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Java基础知识从0.1到0.000001

发布于: 2021-07-04
更新于: 2023-07-09

Java 基础知识

类

抽象类与接口

抽象类与接口

ArrayList

扩容方法

ArrayList扩容


    /**
     * Increases the capacity to ensure that it can hold at least the
     * number of elements specified by the minimum capacity argument.
     *
     * @param minCapacity the desired minimum capacity
     * @throws OutOfMemoryError if minCapacity is less than zero
     */
    private Object[] grow(int minCapacity) {
        int oldCapacity = elementData.length;
        if (oldCapacity > 0 || elementData != DEFAULTCAPACITY_EMPTY_ELEMENTDATA) {
            // 正常是旧容量的1.5倍
            // 通过oldCapacity+ (oldCapacity>>1) 实现

            // 如果1.5倍仍不够,则将最小需要容量作为数组的新容量
            int newCapacity = ArraysSupport.newLength(oldCapacity,
                    minCapacity - oldCapacity, /* minimum growth */
                    oldCapacity >> 1           /* preferred growth */);
            return elementData = Arrays.copyOf(elementData, newCapacity);
        } else {
            return elementData = new Object[Math.max(DEFAULT_CAPACITY, minCapacity)];
        }
    }

HashMap

数据结构


    /**
     * The table, initialized on first use, and resized as
     * necessary. When allocated, length is always a power of two.
     * (We also tolerate length zero in some operations to allow
     * bootstrapping mechanics that are currently not needed.)
     */
    transient Node<K,V>[] table;

    /**
     * Holds cached entrySet(). Note that AbstractMap fields are used
     * for keySet() and values().
     */
    transient Set<Map.Entry<K,V>> entrySet;

基本结构是Node数组,也就是Hash桶,用来存放hash后的相应值,如果发生了哈希冲突,则以链表形式存储[链地址法],假如说元素大于红黑树定义的上线(8),则链表会转为红黑树,提高命中效率。

开放地址法:记录空位,插入时放到空位,比如ThreadLocal这种

hashMap数据结构

resize

    // 主要涉及了负载因子计算容量大小,扩容时double,以及扩容时是否需要重新hash key,针对不同的数据结构,要适用不同的hash形式
    /**
     * Initializes or doubles table size.  If null, allocates in
     * accord with initial capacity target held in field threshold.
     * Otherwise, because we are using power-of-two expansion, the
     * elements from each bin must either stay at same index, or move
     * with a power of two offset in the new table.
     *
     * @return the table
     */
    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;
    }
Java基础知识从0.1到0.000001
/archives/94b7e3f2/
作者
tyrantqiao
发布于
2021-07-04
更新于
2023-07-09
许可协议
CC BY-NC-SA 4.0
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