03_HashMap源码剖析

一、 基本原理

  1. HashMap底层基于数组+链表的数据结构,当出现hash冲突的时候,就将冲突的节点挂在链表尾部
  2. JDK8以后,为了提高性能,解决hash冲突采用了链表+红黑树,如果只有链表的话,他的查询时间复杂度为O(n),而红黑树时间复杂度为O(log(n)

二、红黑树简述

  1. 红黑树是二叉查找树,左小右大,根据这个规则可以快速查找某个值
  2. 普通的二叉查找树,是有可能出现瘸子的情况,只有一条腿,不平衡了,导致查询性能变成O(n),线性查询了
  3. 红黑树,红色和黑色两种节点,会有条件限制去保证树是平衡的,不会出现瘸腿的情况
  4. 如果插入节点的时候破坏了红黑树的规则和平衡,会自动重新平衡,变色(红 <-> 黑),旋转,左旋转,右旋转
  1. 如果要完全搞得红黑树,还是需要花点时间和精力的,我们研究HashMap的话,重点放在源码上

三、核心成员变量

/**
* HashMap里的数组默认大小,16
 * 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;


/**
* 默认加载因子,0.75f,当数组里的元素达到 16 *0.75 = 12的时候,就会进行扩容
* 这个参数我们一般不会去修改,采用默认的就好
 * 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;


/**
* 这个Node其实就是代表数组里的key-value对,key的hash值,key,vlue,以及链表指向的下一个指针
 * Basic hash bin node, used for most entries.  (See below for
 * TreeNode subclass, and in LinkedHashMap for its Entry subclass.)
 */
static class Node implements Map.Entry {
    final int hash;
    final K key;
    V value;
    Node next;


    Node(int hash, K key, V value, Node next) {
        this.hash = hash;
        this.key = key;
        this.value = value;
        this.next = next;
    }


/**
* 代表map的底层的数组
 * 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[] table;


/**
 * Holds cached entrySet(). Note that AbstractMap fields are used
 * for keySet() and values().
 */
transient Set> entrySet;


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

四、hashmap如何降低hash冲突的算法

/**
* 将key-value放入到map中,如果这个key已经存在的话,就会将原来的值替换掉
 * 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);
}

/**
 * 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);
}
  1. 针对这个hash算法(h = key.hashCode()) ^ (h >>> 16),首先h为一个Int类型的变量
  2. 假设这个hashCode为1111 1111 1111 1111 1111 1010 0111 1100,那么h>>>16,就是将1111 1111 1111 1111 1111 1010 0111 1100右移16位

右移16位后为0000 0000 0000 0000 1111 1111 1111 1111
然后将右移16位的h和原来的h进行异或运算

1111 1111 1111 1111 1111 1010 0111 1100
^0000 0000 0000 0000 1111 1111 1111 1111
1111 1111 1111 1111 0000 0101 1000 0011

这样计算,其实就是将h的高16位和低16位进行一个异或运算,保证同时将高16位和低16位的特征同时纳入运算。通过这样的方式算出来的hash值,可以降低hash冲突的概率

五、put操作以及hash寻址算法

  1. 这里的源码细节中的一些参数属于核心,捋清楚这些参数是读懂源码的关键
  2. 我们知道,hashmap底层是基于数组和链表实现的。当出现hash冲突的时候,用链表来解决hash冲突,但是链表的get时间复杂度是O(n),正常来说,table[i]数组索引直接定位的方式的话,O(1)
  3. 如果链表,大量的key冲突,会导致get()操作,性能急剧下降,导致很多的问题
  4. JDK 1.8以后人家优化了这块东西,会判断,如果链表的长度达到8的时候,那么就会将链表转换为红黑树,如果用红黑树的话,get()操作,即使对一个很大的红黑树进行二叉查找,那么时间复杂度会变成O(logn),性能会比链表的O(n)得到大幅度的提升
  5. 新的数组是老数组的大小的两倍
  6. 扩容过以后,会判断一下,如果是一个链表里的元素的话,那么要么是直接放在新数组的原来的那个index,要么就是原来的index + oldCap

···
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;
    // 新增一个map的table是空的,所以会走if逻辑
    if ((tab = table) == null || (n = tab.length) == 0)
    // 这里会走resize方法,resize第一次进来创建一个默认大小16的空数组
    // 这样n就是16
    n = (tab = resize()).length;
    // 这行代码是计算key在数组中的位置的关键
    // 16-1 = 15 用15和上面计算的hash值进行与运算
    // 上面的hash:1111 1111 1111 1111 0000 0101 1000 0011
    // 15的二进制:0000 0000 0000 0000 0000 0000 0000 1111
    // 所以结果为:index = 3
    // 这样的话,其实就是在数组的第三个位置放入这个key-value对
    // 这里采用的是二进制的与运算,而不是取模,是因为性能比取模运算要高很多,而且只有每次扩容的时候
    // 数组的大小是2的n次方就可以保证二进制和取模运算的结果一样了
    if ((p = tab[i = (n - 1) & hash]) == null)
    // 这里的逻辑,就是通过hash寻找之后定位到的index位置上是空的,那么就可以直接将元素放到index的数组中
    tab[i] = newNode(hash, key, value, null);
    else {
    // 走到else逻辑后,说明出现了hash冲突
    Node e; K k;
    if (p.hash == hash &&
    ((k = p.key) == key || (key != null && key.equals(k))))
    // 这里的条件,说明key相同,那么直接覆盖原来的value,而这里会把e指向index这个位置的node
    e = p;
    else if (p instanceof TreeNode)
    e = ((TreeNode)p).putTreeVal(this, tab, hash, key, value);
    else {
    for (int binCount = 0; ; ++binCount) {
    if ((e = p.next) == null) {
    // 这里是说,如果链表的长度大于了8,达到9时,那么就要将这个链表转换为一个红黑树的数据结构
    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
    // oldValue原来老的值
    V oldValue = e.value;
    if (!onlyIfAbsent || oldValue == null)
    e.value = value;
    // value代表新的值,也就是将数组那个位置的Node的value设置为了新的value
    afterNodeAccess(e);

         return oldValue;
     }
    

    }
    ++modCount;
    if (++size > threshold)
    resize();
    afterNodeInsertion(evict);
    return null;
    }

/**

  • 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() {
    // 第一次进来,table为空
    // 我们不用一行一行的分析,其实这里第一次put方法进来的时候,就是创建一个空的
    // 默认大小16的空数组
    Node[] 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[] newTab = (Node[])new Node[newCap];
    table = newTab;
    if (oldTab != null) {
    for (int j = 0; j < oldCap; ++j) {
    Node 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)e).split(this, newTab, j, oldCap);
    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;
    }
    ···

五、get和remove

get和remove的逻辑思路其实是类似的

···
public V get(Object key) {
Node e;
// hash(key)首先找到key对应的index,然后使用getNode方法读取数据
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;
    // 如果说通过hash寻址算法找到的index的数据不为空的话
    if ((tab = table) != null && (n = tab.length) > 0 &&
    (first = tab[(n - 1) & hash]) != null) {
    // 先检查是不是链表第一个元素,是的话就直接返回
    if (first.hash == hash && // always check first node
    ((k = first.key) == key || (key != null && key.equals(k))))
    return first;
    if ((e = first.next) != null) {
    if (first instanceof TreeNode)
    // 如果链表是红黑树的话,使用红黑树的二分查找读取数据
    return ((TreeNode)first).getTreeNode(hash, key);
    do {
    // 否则遍历链表
    if (e.hash == hash &&
    ((k = e.key) == key || (key != null && key.equals(k))))
    return e;
    } while ((e = e.next) != null);
    }
    }
    return null;
    }

/**

  • Removes the mapping for the specified key from this map if present.
  • @param key key whose mapping is to be removed from the map
  • @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 remove(Object key) {
Node e;
return (e = removeNode(hash(key), key, null, false, true)) == null ?
null : e.value;
}

/**

  • 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 removeNode(int hash, Object key, Object value,
    boolean matchValue, boolean movable) {
    Node[] tab; Node p; int n, index;
    if ((tab = table) != null && (n = tab.length) > 0 &&
    (p = tab[index = (n - 1) & hash]) != null) {
    Node node = null, e; K k; V v;
    if (p.hash == hash &&
    ((k = p.key) == key || (key != null && key.equals(k))))
    node = p;
    else if ((e = p.next) != null) {
    if (p instanceof TreeNode)
    node = ((TreeNode)p).getTreeNode(hash, key);
    else {
    do {
    if (e.hash == hash &&
    ((k = e.key) == key ||
    (key != null && key.equals(k)))) {
    node = e;
    break;
    }
    p = e;
    } while ((e = e.next) != null);
    }
    }
    if (node != null && (!matchValue || (v = node.value) == value ||
    (value != null && value.equals(v)))) {
    if (node instanceof TreeNode)
    ((TreeNode)node).removeTreeNode(this, tab, movable);
    else if (node == p)
    tab[index] = node.next;
    else
    p.next = node.next;
    ++modCount;
    --size;
    afterNodeRemoval(node);
    return node;
    }
    }
    return null;
    }

···

总结

  1. hash算法:为什么要高位和低位做异或运算,这样可以保证说高16位和低16位都参与了hash寻址
  2. hash寻址没有使用取模而是使用了位运算,因为位运算的性能要远远的高于取模
  3. hash冲突后数据挂在链表上,当链表的数量达到8以上后,就会将链表升级为红黑树,避免读取数据的时候,遍历整个链表。

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