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  • hashmap

    1.前言

    2.公共方法

    3.全局变量&&公共方法

    
    
    /*
         * hashmap中同一个桶中的多个节点默认以链表的形式连接,超过一定数量后会以红黑树的形式连接,并将Node转化为TreeNode
         * 由于TreeNode的大小是Node的两倍,所以当bins(链表)数量下降到一定值的时候会再转化回Node
         * 术语描述:
         *  table hashmap中存放数据的数组,所有的操作都是以这个table为入口
         *  hash桶:table数组中的每一个下标位置对应一个hash桶
         *  bins  同一个桶中的所有子节点集合,也就是node的集合
         *  Node  hash桶中的节点
         *  
         */

    1 /** 2 * 默认初始化容量,必须是2的指数倍 3 */ 4 static final int DEFAULT_INITIAL_CAPACITY = 1 << 4; // 16 5 6 /** 7 * 最大容量 8 */ 9 static final int MAXIMUM_CAPACITY = 1 << 30; 10 11 /** 12 * 默认的加载因子. 13 */ 14 static final float DEFAULT_LOAD_FACTOR = 0.75f; 15 16 /** 17 * 横向链表bin节点的阈值,超过这个值会进行树化(将链表转化为tree) 18 */ 19 static final int TREEIFY_THRESHOLD = 8; 20 21 /** 22 * 将树化结构转回链表的阈值 23 */ 24 static final int UNTREEIFY_THRESHOLD = 6; 25 26 /** 27 * 触发树化时的表的最小容量,即当链表长度过大时,优先选择扩容,容量超过64之后,才会对链表进行树化 28 */ 29 static final int MIN_TREEIFY_CAPACITY = 64; 30 31 /** 32 * 存储hash entry的数组,第一次使用的时候初始化,在必要的时候进行扩容 33 */ 34 transient Node<K,V>[] table; 35 36 /** 37 * 保存缓存的entry 集合 38 */ 39 transient Set<Map.Entry<K,V>> entrySet; 40 41 /** 42 * 当前map中的entry set的数量(放入的对象的数量) 43 */ 44 transient int size; 45 46 /** 47 * MyHashMap被结构性修改的次数,如:增减map中entry的数量,rehash。 这个字段主要用于iterators遍历时使用,判断是否存在并发操作问题 48 */ 49 transient int modCount; 50 51 /** 52 * 进行扩容的阈值,= (capacity * load factor). 53 */ 54 int threshold; 55 56 /** 57 * 加载因子. 58 */ 59 final float loadFactor;
     1  /**
     2      * hash值计算
     3      */
     4     static final int hash(Object key) {
     5         int h;
     6         return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
     7     }
     8 
     9     /**
    10      * 返回2的指数倍的数组大小
    11      */
    12     static final int tableSizeFor(int cap) {
    13         int n = cap - 1;
    14         n |= n >>> 1;
    15         n |= n >>> 2;
    16         n |= n >>> 4;
    17         n |= n >>> 8;
    18         n |= n >>> 16;
    19         return (n < 0) ? 1 : (n >= MAXIMUM_CAPACITY) ? MAXIMUM_CAPACITY : n + 1;
    20     }
    21     /**
    22      *返回hash桶中节点的总数量
    23      */
    24     public int size() {
    25         return size;
    26     }
    27 
    28     /**
    29      * 判断hash桶是否为空
    30      */
    31     public boolean isEmpty() {
    32         return size == 0;
    33     }

    4.数据结构

    /**
         * bin节点的数据结构
         */
        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;
            }
    
            public final K getKey()        { return key; }
            public final V getValue()      { return value; }
            public final String toString() { return key + "=" + value; }
    
            public final int hashCode() {
                return Objects.hashCode(key) ^ Objects.hashCode(value);
            }
    
            public final V setValue(V newValue) {
                V oldValue = value;
                value = newValue;
                return oldValue;
            }
    
            public final boolean equals(Object o) {
                if (o == this)
                    return true;
                if (o instanceof Map.Entry) {
                    Map.Entry<?,?> e = (Map.Entry<?,?>)o;
                    if (Objects.equals(key, e.getKey()) &&
                        Objects.equals(value, e.getValue()))
                        return true;
                }
                return false;
            }
        }

    5.插入元素

     /**
         * 将对应的key,value放到hashmap中,如果之前存在这个key则直接更新value
         */
        public V put(K key, V value) {
            return putVal(hash(key), key, value, false, true);
        }
       
        /**
         * 将一个map添加到当前hashmap中
         * @param m 要添加的map
         * @param evict 当第一次构造这个map时设置为false
         */
        final void putMapEntries(Map<? extends K, ? extends V> m, boolean evict) {
            int s = m.size();
            if (s > 0) {
                if (table == null) { // pre-size
                    //计算 map的capacity
                    float ft = ((float)s / loadFactor) + 1.0F;
                    int t = ((ft < (float)MAXIMUM_CAPACITY) ?
                             (int)ft : MAXIMUM_CAPACITY);
                    if (t > threshold)
                        threshold = tableSizeFor(t);        //更新阈值
                }
                else if (s > threshold)
                    resize();       //如果map的大小超过了当前map的阈值,则进行扩容
                for (Map.Entry<? extends K, ? extends V> e : m.entrySet()) {
                    K key = e.getKey();
                    V value = e.getValue();
                    putVal(hash(key), key, value, false, evict);    //遍历当前map,将map中的数据加到当前hashmap中
                }
            }
        }
    
        /**
         * @param hash key的hash值
         * @param key key
         * @param value value
         * @param onlyIfAbsent 是否只有不存在这个key的时候才put
         * @param evict if false, the table is in creation mode.
         * @return 改key中以前的值,或者null 如果这是第一次添加的话
         */
        final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
                       boolean evict) {
            Node<K,V>[] tab;     //指向当前的hash数组
            Node<K,V> p;     //指向当前hash对应桶中的首节点
            int n, i;       //n表示数组的长度
            if ((tab = table) == null || (n = tab.length) == 0)
                n = (tab = resize()).length;     //初始化数组
            if ((p = tab[i = (n - 1) & hash]) == null)      //判断这个hash所对应的hash桶是否已经有Node节点
                tab[i] = newNode(hash, key, value, null);   //hash桶中无其它数据,直接在hash桶中添加这个节点即可
            else {
                Node<K,V> e;   //保存当前桶中该key对应的数据节点(可能这个key之前插入过) 
                K k;
                if (p.hash == hash &&
                    ((k = p.key) == key || (key != null && key.equals(k))))   //当前key存在
                    e = p;
                else if (p instanceof TreeNode)     //当前桶中的节点为tree节点
                    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) // 如果当前桶中的节点数量大于树化阈值,则进行树化
                                treeifyBin(tab, hash);
                            break;
                        }
                        if (e.hash == hash &&
                            ((k = e.key) == key || (key != null && key.equals(k))))
                            break;
                        p = e;
                    }
                }
                if (e != null) { // 当前key已经在map中存在,根据入参决定是否更新key的值
                    V oldValue = e.value;
                    if (!onlyIfAbsent || oldValue == null)
                        e.value = value;
                    afterNodeAccess(e);
                    return oldValue;
                }
            }
            ++modCount;         //修改modCount,用于iterator判断当前是否存在并发
            if (++size > threshold)
                resize();
            afterNodeInsertion(evict);
            return null;
        }
    /**
         *将map中的数据放到当前 hashmap中
         */
        public void putAll(Map<? extends K, ? extends V> m) {
            putMapEntries(m, true);
        }
    
    
    
     

    6.获取元素

    /**
         * 返回这个key对应的值
         * 找不到这个key或者这个key的值为null时返回null
         */
        public V get(Object key) {
            Node<K,V> e;
            return (e = getNode(hash(key), key)) == null ? null : e.value;
        }
        
        /**
         * 判断当前hashmap中是否有这个key
         */
        public boolean containsKey(Object key) {
            return getNode(hash(key), key) != null;
        }
    
        /**
         * 查找方式:先根据hash值找到hash桶,然后遍历hash桶中的Node链表(或者红黑树),找到这个节点
         * @param hash key的hash值
         * @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;     //first保存相应hash桶中的第一个节点
            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;
        }
    /**
         * 判断当前map中是否存在这个值
         */
        public boolean containsValue(Object value) {
            Node<K,V>[] tab; V v;
            if ((tab = table) != null && size > 0) {
                for (int i = 0; i < tab.length; ++i) {
                    for (Node<K,V> e = tab[i]; e != null; e = e.next) {
                        if ((v = e.value) == value ||
                            (value != null && value.equals(v)))
                            return true;
                    }
                }
            }
            return false;
        }
    
    
    
     

    7.移除元素

    /**
         * 移除指定的节点,返回节点对应的值
         */
        public V remove(Object key) {
            Node<K,V> e;
            return (e = removeNode(hash(key), key, null, false, true)) == null ?
                null : e.value;
        }
    
        /**
         * 移除指定的数据,实现方式是在链表中跳过当前节点
         * @param matchValue 是否需要匹配值
         */
        final Node<K,V> removeNode(int hash, Object key, Object value,
                                   boolean matchValue, boolean movable) {
            Node<K,V>[] tab; Node<K,V> p; int n, index;
            if ((tab = table) != null && (n = tab.length) > 0 &&
                (p = tab[index = (n - 1) & hash]) != null) {
                Node<K,V> 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<K,V>)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<K,V>)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;
        }
    /**
         * 清空table中的数据
         */
        public void clear() {
            Node<K,V>[] tab;
            modCount++;
            if ((tab = table) != null && size > 0) {
                size = 0;
                for (int i = 0; i < tab.length; ++i)
                    tab[i] = null;
            }
        }
    
    
    
     

    7.扩容

    /**
         * 扩容或者初始化表。
         * 扩容是double table,这里注意一个hash一致性问题(2的幂)
         * @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) {   //扩容流程
                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 扩容table的流程,修改容量值和阈值
            }
            else if (oldThr > 0) // initial capacity was placed in threshold
                newCap = oldThr;
            else {               // 初始化表的流程,使用默认的初始容量和加载因子
                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];         //使用数组容量初始化空的数组
            table = newTab;                 //将当前新初始化的空数组置为全局变量 table(hashmap存放node的数组)
            if (oldTab != null) {           //如果这次是扩容操作,需要将原表中的数组拷贝到新的数组中(table)
                for (int j = 0; j < oldCap; ++j) {
                    Node<K,V> e;
                    if ((e = oldTab[j]) != null) {  //遍历处理每一个hash桶
                        oldTab[j] = null;
                        if (e.next == null)         //如果这个hash桶中只有一个节点,则直接将这条链移到相应位置即可
                            newTab[e.hash & (newCap - 1)] = e;
                        else if (e instanceof TreeNode)     //如果子链是树节点,则将树节点进行拆分处理
                            ((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
                        else { // preserve order
                            Node<K,V> loHead = null, loTail = null;        //存放的是要保留在当前桶中的Node节点链表
                            Node<K,V> hiHead = null, hiTail = null;         //存放的是要放在扩容后新产生的桶(j + oldCap)中的链表
                            Node<K,V> next;
                            do {
                                next = e.next;
                                if ((e.hash & oldCap) == 0) {   //根据高位是否为1(扩容正好是2倍扩容,高位为1的话说明hash值会落到新桶中),判断数据迁移
                                    if (loTail == null)
                                        loHead = e;
                                    else
                                        loTail.next = e;
                                    loTail = e;
                                }
                                else {                  //存放要移到新桶中的Node节点链表
                                    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;
        }

    8. map视图(keyset,valueset,entryset)

     /**
         * 返回当前map中所有key集合的视图
         */
        public Set<K> keySet() {
            Set<K> ks = keySet;
            if (ks == null) {
                ks = new KeySet();
                keySet = ks;
            }
            return ks;
        }
    
        /**
         * hashmap 的key视图
         */
        final class KeySet extends AbstractSet<K> {
            public final int size()                 { return size; }
            public final void clear()               { HashMap_tmp.this.clear(); }
            public final Iterator<K> iterator()     { return new KeyIterator(); }
            public final boolean contains(Object o) { return containsKey(o); }
            public final boolean remove(Object key) {
                return removeNode(hash(key), key, null, false, true) != null;
            }
            public final Spliterator<K> spliterator() {
                return new KeySpliterator<>(HashMap_tmp.this, 0, -1, 0, 0);
            }
            public final void forEach(Consumer<? super K> action) {
                Node<K,V>[] tab;
                if (action == null)
                    throw new NullPointerException();
                if (size > 0 && (tab = table) != null) {
                    int mc = modCount;
                    for (int i = 0; i < tab.length; ++i) {
                        for (Node<K,V> e = tab[i]; e != null; e = e.next)
                            action.accept(e.key);
                    }
                    if (modCount != mc)
                        throw new ConcurrentModificationException();
                }
            }
        }
    
        /**
         * 返回当前map的values的集合视图
         */
        public Collection<V> values() {
            Collection<V> vs = values;
            if (vs == null) {
                vs = new Values();
                values = vs;
            }
            return vs;
        }
        
        /**
         * hashmap的values视图
         */
        final class Values extends AbstractCollection<V> {
            public final int size()                 { return size; }
            public final void clear()               { HashMap_tmp.this.clear(); }
            public final Iterator<V> iterator()     { return new ValueIterator(); }
            public final boolean contains(Object o) { return containsValue(o); }
            public final Spliterator<V> spliterator() {
                return new ValueSpliterator<>(HashMap_tmp.this, 0, -1, 0, 0);
            }
            public final void forEach(Consumer<? super V> action) {
                Node<K,V>[] tab;
                if (action == null)
                    throw new NullPointerException();
                if (size > 0 && (tab = table) != null) {
                    int mc = modCount;
                    for (int i = 0; i < tab.length; ++i) {
                        for (Node<K,V> e = tab[i]; e != null; e = e.next)
                            action.accept(e.value);
                    }
                    if (modCount != mc)
                        throw new ConcurrentModificationException();
                }
            }
        }
    
        /**
         * 返回hashmap的key,value视图
         */
        public Set<Map.Entry<K,V>> entrySet() {
            Set<Map.Entry<K,V>> es;
            return (es = entrySet) == null ? (entrySet = new EntrySet()) : es;
        }
    
        /**
         * hashmap的key value视图
         * 
         */
        final class EntrySet extends AbstractSet<Map.Entry<K,V>> {
            public final int size()                 { return size; }
            public final void clear()               { HashMap_tmp.this.clear(); }
            public final Iterator<Map.Entry<K,V>> iterator() {
                return new EntryIterator();
            }
            public final boolean contains(Object o) {
                if (!(o instanceof Map.Entry))
                    return false;
                Map.Entry<?,?> e = (Map.Entry<?,?>) o;
                Object key = e.getKey();
                Node<K,V> candidate = getNode(hash(key), key);
                return candidate != null && candidate.equals(e);
            }
            public final boolean remove(Object o) {
                if (o instanceof Map.Entry) {
                    Map.Entry<?,?> e = (Map.Entry<?,?>) o;
                    Object key = e.getKey();
                    Object value = e.getValue();
                    return removeNode(hash(key), key, value, true, true) != null;
                }
                return false;
            }
            public final Spliterator<Map.Entry<K,V>> spliterator() {
                return new EntrySpliterator<>(HashMap_tmp.this, 0, -1, 0, 0);
            }
            public final void forEach(Consumer<? super Map.Entry<K,V>> action) {
                Node<K,V>[] tab;
                if (action == null)
                    throw new NullPointerException();
                if (size > 0 && (tab = table) != null) {
                    int mc = modCount;
                    for (int i = 0; i < tab.length; ++i) {
                        for (Node<K,V> e = tab[i]; e != null; e = e.next)
                            action.accept(e);
                    }
                    if (modCount != mc)
                        throw new ConcurrentModificationException();
                }
            }
        }

    9.树化

     1 /**
     2      * 将Node链表进行树化,前提是满足树化条件(默认是 桶的数量要达到64),否则进行扩容
     3      */
     4     final void treeifyBin(Node<K,V>[] tab, int hash) {
     5         int n, index; 
     6         Node<K,V> e;    //存放链表的首节点
     7         if (tab == null || (n = tab.length) < MIN_TREEIFY_CAPACITY)
     8             resize();
     9         else if ((e = tab[index = (n - 1) & hash]) != null) {
    10             TreeNode<K,V> hd = null, tl = null;     //hd存放头节点,tl存放尾节点,建立双向链表
    11             do {
    12                 TreeNode<K,V> p = replacementTreeNode(e, null);   //将node转化为TreeNode
    13                 if (tl == null)
    14                     hd = p;
    15                 else {
    16                     p.prev = tl;
    17                     tl.next = p;
    18                 }
    19                 tl = p;
    20             } while ((e = e.next) != null);
    21             if ((tab[index] = hd) != null)
    22                 hd.treeify(tab);
    23         }
    24     }
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  • 原文地址:https://www.cnblogs.com/gc65/p/9671226.html
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