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Android 开发也要懂得数据结构 - HashMap源码

Android 开发也要懂得数据结构 - HashMap源码

作者: 进击的包籽 | 来源:发表于2020-12-25 12:50 被阅读0次
    • HashMap不仅是Android开发中常用的数据结构,面试也是高频出现,所以了解一下源码还是非常必要的。
    • 本文章使用的是 JDK1.8 ,不同版本源码有差异。
    • 文章里面的图片来自 极客时间,王争老师的数据结构与算法课
    • 极客时间 - 数据结构与算法

    1.HashMap特点

    • Collection 是集合,有数组(ArrayList)查找快增删慢,有链表(LinkList)增删快查找慢,Map 就是数组与链表的结合体,结合了两的优点。
    • HashMap 的数据关系是 keyvalue 的映射关系,key 是唯一的,value 是可以重复的。
    • HashMapHash , 是因为 key 是需要计算哈希值,这种数组就是散列表
    • HashMap 可以理解为 key计算后的位置用 数组 保存,数组里面的内容放着 链表 ,链表的节点是 key-value 一个个保存起来,查找的时候,快速找到数组中对应的位置,然后遍历链表。
    • HashMap 非线程安全,可以用 HashTable
    • HashMap 数据是无序的,需要有序的使用 LinkedHashMap

    2.HashMap 的继承关系

    • HashMap继承于 Map,而LinkedHashMap 是继承于HashMap

    3.HashMap常用方法

    3.1 构造方法

    • 默认构造方法,只初始化了扩容因数,0.75,就是HashMap的数组容量使用了75%,就要进行扩容操作了,注释还说明初始容量为16
        /**
         * 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
        }
    
    • 自定义初始容量和扩容因子的构造方法,initialCapacity为初始容量,最大不超过2的30次方,loadFactor为扩容因子,必需为大于0的浮点数。
        /**
         * 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;
    
        /**
         * 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) {
            //小于0走异常处理。
            if (initialCapacity < 0)
                throw new IllegalArgumentException("Illegal initial capacity: " +
                                                   initialCapacity);
            //最大不能超过1左移30位,也就是2的30次方,非常大的数。  
            if (initialCapacity > MAXIMUM_CAPACITY)
                initialCapacity = MAXIMUM_CAPACITY;
            //如果扩容因子小于0,或者不是浮点数,报异常处理。
            if (loadFactor <= 0 || Float.isNaN(loadFactor))
                throw new IllegalArgumentException("Illegal load factor: " +
                                                   loadFactor);
            this.loadFactor = loadFactor;
            this.threshold = tableSizeFor(initialCapacity);
        }
    
    • 传入Map的构造方法,默认扩容因子也是0.75,可以将Map转化为HashMap。
        /**
         * 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);
        }
    

    3.2 放入元素 put(K key, V value)

    • 首先看 hash(Object key) 方法,就是判断元素存放在数组的位置,如果空就返回 0,否则 key 的哈希值用临时变量 h 保存,再和 h 无符号右移16位的结果(>>>是无符号右移,高位补0),做异或操作(^是异或),算出存放在数组的位置,这种哈希计算过的数组其实就是 散列表。如果计算出来的结果一样,也就是哈希碰撞,那数据后面会用链表存放。
    • 下图就是散列表。
    散列表
    • putVal(hash(key), key, value, false, true),方法有五个参数,第一个是计算的位置,第二个是 key , 第三个是value , 第四个是否修改已存在的值,最后一个参数看意思是创建表格。
    • 如果key 为空,null 是无法计算哈希值的,就返回0,所以 HashMap 是可以放一个 key 为空的元素的。
        /**
         * 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);
        }
    
        /**
         * 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;
            //如果key为空,就返回0
            //否者key的哈希码 异或 key无符号右移16位的结果
            return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
        }
    
    • Node<K,V> 节点类,我们可以看到,是单链表结构,还重写了equals(Object o)
        /**
         * Basic hash bin node, used for most entries.  (See below for
         * TreeNode subclass, and in LinkedHashMap for its Entry subclass.)
         */
        static class Node<K,V> implements Map.Entry<K,V> {
            final int hash;
            final K key;
            V value;
            //单链表结构
            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;
            }
    
            //重写equals
            public final boolean equals(Object o) {
                if (o == this)
                    return true;
                if (o instanceof Map.Entry) {
                    Map.Entry<?,?> e = (Map.Entry<?,?>)o;
                    //对比key和value
                    if (Objects.equals(key, e.getKey()) &&
                        Objects.equals(value, e.getValue()))
                        return true;
                }
                return false;
            }
        }
    

    3.3 HashMap核心 putVal(int hash, K key, V value, boolean onlyIfAbsent,boolean evict)

    • 这个方法看到Node<K,V>[],HashMap的结构就能理解了吧。

    • HashMap为了解决散列冲突,就用了链表法。

      链表法
    • 存放数据的链表,在长度在 8以下 的时候,是 链表 存储,8以上 或者数组长度大于64时就红黑树存储。

    • 由于方法细节太多,直接写注释一步步好理解。

       /**
         * 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;
    
        /**
         * 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) {
            //分别是散列表,节点,散列表长度,索引位置
            Node<K,V>[] tab; Node<K,V> p; int n, i;
            //如果散列表为空或者散列表长度为0
            if ((tab = table) == null || (n = tab.length) == 0)
                //resize()是创建哈希表,长度为16,并且将长度赋值给n
                n = (tab = resize()).length;
            //找到hash值在当前哈希表中的位置,该位置的节点赋值给p,且判断该位置是否为空
            if ((p = tab[i = (n - 1) & hash]) == null)
                //如果为空就把这个元素放在这个位置
                tab[i] = newNode(hash, key, value, null);
            else {
                Node<K,V> e; K k;
                //如果hash值相同,key也相同
                if (p.hash == hash &&
                    ((k = p.key) == key || (key != null && key.equals(k))))
                    //将节点p赋值给e              
                    e = p;
                //如果p是树节点
                else if (p instanceof TreeNode)
                    //创建一个树节点赋值给e
                    e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
                else {
                    //链表节点,就遍历链表
                    for (int binCount = 0; ; ++binCount) {
                        //将p的下一节点赋值给e,且为空
                        if ((e = p.next) == null) {
                            //找到链表尾部,插入新的节点
                            p.next = newNode(hash, key, value, null);
                            //如果链表的长度大于8的时候
                            if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
                                //链表转树结构
                                treeifyBin(tab, hash);
                            break;
                        }
                        //遍历到的位置已经有元素了,这里就是遍历链表一直循环next,直到为空停止
                        if (e.hash == hash &&
                            ((k = e.key) == key || (key != null && key.equals(k))))
                            break;
                        p = e;
                    }
                }
                //如果当前节点不为空,前面的操作除了最后一个else,其他就是找到已存在的节点
                if (e != null) { // existing mapping for key
                    //当前节点的值赋值给oldValue 
                    V oldValue = e.value;
                    //如果不修改值,或者oldValue 为空(存储value为空)
                    if (!onlyIfAbsent || oldValue == null)
                        //就修改当前节点的值
                        e.value = value;
                    afterNodeAccess(e);
                    //修改值,在这return,不再增加数据的长度
                    return oldValue;
                }
            }
            ++modCount;
            //添加好元素,长度+1
            if (++size > threshold)
                //扩容操作
                resize();
            afterNodeInsertion(evict);
            return null;
        }
        
        /**
         * Replaces all linked nodes in bin at index for given hash unless
         * table is too small, in which case resizes instead.
         */
        final void treeifyBin(Node<K,V>[] tab, int hash) {
            int n, index; Node<K,V> e;
            //如果哈希表为空,或者哈希表长度小于64,优先扩容,而不是转为红黑树
            if (tab == null || (n = tab.length) < MIN_TREEIFY_CAPACITY)
                //扩容
                resize();
            //否则判断不为空就转为树节点
            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);
            }
        }
    

    3.4 扩容 resize()

    • 扩容为原来的2倍大小,扩容完,需要重写计算位置,重写排位置。
        /**
         * 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() {
            Node<K,V>[] oldTab = table;
            int oldCap = (oldTab == null) ? 0 : oldTab.length;
            int oldThr = threshold;
            int newCap, newThr = 0;
            if (oldCap > 0) {
                //最大不能超过Integer.MAX_VALUE
                if (oldCap >= MAXIMUM_CAPACITY) {
                    threshold = Integer.MAX_VALUE;
                    return oldTab;
                }
                //扩容为原来的2倍
                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
                //创建默认大小,长度16
                newCap = DEFAULT_INITIAL_CAPACITY;
                //阈值0.75 * 16
                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;
        }
    

    3.5 查找元素 get(Object key)

    • 查找比较简单,key为空,就是hash为0,有就返回,没有就返回null。
    • key 不为空,找到就返回 value,找不到就返回null。
        /**
         * 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;
            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;
        }
    

    3.6 移除元素 remove(Object key)

        /**
         * 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 <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 remove(Object key) {
            Node<K,V> 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<K,V> removeNode(int hash, Object key, Object value,
                                   boolean matchValue, boolean movable) {
            Node<K,V>[] tab; Node<K,V> p; int n, index;
            //如果散列表不为空且hash对应位置有元素,找到赋值给p
            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;
                //如果p就是要找的元素,赋值给node
                if (p.hash == hash &&
                    ((k = p.key) == key || (key != null && key.equals(k))))
                    node = p;
                //如果p不是要找的元素
                else if ((e = p.next) != null) {
                    //如果是树节点,遍历树,找到节点赋值给node
                    if (p instanceof TreeNode)
                        node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
                    else {
                        //如果是链表,就不停的next,找节点赋值给node
                        do {
                            if (e.hash == hash &&
                                ((k = e.key) == key ||
                                 (key != null && key.equals(k)))) {
                                node = e;
                                break;
                            }
                            p = e;
                        } while ((e = e.next) != null);
                    }
                }
                //如果找的节点不为空,且value值符合
                if (node != null && (!matchValue || (v = node.value) == value ||
                                     (value != null && value.equals(v)))) {
                    //树节点就用树的删除
                    if (node instanceof TreeNode)
                        ((TreeNode<K,V>)node).removeTreeNode(this, tab, movable);
                    //链表只有一个的情况next为空,相当于置空
                    else if (node == p)
                        tab[index] = node.next;
                    //链表后面有元素,直接指向next
                    else
                        p.next = node.next;
                    ++modCount;
                    --size;
                    afterNodeRemoval(node);
                    return node;
                }
            }
            return null;
        }
    

    3.7 清空 clear()

    • 直接遍历散列表,置空,引用断开GC时自己会回收。
        /**
         * Removes all of the mappings from this map.
         * The map will be empty after this call returns.
         */
        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;
            }
        }
    

    3.8 长度 size()

    • 这个长度是元素的数量。
        /**
         * Returns the number of key-value mappings in this map.
         *
         * @return the number of key-value mappings in this map
         */
        public int size() {
            return size;
        }
    

    3.9 获取所有元素的集合 entrySet()

    • 这个方法可以获取所有 Entry
        /**
         * Returns a {@link Set} view of the mappings contained in this map.
         * The set is backed by the map, so changes to the map are
         * reflected in the set, and vice-versa.  If the map is modified
         * while an iteration over the set is in progress (except through
         * the iterator's own <tt>remove</tt> operation, or through the
         * <tt>setValue</tt> operation on a map entry returned by the
         * iterator) the results of the iteration are undefined.  The set
         * supports element removal, which removes the corresponding
         * mapping from the map, via the <tt>Iterator.remove</tt>,
         * <tt>Set.remove</tt>, <tt>removeAll</tt>, <tt>retainAll</tt> and
         * <tt>clear</tt> operations.  It does not support the
         * <tt>add</tt> or <tt>addAll</tt> operations.
         *
         * @return a set view of the mappings contained in this map
         */
        public Set<Map.Entry<K,V>> entrySet() {
            Set<Map.Entry<K,V>> es;
            return (es = entrySet) == null ? (entrySet = new EntrySet()) : es;
        }
    
    • 遍历使用方法
    HashMap<String, String> hashMap = new HashMap<>();
    Iterator iterator = hashMap.entrySet().iterator();
    while (iterator.hasNext()) {
        Map.Entry<String, String> entry = (Map.Entry<String, String>) iterator.next();
        String key = entry.getKey();
        String value = entry.getValue();
    }
    

    3.10 获取key的集合 keySet()

        /**
         * Returns a {@link Set} view of the keys contained in this map.
         * The set is backed by the map, so changes to the map are
         * reflected in the set, and vice-versa.  If the map is modified
         * while an iteration over the set is in progress (except through
         * the iterator's own <tt>remove</tt> operation), the results of
         * the iteration are undefined.  The set supports element removal,
         * which removes the corresponding mapping from the map, via the
         * <tt>Iterator.remove</tt>, <tt>Set.remove</tt>,
         * <tt>removeAll</tt>, <tt>retainAll</tt>, and <tt>clear</tt>
         * operations.  It does not support the <tt>add</tt> or <tt>addAll</tt>
         * operations.
         *
         * @return a set view of the keys contained in this map
         */
        public Set<K> keySet() {
            Set<K> ks = keySet;
            if (ks == null) {
                ks = new KeySet();
                keySet = ks;
            }
            return ks;
        }
    
    • 使用
    HashMap<String, String> hashMap = new HashMap<>();
    Iterator iterator = hashMap.keySet().iterator();
    while (iterator.hasNext()){
        String key = (String) iterator.next();
        String value = hashMap.get(key);
    }
    

    最后,如果有错误,欢迎大家指出,我会继续学习修改,谢谢~~

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