingByValue(Comparator<? super V> cmp) {
            Objects.requireNonNull(cmp);
            return (Comparator<Map.Entry<K, V>> & Serializable)
                (c1, c2) -> cmp.compare(c1.getValue(), c2.getValue());
        }

HashMap

HashMap 底层是一个长度为 $2^k$ 的数组，数组元素是key-value 的Entry

    /**
     * The table, initialized on first use, and resized as
     * necessary. When allocated, length is always a power of two.
     * (We also tolerate length zero in some operations to allow
     * bootstrapping mechanics that are currently not needed.)
     */
    transient Node<K,V>[] table;

这在取模运算的时候有一个非常好的优势，可以用 与 运算代替代替模运算。带来常数的优化。可是同时的缺点就是这个 hash 值在取模的时候仅会保留低位，高位的不同完全被忽略，所以真正计算hash的时候，他是将key 异或上了(h>>16)

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

冲突解决策略

有趣的是这里解决冲突的时候不是在这个 table 中用线性或者平方探测找下一个位置，而将每一个 bin（桶，table的单元）做成一个 链表，并且为了解决链表元素过高插入和寻找比较难的问题，他会将在链表的size大于某个特定值(8)的时候，组织成一颗红黑树, 保证 $O(logN)$ 的时间优化，$N$ 是冲突的元素个数。

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    public V put(K key, V value) {
        return putVal(hash(key), key, value, false, true);
    }
    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)// 无冲突，直接添加
            tab[i] = newNode(hash, key, value, null);
        else {// 冲突
            Node<K,V> e; K k;
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))//bin首元素，直接修改
                e = p;
            else if (p instanceof TreeNode)// 是一颗红黑树
                e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
            else {// 是链表，直接查找
                for (int binCount = 0; ; ++binCount) {
                    if ((e = p.next) == null) {//没有找到,添加到后面
                        p.next = newNode(hash, key, value, null);
                        if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st// 超过特定值，变化为红黑树
                            treeifyBin(tab, hash);
                        break;
                    }
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))// 找到
                        break;//相等
                    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;
    }

resize 的策略

1. risize就是将 table 的大小乘以2
2. risize 的时候会涉及再hash的问题，这在这里面也很简单，映射到同一个桶的元素仅会被hash到同一个方，或者原来的桶的位置加上原来的容量。即在最高位发生变化
3. 还有一个问题是当在hash之后链上元素小于固定值的时候会将它从红黑树展开为链表

    /**
     * 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() {
        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;
        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;// 重新hash分为两部分高位j+oldCap，和低位j
                        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 {// e idx of e + oldCap, 属于重新hash后的高位bin
                                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;
    }

迭代器策略

正如你所想的一样，当做 for-each 的遍历的时候它所做的遍历次数并不是恰好等于元素个数，而是等于箱子的个数，即会浪费一定的变量时间，不过好在默认情况下装箱的因子 loadFactor = 0.75 相当于浪费$1//3$ 的时间

    abstract class HashIterator {
        Node<K,V> next;        // next entry to return
        Node<K,V> current;     // current entry
        int expectedModCount;  // for fast-fail
        int index;             // current slot

        HashIterator() {
            expectedModCount = modCount;
            Node<K,V>[] t = table;
            current = next = null;
            index = 0;
            if (t != null && size > 0) { // advance to first entry
                do {} while (index < t.length && (next = t[index++]) == null);//遍历数组，找到一个非空的桶
            }
        }

        public final boolean hasNext() {
            return next != null;
        }

        final Node<K,V> nextNode() {
            Node<K,V>[] t;
            Node<K,V> e = next;
            if (modCount != expectedModCount)
                throw new ConcurrentModificationException();
            if (e == null)
                throw new NoSuchElementException();
            if ((next = (current = e).next) == null && (t = table) != null) {
                do {} while (index < t.length && (next = t[index++]) == null);//遍历数组，找到一个非空的桶
            }
            return e;
        }

        public final void remove() {
            Node<K,V> p = current;
            if (p == null)
                throw new IllegalStateException();
            if (modCount != expectedModCount)
                throw new ConcurrentModificationException();
            current = null;
            K key = p.key;
            removeNode(hash(key), key, null, false, false);
            expectedModCount = modCount;
        }
    }

LinkedHashMap

LinkedHashMap 就是继承自 HashMap 的类了 它并没有重写put, 而是选择了重写 Node, 并且改写自己在 HashMap 调用put 方法时和 node相关的方法就行了，因为hashMap本来就是设计成的list解决冲突不过在元素多的时候改成了红黑树节点而已。