blocks|key|366136|text|使用构造函数ArrayList(int+initialCapacity)。这样，您就可以设置初始容量。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|366137|entityMap^0|6|U|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

Use constructor <code>ArrayList(int initialCapacity)</code>. This way you can set an initial capacity.

blocks|key|366145|text|不，你不能：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|366146|http://docs.oracle.com/javase/6/docs/api/java/util/ArrayList.html#add(int|offset|length|366147|366148|抛出:+IndexOutOfBoundsException+-如果索引超出范围(索引<0+%7C%7C索引>+size())|blockquote|366149|366150|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/6/docs/api/java/util/ArrayList.html#add(int),%2520E^0|0|0|21|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@$D|U|E|V|1|W]]|A|$]]|$1|F|3|-4|5|6|7|X|8|@]|9|@]|A|$]]|$1|G|3|H|5|I|7|Y|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

No, you can't:

<a href="http://docs.oracle.com/javase/6/docs/api/java/util/ArrayList.html#add(int" rel="nofollow">http://docs.oracle.com/javase/6/docs/api/java/util/ArrayList.html#add(int</a>, E)

<blockquote>
 Throws: IndexOutOfBoundsException - if the index is out of range
 (index &lt; 0 || index > size())
</blockquote>

blocks|key|4307057|text|我会考虑在这里使用SortedMap而不是List。这将允许索引不存在：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4307058|SorteMap<Integer,+Item>+myMap+=+new+TreeMap<Integer,+Map>();
int+i=0;
myMap.put(i%2B%2B,+first);
myMap.put(i%2B%2B,+second);
myMap.put(i%2B%2B,+third);
myMap.put(10,+other);|code-block|syntax|javascript|4307059|如果Map真的不能工作，就像你说的那样。然后我会建议在ArrayList周围创建一个装饰器。在insert方法中，添加空值以填充空位置。我建议使用Guava的ForwardingList来简化类的创建。这样你就只需要实现一个方法。|4307060|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/6/docs/api/java/util/SortedMap.html|1|http://docs.guava-libraries.googlecode.com/git-history/v11.0.1/javadoc/com/google/common/collect/ForwardingList.html^0|9|9|0|0|0|27|E|1|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@$A|10|B|11|1|12]]|C|$]]|$1|K|3|-4|5|6|7|13|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|T]]]]

I would consider using a <a href="http://docs.oracle.com/javase/6/docs/api/java/util/SortedMap.html" rel="nofollow">SortedMap</a> instead of a List here. This will allow for indexes to not exist:

<pre><code>SorteMap&lt;Integer, Item&gt; myMap = new TreeMap&lt;Integer, Map&gt;();
int i=0;
myMap.put(i++, first);
myMap.put(i++, second);
myMap.put(i++, third);
myMap.put(10, other);
</code></pre>

If a Map truly will not work, as you stated. I would then suggest creating a Decorator around ArrayList. In the insert method, add nulls to fill the empty locations. I would suggest using Guava's <a href="http://docs.guava-libraries.googlecode.com/git-history/v11.0.1/javadoc/com/google/common/collect/ForwardingList.html" rel="nofollow">ForwardingList</a> to ease the creation of the class. This way you would only have to implement one method.

blocks|key|365148|text|不，您不能这样做，但如果您希望这样做，那么在剩余的索引中添加空对象，例如..|type|unstyled|depth|inlineStyleRanges|entityRanges|data|365149|++++ArrayList<Object>+items+=+new+ArrayList<Object>();
++++items.add(new+Object());
++++items.add(new+Object());
++++items.add(new+Object());
++++items.add(3,new+Object());|code-block|syntax|javascript|365150|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

No, you can't do this,
But if you wish to do that, then add empty object in remaining index such as..

<pre><code> ArrayList&lt;Object&gt; items = new ArrayList&lt;Object&gt;();
 items.add(new Object());
 items.add(new Object());
 items.add(new Object());
 items.add(3,new Object());
</code></pre>

blocks|key|266474|text|我所能做的最好的事情就是items.addAll(Collections.nCopies(6,+null))，并希望ArrayList实现一些行为来在内部加速这一过程|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|266475|entityMap^0|C|16|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

imho the best thing you can do is <code>items.addAll(Collections.nCopies(6, null))</code> and hope, that ArrayList implements some behaviour to internally fasten this up

blocks|key|365174|text|这个怎么样？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|365175|ArrayList<Item>+items+=+new+ArrayList<Item>();

items.add(new+Item(0));
items.add(new+Item(1));
items.add(new+Item(2));

items.addAll(Collections.<Item>nCopies(7,+null));
items.add(10,new+Item(10));

System.out.println(items);|code-block|syntax|javascript|365176|打印|365177|[0,+1,+2,+null,+null,+null,+null,+null,+null,+null,+10]|365178|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

How about this?

<pre><code>ArrayList&lt;Item&gt; items = new ArrayList&lt;Item&gt;();

items.add(new Item(0));
items.add(new Item(1));
items.add(new Item(2));

items.addAll(Collections.&lt;Item&gt;nCopies(7, null));
items.add(10,new Item(10));

System.out.println(items);
</code></pre>

prints

<pre><code>[0, 1, 2, null, null, null, null, null, null, null, 10]
</code></pre>

blocks|key|365191|text|如果内存和索引如此重要以至于使用普通数组。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|365192|当它变得很小的时候，使用System.arraycopy就是ArrayList内部的方式。|offset|length|365193|--|365194|即使您使用ArrayList并且有一百万个对象，也建议您使用ArrayList(int+initialCapacity)-Constructor来避免大量的复制操作|365195|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/6/docs/api/java/lang/System.html#arraycopy%2528java.lang.Object,%2520int,%2520java.lang.Object,%2520int,%2520int%2529^0|0|C|G|0|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@]|9|@$D|T|E|U|1|V]]|A|$]]|$1|F|3|G|5|6|7|W|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|X|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|K|$L|$5|M|N|O|A|$P|Q]]]]

If memory and index is so important that use a normal array. 

When it becomes to small use <a href="http://docs.oracle.com/javase/6/docs/api/java/lang/System.html#arraycopy%28java.lang.Object,%20int,%20java.lang.Object,%20int,%20int%29" rel="nofollow">System.arraycopy</a> thats the way ArrayList does it internal. 

--

Even if you use the ArrayList and have a Million Objects it is advisable to use the ArrayList(int initialCapacity)-Constructor to avoid a lot of copy-operations

blocks|key|1383167|text|请改用TreeMap。这里有一个简单的例子来检查内存消耗。分别运行第一个和第二个测试，并使用jvisualvm检查堆大小。记住要多次执行GC。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1383168|++++public+class+Test+{


++++++++++++public+static+void+main(String[]+args)+throws+InterruptedException+{
++++++++++++++++String+s+=+"Lorem+ipsum+dolor+sit+amet,+consectetur+adipiscing+elit.+Quisque+metus.";


++++++++++++++++//Test+1
++++++++++++++++ArrayList<String>+l+=+new+ArrayList<String>();

++++++++++++++++for+(int+i+=+0;+i+<+1000000;+i%2B%2B)+{
++++++++++++++++++++l.add(s+%2B+"+"+%2B+i);
++++++++++++++++++++l.addAll(Collections.nCopies(i+%25+10,+(String)null));+//Add+some+nulls
++++++++++++++++}
++++++++++++++++//Heap+is+>+5MB

++++++++++++++++//Test+2+uncomment+and+comment+test+1
++++//++++++++++SortedMap<Integer,+String>+map+=+new+TreeMap<Integer,+String>();
++++//++++++++++for+(int+i+=+0;+i+<+1000000;+i%2B%2B)+{
++++//++++++++++++++map.put(i,s+%2B+"+"+%2B+i);
++++//++++++++++}
++++++++++++++++//Heap+is+<+5MB

++++++++++++++++Thread.sleep(100000);

++++++++++++}
++++}|code-block|syntax|javascript|1383169|看起来TreeMap版本比ArrayList版本占用的内存更少。检查一下你自己。|offset|length|style|CODE|1383170|entityMap^0|0|0|3|7|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Use TreeMap instead. Here is simple example to check memony consuption. Run first and second test separatly and use jvisualvm to check heap size. Remember to Perform GC several times.

<pre><code> public class Test {


 public static void main(String[] args) throws InterruptedException {
 String s = "Lorem ipsum dolor sit amet, consectetur adipiscing elit. Quisque metus.";


 //Test 1
 ArrayList&lt;String&gt; l = new ArrayList&lt;String&gt;();

 for (int i = 0; i &lt; 1000000; i++) {
 l.add(s + " " + i);
 l.addAll(Collections.nCopies(i % 10, (String)null)); //Add some nulls
 }
 //Heap is &gt; 5MB

 //Test 2 uncomment and comment test 1
 // SortedMap&lt;Integer, String&gt; map = new TreeMap&lt;Integer, String&gt;();
 // for (int i = 0; i &lt; 1000000; i++) {
 // map.put(i,s + " " + i);
 // }
 //Heap is &lt; 5MB

 Thread.sleep(100000);

 }
 }
</code></pre>

It looks like <code>TreeMap</code> version is even less memory consuming than ArrayList version. Check yourself.

blocks|key|366286|text|@icCube-你说过，这个列表应该有90%25是满的。我对这个解决方案的想法是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|366287|如果您确切知道目标大小，请使用|366288|-如果您知道目标大小，请使用普通数组|unordered-list-item|366289|-使用初始容量尽可能接近目标大小的ArrayList。|366290|如果你不知道目标大小，你的ArrayList会被多次调整大小。调整大小意味着复制整个底层数组(它使用Arrays.copyOf)。你可以想象如果数组被复制会发生什么-+GC有很多工作要做。然后使用TreeMap。|366291|366292|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|P|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|F|7|R|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|S|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|M|$]]

@icCube- you said, that list should be about 90% full. 
My idea for this solution is:

<ul>
<li>If you exactly know target size - use plain array</li>
<li>If you are aware of target size - use ArrayList with initial capacity as close as possible to target size. Put nulls with <code>l.addAll(Collections.nCopies(n, (String)null));</code> as people said.</li>
<li>If you don't know target size - your ArrayList will be resized many times. Resizing means copying whole underlying array (it uses Arrays.copyOf). You can imagine what happens if array is copied around - GC has lots of work. Use TreeMap then.</li>
</ul>

blocks|key|266640|text|这是一个较老的问题，但您现在可以使用SparseArray作为ArrayList的(几乎)直接替代。它允许不连续的整数键值，如果没有为键设置值，则返回null。在性能方面，它正是为满足您的需求而设计的。您可以使用append而不是add，它更具描述性，因为无论最大键是什么，您都会将其附加到末尾，即使有间隙也是如此。您还可以在任何键值上执行set操作，即使它超出了最大键值。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|266641|entityMap^0|I|B|V|9|2Y|6|37|3|4Q|3|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|G|$]]

This is an older question, but you can now use <code>SparseArray</code> as an (almost) direct drop-in replacement for <code>ArrayList</code>. It allows non-contiguous integer key values, and returns null if a value has not been set for a key. Performance-wise it was designed exactly for your needs. Instead of <code>add</code> you use <code>append</code>, which is more descriptive in that you are appending to the end of whatever the max key is, even if there is gaps. You can also <code>set</code> at any key value you'd like, even if it is beyond the maximum key.

Wondering whether there is an efficient way to add an item to Java's ArrayList at a bigger position than its current size:

Scenario:

<pre><code> ArrayList&lt;Item&gt; items = new ArrayList&lt;Item&gt;;
 ... let's say I add three elements
</code></pre>

Now I would like to add an item at position 10 (leaving items from 3 to 10 to null)

<pre><code> items.add(10,newItem); // item.size() == 3 
</code></pre>

Is there an efficient way resizing/filling an ArrayList with nulls?

Java's implementation makes size field private :-(..

Java ArrayList add item outside current size

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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想知道是否有一种有效的方法可以将项目添加到比当前大小更大的位置上的ArrayList中：场景： ArrayList<Item> items = new ArrayList<Item>; ... let's say I add three elements现在我想在位置10添加一个项目(将3到10的项目保留为空)...

问Java ArrayList添加超出当前大小的项
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