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

构造方法

/**     * Constructs an empty HashMap 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 HashMap 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 HashMap 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 HashMap with the same mappings as the     * specified Map.  The HashMap is created with     * default load factor (0.75) and an initial capacity sufficient to     * hold the mappings in the specified Map.     *     * @param   m the map whose mappings are to be placed in this map     * @throws  NullPointerException if the specified map is null     */    public HashMap(Map
      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 --->>1629 --->>3244 --->>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 key, or     *         null if there was no mapping for key.     *         (A null return can also indicate that the map     *         previously associated null with key.)     */    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
     
      [] tab; Node
      
        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
       
         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
        
         )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.     *     * 
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.)     *     * 
A return value of {
     @code null} does not necessarily     * 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
      
        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
       
         getNode(int hash, Object key) {        Node
        
         [] tab; Node
         
           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
          
           )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
     
      [] resize() {         //保存旧的        Node
      
       [] 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
       
        [] newTab = (Node
        
         [])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
         
           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
          
           )e).split(this, newTab, j, oldCap); //只能是链表了 else { // preserve order Node
           
             loHead = null, loTail = null; Node
            
              hiHead = null, hiTail = null; Node
             
               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。