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社区首页 >专栏 >【重点,要考的】数据结构及算法基础--哈希图(HashMap)

【重点,要考的】数据结构及算法基础--哈希图(HashMap)

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用户5640963
发布2019-07-26 14:09:56
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发布2019-07-26 14:09:56
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文章被收录于专栏:卯金刀GG卯金刀GG

HashMap可以说是java中最常见的几种集合了。

在了解HashMap前我们要先了解Object的两个方法:Equals和hashCode()

首先我们来看一下object内的源码是怎样实现的:

hashcode():

代码语言:javascript
复制
/**
     * Returns a hash code value for the object. This method is
     * supported for the benefit of hash tables such as those provided by
     * {@link java.util.HashMap}.
     * <p>
     * The general contract of {@code hashCode} is:
     * <ul>
     * <li>Whenever it is invoked on the same object more than once during
     *     an execution of a Java application, the {@code hashCode} method
     *     must consistently return the same integer, provided no information
     *     used in {@code equals} comparisons on the object is modified.
     *     This integer need not remain consistent from one execution of an
     *     application to another execution of the same application.
     * <li>If two objects are equal according to the {@code equals(Object)}
     *     method, then calling the {@code hashCode} method on each of
     *     the two objects must produce the same integer result.
     * <li>It is <em>not</em> required that if two objects are unequal
     *     according to the {@link java.lang.Object#equals(java.lang.Object)}
     *     method, then calling the {@code hashCode} method on each of the
     *     two objects must produce distinct integer results.  However, the
     *     programmer should be aware that producing distinct integer results
     *     for unequal objects may improve the performance of hash tables.
     * </ul>
     * <p>
     * As much as is reasonably practical, the hashCode method defined by
     * class {@code Object} does return distinct integers for distinct
     * objects. (This is typically implemented by converting the internal
     * address of the object into an integer, but this implementation
     * technique is not required by the
     * Java&trade; programming language.)
     *
     * @return  a hash code value for this object.
     * @see     java.lang.Object#equals(java.lang.Object)
     * @see     java.lang.System#identityHashCode
     */
    public native int hashCode();

但是这个方法没有实现!注意上面这句话:

代码语言:javascript
复制
but this implementation technique is not required by the Java&trade; programming language.
我们不需要知道具体怎样实现的hashCode的运行过程,我们需要知道的是它返回这个对象的特定的类型为整数的hashcode
代码语言:javascript
复制
equals():
代码语言:javascript
复制
/**
     * Indicates whether some other object is "equal to" this one.
     * <p>
     * The {@code equals} method implements an equivalence relation
     * on non-null object references:
     * <ul>
     * <li>It is <i>reflexive</i>: for any non-null reference value
     *     {@code x}, {@code x.equals(x)} should return
     *     {@code true}.
     * <li>It is <i>symmetric</i>: for any non-null reference values
     *     {@code x} and {@code y}, {@code x.equals(y)}
     *     should return {@code true} if and only if
     *     {@code y.equals(x)} returns {@code true}.
     * <li>It is <i>transitive</i>: for any non-null reference values
     *     {@code x}, {@code y}, and {@code z}, if
     *     {@code x.equals(y)} returns {@code true} and
     *     {@code y.equals(z)} returns {@code true}, then
     *     {@code x.equals(z)} should return {@code true}.
     * <li>It is <i>consistent</i>: for any non-null reference values
     *     {@code x} and {@code y}, multiple invocations of
     *     {@code x.equals(y)} consistently return {@code true}
     *     or consistently return {@code false}, provided no
     *     information used in {@code equals} comparisons on the
     *     objects is modified.
     * <li>For any non-null reference value {@code x},
     *     {@code x.equals(null)} should return {@code false}.
     * </ul>
     * <p>
     * The {@code equals} method for class {@code Object} implements
     * the most discriminating possible equivalence relation on objects;
     * that is, for any non-null reference values {@code x} and
     * {@code y}, this method returns {@code true} if and only
     * if {@code x} and {@code y} refer to the same object
     * ({@code x == y} has the value {@code true}).
     * <p>
     * Note that it is generally necessary to override the {@code hashCode}
     * method whenever this method is overridden, so as to maintain the
     * general contract for the {@code hashCode} method, which states
     * that equal objects must have equal hash codes.
     *
     * @param   obj   the reference object with which to compare.
     * @return  {@code true} if this object is the same as the obj
     *          argument; {@code false} otherwise.
     * @see     #hashCode()
     * @see     java.util.HashMap
     */
    public boolean equals(Object obj) {
        return (this == obj);
    }

这里我将jdk源码中所有相关信息都给出来了,希望在某些地方理解的时候,会提供一定的帮助。

当然我们可以重写这两个函数,但是在java1.8中定义的函数最好不要进行重写,不然对hashmap的性能产生很大的影响;

1)HashMap概述

HashMap是基于哈希表的map接口的非同步实现,此实现提供所有可选的映射操作,并允许使用null值和null键。此类不保证映射的顺序,特别是它不保证该顺序恒久不变。

2)HashMap数据结构

在java语言编程中,最基本的数据结构就两种:数组和引用,其他所有的数据结构都可以通过这两个基本的数据结构来实现,在jkd 1.7以前,hashmap就是一个链表散列的结构,但是在jdk1.8发布后,hashmap的链表长度大于一定值过后,变编程红黑树,关于红黑树的概念,在上篇文章中进行了讲解:

java中采用的便是链地址法,便是每个数组元素上都是一个链表。当数据被hash后,得到数组下标,将数据放在对应数组下标的链表上

其中每个元素都用node节点表示:

代码语言:javascript
复制
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;
        }
}

node是hashmap的一个内部类,用来储存数据和保持链表结构的。它的本质就是一个映射(键值对)。

当然,会产生两个key值产生同一个位置,(最主要的便是因为index的产生原理,当然也有可能是产生了一样的hash值)这种情况叫哈希碰撞。当然hash算法计算结果越分散均匀,发生hash碰撞的机率就越小,map的存储效率就越高。

hashmap中又一个很重要的字段就是Node[] table。如上图所示,这就是hashmap的基本结构,构成链表的数组。

如果哈希桶数组很大,即使较差的Hash算法也会比较分散,如果哈希桶数组数组很小,即使好的Hash算法也会出现较多碰撞,所以就需要在空间成本和时间成本之间权衡,其实就是在根据实际情况确定哈希桶数组的大小,并在此基础上设计好的hash算法减少Hash碰撞。那么通过什么方式来控制map使得Hash碰撞的概率又小,哈希桶数组(Node[] table)占用空间又少呢?答案就是好的Hash算法和扩容机制。

在此之前,我们先来了解一下hashmap一些非常非常重要的参数。源代码中如下:

代码语言:javascript
复制
/**
     * 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;
    
    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;
    /**
     * The number of key-value mappings contained in this map.
     */
代码语言:javascript
复制
上面这些参数的是非常非常重要的,其重要性相当于hashmap的数据结构的重要性。在本篇中,我们运用到并重点讲解的为一下几个参数
代码语言:javascript
复制
static final int DEFAULT_INITIAL_CAPACITY = 1 << 4;
static final float DEFAULT_LOAD_FACTOR = 0.75f;
transient int size;
transient int modCount;
int threshold;
final float loadFactor;

首先可以单刀,Node[] table的默认长度是16,loadFactor的默认大小为0.75,threshold是hashmap所能容纳的最大数据量的Node个数,默认为0.75,threshold=DEFAULT_INITIAL_CAPACITY*loadFactor;当添加元素数量超过这个数量过后,就要进行扩容,扩容后hashmap的容量是之前的两倍。对于0.75,建议大家不要轻易修改。除非在时间和空间比较特殊的情况下,如果内存空间很多而又对时间效率要求很高,可以降低负载因子Load factor的值;相反,如果内存空间紧张而对时间效率要求不高,可以增加负载因子loadFactor的值,这个值可以大于1。

size就是在这个hashmpa中实际存在的node数量。modCount便是hashmap结构修改的次数。在之前对iterator(迭代器)进行讲解的时候我已经进行了说明,需要注意的是在hashmap中modcount指的是结构更改的次数,例如添加新的node,但是如果是替换原有node的value,modcount是不变的,因为它不属于结构变化。

有兴趣可以了解下: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。

3)确认hashmap索引位置

代码:

代码语言:javascript
复制
/**
     * 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);
    }

这里的Hash算法本质上就是三步:取key的hashCode值、高位运算、取模运算。

对于任意给定的对象,只要它的hashCode()返回值相同,那么程序调用方法一所计算得到的Hash码值总是相同的。我们首先想到的就是把hash值对数组长度取模运算,这样一来,元素的分布相对来说是比较均匀的。但是,模运算的消耗还是比较大的,在HashMap中是这样做的:我们通过h & (table.length -1)来计算该对象应该保存在table数组的哪个索引处。

这个方法非常巧妙,它通过h & (table.length -1)来得到该对象的保存位,而HashMap底层数组的长度总是2的n次方,这是HashMap在速度上的优化。当length总是2的n次方时,h& (length-1)运算等价于对length取模,也就是h%length,但是&比%具有更高的效率。

在JDK1.8的实现中,优化了高位运算的算法,通过hashCode()的高16位异或低16位实现的:(h = k.hashCode()) ^ (h >>> 16),主要是从速度、功效、质量来考虑的,这么做可以在数组table的length比较小的时候,也能保证考虑到高低Bit都参与到Hash的计算中,同时不会有太大的开销。

我们举个栗子:

大概的得到索引的流程就是上面所示。

4)hashmap的put实现方法(划重点,要考):

put函数大致的思路为:

  1. 对key的hashCode()做hash,然后再计算index;
  2. 如果没碰撞直接放到bucket里;
  3. 如果碰撞了,以链表的形式存在buckets后;
  4. 如果碰撞导致链表过长(大于等于TREEIFY_THRESHOLD),就把链表转换成红黑树;
  5. 如果节点已经存在就替换old value(保证key的唯一性)
  6. 如果bucket满了(超过load factor*current capacity),就要resize。

代码如下:

代码语言:javascript
复制
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;
        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;
    }

5)hashmap的get方法:

思路如下:

  1. bucket里的第一个节点,直接命中;
  2. 如果有冲突,则通过key.equals(k)去查找对应的entry 若为树,则在树中通过key.equals(k)查找,O(logn); 若为链表,则在链表中通过key.equals(k)查找,O(n)。

代码如下:

代码语言:javascript
复制
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))))
                return first;
            if ((e = first.next) != null) {
                if (first instanceof TreeNode)
                    return ((TreeNode<K,V>)first).getTreeNode(hash, key);
                do {
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        return e;
                } while ((e = e.next) != null);
            }
        }
        return null;
    }

注意(重点,要考的):上述put的思路从putval的方法中是正确的,但是如果将putval方法打碎了分析,这个思路是不完全的,这就涉及到了hashmap的扩容机制,我会在下一篇hashmap的讲解中来具体讲解,putval在不同情况下是怎么运行的,以及扩容机制中最重要的函数,resize();

jdk1.8中对hashmap有着非常棒的扩容机制,我们在上一篇文章提到了当链表长度大于某个值的时候,hashmap中的链表会变成红黑树结构,但是实际上真的是这样么?我们来看一下树化的函数是怎样进行的:

代码语言:javascript
复制
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)
            resize();
        else if ((e = tab[index = (n - 1) & hash]) != null) {
            TreeNode<K,V> hd = null, tl = null;
            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);
            if ((tab[index] = hd) != null)
                hd.treeify(tab);
        }
    }

我们从第一个判断语句就发现,如果hashmap中table的长度小于64(MIN_TREEIFY_CAPACITY)的时候,其实是不会进行树化的,而是对这个hashmap进行扩容。所以我们发现,扩容不仅仅用于node的个数超过threshold的时候。

这个树化函数的设计便是想保持算法设计中的相对较好。

要了解扩容机制,我们先来看看jdk1.7是怎么设计的,因为我用的是jdk1.8,所以一下代码是从网上摘取,如果和源码有区别,请各位告知:

代码语言:javascript
复制
void resize(int newCapacity) {   //传入新的容量  
    Entry[] oldTable = table;    //引用扩容前的Entry数组  
    int oldCapacity = oldTable.length;  
    if (oldCapacity == MAXIMUM_CAPACITY) {  //扩容前的数组大小如果已经达到最大(2^30)了  
        threshold = Integer.MAX_VALUE; //修改阈值为int的最大值(2^31-1),这样以后就不会扩容了  
        return;  
    }  
  
    Entry[] newTable = new Entry[newCapacity];  //初始化一个新的Entry数组  
    transfer(newTable);                         //!!将数据转移到新的Entry数组里  
    table = newTable;                           //HashMap的table属性引用新的Entry数组  
    threshold = (int) (newCapacity * loadFactor);//修改阈值  
}

其中transfer方法如下:

代码语言:javascript
复制
void transfer(Entry[] newTable) {  
    Entry[] src = table;                   //src引用了旧的Entry数组  
    int newCapacity = newTable.length;  
    for (int j = 0; j < src.length; j++) { //遍历旧的Entry数组  
        Entry<K, V> e = src[j];             //取得旧Entry数组的每个元素  
        if (e != null) {  
            src[j] = null;//释放旧Entry数组的对象引用(for循环后,旧的Entry数组不再引用任何对象)  
            do {  
                Entry<K, V> next = e.next;  
                int i = indexFor(e.hash, newCapacity); //!!重新计算每个元素在数组中的位置  
                e.next = newTable[i]; //标记[1]  
                newTable[i] = e;      //将元素放在数组上  
                e = next;             //访问下一个Entry链上的元素  
            } while (e != null);  
        }  
    }  
}

我们通过上面代码可以知道,我们其实是遍历这个链表,然后将新的元素位置从头位置插入。这样我们可以知道,我们链表中的先后顺序是会改变的。前后顺序会反过来。下图可以很明白的开出这种变换关系:

那么,关于jdk1.8,我们做了哪些优化呢?

(重点,要考)我们一定要先明确一个很重要的东西!!jdk1.8的table长度一定是2的幂!!

也就是说在jdk1.8中 resize()一定是扩大两倍的容量

jdk1.8中的索引和1.7的原则是一样的,都采用的是:h & (length - 1)作为node的索引

如果我们扩展长度为两倍,那么作为length-1就是尾端为一串1,其余为0的位序列。

那么位运算可以得到下图:

图a是扩展前产生的index,图二为扩展两倍容量的index,java1.8很巧妙的运用扩展2倍产生index这一点,我们直接判断hash值在位中,比n-1高一位的比特是1还是0来移动:

这就是上图中,红点标出的比特位便成了一种标志,我们通过判断它为0为1来进行扩容操作。红圈的16不是定值,而是原hashmap的table的长度。

上面的例子,也说明,我们table长度只有16的时候,有很大的情况能够让index相同,但是扩容后又不在拥有相同的index。

这个设计确实非常的巧妙,既省去了重新计算hash值的时间,而且同时,由于新增的1bit是0还是1可以认为是随机的,因此resize的过程,均匀的把之前的冲突的节点分散到新的bucket了。这一块就是JDK1.8新增的优化点。有一点注意区别,JDK1.7中rehash的时候,旧链表迁移新链表的时候,如果在新表的数组索引位置相同,则链表元素会倒置,但是从上图可以看出,JDK1.8不会倒置,这一点正如之前的代码所示。

我们可以用一张图略微表示一下,下图中蓝色为新增的index位为0,绿色的表示1:

当然,jdk1.8的resize代码复杂了很多,虽然大家都说它写的很好,我还是在判断语句的执行中有很多疑惑,感觉很多判断语句都是相互包含的。具体的我还要继续学习一下,但是jdk1.8中的resize()流程还是很清晰的,怎么扩展,怎么移动链表,代码都很棒的:

代码语言:javascript
复制
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;
            }
            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<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)
                        newTab[e.hash & (newCap - 1)] = e;
                    else if (e instanceof TreeNode)
                        ((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
                    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;
    }

其实说了这么多,hashmap如果只是运用的话,我们只需要了解她的基础函数和结构即可,但是我相信对hashmap的原理有了解肯定能加强对它理解和应用,对不同情况的使用也有理解。

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目录
  • 1)HashMap概述
  • 2)HashMap数据结构
  • 3)确认hashmap索引位置
  • 4)hashmap的put实现方法(划重点,要考):
  • 5)hashmap的get方法:
  • (重点,要考)我们一定要先明确一个很重要的东西!!jdk1.8的table长度一定是2的幂!!
  • 也就是说在jdk1.8中 resize()一定是扩大两倍的容量
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