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当前位置: Java > Java集合高频面试题 > 22.HashMap的扩容操作是怎么实现的?
  • 初始值为16,负载因子为0.75,阈值为负载因子*容量
  • resize()方法是在hashmap中的键值对大于阀值时或者初始化时,就调用resize()方法进行扩容。
  • 每次扩容,容量都是之前的两倍
  • 扩容时有个判断e.hash & oldCap是否为零,也就是相当于hash值对数组长度的取余操作,若等于0,则位置不变,若等于1,位置变为原位置加旧容量。

源码如下:

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) { //如果旧容量已经超过最大值,阈值为整数最大值
            threshold = Integer.MAX_VALUE;
            return oldTab;
        }else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
                 oldCap >= DEFAULT_INITIAL_CAPACITY)
            newThr = oldThr << 1;  //没有超过最大值就变为原来的2倍
    }
    else if (oldThr > 0) 
        newCap = oldThr;

    else {               
        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 { 
                    Node<K,V> loHead = null, loTail = null;//loHead,loTail 代表扩容后在原位置
                    Node<K,V> hiHead = null, hiTail = null;//hiHead,hiTail 代表扩容后在原位置+旧容量
                    Node<K,V> next;
                    do {             
                        next = e.next;
                        if ((e.hash & oldCap) == 0) { //判断是否为零,为零赋值到loHead,不为零赋值到hiHead
                            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;   //loHead放在原位置
                    }
                    if (hiTail != null) {
                        hiTail.next = null;
                        newTab[j + oldCap] = hiHead;  //hiHead放在原位置+旧容量
                    }
                }
            }
        }
    }
    return newTab;
}

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