HashMap/ArrayMap/SpareArray

复杂度 HashMap ArrayMap SpareArray
get时间复杂度 O(logn),红黑树 O(logn)，二分查找 O(logn)，二分查找

ps: 大量数据（1000以上）, 红黑树优于二分查找

构建一个Map需要解决这些问题

  • hash code 冲突
  • 扩容

HashMap

    transient Node<K,V>[] table;
    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;
                }
            }
            ...
        }
        ++modCount;
        if (++size > threshold)
            resize();
        afterNodeInsertion(evict);
        return null;
    }

HashMap由数组+链表实现 , 使用拉链法解决 hash code 冲突，使用红黑树解决过长的链表时间复杂度为O(n)的问题，HashMap扩容时是 2的幂次(由于h & (length-1) 比 h % length 快 所以扩容幂次方)极其浪费空间资源，并且会重新构建 hash 表

ArrayMap

    int[] mHashes;
    Object[] mArray;
    public V put(K key, V value) {
        final int osize = mSize;
        final int hash;
        int index;
        if (key == null) {
            hash = 0;
            index = indexOfNull();
        } else {
            hash = key.hashCode();
            index = indexOf(key, hash);
        }
        if (index >= 0) {
            index = (index<<1) + 1;
            final V old = (V)mArray[index];
            mArray[index] = value;
            return old;
        }

        index = ~index;
        if (osize >= mHashes.length) {
            final int n = osize >= (BASE_SIZE*2) ? (osize+(osize>>1))
                    : (osize >= BASE_SIZE ? (BASE_SIZE*2) : BASE_SIZE);

            if (DEBUG) System.out.println(TAG + " put: grow from " + mHashes.length + " to " + n);

            final int[] ohashes = mHashes;
            final Object[] oarray = mArray;
            allocArrays(n);

            if (CONCURRENT_MODIFICATION_EXCEPTIONS && osize != mSize) {
                throw new ConcurrentModificationException();
            }

            if (mHashes.length > 0) {
                if (DEBUG) System.out.println(TAG + " put: copy 0-" + osize + " to 0");
                System.arraycopy(ohashes, 0, mHashes, 0, ohashes.length);
                System.arraycopy(oarray, 0, mArray, 0, oarray.length);
            }

            freeArrays(ohashes, oarray, osize);
        }

        if (index < osize) {
            if (DEBUG) System.out.println(TAG + " put: move " + index + "-" + (osize-index)
                    + " to " + (index+1));
            System.arraycopy(mHashes, index, mHashes, index + 1, osize - index);
            System.arraycopy(mArray, index << 1, mArray, (index + 1) << 1, (mSize - index) << 1);
        }

        if (CONCURRENT_MODIFICATION_EXCEPTIONS) {
            if (osize != mSize || index >= mHashes.length) {
                throw new ConcurrentModificationException();
            }
        }

        mHashes[index] = hash;
        mArray[index<<1] = key;
        mArray[(index<<1)+1] = value;
        mSize++;
        return null;
    }

ArrayMap由两个数组组成，一个hash 数组，一个 存k,v数组。hash code 冲突时，先向后查找，然后再向前查找。扩容不需要 2 倍，使用System.arraycopy效率更高。SpareArray的 key 为 int，更省内存，内部对采用压缩的方式表示稀疏数据。

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