HashMap

作者: Pillar_Zhong | 来源:发表于2019-05-23 17:53 被阅读0次

put

final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
               boolean evict) {
    Node<K,V>[] tab; Node<K,V> p; int n, i;
    // 如果是空,resize扩容
    if ((tab = table) == null || (n = tab.length) == 0)
        n = (tab = resize()).length;
    // 如果hash求余的位置空余,直接插入
    if ((p = tab[i = (n - 1) & hash]) == null)
        tab[i] = newNode(hash, key, value, null);
    else {
        Node<K,V> e; K k;
        // 能找到对应hash+key匹配的位置
        if (p.hash == hash &&
            ((k = p.key) == key || (key != null && key.equals(k))))
            e = p;
        // 如果是红黑树,直接插入
        else if (p instanceof TreeNode)
            e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
        else {
            // 遍历链表
            for (int binCount = 0; ; ++binCount) {
                if ((e = p.next) == null) {
                    // 插入到链表末尾
                    p.next = newNode(hash, key, value, null);
                    // 如果超过阈值,链表转换成红黑树
                    if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
                        treeifyBin(tab, hash);
                    break;
                }
                // 找到链表中对应的hash+key的位置结果
                if (e.hash == hash &&
                    ((k = e.key) == key || (key != null && key.equals(k))))
                    break;
                p = e;
            }
        }
        // 如果能之前已经存在,则这里用来更新已有的值
        if (e != null) { // existing mapping for key
            V oldValue = e.value;
            if (!onlyIfAbsent || oldValue == null)
                e.value = value;
            afterNodeAccess(e);
            return oldValue;
        }
    }
    // 设置该table已经更新
    ++modCount;
    // 如果size已经超过阈值,立即扩容
    if (++size > threshold)
        resize();
    afterNodeInsertion(evict);
    return null;
}

resize

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) {
        // 如果旧的容量已经over,直接返回
        if (oldCap >= MAXIMUM_CAPACITY) {
            threshold = Integer.MAX_VALUE;
            return oldTab;
        }
        // 否则,在原来的基础上翻倍,且不能超过阈值
        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);
    }
    // 根据新的容量,设置新的threshold
    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) {
        // 遍历原来的table
        for (int j = 0; j < oldCap; ++j) {
            Node<K,V> e;
            if ((e = oldTab[j]) != null) {
                oldTab[j] = null
                // 如果该节点是单一元素,直接移动到新容器
                if (e.next == null)
                    newTab[e.hash & (newCap - 1)] = e;
                // 如果该节点是红黑树,移动到新容器中对应位置的红黑树中
                else if (e instanceof TreeNode)
                    ((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
                // 否则就是链表
                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;
                        // 举例:
                        // 假如 hash 值为 111001
                        // 扩容前 n=16(010000) n-1=15(01111)  (n - 1)&hash=01111 & 111001=001001
                        // 扩容后 n=32(100000) n-1=31(11111)  (n - 1)&hash=11111 & 111001=011001
                        // 最后的位置有变化 001001 -> 011001
                        // 假如 hash 值为 101001
                        // 扩容前  1111 & 101001 = 001001
                        // 扩容后 11111 & 101001 = 001001
                        // 最后的位置没变 001001
                        // hash & oldCap 的作用在于找到哪些扩容后,哪些hash对应的index不用变
                        // 101001 & 010000 = 0 但是 111001 & 010000 = 010000
                        if ((e.hash & oldCap) == 0) {
                            // 旧的部分从loHead开始
                            if (loTail == null)
                                loHead = e;
                            else
                                loTail.next = e;
                            loTail = e;
                        }
                        // 那么这里就是需要移动的部分,从hiHead开始
                        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;
                    }
                    // 新的部分的起始位置是原索引+原数组长度
                    if (hiTail != null) {
                        hiTail.next = null;
                        newTab[j + oldCap] = hiHead;
                    }
                }
            }
        }
    }
    return newTab;
}

get

final Node<K,V> getNode(int hash, Object key) {
    Node<K,V>[] tab; Node<K,V> first, e; int n; K k;
    // 如果hash对应的位置不为空
    if ((tab = table) != null && (n = tab.length) > 0 &&
        (first = tab[(n - 1) & hash]) != null) {
        // 如果hash+key相等,直接返回
        if (first.hash == hash && // always check first node
            ((k = first.key) == key || (key != null && key.equals(k))))
            return first;
        if ((e = first.next) != null) {
            // 如果是红黑树的部分,直接返回
            if (first instanceof TreeNode)
                return ((TreeNode<K,V>)first).getTreeNode(hash, key);
            // 否则遍历链表,直到找到hash+key
            do {
                if (e.hash == hash &&
                    ((k = e.key) == key || (key != null && key.equals(k))))
                    return e;
            } while ((e = e.next) != null);
        }
    }
    return null;
}

remove

final Node<K,V> removeNode(int hash, Object key, Object value,
                           boolean matchValue, boolean movable) {
    Node<K,V>[] tab; Node<K,V> p; int n, index;
    // 或者hash+key对应的值
    if ((tab = table) != null && (n = tab.length) > 0 &&
        (p = tab[index = (n - 1) & hash]) != null) {
        Node<K,V> node = null, e; K k; V v;
        // 直接在主链中找到
        if (p.hash == hash &&
            ((k = p.key) == key || (key != null && key.equals(k))))
            node = p;
        else if ((e = p.next) != null) {
            // 去红黑树找
            if (p instanceof TreeNode)
                node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
            // 去链表找
            else {
                do {
                    if (e.hash == hash &&
                        ((k = e.key) == key ||
                         (key != null && key.equals(k)))) {
                        node = e;
                        break;
                    }
                    p = e;
                } while ((e = e.next) != null);
            }
        }
        if (node != null && (!matchValue || (v = node.value) == value ||
                             (value != null && value.equals(v)))) {
            // 如果是红黑树,去那边删除                             
            if (node instanceof TreeNode)
                ((TreeNode<K,V>)node).removeTreeNode(this, tab, movable);
            else if (node == p)
                tab[index] = node.next;
            else
                // 跳过该节点,也就是删除
                p.next = node.next;
            ++modCount;
            --size;
            afterNodeRemoval(node);
            return node;
        }
    }
    return null;
}

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