JDK 1.7 HashMap 源码分析

Java 面试 About 1,319 words

数据结构

数组 + 单链表

哈希冲突解决方法

头插方式

void createEntry(int hash, K key, V value, int bucketIndex) {
    Entry<K,V> e = table[bucketIndex];
    table[bucketIndex] = new Entry<>(hash, key, value, e);
    size++;
}

Entry(int h, K k, V v, Entry<K,V> n) {
    value = v;
    next = n;
    key = k;
    hash = h;
}

null key

空键插入到数组索引0的位置上。

private V putForNullKey(V value) {
    for (Entry<K,V> e = table[0]; e != null; e = e.next) {
        if (e.key == null) {
            V oldValue = e.value;
            e.value = value;
            e.recordAccess(this);
            return oldValue;
        }
    }
    modCount++;
    addEntry(0, null, value, 0);
    return null;
}

扩容

扩容为原始数组的2倍。

if ((size >= threshold) && (null != table[bucketIndex])) {
    resize(2 * table.length);
    hash = (null != key) ? hash(key) : 0;
    bucketIndex = indexFor(hash, table.length);
}

void resize(int newCapacity) {
    Entry[] oldTable = table;
    int oldCapacity = oldTable.length;
    if (oldCapacity == MAXIMUM_CAPACITY) {
        threshold = Integer.MAX_VALUE;
        return;
    }

    Entry[] newTable = new Entry[newCapacity];
    // 赋值给新的数组
    transfer(newTable, initHashSeedAsNeeded(newCapacity));
    table = newTable;
    threshold = (int)Math.min(newCapacity * loadFactor, MAXIMUM_CAPACITY + 1);
}
Views: 2,690 · Posted: 2021-10-14

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