HashMap 源码

概述

HashMap 是一个散列表，是 key/value 的存储结构，每个 key 对应唯一的一个 value，查询和修改的速度都很快；不能保证元素的存储顺序。
HashMap 的实现不是同步的，非线程安全。key 和 value 都可以为 null。
HashMap 的实例有两个参数影响性能：初始容量 和 加载因子。
容量 是哈希表中桶的数量； 初始容量 是哈希表在创建时的容量；加载因子 是哈希表在其容量自动增加之前可以达到多满的一种尺度。

继承结构

• HashMap 继承 AbstractMap<K,V>，实现 Map 接口。

存储结构

• HashMap 的实现采用了 数组 + 链表 + 红黑树 的复杂结构；
• 添加元素时，对 key 进行 hash 计算，根据 hash 值散列存储在数组中；
• 如果要存储的位置已经有元素存在了，则以链表的形式把元素存放在尾部；
• 如果数组的长度达到 64，且 链表 的长度也达到 8，则进行树化；
• 链表树化能大大地提高查询性能；

  数组的查询复杂度是 O(1)，链表的查询复杂度是 O(n)，红黑树的查询复杂度是 O(log n)；

源码实现

属性

View Code

/**
 * The default initial capacity - MUST be a power of two.
 */
static final int DEFAULT_INITIAL_CAPACITY = 1 << 4; // aka 16

/**
 * The maximum capacity, used if a higher value is implicitly specified
 * by either of the constructors with arguments.
 * MUST be a power of two <= 1<<30.
 */
static final int MAXIMUM_CAPACITY = 1 << 30;

/**
 * The load factor used when none specified in constructor.
 */
static final float DEFAULT_LOAD_FACTOR = 0.75f;

/**
 * The bin count threshold for using a tree rather than list for a
 * bin.  Bins are converted to trees when adding an element to a
 * bin with at least this many nodes. The value must be greater
 * than 2 and should be at least 8 to mesh with assumptions in
 * tree removal about conversion back to plain bins upon
 * shrinkage.
 */
static final int TREEIFY_THRESHOLD = 8;

/**
 * The bin count threshold for untreeifying a (split) bin during a
 * resize operation. Should be less than TREEIFY_THRESHOLD, and at
 * most 6 to mesh with shrinkage detection under removal.
 */
static final int UNTREEIFY_THRESHOLD = 6;

/**
 * The smallest table capacity for which bins may be treeified.
 * (Otherwise the table is resized if too many nodes in a bin.)
 * Should be at least 4 * TREEIFY_THRESHOLD to avoid conflicts
 * between resizing and treeification thresholds.
 */
static final int MIN_TREEIFY_CAPACITY = 64;

/**
 * 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;

/**
 * Holds cached entrySet(). Note that AbstractMap fields are used
 * for keySet() and values().
 */
transient Set<Map.Entry<K,V>> entrySet;

/**
 * The number of key-value mappings contained in this map.
 */
transient int size;

/**
 * The number of times this HashMap has been structurally modified
 * Structural modifications are those that change the number of mappings in
 * the HashMap or otherwise modify its internal structure (e.g.,
 * rehash).  This field is used to make iterators on Collection-views of
 * the HashMap fail-fast.  (See ConcurrentModificationException).
 */
transient int modCount;

/**
 * The next size value at which to resize (capacity * load factor).
 *
 * @serial
 */
// (The javadoc description is true upon serialization.
// Additionally, if the table array has not been allocated, this
// field holds the initial array capacity, or zero signifying
// DEFAULT_INITIAL_CAPACITY.)
int threshold;

/**
 * The load factor for the hash table.
 *
 * @serial
 */
final float loadFactor;

• 默认初始化容量 16（1 << 4），最大容量为 2^30；
• 默认装载因子为 0.75，容量达到装载因子，则进行扩容；
• 当数组的容量达到 64 且链表的长度达到 8 的时候进行树化，当链表的长度小于 6 的时侯反树化；

Node内部类

Node 类是一个单向链表。

View Code

/**
 * Basic hash bin node, used for most entries.  (See below for
 * TreeNode subclass, and in LinkedHashMap for its Entry subclass.)
 */
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;
    }

    public final int hashCode() {
        return Objects.hashCode(key) ^ Objects.hashCode(value);
    }

思考

为什么 hashCode 分别取 key 和 value 的 hashCode 求异或？（异或运算的妙用）
相关文章

TreeNode 内部类

TreeNode是一个树型节点，其中，prev是链表中的节点，用于在删除元素的时候可以快速找到它的前置节点。

View Code

/**
 * Entry for Tree bins. Extends LinkedHashMap.Entry (which in turn
 * extends Node) so can be used as extension of either regular or
 * linked node.
 */
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);
    }
    /**
     * Returns root of tree containing this node.
     */
    final TreeNode<K,V> root() {
        for (TreeNode<K,V> r = this, p;;) {
            if ((p = r.parent) == null)
                return r;
            r = p;
        }
    }
}

构造方法

HashMap 提供多个构造方法，供调用者设置初始化容量和加载因子。

View Code

/**
 * Constructs an empty <tt>HashMap</tt> with the specified initial
 * capacity and load factor.
 *
 * @param  initialCapacity the initial capacity
 * @param  loadFactor      the load factor
 * @throws IllegalArgumentException if the initial capacity is negative
 *         or the load factor is nonpositive
 */
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);
}

/**
 * Constructs an empty <tt>HashMap</tt> with the specified initial
 * capacity and the default load factor (0.75).
 *
 * @param  initialCapacity the initial capacity.
 * @throws IllegalArgumentException if the initial capacity is negative.
 */
public HashMap(int initialCapacity) {
    this(initialCapacity, DEFAULT_LOAD_FACTOR);
}

/**
 * Constructs an empty <tt>HashMap</tt> with the default initial capacity
 * (16) and the default load factor (0.75).
 */
public HashMap() {
    this.loadFactor = DEFAULT_LOAD_FACTOR; // all other fields defaulted
}

/**
 * Constructs a new <tt>HashMap</tt> with the same mappings as the
 * specified <tt>Map</tt>.  The <tt>HashMap</tt> is created with
 * default load factor (0.75) and an initial capacity sufficient to
 * hold the mappings in the specified <tt>Map</tt>.
 *
 * @param   m the map whose mappings are to be placed in this map
 * @throws  NullPointerException if the specified map is null
 */
public HashMap(Map<? extends K, ? extends V> m) {
    this.loadFactor = DEFAULT_LOAD_FACTOR;
    putMapEntries(m, false);
}

/**
 * Returns a power of two size for the given target capacity.
 */
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;
}

• 调用者没有指定初始化容量和加载因子，则使用默认的参数；
• 调用者初始化容量，通过 tableSizeFor(int cap) 方法把初始化容量设置为最近的 2^n；

思考

给定一个数，如何求出大于等于这个数，且最小的 2 的 n 次幂；

/**
 * Returns a power of two size for the given target capacity.
 */
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;
}

操作方法

put(K key, V value)

View Code

/**
 * Associates the specified value with the specified key in this map.
 * If the map previously contained a mapping for the key, the old
 * value is replaced.
 *
 * @param key key with which the specified value is to be associated
 * @param value value to be associated with the specified key
 * @return the previous value associated with <tt>key</tt>, or
 *         <tt>null</tt> if there was no mapping for <tt>key</tt>.
 *         (A <tt>null</tt> return can also indicate that the map
 *         previously associated <tt>null</tt> with <tt>key</tt>.)
 */
public V put(K key, V value) {
    return putVal(hash(key), key, value, false, true);
}

/**
 * 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);
}

/**
 * Implements Map.put and related methods
 *
 * @param hash hash for key
 * @param key the key
 * @param value the value to put
 * @param onlyIfAbsent if true, don't change existing value
 * @param evict if false, the table is in creation mode.
 * @return previous value, or null if none
 */
final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
               boolean evict) {
    Node<K,V>[] tab; Node<K,V> p; int n, i;
    // 数组容量为0，初始化数组
    if ((tab = table) == null || (n = tab.length) == 0)
        // resize 初始化
        n = (tab = resize()).length;
        // (n - 1) & hash 计算元素在哪个桶中，如果桶没有元素，则直接保存
    if ((p = tab[i = (n - 1) & hash]) == null)
        tab[i] = newNode(hash, key, value, null);
    else {
        // 桶中已经有元素了
        Node<K,V> e; K k;
        // 如果桶中的 key 与待插入的 key 相等，则修改 value 值
        if (p.hash == hash &&
            ((k = p.key) == key || (key != null && key.equals(k))))
            e = p;
        else if (p instanceof TreeNode)
            // 如果是树节点，调用树的插入方法插入元素
            e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
        else {
            // 如果是链表，遍历链表，存在相同的 key 就修改 value，否则追加在链表末尾；如果复合树化规则，则树化
            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
                        // 触发树化检查，不一定就树化，还要检查数组的大小
                        treeifyBin(tab, hash);
                    break;
                }
                if (e.hash == hash &&
                    ((k = e.key) == key || (key != null && key.equals(k))))
                    break;
                p = e;
            }
        }
        // 找到相同 key 的元素，执行 value 的修改
        if (e != null) { // existing mapping for key
            V oldValue = e.value;
            if (!onlyIfAbsent || oldValue == null)
                e.value = value;
            // 移动元素节点到最后
            afterNodeAccess(e);
            return oldValue;
        }
    }
    ++modCount;
    if (++size > threshold)
        resize();
    // 插入元素后，转后续流程，可能移动到最大的位置
    afterNodeInsertion(evict);
    return null;
}

• 计算 key 的 hash 值：(h = key.hashCode()) ^ (h >>> 16) ；
• 如果数组的容量为 0，则初始化数组；元素在数组中的位置位置：hash & (size - 1) ；
• 如果 key 不存在，则直接插入；如果插入的位置是数组，检查是否需要扩容；如果插入的位置是链表，检查是否需要树化；如果插入的位置是树，则检查是否需要做平衡；
• 如果 key 存在，则执行 value 修改；

  三种可能：

  1. key 存在于桶中，直接修改
  2. key 存在于链表中，修改 value
  3. key 存在于树中，修改 value，检查是否做平衡

思考

为什么 hashCode 如此设计？
相关文档

resize()

扩容。

View Code

/**
 * 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) {
            // 达到最大容量，不再扩容
            threshold = Integer.MAX_VALUE;
            return oldTab;
        }
        else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
                 oldCap >= DEFAULT_INITIAL_CAPACITY)
            // 扩容后的容量没有超越最大容量，且大于默认初始化容量，则扩容为原来的2倍
            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) {
        // 如果扩容门槛被设置为0，则计算为容量 * 装载因子，当不能超过最大容量
        float ft = (float)newCap * loadFactor;
        newThr = (newCap < MAXIMUM_CAPACITY && ft < (float)MAXIMUM_CAPACITY ?
                  (int)ft : Integer.MAX_VALUE);
    }
    // 重置扩容门槛
    threshold = newThr;
    @SuppressWarnings({"rawtypes","unchecked"})
    // 创建新数组，开始复制操作
        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)
                    // 复制的元素没有下一个节点（不是链表和树），直接赋值
                    newTab[e.hash & (newCap - 1)] = e;
                else if (e instanceof TreeNode)
                    // 如果是树，则拆分成两棵树存放到桶中
                    ((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
                else { // preserve order
                    // 如果是链表
                    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);

                    // 拆分成了两链表
                    // 低位链表在新桶中的位置一致
                    if (loTail != null) {
                        loTail.next = null;
                        newTab[j] = loHead;
                    }

                    // 高位链表在新桶中的位置正好是原来的位置加上旧容量
                    if (hiTail != null) {
                        hiTail.next = null;
                        newTab[j + oldCap] = hiHead;
                    }
                }
            }
        }
    }
    return newTab;
}

// TreeNode 类中拆分树的方法
/**
    * Splits nodes in a tree bin into lower and upper tree bins,
    * or untreeifies if now too small. Called only from resize;
    * see above discussion about split bits and indices.
    *
    * @param map the map
    * @param tab the table for recording bin heads
    * @param index the index of the table being split
    * @param bit the bit of hash to split on
    */
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) {
        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) {
        if (hc <= UNTREEIFY_THRESHOLD)
            tab[index + bit] = hiHead.untreeify(map);
        else {
            tab[index + bit] = hiHead;
            if (loHead != null)
                hiHead.treeify(tab);
        }
    }
}

• 数组的容量总是 2^n，每次扩容是原来的 2 倍；
• 根据加载因子计算扩容门槛，达到扩容门槛，则扩容；
• 最大容量不超过 2^30；
• 扩容中的数据复制分三种情况处理；

  • 如果要复制的元素不是链表和树，直接引用；
  • 如果要复制的元素是树，则拆分为两棵树放到桶中，低位树的位置跟旧数组中的位置一致，高位树的位置为 旧数组的容量 + 旧数组中的位置；
  • 如果要复制的元素是链表，则拆分成两个链表，低位链表的位置跟旧数组中的一致，高位链表的位置为 旧数组的容量 + 旧数组中的位置；
    JDK8，这里顺序遍历，追加到链表尾部。JDK7没有使用红黑树，顺序遍历，添加在头部；并发场景下，转移数据扩容后，可能出现链表逆序，从而形成环型链表，造成死循环。

思考

一个链表拆分成两个的算法为什么是 (e.hash & oldCap) == 0 ?
相关文章

TreeNode.putTreeVal(HashMap<K,V> map, Node<K,V>[] tab, int h, K k, V v)

向红黑树中插入元素。

View Code

/**
* Tree version of putVal.
*/
final TreeNode<K,V> putTreeVal(HashMap<K,V> map, Node<K,V>[] tab,
                                int h, K k, V v) {
    Class<?> kc = null;
    // 标记是否找到
    boolean searched = false;
    // 找到根节点
    TreeNode<K,V> root = (parent != null) ? root() : this;
    for (TreeNode<K,V> p = root;;) {
        // dir = direction , 标记是左边还是右边
        // ph = p.hash , 当前节点的 hash 值
        // pk = p.key , 当前节点的 key 值
        int dir, ph; K pk;
        if ((ph = p.hash) > h)
            // 当前 hash 比目标 hash 大，在左边
            dir = -1;
        else if (ph < h)
            // 当前 hash 比目标 hash 小，在右边
            dir = 1;
        else if ((pk = p.key) == k || (k != null && k.equals(pk)))
            // 两者 hash 值相等，且 key 相等，说明找到该节点
            return p;
        else if ((kc == null &&
                    (kc = comparableClassFor(k)) == null) ||
                    (dir = compareComparables(kc, k, pk)) == 0) {
            if (!searched) {
                TreeNode<K,V> q, ch;
                searched = true;
                if (((ch = p.left) != null &&
                        (q = ch.find(h, k, kc)) != null) ||
                    ((ch = p.right) != null &&
                        (q = ch.find(h, k, kc)) != null))
                    return q;
            }

            // 如果两者类型相同，再根据它们的内存地址计算hash值进行比较
            dir = tieBreakOrder(k, pk);
        }

        TreeNode<K,V> xp = p;
        if ((p = (dir <= 0) ? p.left : p.right) == null) {
            // 如果最后确实没找到对应key的元素，则新建一个节点
            Node<K,V> xpn = xp.next;
            TreeNode<K,V> x = map.newTreeNode(h, k, v, xpn);
            if (dir <= 0)
                xp.left = x;
            else
                xp.right = x;
            xp.next = x;
            x.parent = x.prev = xp;
            if (xpn != null)
                ((TreeNode<K,V>)xpn).prev = x;
            // 插入树节点后平衡
            // 把root节点移动到链表的第一个节点
            moveRootToFront(tab, balanceInsertion(root, x));
            return null;
        }
    }
}

• 寻找根节点；
• 从根节点查找；
• 比较 hash 值及 key 值，如果相同，直接返回。在 putVal() 方法中决定是否要替换 value；
• 根据 hash 值及 key 值确定在树的左子树还是右子树查找，找到了直接返回；
• 如果最后没有找到则在树的相应位置插入元素，并做平衡；

treeifyBin(Node<K,V>[] tab, int hash)

如果插入元素后链表的长度大于等于 8 则判断是否需要树化。

View Code

/**
 * Replaces all linked nodes in bin at index for given hash unless
 * table is too small, in which case resizes instead.
 */
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)
        // 如果桶数量小于64，扩容
        resize();
    else if ((e = tab[index = (n - 1) & hash]) != null) {
        TreeNode<K,V> hd = null, tl = null;
        // 把所有节点换成树节点
        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);

        // 如果进入过上面的循环，则从头节点开始树化
        if ((tab[index] = hd) != null)
            hd.treeify(tab);
    }
}

TreeNode.treeify(Node<K,V>[] tab)

树化。

View Code

/**
* Forms tree of the nodes linked from this node.
* @return root of tree
*/
final void treeify(Node<K,V>[] tab) {
    TreeNode<K,V> root = null;
    for (TreeNode<K,V> x = this, next; x != null; x = next) {
        next = (TreeNode<K,V>)x.next;
        x.left = x.right = null;
        // 第一个元素作为根节点，后面在做平衡
        if (root == null) {
            x.parent = null;
            x.red = false;
            root = x;
        }
        else {
            K k = x.key;
            int h = x.hash;
            Class<?> kc = null;
            for (TreeNode<K,V> p = root;;) {
                int dir, ph;
                K pk = p.key;
                if ((ph = p.hash) > h)
                    dir = -1;
                else if (ph < h)
                    dir = 1;
                else if ((kc == null &&
                            (kc = comparableClassFor(k)) == null) ||
                            (dir = compareComparables(kc, k, pk)) == 0)
                    dir = tieBreakOrder(k, pk);

                TreeNode<K,V> xp = p;
                if ((p = (dir <= 0) ? p.left : p.right) == null) {
                    x.parent = xp;
                    if (dir <= 0)
                        xp.left = x;
                    else
                        xp.right = x;
                    root = balanceInsertion(root, x);
                    break;
                }
            }
        }
    }
    moveRootToFront(tab, root);
}

TreeNode.removeTreeNode(HashMap<K,V> map, Node<K,V>[] tab, boolean movable)

View Code

/**
* Removes the given node, that must be present before this call.
* This is messier than typical red-black deletion code because we
* cannot swap the contents of an interior node with a leaf
* successor that is pinned by "next" pointers that are accessible
* independently during traversal. So instead we swap the tree
* linkages. If the current tree appears to have too few nodes,
* the bin is converted back to a plain bin. (The test triggers
* somewhere between 2 and 6 nodes, depending on tree structure).
*/
final void removeTreeNode(HashMap<K,V> map, Node<K,V>[] tab,
                            boolean movable) {
    int n;
    if (tab == null || (n = tab.length) == 0)
        return;
    //  节点在桶中的索引  
    int index = (n - 1) & hash;
    TreeNode<K,V> first = (TreeNode<K,V>)tab[index], root = first, rl;
    // 前置节点，后继节点
    TreeNode<K,V> succ = (TreeNode<K,V>)next, pred = prev;
    if (pred == null)
        // 前置节点为空，当前位置是根节点。后继节点替换当前节点，删除了当前节点
        tab[index] = first = succ;
    else
        // 否则把前置节点的下一个节点设置为当前节点的后继节点。相当于删除了当前节点。
        pred.next = succ;
        
        // 后继节点不为空，则后继节点的前置节点指向当前节点的前置节点，相当于删除了当前节点
    if (succ != null)
        succ.prev = pred;
        // 无后继节点，直接返回
    if (first == null)
        return;
    if (root.parent != null)
        root = root.root();
    if (root == null || root.right == null ||
        (rl = root.left) == null || rl.left == null) {
        tab[index] = first.untreeify(map);  // too small
        return;
    }

    // 删除红黑树中的节点
    TreeNode<K,V> p = this, pl = left, pr = right, replacement;
    if (pl != null && pr != null) {
        TreeNode<K,V> s = pr, sl;
        while ((sl = s.left) != null) // find successor
            s = sl;
        boolean c = s.red; s.red = p.red; p.red = c; // swap colors
        TreeNode<K,V> sr = s.right;
        TreeNode<K,V> pp = p.parent;
        if (s == pr) { // p was s's direct parent
            p.parent = s;
            s.right = p;
        }
        else {
            TreeNode<K,V> sp = s.parent;
            if ((p.parent = sp) != null) {
                if (s == sp.left)
                    sp.left = p;
                else
                    sp.right = p;
            }
            if ((s.right = pr) != null)
                pr.parent = s;
        }
        p.left = null;
        if ((p.right = sr) != null)
            sr.parent = p;
        if ((s.left = pl) != null)
            pl.parent = s;
        if ((s.parent = pp) == null)
            root = s;
        else if (p == pp.left)
            pp.left = s;
        else
            pp.right = s;
        if (sr != null)
            replacement = sr;
        else
            replacement = p;
    }
    else if (pl != null)
        replacement = pl;
    else if (pr != null)
        replacement = pr;
    else
        replacement = p;
    if (replacement != p) {
        TreeNode<K,V> pp = replacement.parent = p.parent;
        if (pp == null)
            root = replacement;
        else if (p == pp.left)
            pp.left = replacement;
        else
            pp.right = replacement;
        p.left = p.right = p.parent = null;
    }

    TreeNode<K,V> r = p.red ? root : balanceDeletion(root, replacement);

    if (replacement == p) {  // detach
        TreeNode<K,V> pp = p.parent;
        p.parent = null;
        if (pp != null) {
            if (p == pp.left)
                pp.left = null;
            else if (p == pp.right)
                pp.right = null;
        }
    }
    if (movable)
        moveRootToFront(tab, r);
}

• TreeNode 既是链表，也是红黑树；
• 先删除链表；
• 再删除红黑树，最后做平衡；

get(Object key)

View Code

/**
 * 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) {
        if (first.hash == hash && // always check first node
            ((k = first.key) == key || (key != null && key.equals(k))))
            // 在数组中查找的第一个元素匹配上，则返回
            return first;
        if ((e = first.next) != null) {
            if (first instanceof TreeNode)
                // 在树中查找
                return ((TreeNode<K,V>)first).getTreeNode(hash, key);
            do {
                if (e.hash == hash &&
                    ((k = e.key) == key || (key != null && key.equals(k))))
                    // 在链表中查找
                    return e;
            } while ((e = e.next) != null);
        }
    }
    return null;
}

• 计算 key 的 hash 值；
• 找到 key 所在的桶的第一个元素；
• 如果第一个元素的 key 等于待查找的 key，直接返回；
• 如果第一个元素是树节点，就按树的方式查找，否则按链表的方式查找；

TreeNode.getTreeNode(int h, Object k)

View Code

/**
* Calls find for root node.
*/
final TreeNode<K,V> getTreeNode(int h, Object k) {
    return ((parent != null) ? root() : this).find(h, k, null);
}

/**
* Finds the node starting at root p with the given hash and key.
* The kc argument caches comparableClassFor(key) upon first use
* comparing keys.
*/
final TreeNode<K,V> find(int h, Object k, Class<?> kc) {
    TreeNode<K,V> p = this;
    do {
        int ph, dir; K pk;
        TreeNode<K,V> pl = p.left, pr = p.right, q;
        if ((ph = p.hash) > h)
            // 左子树
            p = pl;
        else if (ph < h)
            // 右子树
            p = pr;
        else if ((pk = p.key) == k || (k != null && k.equals(pk)))
            // 找到则返回
            return p;
        else if (pl == null)
            // hash相同，但key不同，左子树为空，查右子树
            p = pr;
        else if (pr == null)
            // 右子树为空，查左子树
            p = pl;
        else if ((kc != null ||
                    (kc = comparableClassFor(k)) != null) &&
                    (dir = compareComparables(kc, k, pk)) != 0)
            // 比较 key 值大小，判断使用左子树还是右子树
            p = (dir < 0) ? pl : pr;
        else if ((q = pr.find(h, k, kc)) != null)
            // 以上条件都不通过，则尝试右子树查找
            return q;
        else
            // 都没有找到，左子树查找
            p = pl;
    } while (p != null);
    return null;
}

• 红黑树查找，先根据 hash 值比较，再根据 key 值比较决定查左子树还是右子树。

remove(Object key)

View Code

@Override
public boolean remove(Object key, Object value) {
    return removeNode(hash(key), key, value, true, true) != null;
}

/**
 * Implements Map.remove and related methods
 *
 * @param hash hash for key
 * @param key the key
 * @param value the value to match if matchValue, else ignored
 * @param matchValue if true only remove if value is equal
 * @param movable if false do not move other nodes while removing
 * @return the node, or null if none
 */
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))))
            // 第一个元素是要找的目标元素，保存，然后删除使用
            node = p;
        else if ((e = p.next) != null) {
            if (p instanceof TreeNode)
                // 第一个元素是树节点，以树的方式查找
                node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
            else {
                // 如果是链表，则遍历查找
                do {
                    if (e.hash == hash &&
                        ((k = e.key) == key ||
                         (key != null && key.equals(k)))) {
                        node = e;
                        break;
                    }
                    p = e;
                } while ((e = e.next) != null);
            }
        }
        // 查找到元素，匹配删除
        if (node != null && (!matchValue || (v = node.value) == value ||
                             (value != null && value.equals(v)))) {
            if (node instanceof TreeNode)
                // 树的方式删除
                ((TreeNode<K,V>)node).removeTreeNode(this, tab, movable);
            else if (node == p)
                // 链表删除的，下一个个元素上移
                tab[index] = node.next;
            else
                // 数组删除
                p.next = node.next;
            ++modCount;
            --size;
            // 删除节点后处理
            afterNodeRemoval(node);
            return node;
        }
    }
    return null;
}

总结

• HashMap 是一种散列表，采用 数组 + 链表 + 红黑树 的存储结构；
• HashMap 默认初始容量是 16，默认装载因子是 0.75f，容量总是 2 的 n 次方；
• HashMap 每次扩容时，容量变为原来的两倍；
• 当数组的容量大于等于 64 且链表的长度大于 8 时，进行树化；
• 当单个桶的数据小于 6 时，进行反树化；

FAQ

Q：HashMap 为什么用 数组 + 链表 + 红黑树 ？（实现原理）

• 数组用来确定桶的位置；
  key 的 hash 值：(hash = key.hashCode()) ^ (hash >>> 16) ，桶在数组中的位置：hash & (size - 1)；

• 链表是解决 hash 冲突的一种方式；
  如果元素所在的桶已经有元素了，以链表的方式追加在末尾；

• 红黑树是解决链表查询性能的问题；
  链表的查询时间复杂度是 O(n)，红黑树的查询时间复杂度是 O(log n)；当链表的长度大于 8，转成红黑树能大大提高查询性能；

Q：HashMap 什么时候触发扩容？每次扩容的长度总是 2 的 n 次幂？

• 如果数组大小达到 (float)capacity * loadFactor，就进行扩容；
  默认加载因子为 0.75，最大容量不超过 2^30，每次扩容为原来的 2 倍，数组容量总是 2^n；
• 每次扩容是 2 的 n 次幂；
  桶在数组中的位置为：hash & (size - 1)，而 size - 1 转二进制，最高位为 0，低位全是 1，按位取与能减少 hash 碰撞；
• 干扰函数 混合低位和高位，减少 hash 冲突：(hash = key.hashCode()) ^ (hash >>> 16)；

Q：String 中 hashcode 的实现？

用自然溢出等效取模：s[0]*31^(n-1) + s[1]*31^(n-2) + ... + s[n-1]；

Q：HashMap 扩容时每个 entry 需要再计算一次 Hash 吗？

不用 rehash，因为 HashMap 在扩容量、HashCode 和 数组索引位置计算上做了巧妙的设计，扩容后的元素，分高低位，低位保持原来数组索引的位置，而高位为原来的位置+扩容 size，所以不用 rehash；

Q：HashMap 中元素的位置计算应用了 key.hashCode()，重写 hashcode 和 equals 要方法注意什么?

四个原则（存在 hash 碰撞）：

• 两个对象相等，hashcode 一定相等
• 两个对象不等，hashcode 不一定不等
• hashcode 相等，两个对象不一定相等
• hashcode 不等，两个对象一定不等

Q：HashMap 的同步实现？

Map m = Collections.synchronizeMap(hashMap)