【HashMap】源码解读

最新推荐文章于 2025-05-02 22:11:26 发布

north_china

最新推荐文章于 2025-05-02 22:11:26 发布

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分类专栏：常用工具类文章标签：哈希算法散列表数据结构

本文链接：https://blog.youkuaiyun.com/kebo_china/article/details/126630560

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本文深入解析了HashMap的数据结构和核心操作，包括默认容量、负载因子、扩容策略等属性，详细阐述了put方法的插入逻辑，get方法的查找过程，以及扩容resize方法的实现，同时介绍了remove方法如何删除节点。这些内容有助于理解HashMap的内部工作原理。

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属性

    //数组的默认长度 16
    static final int DEFAULT_INITIAL_CAPACITY = 1 << 4;
    //数组最大长度
    static final int MAXIMUM_CAPACITY = 1 << 30;
    //负载因子大小
    static final float DEFAULT_LOAD_FACTOR = 0.75f;
    //树化阈值 >=8 链表转红黑树
    static final int TREEIFY_THRESHOLD = 8;
    //树降阀值 <=6 红黑树转为链表
    static final int UNTREEIFY_THRESHOLD = 6;
    //数组长度大于64 且满足树化>=8 才会将链表转红黑树
    static final int MIN_TREEIFY_CAPACITY = 64;

put方法

final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
                   boolean evict) {
        Node<K,V>[] tab; //数组
        Node<K,V> p;  //开始头部指针
        int n, i;//n 数组长度 i为数组下标
        
        //数组为空或者长度为0，对数组进行扩容
        if ((tab = table) == null || (n = tab.length) == 0)
            n = (tab = resize()).length;
        //i = (n - 1) & hash n-1&hash 相当于取模
        //tab[i]没有元素，直接插入结点到tab[i]
        if ((p = tab[i = (n - 1) & hash]) == null)
            tab[i] = newNode(hash, key, value, null);
        else {
        //当前数组位置有值
            Node<K,V> e;//记录需要插入的节点位置
            K k;
            //key和头节点key相同，hash相同
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                e = p;
            //头节点为红黑树
            else if (p instanceof TreeNode)
                e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
            //为连表
            else {
                for (int binCount = 0; ; ++binCount) {
                    //从第二节点开始遍历
                    if ((e = p.next) == null) {
                    //下一个节点为空，添加数据
                        p.next = newNode(hash, key, value, null);
                        //链表长度>=8 链表转变为红黑树(双向连标)
                        if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
                            //此方法中，判断数组长度是否大于64，如果数组的长度不大，不转变，性能不好
                            treeifyBin(tab, hash);
                        break;
                    }
                    //后面的节点，有相同key，覆盖
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        break;
                    p = e;
                }
            }
            //找到了存放节点位置
            if (e != null) { // existing mapping for key
                V oldValue = e.value;
                if (!onlyIfAbsent || oldValue == null)
                    e.value = value;
                afterNodeAccess(e);
                return oldValue;
            }
        }
        //不存在相同的key,新插入元素才执行下面的代码
        ++modCount;
        //数组长度大于临界值
        if (++size > threshold)
            resize();
        afterNodeInsertion(evict);
        return null;
    }

get方法

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) {
            //hash取模，找到数组桶的位置，key和头节点一样的，返回
            if (first.hash == hash &&
                ((k = first.key) == key || (key != null && key.equals(k))))
                return first;
                //下一个节点
            if ((e = first.next) != null) {
                //如果是红黑树，while双向链表遍历
                if (first instanceof TreeNode)
                    return ((TreeNode<K,V>)first).getTreeNode(hash, key);
                do {
                //链表，while遍历查询key返回
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        return e;
                } while ((e = e.next) != null);
            }
        }
        return null;
    }

扩容

HashMap在进行扩容时，使用的rehash方式非常巧妙，因为每次扩容都是翻倍，与原来计算的 (n-1)&hash的结果相比，只是多了一个bit位，所以节点要么就在原来的位置，要么就被分配到"原位置+旧容量"这个位置。当然resize非常消耗性能，要尽量避免

 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;
    }

remove方法

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) {
            //找到数组对应桶
            Node<K,V> node = null, e; K k; V v;
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                //头节点
                node = p;
            else if ((e = p.next) != null) {
            //next节点
                if (p instanceof TreeNode)
                    //双链表 while双向遍历 红黑树
                    node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
                else {
                    //遍历
                    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)
                    //双向链表删除
                    ((TreeNode<K,V>)node).removeTreeNode(this, tab, movable);
                else if (node == p)
                     //头节点替换
                    tab[index] = node.next;
                else
                   // next节点替换
                    p.next = node.next;
                ++modCount;
                --size;
                afterNodeRemoval(node);
                return node;
            }
        }
        return null;
    }