java.util.HashMap

一般深入阅读前，我习惯先扫下开头的类的Java doc。里面有如下信息点：

HashMap是Java集合框架中的一员。它是基于Map接口实现的一个哈希表。
它允许插入null的key和null的value。它的存储结点的结构叫bucket。
两个影响到性能的参数：初始容量和加载因子。
当哈希表中的条目数超出了加载因子与当前容量的乘积时，则要对该哈希表进行 rehash 操作（即重建内部数据结构），从而哈希表将具有大约两倍的桶数。
该实现不是同步的。可使用Collections.synchronizedMap(new HashMap(...));“封装”一个简单的同步的map，但需注意一点，这个返回的迭代器都是快速失败的。
作者是四位神： Doug Lea、Josh Bloch、Arthur van Hoff、Neal Gafter。

是什么？为了解决什么问题？

从目的来讲，HashMap是一个存储键值对的集合，也就是符号表。往细里说，如上叙，是基于Map接口实现的一个哈希表。

内部实现

先简单说说存储结构，在说下重要属性，最后说下重点方法。

存储结构

重要属性


transient Node<K,V>[] table; //存储Node bucket的数组

transient int size; //当前存储的key-value对的总数

int modCount;  //记录集合结构变化的次数，用于快速失败等等

int threshold; //所能容纳的key-value对总数，当集合的key-value对大于该值resize

final float loadFactor; //加载因子，默认0.75，一般不要改

关于table这数组的length为什么必须为2的n次方(JavaDoc中有capacity MUST be a power of two.)？摘自美团点评技术团队 | Java 8系列之重新认识HashMap:

在HashMap中，哈希桶数组table的长度length大小必须为2的n次方(一定是合数)，这是一种非常规的设计，常规的设计是把桶的大小设计为素数。相对来说素数导致冲突的概率要小于合数，具体证明可以参考http://blog.csdn.net/liuqiyao_01/article/details/14475159， Hashtable初始化桶大小为11，就是桶大小设计为素数的应用（Hashtable扩容后不能保证还是素数）。HashMap采用这种非常规设计，主要是为了在取模和扩容时做优化，同时为了减少冲突，HashMap定位哈希桶索引位置时，也加入了高位参与运算的过程。

当自定义设置初始容量和负载因子时，有个有趣的方法tableSizeFor(int cap):

    public HashMap(int initialCapacity, float loadFactor) {
        ……(省略)
        this.loadFactor = loadFactor;
        this.threshold = tableSizeFor(initialCapacity);
    }

    /**
     * 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;
    }

可自行模拟一下方法是如何执行的。主要功能是返回一个比给定整数大且最接近的2的n次方整数，如给定12，返回2的4次方16。

方法

put

    /**
     * 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);
    }

首先看下，hash方法：

    /**
     * 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);
    }

为什么要多算这一步hash呢？

因为通过h >>> 16并异或h能使，在HashMap的Node<K,V>[] table的length还小时， key的哈希值的高位更充分参与到定位bucket的过程。(注意下文中tab[i = (n - 1) & hash])。

拓展阅读：美团点评技术团队 | Java 8系列之重新认识HashMap的"确定哈希桶数组索引位置"一小节。

    /**
     * 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;
        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 {
                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
                            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;
    }

感觉没什么可说，源码本身也有注释。

resize

扩容方法。挺赞的代码。

   /**
     * 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;
            }
            //没超过最大值，大小扩大为2倍
            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) {//初始化threshold
            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) {
            //重整每个node的位置
            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); //红黑树重构
                    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;//下标位置移动原来容量大小
                        }
                    }
                }
            }
        }
        return newTab;
    }

以上代码做了点注释。

btw， 思考一下每个&按位运算的巧妙处。

核心算法及性能

拉链法哈希实现
红黑树

由于红黑树实现，所以key-value对的value必须实现了Comparable接口。

性能相关点

由上算法可知，n次插入后单个查询的时间复杂度，最好O(1)，最差O(n)，如果是红黑O(log n)。
key计算hashcode的算法是否复杂。
初始容量和加载因子的设置。

设置初始容量前，最好是对新建的HashMap对象存储多少键值对有一个估量。因为resize会浪费性能。

加载因子个人认为一般使用默认的0.75，当然，认为系统有很多空闲空间，可以设置小一点。