blocks|key|564939|text|String[]+mString+=+new+String[]+{"B",+"D",+"F"};

for+(int+j+=+0;+j+<+mString.length-1;+j%2B%2B)+{
++++++++List_Of_Array.remove(mString[j]);
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|564940|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>String[] mString = new String[] {"B", "D", "F"};

for (int j = 0; j &lt; mString.length-1; j++) {
 List_Of_Array.remove(mString[j]);
}
</code></pre>

blocks|key|564969|text|这个怎么样？好好想一想-|type|unstyled|depth|inlineStyleRanges|entityRanges|data|564970|import+java.util.ArrayList;

class+Solution
{
++++++++public+static+void+main+(String[]+args){

+++++++++++++ArrayList<String>+List_Of_Array+=+new+ArrayList<String>();
+++++++++++++List_Of_Array.add("A");
+++++++++++++List_Of_Array.add("B");
+++++++++++++List_Of_Array.add("C");
+++++++++++++List_Of_Array.add("D");
+++++++++++++List_Of_Array.add("E");
+++++++++++++List_Of_Array.add("F");
+++++++++++++List_Of_Array.add("G");
+++++++++++++List_Of_Array.add("H");

+++++++++++++int+i[]+=+{1,3,5};

+++++++++++++for+(int+j+=+0;+j+<+i.length;+j%2B%2B)+{
+++++++++++++++++List_Of_Array.remove(i[j]-j);
+++++++++++++}

+++++++++++++System.out.println(List_Of_Array);

++++++++}


}|code-block|syntax|javascript|564971|输出结果是-|564972|[A,+C,+E,+G,+H]|564973|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? Just give it a thought-

<pre><code>import java.util.ArrayList;

class Solution
{
 public static void main (String[] args){

 ArrayList&lt;String&gt; List_Of_Array = new ArrayList&lt;String&gt;();
 List_Of_Array.add("A");
 List_Of_Array.add("B");
 List_Of_Array.add("C");
 List_Of_Array.add("D");
 List_Of_Array.add("E");
 List_Of_Array.add("F");
 List_Of_Array.add("G");
 List_Of_Array.add("H");

 int i[] = {1,3,5};

 for (int j = 0; j &lt; i.length; j++) {
 List_Of_Array.remove(i[j]-j);
 }

 System.out.println(List_Of_Array);

 }


}
</code></pre>

And the output was-

<pre><code>[A, C, E, G, H]
</code></pre>

blocks|key|348927|text|您可以使用ListIterator从ArrayList中删除元素，|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|348928|ListIterator+listIterator+=+List_Of_Array.listIterator();

+/*+Use+void+remove()+method+of+ListIterator+to+remove+an+element+from+List.
+++++It+removes+the+last+element+returned+by+next+or+previous+methods.
+*/
listIterator.next();

//remove+element+returned+by+last+next+method
listIterator.remove();//remove+element+at+1st+position
listIterator.next();
listIterator.next();
listIterator.remove();//remove+element+at+3rd+position
listIterator.next();
listIterator.next();
listIterator.remove();//remove+element+at+5th+position|code-block|syntax|javascript|348929|entityMap^0|5|C|I|9|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

You can remove elements from <code>ArrayList</code> using <code>ListIterator</code>,

<pre><code>ListIterator listIterator = List_Of_Array.listIterator();

 /* Use void remove() method of ListIterator to remove an element from List.
 It removes the last element returned by next or previous methods.
 */
listIterator.next();

//remove element returned by last next method
listIterator.remove();//remove element at 1st position
listIterator.next();
listIterator.next();
listIterator.remove();//remove element at 3rd position
listIterator.next();
listIterator.next();
listIterator.remove();//remove element at 5th position
</code></pre>

blocks|key|348954|text|试着这样做，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|348955|ArrayList<String>+List_Of_Array+=+new+ArrayList<String>();
List_Of_Array.add("A");
List_Of_Array.add("B");
List_Of_Array.add("C");
List_Of_Array.add("D");
List_Of_Array.add("E");
List_Of_Array.add("F");
List_Of_Array.add("G");
List_Of_Array.add("H");

int+i[]+=+{5,3,1};

for+(int+j+=+0;+j+<+i.length;+j%2B%2B)+{
++++List_Of_Array.remove(i[j]);
}|code-block|syntax|javascript|348956|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|$]]

Try it this way,

<pre><code>ArrayList&lt;String&gt; List_Of_Array = new ArrayList&lt;String&gt;();
List_Of_Array.add("A");
List_Of_Array.add("B");
List_Of_Array.add("C");
List_Of_Array.add("D");
List_Of_Array.add("E");
List_Of_Array.add("F");
List_Of_Array.add("G");
List_Of_Array.add("H");

int i[] = {5,3,1};

for (int j = 0; j &lt; i.length; j++) {
 List_Of_Array.remove(i[j]);
}
</code></pre>

blocks|key|565020|text|+public+void+DeleteUserIMP(UserIMP+useriamp)+{
+++++++synchronized+(ListUserIMP)+{
++++++++++++if+(ListUserIMP.isEmpty())+{
++++++++++++System.out.println("user+is+empty");
++++++++}++else+{
++++++++++++Iterator<UserIMP>+it+=+ListUserIMP.iterator();
++++++++++++while+(it.hasNext())+{
++++++++++++++++UserIMP+user+=+it.next();
++++++++++++++++if+(useriamp.getMoblieNumber().equals(user.getMoblieNumber()))+{
++++++++++++++++++++it.remove();
++++++++++++++++++++System.out.println("remove+it");
++++++++++++++++}
++++++++++++}
++++++++++++//+ListUserIMP.remove(useriamp);

++++++++++++System.out.println("+this+user+removed");
++++++++}
++++++++Constants.RESULT_FOR_REGISTRATION+=+Constants.MESSAGE_OK;
++++++++//+System.out.println("This+user+Deleted+"+%2B+Constants.MESSAGE_OK);

++++}
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|565021|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code> public void DeleteUserIMP(UserIMP useriamp) {
 synchronized (ListUserIMP) {
 if (ListUserIMP.isEmpty()) {
 System.out.println("user is empty");
 } else {
 Iterator&lt;UserIMP&gt; it = ListUserIMP.iterator();
 while (it.hasNext()) {
 UserIMP user = it.next();
 if (useriamp.getMoblieNumber().equals(user.getMoblieNumber())) {
 it.remove();
 System.out.println("remove it");
 }
 }
 // ListUserIMP.remove(useriamp);

 System.out.println(" this user removed");
 }
 Constants.RESULT_FOR_REGISTRATION = Constants.MESSAGE_OK;
 // System.out.println("This user Deleted " + Constants.MESSAGE_OK);

 }
}
</code></pre>

blocks|key|349001|text|如果您使用"="，则会为第二个列表中的原始arraylist创建一个副本，但引用是相同的，因此如果您在一个列表中更改，另一个列表也会被修改。使用这个而不是"=“|type|unstyled|depth|inlineStyleRanges|entityRanges|data|349002|++++++++List_Of_Array1.addAll(List_Of_Array);|code-block|syntax|javascript|349003|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|$]]

If you use "=", a replica is created for the original arraylist in the second one, but the reference is same so if you change in one list , the other one will also get modified. Use this instead of "="

<pre><code> List_Of_Array1.addAll(List_Of_Array);
</code></pre>

blocks|key|349131|text|如前所述|type|unstyled|depth|inlineStyleRanges|entityRanges|data|349132|iterator.remove()|code-block|syntax|javascript|349133|可能是在循环期间删除列表项的唯一安全方法。|349134|要更深入地理解使用迭代器删除项，请尝试查看此thread|offset|length|349135|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/15993356/how-iterators-remove-method-actually-remove-an-object^0|0|0|0|M|6|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|X|8|@]|9|@$K|Y|L|Z|1|10]]|A|$]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

As mentioned before

<pre><code>iterator.remove()
</code></pre>

is maybe the only safe way to remove list items during the loop.

For deeper understanding of items removal using the iterator, try to look at this <a href="https://stackoverflow.com/questions/15993356/how-iterators-remove-method-actually-remove-an-object">thread</a>

blocks|key|565112|text|arraylist的remove(int+index)方法删除列表中指定位置(Index)的元素。移除arraylist项后，将所有后续元素向左移动。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|565113|表示数组列表是否包含{20,15,30,40}|565114|我已经调用了这个方法:+arraylist.remove(1)|565115|那么数据15将被删除，并且30+&+40这两个项目将左移1。|565116|为此，您必须先删除arraylist的较高索引项。|565117|So..for您给定的situation..the代码将是..|565118|ArrayList<String>+list+=+new+ArrayList<String>();
list.add("A");
list.add("B");
list.add("C");
list.add("D");
list.add("E");
list.add("F");
list.add("G");
list.add("H");

int+i[]+=+{1,3,5};

for+(int+j+=+i.length-1;+j+>=+0;+j--)+{
++++list.remove(i[j]);
}|code-block|syntax|javascript|565119|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|U|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|V|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|W|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|X|8|@]|9|@]|A|$]]|$1|L|3|M|5|N|7|Y|8|@]|9|@]|A|$O|P]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

remove(int index) method of arraylist removes the element at the specified position(index) in the list. After removing arraylist items shifts any subsequent elements to the left.

Means if a arraylist contains {20,15,30,40}

I have called the method: arraylist.remove(1)

then the data 15 will be deleted and 30 &amp; 40 these two items will be left shifted by 1.

For this reason you have to delete higher index item of arraylist first.

So..for your given situation..the code will be..

<pre><code>ArrayList&lt;String&gt; list = new ArrayList&lt;String&gt;();
list.add("A");
list.add("B");
list.add("C");
list.add("D");
list.add("E");
list.add("F");
list.add("G");
list.add("H");

int i[] = {1,3,5};

for (int j = i.length-1; j &gt;= 0; j--) {
 list.remove(i[j]);
}
</code></pre>

blocks|key|349199|text|我假设数组i是升序排序的，这是另一个使用迭代器的解决方案，它更通用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|349200|ArrayList<String>+list+=+new+ArrayList<String>();
list.add("A");
list.add("B");
list.add("C");
list.add("D");
list.add("E");
list.add("F");
list.add("G");
list.add("H");

int+i[]+=+{1,3,5};

Iterator<String>+itr+=+list.iterator();
int+pos+=+0;
int+index+=+0;
while(+itr.hasNext()+){
++++itr.next();
++++if(+pos+>=+i.length+){
++++++++break;
++++}
++++if(+i[pos]+==+index+){
++++++++itr.remove();
++++++++pos%2B%2B;
++++}

++++index%2B%2B;
}|code-block|syntax|javascript|349201|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|$]]

I assume the array i is ascend sorted, here is another solution with Iterator, it is more generic:

<pre><code>ArrayList&lt;String&gt; list = new ArrayList&lt;String&gt;();
list.add("A");
list.add("B");
list.add("C");
list.add("D");
list.add("E");
list.add("F");
list.add("G");
list.add("H");

int i[] = {1,3,5};

Iterator&lt;String&gt; itr = list.iterator();
int pos = 0;
int index = 0;
while( itr.hasNext() ){
 itr.next();
 if( pos &gt;= i.length ){
 break;
 }
 if( i[pos] == index ){
 itr.remove();
 pos++;
 }

 index++;
}
</code></pre>

blocks|key|349224|text|如果您需要删除列表中的成品，可以使用此代码|type|unstyled|depth|inlineStyleRanges|entityRanges|data|349225|+if(yourList.size()+!=+0){
+++yourList.remove(yourList.size()-1);
+}|code-block|syntax|javascript|349226|或者您可能需要在选择列表中的项目后删除某个时间|349227|if(selected){//+boolean+for+checked+add+or+remove
+++yourList.add(position+,+id);+//+add+to+list+by+position+can+use+your+list+position
}else{
+++if(yourList.size()+!=+0){+
+++yourList.remove(position);
+}
}|349228|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|$]]

if you need to remove end item in list can use this code
<pre><code> if(yourList.size() != 0){
 yourList.remove(yourList.size()-1);
 }
</code></pre>
or maybe you need to remove a time after select item in list
<pre><code>if(selected){// boolean for checked add or remove
 yourList.add(position , id); // add to list by position can use your list position
}else{
 if(yourList.size() != 0){ 
 yourList.remove(position);
 }
}
</code></pre>

I have an <code>ArrayList</code> suppose <code>list</code>, and it has 8 items A-H and now I want to delete 1,3,5 position Item stored in int array from the <code>list</code> how can I do this.

I am trying to do this with 

<pre><code>ArrayList&lt;String&gt; list = new ArrayList&lt;String&gt;();
list.add("A");
list.add("B");
list.add("C");
list.add("D");
list.add("E");
list.add("F");
list.add("G");
list.add("H");

int i[] = {1,3,5};

for (int j = 0; j &lt; i.length; j++) {
 list.remove(i[j]);
}
</code></pre>

But after first item deleted positioned of array is changed and in the next iterate it deletes wrong element or give exception.

Remove Item 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ArrayList假设list，它有8个项A-H，现在我想从list中删除存储在整数数组中的1,3,5个位置项，我该怎么做呢？我正试着这样做ArrayList<String> list = new ArrayList<String>();list.add("A");list.add("B");list.add(...

问从ArrayList中删除项目
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