你要的hashMap解读来了

最近快到金三银四了,不少小伙伴又开始忙忙碌碌找工作了,我也是,秉着为自己也为大家,整理了下hashMap的知识点,希望能对你有所帮助!

这里主要是对jdk1.7和jdk1.8的分析

JDK1.7 hashMap结构图

img

1.7 hashMap变量

put函数过程

 public V put(K key, V value) {
     
      // 容量为空 则初始化
        if (table == EMPTY_TABLE) {
            inflateTable(threshold);
        }
        // key为null 则进行put key为null的操作
        if (key == null)
            return putForNullKey(value);
     
      // 求hash值 根据hashCode 再进行一些位运算 降低hash冲突的概率
        int hash = hash(key);
     
      // 根据元素hash值 求索引
        int i = indexFor(hash, table.length);
        for (Entry e = table[i]; e != null; e = e.next) {
            Object k;
            // 如果元素hash值和key都和桶上的第一个元素相同,则替换value
            if (e.hash == hash && ((k = e.key) == key || key.equals(k))) {
                V oldValue = e.value;
                e.value = value;
                e.recordAccess(this);
                return oldValue;
            }
        }

        modCount++;
      // 添加元素 头插法(可能会造成死循环)
        addEntry(hash, key, value, i);
        return null;
    }

hash函数过程

hash冲突的解决办法:

开放定址法

再哈希法

链地址法(拉链法,hashMap使用)

    /**
     * Retrieve object hash code and applies a supplemental hash function to the
     * result hash, which defends against poor quality hash functions.  This is
     * critical because HashMap uses power-of-two length hash tables, that
     * otherwise encounter collisions for hashCodes that do not differ
     * in lower bits. Note: Null keys always map to hash 0, thus index 0.
     */
    final int hash(Object k) {
        int h = hashSeed;
        if (0 != h && k instanceof String) {
            return sun.misc.Hashing.stringHash32((String) k);
        }

        h ^= k.hashCode();

        // This function ensures that hashCodes that differ only by
        // constant multiples at each bit position have a bounded
        // number of collisions (approximately 8 at default load factor).
        h ^= (h >>> 20) ^ (h >>> 12);
        return h ^ (h >>> 7) ^ (h >>> 4);
    }

求索引位置 indexFor函数

// jdk1.7  求索引位置函数   
static int indexFor(int h, int length) {
        // assert Integer.bitCount(length) == 1 : "length must be a non-zero power of 2";
        return h & (length-1);
    }

扩容过程

1.addEntry函数

    /**
     * Adds a new entry with the specified key, value and hash code to
     * the specified bucket.  It is the responsibility of this
     * method to resize the table if appropriate.
     * 添加元素
     * Subclass overrides this to alter the behavior of put method.
     */
    void addEntry(int hash, K key, V value, int bucketIndex) {
        // 当前容量大于等于负载数 且当前位置首个元素不为空
        if ((size >= threshold) && (null != table[bucketIndex])) {
            // 扩容为原来的两倍
            resize(2 * table.length);
            hash = (null != key) ? hash(key) : 0;
            bucketIndex = indexFor(hash, table.length);
        }

        createEntry(hash, key, value, bucketIndex);
    }

2.resize函数

    /**
     * Rehashes the contents of this map into a new array with a
     * larger capacity.  This method is called automatically when the
     * number of keys in this map reaches its threshold.
     *
     * If current capacity is MAXIMUM_CAPACITY, this method does not
     * resize the map, but sets threshold to Integer.MAX_VALUE.
     * This has the effect of preventing future calls.
     *
     * @param newCapacity the new capacity, MUST be a power of two;
     *        must be greater than current capacity unless current
     *        capacity is MAXIMUM_CAPACITY (in which case value
     *        is irrelevant).
     */
    void resize(int newCapacity) {
        Entry[] oldTable = table;
        int oldCapacity = oldTable.length;
        if (oldCapacity == MAXIMUM_CAPACITY) {
            threshold = Integer.MAX_VALUE;
            return;
        }
  
        // 创建新数组
        Entry[] newTable = new Entry[newCapacity];
        
        // 元素重新求索引位置 元素从旧数组移到新数组上
        transfer(newTable, initHashSeedAsNeeded(newCapacity));
        table = newTable;
        
        // 设置新的负载数
        threshold = (int)Math.min(newCapacity * loadFactor, MAXIMUM_CAPACITY + 1);
    }

    /**
     * Transfers all entries from current table to newTable.
     */
    void transfer(Entry[] newTable, boolean rehash) {
        int newCapacity = newTable.length;
        for (Entry e : table) {
            while(null != e) {
                Entry next = e.next;
                if (rehash) {
                    e.hash = null == e.key ? 0 : hash(e.key);
                }
                int i = indexFor(e.hash, newCapacity);
                e.next = newTable[i];
                newTable[i] = e;
                e = next;
            }
        }
    }

移除

    /**
     * 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) {
        
        // 移除key对应的元素 并返回entry
        Entry e = removeEntryForKey(key);
        return (e == null ? null : e.value);
    }
    
        /**
     * Removes and returns the entry associated with the specified key
     * in the HashMap.  Returns null if the HashMap contains no mapping
     * for this key.
     */
    final Entry removeEntryForKey(Object key) {
        if (size == 0) {
            return null;
        }
        int hash = (key == null) ? 0 : hash(key);
        
        // 根据元素hash值求索引
        int i = indexFor(hash, table.length);
        
        // 桶的首个节点
        Entry prev = table[i];
        Entry e = prev;

        while (e != null) {
            Entry next = e.next;
            Object k;
            
            // 判断元素是否和移除的元素key和hash值相同 ,相同则进行移除操作
            if (e.hash == hash &&
                ((k = e.key) == key || (key != null && key.equals(k)))) {
                modCount++;
                size--;
                
                // 如果 桶的首个元素和被移除的元素相同 则将next 置为桶的首个元素
                if (prev == e)
                    table[i] = next;
                else
                    prev.next = next;
                e.recordRemoval(this);
                return e;
            }
            prev = e;
            e = next;
        }

        return e;
    }

get函数(根据key获取entry)

public V get(Object key) {
    if (key == null)
        return getForNullKey();
    Entry entry = getEntry(key);

    return null == entry ? null : entry.getValue();
}

final Entry getEntry(Object key) {
    if (size == 0) {
        return null;
    }

    int hash = (key == null) ? 0 : hash(key);
    
    // 遍历索引上的链表
    for (Entry e = table[indexFor(hash, table.length)];
         e != null;
         e = e.next) {
        Object k;
        // hash值相同 key也相等 则返回entry元素
        if (e.hash == hash &&
            ((k = e.key) == key || (key != null && key.equals(k))))
            return e;
    }
    return null;
}

缩容

JDK1.8 hashMap

1.8 hashMap 变量

put函数过程

    /**
     * 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;
        // 数组为空 或长度为0 初始化数组
        if ((tab = table) == null || (n = tab.length) == 0)
            n = (tab = resize()).length;
        
        // 索引位置的桶为空 创建节点
        if ((p = tab[i = (n - 1) & hash]) == null)
            tab[i] = newNode(hash, key, value, null);
        else {
            Node e; K k;
            
            // 如果put的元素为桶的首个节点,e赋值
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                e = p;
            
            // 如果是红黑树 将元素插入
            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) {
                        p.next = newNode(hash, key, value, null);
                        
                        // 链表长度大于 TREEIFY_THRESHOLD 链表转红黑树
                        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;
                }
            }
            
            // 统一处理,value值
            if (e != null) { // existing mapping for key
                V oldValue = e.value;
                if (!onlyIfAbsent || oldValue == null)
                    e.value = value;
                afterNodeAccess(e);
                return oldValue;
            }
        }
        ++modCount;
        // size大于负载*容量时,扩容
        if (++size > threshold)
            // 扩容
            resize();
        afterNodeInsertion(evict);
        return null;
    }

hash函數

    static final int hash(Object key) {
        int h;
        
        // hashCode值 与 hashCode值右移16位做异或 得出来的值 高低位特征都有保留 扰动效果更好
        return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
    }

扩容过程 函数resize

 final Node[] resize() {
        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;
    }

1.8的resize和1.7最大不同就是,元素重新求索引位置,不是单纯求 hash & (length-1),而是看元素key的hash值在newCap-1 的高位是1还是0,如果是1 元素索引位置+oldCap,如果是0元素索引位置保持不变。

get函数(和1.7类似)

    final Node getNode(int hash, Object key) {
        Node[] tab; Node 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)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;
    }

缩容

注意当链表长度小于6时,红黑树会变为链表

面试考点

JDK1.7,hashMap使用头插法为什么会造成死循环?

 void transfer(Entry[] newTable, boolean rehash) {
        int newCapacity = newTable.length;
        for (Entry e : table) {
            while(null != e) {
                // 第一处
                Entry next = e.next;
                if (rehash) {
                    e.hash = null == e.key ? 0 : hash(e.key);
                }
                int i = indexFor(e.hash, newCapacity);
                e.next = newTable[i];
                newTable[i] = e;
                e = next;
            }
        }
    }

原因是 多线程并发扩容时,假设A,B两个线程,当A线程执行完第一处,时间片耗尽,线程B按照头插法,完成了完整扩容,此时链表相对于原来是逆序,A线程再继续顺序执行,会造成指针混乱,于是出现死循环。

为什么容量是2的指数幂?

有两点,更好的得到新索引和 搭配数组的length-1 可以使 hash%(length) == hash & (length-1) 作用相等

  1. 更好的得到新索引,打个比方 容量如果一开始时 8 ,后面扩容成16 从二进制上看只有最高位不同(length最高位后面都是0),所以在扩容的时候,需要重新用元素的hash值和length-1求余 实际上只有最高位不同,其他位不变,修改的数据少了,提高了代码的执行效率(1.7这样写,但是没利用到,但是1.8利用了这个特点) image-20230201115404578
  2. 如果扩容不是2的指数幂,那么扩容后的长度低位不会那么均匀,(试想下,如果低位有0的话,是不是每个hash值在这个位置都是0,大大提高了hash冲突的概率,是2的指数幂低位,length-1 低位全是1) 也能更好的降低hash冲突的概率
  3. 搭配数组的length-1 可以使 hash%(length) == hash & (length-1) 作用相等

如果使用自定义对象作为hashMap的key 为什么一定需要重写hashCode和equals方法?

因为hashMap插入元素,会用到hashCode计算hash值,如果没有重写的话,默认使用Object的实现,即对象的内存地址

HashMapKey k1 = new HashMapKey(1);
HashMapKey k2 = new HashMapKey(1);
HashMap map = new HashMap<>();
map.put(k1, "test");
System.out.println("map.get(k2) : " + map.get(k2));

上面的例子,如果沒有重写hashCode的话,k1和k2元素肯定不在数组的同一个位置上,因为hashCode使用的是内存地址,所以map.get(k2)=null

那为什么要重写equals呢?

因为获取元素需要用到equals方法

HashMapKey k1 = new HashMapKey(1);
HashMapKey k2 = new HashMapKey(1);
HashMap map = new HashMap<>();
map.put(k1, "test");
System.out.println("map.get(k2) : " + map.get(k2));

还是上面的例子,k1和k2,不管有没有重写hashCode 都有可能hash值相等(不同元素的hash可能相等),假设k1和k2的hash值相等,当根据key获取元素时,不但会判断元素的hash值还会判断key之间是否equals,如果没有重写equals方法(equals默认对象的内存地址),k1和k2的内存地址肯定不相同,所以map.get(k2)=null

总结:使用自定义对象作为hashMap的key 一定需要重写hashCode和equals方法,set集合比较,去重时(先比较hash,hash相同在比较equals),所以一样需要重写。

HashMap线程不安全的体现:

思考问题

谈谈你理解的 HashMap,讲讲其中的 get put 过程。 1.8 做了什么优化? 是线程安全的嘛? 不安全会导致哪些问题? 如何解决?有没有线程安全的并发容器? ConcurrentHashMap 是如何实现的? 1.7、1.8 实现有何不同?为什么这么做?

引用

https://crossoverjie.top/2018/07/23/java-senior/ConcurrentHashMap/#HashMap

有问题,欢迎私信我哦

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页面更新:2024-04-20

标签:低位   重写   数组   节点   负载   线程   函数   索引   元素   容量

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