文章目录

• Lsit中数据复制问题
• • 1.1浅拷贝
  • 1.2深拷贝
  • 1.3 最终结论

Lsit中数据复制问题

这是由一道开放式面试题引发的文章，题目：加入内存足够大，一个集合中有100万条数据，怎么高效的把集合中的数据复制到另外一个集合

1.1浅拷贝

java 中复制分为浅拷贝和深拷贝

如果考察浅拷贝：直接使用clone方法

System.out.println("测试开始时");

        List<String> a=new ArrayList<String>(10000000);
        int i=1;
        for(int x = 1; x < 10000000; x = x+1){ 
            a.add(Integer.toString(x));
        }
        List<String> b=new ArrayList<String>(10000000);
        List<String> c=new ArrayList<String>(10000000);
        long begintime = System.nanoTime();

        b= (List<String>) ((ArrayList<String>) a).clone();  //浅拷贝

明显，问题没有这么简单

深度拷贝的话最简单就是遍历，这个基本都知道，自己实现遍历性能不会是最高的，开始思考
collections包含的方法

Collections.copy(c,a);

但这种方法具有局限，List c的size 要>= a.size()
并且根据代码这种方式也是浅拷贝

接下来想，如果是List
可以使用addall()
原理是数据拷贝，使用了java的native方法复制内存
同样也是浅拷贝

经过对比集中浅拷贝

java中提供了流式计算

 d = a.stream().collect(Collectors.toList());

经过实际验证：
addall()和collections.copy(a,b)
消耗时间接近，是比较好的复制方法，其中addall()略微好于collections

一般java的高效的集合类都在

java.util.concurrent.*;

分析代码：

long begintime5 = System.nanoTime();
        CopyOnWriteArrayList m=new CopyOnWriteArrayList(a);
        long endtime5 = System.nanoTime();
        long costTime5 = (endtime5 - begintime5)/1000;
        System.out.println("测试结束5: "+costTime5);

底层原理分析，看代码

 /** * Creates a list containing the elements of the specified * collection, in the order they are returned by the collection's * iterator. * * @param c the collection of initially held elements * @throws NullPointerException if the specified collection is null */
    public CopyOnWriteArrayList(Collection<? extends E> c) { 
        Object[] elements;
        if (c.getClass() == CopyOnWriteArrayList.class)
            elements = ((CopyOnWriteArrayList<?>)c).getArray();
        else { 
            elements = c.toArray();
            // c.toArray might (incorrectly) not return Object[] (see 6260652)
            if (elements.getClass() != Object[].class)
                elements = Arrays.copyOf(elements, elements.length, Object[].class);
        }
        setArray(elements);
    }

这段显示，调用了Arrays.copuOf()

/** * Copies the specified array, truncating or padding with nulls (if necessary) * so the copy has the specified length. For all indices that are * valid in both the original array and the copy, the two arrays will * contain identical values. For any indices that are valid in the * copy but not the original, the copy will contain <tt>null</tt>. * Such indices will exist if and only if the specified length * is greater than that of the original array. * The resulting array is of the class <tt>newType</tt>. * * @param <U> the class of the objects in the original array * @param <T> the class of the objects in the returned array * @param original the array to be copied * @param newLength the length of the copy to be returned * @param newType the class of the copy to be returned * @return a copy of the original array, truncated or padded with nulls * to obtain the specified length * @throws NegativeArraySizeException if <tt>newLength</tt> is negative * @throws NullPointerException if <tt>original</tt> is null * @throws ArrayStoreException if an element copied from * <tt>original</tt> is not of a runtime type that can be stored in * an array of class <tt>newType</tt> * @since 1.6 */
    public static <T,U> T[] copyOf(U[] original, int newLength, Class<? extends T[]> newType) { 
        @SuppressWarnings("unchecked")
        T[] copy = ((Object)newType == (Object)Object[].class)
            ? (T[]) new Object[newLength]
            : (T[]) Array.newInstance(newType.getComponentType(), newLength);
        System.arraycopy(original, 0, copy, 0,
                         Math.min(original.length, newLength));
        return copy;
    }

接下来调用了

System.arraycopy(original, 0, copy, 0,
Math.min(original.length, newLength));

看看最后：

public static native void arraycopy(Object src,  int  srcPos,
                                        Object dest, int destPos,
                                        int length);

java的native 方法，效率最高
本质和collection的clone的方法逻辑一致

1.2深拷贝

集合的深度拷贝，除了遍历还有一种是序列化和反序列，这种首先排除在外，接下来看看还有没有其他方式

查询了资料，木有发现更好的深拷贝方法

经过实践对比，序列化、反序列化是效率最高的办法

1.3 最终结论

  public static <T> List<T> depCopy(List<T> srcList) { 
        ByteArrayOutputStream byteOut = new ByteArrayOutputStream();
        try { 
            ObjectOutputStream out = new ObjectOutputStream(byteOut);
            out.writeObject(srcList);

            ByteArrayInputStream byteIn = new ByteArrayInputStream(byteOut.toByteArray());
            ObjectInputStream inStream = new ObjectInputStream(byteIn);
            List<T> destList = (List<T>) inStream.readObject();
            return destList;
        } catch (IOException e) { 
            e.printStackTrace();
        } catch (ClassNotFoundException e) { 
            e.printStackTrace();
        }
        return null;
    }

以上针对问题，初步研究。

参考1：https://blog.csdn.net/u010648159/article/details/79144154
参考2：https://blog.csdn.net/demonliuhui/article/details/54572908