hashMap的源码实现

1、初步认识hashMap

public static void main(String[] args) {
    HashMap<String, Integer> map = new HashMap<>(5);
    map.put("中国", 1);
    map.put("美国", 2);
    map.put("俄罗斯", 3);
    map.put("英国", 4);
    map.put("法国", 5);
    for(Map.Entry<String, Integer> entry : map.entrySet()) {
        System.out.println(entry.getKey() + ": " + entry.getValue());
    }
}

debug模式,从数据结构上认知HashMap:

JDK8中HashMap的数据结构源码:

 static class Node<K,V> implements Map.Entry<K,V> {
        final int hash;
        final K key;
        V value;
        Node<K,V> next;

        Node(int hash, K key, V value, Node<K,V> next) {
            this.hash = hash;
            this.key = key;
            this.value = value;
            this.next = next;
        }

        public final K getKey()        { return key; }
        public final V getValue()      { return value; }
        public final String toString() { return key + "=" + value; }

        public final int hashCode() {
            return Objects.hashCode(key) ^ Objects.hashCode(value);
        }

        public final V setValue(V newValue) {
            V oldValue = value;
            value = newValue;
            return oldValue;
        }

        public final boolean equals(Object o) {
            if (o == this)
                return true;
            if (o instanceof Map.Entry) {
                Map.Entry<?,?> e = (Map.Entry<?,?>)o;
                if (Objects.equals(key, e.getKey()) &&
                    Objects.equals(value, e.getValue()))
                    return true;
            }
            return false;
        }
    }

2、HashMap的两个重要参数

/**
* The default initial capacity - MUST be a power of two.
* table的默认初始容量
*/
static final int DEFAULT_INITIAL_CAPACITY = 1 << 4; // aka 16
/** * The load factor used when none specified in constructor.(负载因子) */
static final float DEFAULT_LOAD_FACTOR = 0.75f;
  • capacity 就是初始化HashMap时的数组容量,load factor 指负载因子;
  • 当我们对迭代性能要求比较高时,不能把capacity设置的太大;同时load factor不要超过0.75,否则会明显增加冲突几率,降低HashMap性能;
  •  hashMap中元素数量( put 的元素个数) > (负载因子 * 容量) 时,就需要扩容为原来的2倍

3、HashMap的put(Key k,Value v)的原理

数据存储的步骤

  1.  当在第一次put时,先对table初始化,通过hash计算得到存放位置table[i],存放。
  2.  当再次put时,同样经过hash计算得到位置,则采用链表法解决冲突存放在相同位置的next区域
  3.  在JDK8中设置了链表的默认阈值为8,如果超过这个值,则进行树化。
  4.  如果节点已经存在就替换old value(保证key的唯一性)。
  5.  如果bucket满了(超过load factor*current capacity),就要resize,变为原来2倍。

面试题:解释HashMap的原理,数据量增大时,数据结构是什么样的?
       在数据量小的时候,HashMap是按照链表的模式存储的。当数据量变大之后,为了进行快速的查找,会将这个链表变成红黑树(均衡二叉树),用hash码作为数据的定位来进行保存。

/**
     * Associates the specified value with the specified key in this map.
     * If the map previously contained a mapping for the key, the old
     * value is replaced.
     *
     * @param key key with which the specified value is to be associated
     * @param value value to be associated with the specified key
     * @return the previous value associated with <tt>key</tt>, or
     *         <tt>null</tt> if there was no mapping for <tt>key</tt>.
     *         (A <tt>null</tt> return can also indicate that the map
     *         previously associated <tt>null</tt> with <tt>key</tt>.)
     */
    public V put(K key, V value) {
        return putVal(hash(key), key, value, false, true);
    }

    /**
     * Implements Map.put and related methods
     *
     * @param hash hash for key
     * @param key the key
     * @param value the value to put
     * @param onlyIfAbsent if true, don't change existing value
     * @param evict if false, the table is in creation mode.
     * @return previous value, or null if none
     */
    final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
                   boolean evict) {
        Node<K,V>[] tab; Node<K,V> p; int n, i;
        if ((tab = table) == null || (n = tab.length) == 0)
            n = (tab = resize()).length;
        if ((p = tab[i = (n - 1) & hash]) == null)
            tab[i] = newNode(hash, key, value, null);
        else {
            Node<K,V> e; K k;
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                e = p;
            else if (p instanceof TreeNode)
                e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
            else {
                for (int binCount = 0; ; ++binCount) {
                    if ((e = p.next) == null) {
                        p.next = newNode(hash, key, value, null);
                        if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
                            treeifyBin(tab, hash);
                        break;
                    }
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        break;
                    p = e;
                }
            }
            if (e != null) { // existing mapping for key
                V oldValue = e.value;
                if (!onlyIfAbsent || oldValue == null)
                    e.value = value;
                afterNodeAccess(e);
                return oldValue;
            }
        }
        ++modCount;
        if (++size > threshold)
            resize();
        afterNodeInsertion(evict);
        return null;
    }
原文地址:https://www.cnblogs.com/luao/p/10931769.html