blocks|key|2877053|text|list.remove(Integer.valueOf(value%2B1));|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2877054|应该能正常工作。|unstyled|2877055|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>list.remove(Integer.valueOf(value+1));
</code></pre>

should work fine.

blocks|key|478451|text|list.remove(value%2B1)将在给定索引处删除。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|478452|您可以使用list.remove(Integer.valueOf(value%2B1))删除Integer。|478453|Integer.valueOf(v)被重新命令，而不是new+Integer(v)，因为它允许为特殊值重用Integer实例(参见javadoc)。|478454|编辑:在装箱/取消装箱方面，可以使用像trove4j这样的库来完全消除这个问题，它定义了一个动态的整数数组来存储原语值，而不是数字类(参见TIntArrayList)。没有装箱，没有取消装箱，更低的GC使用和更好的性能。|478455|entityMap|0|LINK|mutability|MUTABLE|url|http://trove4j.sourceforge.net/javadocs/gnu/trove/list/array/TIntArrayList.html^0|0|K|0|5|11|18|7|0|0|I|R|E|1H|7|0|1X|D|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@$9|U|A|V|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]|$9|16|A|17|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|18|8|@]|D|@$9|19|A|1A|1|1B]]|E|$]]|$1|L|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|M|$N|$5|O|P|Q|E|$R|S]]]]

<code>list.remove(value+1)</code> will remove at given index.

You can use <code>list.remove(Integer.valueOf(value+1))</code> to remove the <code>Integer</code>.

<code>Integer.valueOf(v)</code> is recommanded instead of <code>new Integer(v)</code> because it allows reuse of <code>Integer</code> instances for special values (see javadoc).

EDIT :
In term of boxing/unboxing, it is possible to completely eliminate the problem using libraries like trove4j which define an dynamic array of integer storing the primitive values instead of numeric class (see <a href="http://trove4j.sourceforge.net/javadocs/gnu/trove/list/array/TIntArrayList.html" rel="nofollow">TIntArrayList</a>). No boxing, no unboxing, lower GC use and better performance.

blocks|key|2877121|text|你在找这个吗。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2877122|Collections.sort(list);
int+index+=+list.indexOf(5);
list+=+list.subList(0,+index-1);
System.out.println(list);|code-block|syntax|javascript|2877123|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|$]]

Do you looking for this one.

<pre><code>Collections.sort(list);
int index = list.indexOf(5);
list = list.subList(0, index-1);
System.out.println(list);
</code></pre>

blocks|key|476990|text|您可以使用list.indexOf(值%2B+1)，因为该方法只接受对象。然后使用该索引删除元素。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|476991|++++int+i+=+list.indexOf(value+%2B+1);
++++list.remove(i);|code-block|syntax|javascript|476992|如果需要删除值%2B1的所有实例，则继续到I为-1+(未找到)。|476993|或者，只要迭代一遍，找到它们就移除。|476994|++++for+(Iterator<Integer>+iterator+=+list.iterator();+iterator.hasNext();+)+{
++++++++Integer+integer+=+iterator.next();
++++++++if+(integer+==+value+%2B+1)+{
++++++++++++iterator.remove();
++++++++}
++++}|476995|entityMap^0|0|0|0|0|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]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

You can use list.indexOf(value + 1) since that method only takes the object. Then use that index to remove the element.

<pre><code> int i = list.indexOf(value + 1);
 list.remove(i);
</code></pre>

If you need to remove all instances of value + 1 then continue until i is -1 (not found).

Or, just iterate over it and remove as you find them

<pre><code> for (Iterator&lt;Integer&gt; iterator = list.iterator(); iterator.hasNext(); ) {
 Integer integer = iterator.next();
 if (integer == value + 1) {
 iterator.remove();
 }
 }
</code></pre>

blocks|key|1798387|text|更有效的方法可能是使用来自Trove4j库的Trove4j，它包装int[]数组，而不使用包装器类型。因此，它占用的内存减少了3倍，而且速度要快得多。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1798388|entityMap|0|LINK|mutability|MUTABLE|url|http://trove4j.sourceforge.net/javadocs/gnu/trove/list/array/TIntArrayList.html|1^0|X|5|D|7|0|M|7|1|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@$9|R|A|S|1|T]|$9|U|A|V|1|W]]|E|$]]|$1|F|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]|N|$5|I|J|K|E|$L|M]]]]

A more effective way might be to use a <code>TIntArrayList</code> from <a href="http://trove4j.sourceforge.net/javadocs/gnu/trove/list/array/TIntArrayList.html" rel="nofollow">Trove4j</a> library which wraps <code>int[]</code> array and doesn't use wrapper types. For this reason it takes 3x times less memory and is much faster.

blocks|key|2877241|text|在第二种情况下，我是否可以确保删除正确的值？|type|blockquote|depth|inlineStyleRanges|entityRanges|data|2877242|是的你可以。|unstyled|2877243|remove方法将使用equals(Object)标识要从列表中删除的对象。由于Integer.equals根据值(而不是对象标识)进行比较，所以如何创建Integer实例并不重要.如果包装的int值相同的话。|offset|length|style|CODE|2877244|但是..。您的第二个版本不如第一个版本：|2877245|它很麻烦:3条语句而不是1条。|unordered-list-item|2877246|您使用的是new+Integer(...)而不是Integer.valueOf(...)，后者总是要创建一个新实例。相比之下，自动装箱(或显式调用Integer.valueOf(...)+)使用Integer类实例缓存，这通常会避免创建新实例。|2877247|entityMap^0|0|0|0|6|B|E|14|E|25|7|2O|3|0|0|0|5|G|O|K|21|K|2P|7|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|U|8|@]|9|@]|A|$]]|$1|E|3|F|5|D|7|V|8|@$G|W|H|X|I|J]|$G|Y|H|Z|I|J]|$G|10|H|11|I|J]|$G|12|H|13|I|J]|$G|14|H|15|I|J]]|9|@]|A|$]]|$1|K|3|L|5|D|7|16|8|@]|9|@]|A|$]]|$1|M|3|N|5|O|7|17|8|@]|9|@]|A|$]]|$1|P|3|Q|5|O|7|18|8|@$G|19|H|1A|I|J]|$G|1B|H|1C|I|J]|$G|1D|H|1E|I|J]|$G|1F|H|1G|I|J]]|9|@]|A|$]]|$1|R|3|-4|5|D|7|1H|8|@]|9|@]|A|$]]]|S|$]]

<blockquote>
 In the second case, can I be sure the right value will be deleted?
</blockquote>

Yes you can.

The <code>remove</code> method will use <code>equals(Object)</code> to identify the object to be removed from the list. Since <code>Integer.equals</code> compares by value (not by object identity), it doesn't matter how you created the <code>Integer</code> instances ... provided the wrapped <code>int</code> values are the same.

<hr>

However ... your second version is inferior to the first version:

<ul>
<li>It cumbersome: 3 statements instead of 1.</li>
<li>You are using <code>new Integer(...)</code> rather than <code>Integer.valueOf(...)</code> which is always going to create a new instance. By contrast, autoboxing (or calling <code>Integer.valueOf(...)</code> explicitly) makes use of the <code>Integer</code> classes instance cache, and that will often avoid creating a new instance.</li>
</ul>

I have an array list of integers and would like to delete an Integer value from it. To clarify: Let's say I have a function that takes in an <code>ArrayList</code> of Integers, and specified Integer value. I need to return the list with (value+1) deleted from it.

<code>boolean deleteFromList(ArrayList&lt;Integer&gt; list, Integer value)</code>

Now, if I do this: 

<pre><code>return list.remove(value+1)
</code></pre>

compiler will complain because it will try to invoke the delete method that takes <code>int</code> parameter and deletes an object from specified location, not the actual object.
So what is a proper way to deal with this? Is it better to do:

<pre><code>list.remove((Integer)(value+1))
</code></pre>

or 

<pre><code>int v = value.intValue();
v++;
list.remove(new Integer(v)); 
</code></pre>

? In the second case, can I be sure the right value will be deleted?

Java propery way to delete an Integer from ArrayList

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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我有一个整数数组列表，希望从其中删除一个整数值。为了澄清:假设我有一个函数，它接受整数的ArrayList，并指定整数值。我需要返回(value+1)从其中删除的列表。boolean deleteFromList(ArrayList<Integer> list, Integer value)现在，如果我这么做retur...

问从ArrayList中删除整数的Java方法
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