1.HashMap資料結構

首先先看下HashMap的資料結構圖

我們都知道陣列的儲存方式在是連續的，查詢速度比較快，但插入和刪除資料比較慢

而連結串列的儲存方式是非連續的，所以插入和刪除速度較快，但查詢速度就比較慢，HashMap在資料結構上兩種都採用了。

    /**
     * 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;
 

再看下Node中的變數 （Node中含有hash值、key、value以及指向下個節點的next）

        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;
        }
 

通過這個table變數我們就能理解上面那個資料結構圖，它不僅使用了陣列，而且還採用了連結串列結構。

2.HashMap變數

首先我們看下HashMap中的變數，在看之前思考幾個問題：

1. HashMap資料結構既然使用了陣列，那麼陣列預設記憶體大小是多少？

2.當已使用記憶體大小超過總記憶體的一個百分比後，需要繼續增大記憶體，這個百分比是多少？

3.當連結串列節點長多超過多少時儲存結構變成紅黑樹？

    /**
     * The default initial capacity - MUST be a power of two.
     */
    static final int DEFAULT_INITIAL_CAPACITY = 1 << 4; // aka 16
 

由此可以知道陣列的預設大小是16

    /**
     * The maximum capacity, used if a higher value is implicitly specified
     * by either of the constructors with arguments.
     * MUST be a power of two <= 1<<30.
     */
    static final int MAXIMUM_CAPACITY = 1 << 30;

當陣列不夠大的時候，就會增加記憶體大小，大小的上限為1<<30,也就是2的30次方

    /**
     * The load factor used when none specified in constructor.
     */
    static final float DEFAULT_LOAD_FACTOR = 0.75f;

這個變數就是問題2的答案，也被稱為負載因子，預設為0.75

    /**
     * The bin count threshold for using a tree rather than list for a
     * bin.  Bins are converted to trees when adding an element to a
     * bin with at least this many nodes. The value must be greater
     * than 2 and should be at least 8 to mesh with assumptions in
     * tree removal about conversion back to plain bins upon
     * shrinkage.
     */
    static final int TREEIFY_THRESHOLD = 8;

這個變數就是問題三的答案，當節點數量超過8時，儲存結構轉換成紅黑樹

  /**
     * The bin count threshold for untreeifying a (split) bin during a
     * resize operation. Should be less than TREEIFY_THRESHOLD, and at
     * most 6 to mesh with shrinkage detection under removal.
     */
    static final int UNTREEIFY_THRESHOLD = 6;

當bin的個數小於6時由樹形結構調整為列表

    /**
     * The number of key-value mappings contained in this map.
     */
    transient int size;

當前HashMap中儲存key-value的大小

    /**
     * The number of times this HashMap has been structurally modified
     * Structural modifications are those that change the number of mappings in
     * the HashMap or otherwise modify its internal structure (e.g.,
     * rehash).  This field is used to make iterators on Collection-views of
     * the HashMap fail-fast.  (See ConcurrentModificationException).
     */
    transient int modCount;

HashMap被修改和刪除的次數

3.HashMap的儲存過程

        HashMap hashMap = new HashMap();
        hashMap.put("1","1230");
        System.out.print(hashMap.get("1"));

我們從hashMap.put()方法開始

    public V put(K key, V value) {
        return putVal(hash(key), key, value, false, true);
    }

這裡詳細講下hash()函式，首先先看原始碼：

    static final int hash(Object key) {
        int h;
        return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
    }

如果key的值不為空的話，就把變數h等於key值的hashCode，然後把h的高16位與低16位進行一個異或運算，這樣得到的值和陣列的預設長度16相對應，方便後續計算該Node儲存的陣列下標是多少。

然後回到putVal()方法

 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)//初始化tab
            n = (tab = resize()).length;//n為陣列總長度
        if ((p = tab[i = (n - 1) & hash]) == null)//i為儲存的陣列下標
//傳入的hash值與陣列長度進行位與運算，這樣就可以保證數值不越界，得到結果賦值給i，
//如果i處為空的話，就把node儲存到該下標
            tab[i] = newNode(hash, key, value, null);
        else {//如果下標i處存有值的話，就使用連結串列結構，此時p=tab[i]
            Node<K,V> e; K k;
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                e = p;//key值存在的話，就不存
            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;
    }

未完。。。。待續