大数据问题归纳

final Node<K,V>[] resize() {
    // 当前table保存
    Node<K,V>[] oldTab = table;
    // 保存table大小
    int oldCap = (oldTab == null) ? 0 : oldTab.length;
    // 保存当前阈值 
    int oldThr = threshold;
    int newCap, newThr = 0;
    // 之前table大小大于0
    if (oldCap > 0) {
        // 之前table大于最大容量
        if (oldCap >= MAXIMUM_CAPACITY) {
            // 阈值为最大整形
            threshold = Integer.MAX_VALUE;
            return oldTab;
        }
        // 容量翻倍，使用左移，效率更高
        else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
            oldCap >= DEFAULT_INITIAL_CAPACITY)
            // 阈值翻倍
            newThr = oldThr << 1; // double threshold
    }
    // 之前阈值大于0
    else if (oldThr > 0)
        newCap = oldThr;
    // oldCap = 0并且oldThr = 0，使用缺省值（如使用HashMap()构造函数，之后再插入一个元素会调用resize函数，会进入这一步）
    else {           
        newCap = DEFAULT_INITIAL_CAPACITY;
        newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);
    }
    // 新阈值为0
    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"})
    // 初始化table
    Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];
    table = newTab;
    // 之前的table已经初始化过
    if (oldTab != null) {
        // 复制元素，重新进行hash
        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;
                    // 将同一桶中的元素根据(e.hash & oldCap)是否为0进行分割，分成两个不同的链表，完成rehash
                    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;
}