HashMap

基于哈希表的 Map 接口的实现。此实现提供所有可选的映射操作，并允许使用 null 值和 null 键。（除了非同步和允许使用 null 之外，HashMap 类与 Hashtable 大致相同。）此类不保证映射的顺序，特别是它不保证该顺序恒久不变。 此实现假定哈希函数将元素适当地分布在各桶之间，可为基本操作（get 和 put）提供稳定的性能。迭代 collection 视图所需的时间与 HashMap 实例的“容量”（桶的数量）及其大小（键-值映射关系数）成比例。所以，如果迭代性能很重要，则不要将初始容量设置得太高（或将加载因子设置得太低）。

数据结构

先看下hashmap的数据结构

大概就是如图所示。
table就是数组咯。链表的他们称之为桶。大于阈值就转成红黑树咯，主要是为了提高效率。
使用红黑树来实现。

构造方法

/**
     * 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);
    }

其中最主要的是初始化的大小
还有初始化填充因子
static final float DEFAULT_LOAD_FACTOR = 0.75f;
HashMap的容量超过当前数组长度*加载因子，就会执行resize()算法
比如说向水桶中装水，此时HashMap就是一个桶， 这个桶的容量就是加载容量， 
而加载因子就是你要控制向这个桶中倒的水不超过水桶容量的比例，比如加载因子是0.75 ， 
那么在装水的时候这个桶最多能装到3/4 处，超过这个比例时，桶会自动扩容。 
因此，这个桶最多能装水 = 桶的容量 * 加载因子。

/**     
     * 获取初始值，你输入的初始值，不一定是初始化时所用的初始值。
     * 为什么初始值必须是2得倍数呢，下面代码会给你解释。
     * 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;
    }
    MAXIMUM_CAPACITY = 1<<30;

这样得到的始终是你输入初始值 
小于最小的2的次幂，也就是说 
比如你输入 
15 --->>16
29 --->>32
44 --->>64

重要函数

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;
        // 这一顿操作大概的意思就是保留了高16位的值
        // 其实低16位得值也保留了下来，只要在做一次异或，值就变回来了
        return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
    }

public V put(K key, V value) {}

 /**
     * 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);
    }
    /**
     * 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;
        // table未初始化或者长度为0，进行扩容
        if ((tab = table) == null || (n = tab.length) == 0)
            n = (tab = resize()).length;
        // 看下值放在哪一个table[] 
        // 这里也有一个为什么table的大小为什么必须是2的倍数的原因
        // n 是 tab的长度  那么 (n - 1) & hash 的意思就是？
        // 假如 长度为 16(10000) 那么 15(01111) & 就得到最后hash值相当于 h & (length - 1) == h % length
        // 这样数组也不会越界等 运算得比%运算得快 
        if ((p = tab[i = (n - 1) & hash]) == null)
            tab[i] = newNode(hash, key, value, null);
        // 已经有了，就看下是放在 链表还是红黑树。
        else {
            Node<K,V> e; K k;
            //先比较s是不是在头节点
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                e = p;
            //或者是红黑树
            else if (p instanceof TreeNode)
                e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
            //没办法了，只能是链表了
            else {
                for (int binCount = 0; ; ++binCount) {
                    //直接放在尾部
                    if ((e = p.next) == null) {
                        p.next = newNode(hash, key, value, null);
                        //链表大于8个阈值直接转成红黑树
                        if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
                            treeifyBin(tab, hash);
                        break;
                    }
                    //存在一模一样的key则跳出继续
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        break;
                    //继续遍历
                    p = e;
                }
            }
            //如果找到了存放的位置
            if (e != null) { // existing mapping for key
                V oldValue = e.value;
                 // onlyIfAbsent为false或者旧值为null
                 // onlyIfAbsent是传入的参数 默认w为false直接替换
                if (!onlyIfAbsent || oldValue == null)
                    //用新值替换旧值
                    e.value = value;
                afterNodeAccess(e);
                // 返回旧值
                return oldValue;
            }
        }
        ++modCount;
        // 实际大小大于阈值则扩容
        if (++size > threshold)
            resize();
        afterNodeInsertion(evict);
        return null;
    }

public V get(Object key) {}

相对于put，get就比较简单了。
相对jdk1.7版本 
1.7 ---->1.8 
位桶+链表 ----> 位桶+链表大于阈值（8）后切换成红黑树
大数据下 O(n)->>O(Logn)

/**
     * 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;
         // table已经初始化，长度大于0，根据hash寻找table中的项也不为空
        if ((tab = table) != null && (n = tab.length) > 0 &&
            (first = tab[(n - 1) & hash]) != null) {
            //判断是不是第一个结点 是就返回
            if (first.hash == hash && // always check first node
                ((k = first.key) == key || (key != null && key.equals(k))))
                return first;
            // 节点下面还有东西？
            if ((e = first.next) != null) {
                // 是红黑树吗？
                if (first instanceof TreeNode)
                    return ((TreeNode<K,V>)first).getTreeNode(hash, key);
                //不是红黑树那你肯定是链表咯
                do {
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        return e;
                } while ((e = e.next) != null);
            }
        }
        return null;
    }

resize()

hashmap的扩容方法

 /**
     * 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() {
         //保存旧的
        Node<K,V>[] oldTab = table;
        //保存长度
        int oldCap = (oldTab == null) ? 0 : oldTab.length;
        //保存阈值 需要resize的阈值
        int oldThr = threshold;
        int newCap, newThr = 0;
        // 之前table大小大于0
        if (oldCap > 0) {
            // 之前table大于最大容量
            if (oldCap >= MAXIMUM_CAPACITY) {
                 // 阈值为最大整形
                threshold = Integer.MAX_VALUE;
                return oldTab;
            }
            // 容量翻倍，使用左移，效率更高
            else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
                     oldCap >= DEFAULT_INITIAL_CAPACITY)
                // double threshold 阈值翻倍
                newThr = oldThr << 1;        
        // 之前阈值大于0
        else if (oldThr > 0) // initial capacity was placed in threshold
            newCap = oldThr;
        // oldCap = 0并且oldThr = 0，使用缺省值（如使用HashMap()构造函数，之后再插入一个元素会调用resize函数，会进入这一步）
        else {               // zero initial threshold signifies using defaults
            newCap = DEFAULT_INITIAL_CAPACITY;
            newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);
        }
        // 新阈值为0
        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"})
        // 初始化table
        Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];
        table = newTab;
        // 之前的table已经初始化过
        if (oldTab != null) {
            // 复制元素，重新进行hash
            for (int j = 0; j < oldCap; ++j) {
                Node<K,V> e;
                if ((e = oldTab[j]) != null) {
                    oldTab[j] = null;
                    //如果链表只有一个，则直接赋值
                    if (e.next == null)
                        newTab[e.hash & (newCap - 1)] = e;
                    //红黑树啊
                    else if (e instanceof TreeNode)
                        ((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
                    //只能是链表了  
                    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;
                        }
                    }
                }
            }
        }
        return newTab;
    }

这一顿操作之后大概就是这个过程吧

END

HashMap运用了许多非常巧妙的算法吧，大量的使用到了位运算，让这个结构运行更稳定更巧妙。每次看都有新收获。