hashmap深入分析

hashmap底层数据结构

PUT方法

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) {
        //table全局变量,存储链表头节点数组
        Node<K,V>[] tab; Node<K,V> p; int n, i;
        //如果table数组是空的，则创建一个头结点数据，默认长度是16
        if ((tab = table) == null || (n = tab.length) == 0)
            n = (tab = resize()).length;
        //n是table长度,根据数组长度和key的哈希值，定位当前key在table中的下标位置,如果为空则新建一个node。不为空走else
        if ((p = tab[i = (n - 1) & hash]) == null)
            tab[i] = newNode(hash, key, value, null);
        else {
            Node<K,V> e; K k;
            // 如果新的key与table中索引处取出的头节点的key相等，且hash值一致，则把新的node替换掉旧的node
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                e = p;
            //如果头节点不是空，且头节点的类型是树节点类型，则把当前节点插入当前头节点所在的树中(红黑树，防止链表过长，1.8的优化)
            else if (p instanceof TreeNode)
                e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
            //map的数据结构处理
            else {
                //遍历链表
                for (int binCount = 0; ; ++binCount) {
                    if ((e = p.next) == null) {
                        p.next = newNode(hash, key, value, null);
                        //如果链表长度大于或等于8，则把链表转化为红黑树，重点转化方法(treeifyBin)
                        if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
                            //此方法构建红黑树
                            treeifyBin(tab, hash);
                        break;
                    }
                    //如果当前节点的key和hash均和待插入的节点相等，则退出循环，(注意此时e的值在前一个if时赋值过，因此当前的e值，就是链表中的当前节点值)
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        break;
                    p = e;
                }
            }
            //如果老节点不是空，则将老节点的值替换为新值，并返回老的值
            if (e != null) { // existing mapping for key
                V oldValue = e.value;
                if (!onlyIfAbsent || oldValue == null)
                    e.value = value;
                //此方法实现的逻辑,是把入参的节点放置在链表的尾部，但在HashMap中是空实现，在LinkedHashMap中有具体实现
                afterNodeAccess(e);
                return oldValue;
            }
        }
        //保证并发访问时，若HashMap内部结构发生变化，快速响应失败
        ++modCount;
        //当table[]长度大于临界阈值，调用resize方法进行扩容
        if (++size > threshold)
            resize();
        //此方法在HashMap中是空方法，在LinkedHashMap中有实现
        afterNodeInsertion(evict);
        return null;
    }

hashmap的hash算法和寻址算法如何优化

hashmap如何解决hash碰撞

hashmap如何扩容

final Node<K,V>[] resize() {
        Node<K,V>[] oldTab = table;
        //oldCap---原hashMap的最大容量,oldThr---原hashMap的负载容量
        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;
            }
            //左移一位,就是将原来的容量翻倍，翻倍后的值小于2的30次方，大于原来的容量值
            else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
                     oldCap >= DEFAULT_INITIAL_CAPACITY)
                //原来的负载容量翻倍
                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;
    }