JDK8中HashMap源码分析

╰+哭是因爲堅強的太久メ 2022-05-26 01:17 256阅读 0赞

JDK7及以前，HashMap底层基于数组(Entry数组)和链表实现，计算hash时发生hash冲突将该节点以链表形式存储，当hash冲突过多会导致单个链表过长，查找节点时将耗费O(n)的查找时间

JDK8：底层基于数组(Node数组)、链表、红黑树实现，当同一hash值的节点数超过阈值(8)，链表结构将会被转换为红黑树进行数据存储

底层以Node数组存储，对应TreeNode红黑树的实现

transient Node<K,V>[] table;

当hash冲突节点大于指定阈值8时，将链表结构转换为红黑树进行存储

//由链表转换为红黑树的阈值
    static final int TREEIFY_THRESHOLD = 8;
    
    //由红黑树转换为链表的阈值
    static final int UNTREEIFY_THRESHOLD = 6;
    
    //哈希表的最小树形化阈值
    static final int MIN_TREEIFY_CAPACITY = 64;

执行put操作源码分析

public V put(K key, V value) {
        return putVal(hash(key), key, value, false, true);
    }
    
    final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
                   boolean evict) {
        Node<K,V>[] tab; Node<K,V> p; int n, i;
    //当前map无数据，则执行resize()方法，并返回n
        if ((tab = table) == null || (n = tab.length) == 0)
            n = (tab = resize()).length;
    //当前需要存储键值对的位置没有元素，则将键值对封装为Node存放在该位置即可
        if ((p = tab[i = (n - 1) & hash]) == null)
            tab[i] = newNode(hash, key, value, null);
        else {
        //当前位置已有元素占有
            Node<K,V> e; K k;
        //判断该位置元素的key与要插入元素的key是否一致，一致就直接替换
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                e = p;
        //如果是树形结构，则执行putTreeVal方法
            else if (p instanceof TreeNode)
                e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
            else {
           //遍历链表数据，添加新元素
                for (int binCount = 0; ; ++binCount) {
                    if ((e = p.next) == null) {
                        p.next = newNode(hash, key, value, null);
               //判断是否大于等于需要转换树形结构的阈值，大于则执行treeifyBin方法将链表转换为红黑树结构
                        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;
    //当size大于阈值，就执行resize方法扩容
        if (++size > threshold)
            resize();
        afterNodeInsertion(evict);
        return null;
    }

转换为树形结构的treeifyBin方法分析

final void treeifyBin(Node<K,V>[] tab, int hash) {
        int n, index; Node<K,V> e;
    //如果当前哈希表为空或哈希表元素个数小于树形化的最小阈值 就去新建或扩容
        if (tab == null || (n = tab.length) < MIN_TREEIFY_CAPACITY)
            resize();
    //哈希表元素个数超过了树形化的最小阈值 进行树形化 e为该位置链表第一个节点
        else if ((e = tab[index = (n - 1) & hash]) != null) {
            TreeNode<K,V> hd = null, tl = null;
            do {
            //新建树形节点，将链表节点元素信息添加到树形节点
                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);
        //将桶中第一个元素指向新建红黑树头结点
            if ((tab[index] = hd) != null)
                hd.treeify(tab);
        }
    }

将二叉树转换为红黑树treeify方法分析

final void treeify(Node<K,V>[] tab) {
        TreeNode<K,V> root = null;
        for (TreeNode<K,V> x = this, next; x != null; x = next) {
            next = (TreeNode<K,V>)x.next;
            x.left = x.right = null;
            //第一次进入循环，设置头结点为黑色
            if (root == null) {
                x.parent = null;
                x.red = false;
                root = x;
            }
            else {
                K k = x.key;
                int h = x.hash;
                Class<?> kc = null;
                //遍历所有节点与当前节点x进行比较，调整位置
                for (TreeNode<K,V> p = root;;) {
                    int dir, ph;
                    K pk = p.key;
                    //当前比较节点哈希值比x大 dir = -1
                    if ((ph = p.hash) > h)
                        dir = -1;
                    //当前比较节点哈希值比x小 dir = 1
                    else if (ph < h)
                        dir = 1;
                    else if ((kc == null &&
                              (kc = comparableClassFor(k)) == null) ||
                             (dir = compareComparables(kc, k, pk)) == 0)
                        dir = tieBreakOrder(k, pk);
    
                    TreeNode<K,V> xp = p;
                    if ((p = (dir <= 0) ? p.left : p.right) == null) {
                        //将当前节点设置为x的父节点
                        x.parent = xp;
                        //将当前比较节点的哈希值与x比较 比x大则x为左节点，否则x为右节点
                        if (dir <= 0)
                            xp.left = x;
                        else
                            xp.right = x;
                        root = balanceInsertion(root, x);
                        break;
                    }
                }
            }
        }
        moveRootToFront(tab, root);
    }