blocks|key|6001228|text|最好是根本不使用LinkedList，通常它在所有学科中都较慢，而且不太方便。(主要在前面插入/删除时，尤其是对于大型阵列，LinkedList速度更快)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|6001229|6001230|int+len+=+list.size();
for+(int+i+=+0;+i+<+len;+i%2B%2B)+{
++Element+ele+=+list.get(i);
}|code-block|syntax|javascript|6001231|重置是微不足道的，只要再次循环即可。|6001232|如果您坚持使用迭代器，那么您必须使用新的迭代器：|6001233|iter+=+list.listIterator();|6001234|(我一生中只看到过一次LinkedList的优点:我可以遍历while循环并删除第一个元素)|6001235|entityMap^0|8|A|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]|$1|G|3|H|5|I|7|Z|8|@]|D|@]|E|$J|K]]|$1|L|3|M|5|6|7|10|8|@]|D|@]|E|$]]|$1|N|3|O|5|6|7|11|8|@]|D|@]|E|$]]|$1|P|3|Q|5|I|7|12|8|@]|D|@]|E|$J|K]]|$1|R|3|S|5|6|7|13|8|@]|D|@]|E|$]]|$1|T|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|U|$]]

Best would be not using <code>LinkedList</code> at all, usually it is slower in all disciplines, and less handy. (When mainly inserting/deleting to the front, especially for big arrays LinkedList is faster)

Use <code>ArrayList</code>, and iterate with 

<pre><code>int len = list.size();
for (int i = 0; i &lt; len; i++) {
 Element ele = list.get(i);
}
</code></pre>

Reset is trivial, just loop again. 
If you insist on using an iterator, then you have to use a new iterator:

<pre><code>iter = list.listIterator();
</code></pre>

(I saw only once in my life an advantage of LinkedList: i could loop through whith a while loop and remove the first element)

blocks|key|3791944|text|您可以再次调用listIterator方法来获取指向列表开头的迭代器的实例：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3791945|iter+=+list.listIterator();|code-block|syntax|javascript|3791946|entityMap^0|7|C|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

You can call <code>listIterator</code> method again to get an instance of iterator pointing at beginning of list:

<pre><code>iter = list.listIterator();
</code></pre>

blocks|key|6001562|text|这是一种替代解决方案，但有人可能会说，它并没有增加足够的价值，因此它是值得的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|6001563|import+com.google.common.collect.Iterables;
...
Iterator<String>+iter+=+Iterables.cycle(list).iterator();
if(iter.hasNext())+{
++++str+=+iter.next();
}|code-block|syntax|javascript|6001564|6001565|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|K|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|L|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|M|8|@]|9|@]|A|$]]]|I|$]]

This is an alternative solution, but one could argue it doesn't add enough value to make it worth it:

<pre><code>import com.google.common.collect.Iterables;
...
Iterator&lt;String&gt; iter = Iterables.cycle(list).iterator();
if(iter.hasNext()) {
 str = iter.next();
}
</code></pre>

Calling hasNext() will reset the iterator cursor to the beginning if it's a the end.

blocks|key|3792194|text|如果顺序不重要，我们可以使用相同的迭代器使用hasPrevious()和previous()方法向后重新迭代：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3792195|ListIterator<T>+lit+=+myList.listIterator();+//+create+just+one+iterator|code-block|syntax|javascript|3792196|最初迭代器位于开始处，我们进行前向迭代：|3792197|while+(lit.hasNext())+process(lit.next());+//+begin+->+end|3792198|然后迭代器坐在最后，我们可以进行反向迭代：|BOLD|3792199|while+(lit.hasPrevious())+process2(lit.previous());+//+end+->+begin|3792200|entityMap^0|M|D|10|A|0|0|0|0|G|2|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@$9|W|A|X|B|C]|$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|10|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|11|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|13|8|@$9|14|A|15|B|Q]]|D|@]|E|$]]|$1|R|3|S|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|T|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|U|$]]

If the order doesn't matter, we can re-iterate backward with the same iterator using the <code>hasPrevious()</code> and <code>previous()</code> methods:



<pre class="lang-java prettyprint-override"><code>ListIterator&lt;T&gt; lit = myList.listIterator(); // create just one iterator
</code></pre>

Initially the iterator sits at the beginning, we do forward iteration:

<pre class="lang-java prettyprint-override"><code>while (lit.hasNext()) process(lit.next()); // begin -&gt; end
</code></pre>

Then the iterator sits at the end, we can do backward iteration:

<pre class="lang-java prettyprint-override"><code>while (lit.hasPrevious()) process2(lit.previous()); // end -&gt; begin
</code></pre>

blocks|key|4273091|text|您可能真正想要使用的是一个Iterable，它可以通过调用iterator()多次返回一个新的Iterator。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4273092|//A+function+that+needs+to+iterate+multiple+times+can+be+given+one+Iterable:
public+void+func(Iterable<Type>+ible)+{
++++Iterator<Type>+it+=+ible.iterator();+//Gets+an+iterator
++++while+(it.hasNext())+{
++++++++it.next();
++++}
++++it+=+ible.iterator();+//Gets+a+NEW+iterator,+also+from+the+beginning
++++while+(it.hasNext())+{
++++++++it.next();
++++}
}|code-block|syntax|javascript|4273093|您必须事先只定义一次iterator()方法的作用：|4273094|void+main()+{
++++LinkedList<String>+list;+//This+could+be+any+type+of+object+that+has+an+iterator
++++//Define+an+Iterable+that+knows+how+to+retrieve+a+fresh+iterator
++++Iterable<Type>+ible+=+new+Iterable<Type>()+{
++++++++@Override
++++++++public+Iterator<Type>+iterator()+{
++++++++++++return+list.listIterator();+//Define+how+to+get+a+fresh+iterator+from+any+object
++++++++}
++++};
++++//Now+with+a+single+instance+of+an+Iterable,
++++func(ible);+//you+can+iterate+through+it+multiple+times.
}|4273095|entityMap^0|D|8|T|A|1B|8|0|0|A|A|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|P|$]]

What you may actually want to use is an <code>Iterable</code> that can return a fresh <code>Iterator</code> multiple times by calling <code>iterator()</code>.

<pre><code>//A function that needs to iterate multiple times can be given one Iterable:
public void func(Iterable&lt;Type&gt; ible) {
 Iterator&lt;Type&gt; it = ible.iterator(); //Gets an iterator
 while (it.hasNext()) {
 it.next();
 }
 it = ible.iterator(); //Gets a NEW iterator, also from the beginning
 while (it.hasNext()) {
 it.next();
 }
}
</code></pre>

You must define what the <code>iterator()</code> method does just once beforehand:

<pre><code>void main() {
 LinkedList&lt;String&gt; list; //This could be any type of object that has an iterator
 //Define an Iterable that knows how to retrieve a fresh iterator
 Iterable&lt;Type&gt; ible = new Iterable&lt;Type&gt;() {
 @Override
 public Iterator&lt;Type&gt; iterator() {
 return list.listIterator(); //Define how to get a fresh iterator from any object
 }
 };
 //Now with a single instance of an Iterable,
 func(ible); //you can iterate through it multiple times.
}
</code></pre>

blocks|key|443687|text|在集合impl上调用iterator()，可能会在每次调用时获得一个新的迭代器。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|443688|因此，您可以简单地再次调用iterator()来获得一个新的。|443689|443690|代码|443691|IteratorLearn.java|BOLD|443692|import+org.testng.Assert;
import+org.testng.annotations.Test;

import+java.util.Collection;
import+java.util.HashSet;
import+java.util.Iterator;

/**
+*+Iterator+learn.
+*
+*+@author+eric
+*+@date+12/30/18+4:03+PM
+*/
public+class+IteratorLearn+{
++++@Test
++++public+void+test()+{
++++++++Collection<Integer>+c+=+new+HashSet<>();
++++++++for+(int+i+=+0;+i+<+10;+i%2B%2B)+{
++++++++++++c.add(i);
++++++++}

++++++++Iterator+it;

++++++++//+iterate,
++++++++it+=+c.iterator();
++++++++System.out.println("\niterate:");
++++++++while+(it.hasNext())+{
++++++++++++System.out.printf("\t%25d\n",+it.next());
++++++++}
++++++++Assert.assertFalse(it.hasNext());

++++++++//+consume,
++++++++it+=+c.iterator();
++++++++System.out.println("\nconsume+elements:");
++++++++it.forEachRemaining(ele+->+System.out.printf("\t%25d\n",+ele));
++++++++Assert.assertFalse(it.hasNext());
++++}
}|code-block|syntax|javascript|443693|输出：|443694|iterate:
++++0
++++1
++++2
++++3
++++4
++++5
++++6
++++7
++++8
++++9

consume+elements:
++++0
++++1
++++2
++++3
++++4
++++5
++++6
++++7
++++8
++++9|443695|entityMap^0|A|A|0|D|A|0|0|0|0|I|0|0|0|3|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|14|8|@]|D|@]|E|$]]|$1|I|3|J|5|6|7|15|8|@]|D|@]|E|$]]|$1|K|3|L|5|6|7|16|8|@$9|17|A|18|B|M]]|D|@]|E|$]]|$1|N|3|O|5|P|7|19|8|@]|D|@]|E|$Q|R]]|$1|S|3|T|5|6|7|1A|8|@$9|1B|A|1C|B|M]]|D|@]|E|$]]|$1|U|3|V|5|P|7|1D|8|@]|D|@]|E|$Q|R]]|$1|W|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]]|X|$]]

Calling <code>iterator()</code> on a Collection impl, probably would get a new Iterator on each call.

Thus, you can simply call <code>iterator()</code> again to get a new one.

<hr>

<h2>Code</h2>

IteratorLearn.java

<pre><code>import org.testng.Assert;
import org.testng.annotations.Test;

import java.util.Collection;
import java.util.HashSet;
import java.util.Iterator;

/**
 * Iterator learn.
 *
 * @author eric
 * @date 12/30/18 4:03 PM
 */
public class IteratorLearn {
 @Test
 public void test() {
 Collection&lt;Integer&gt; c = new HashSet&lt;&gt;();
 for (int i = 0; i &lt; 10; i++) {
 c.add(i);
 }

 Iterator it;

 // iterate,
 it = c.iterator();
 System.out.println("\niterate:");
 while (it.hasNext()) {
 System.out.printf("\t%d\n", it.next());
 }
 Assert.assertFalse(it.hasNext());

 // consume,
 it = c.iterator();
 System.out.println("\nconsume elements:");
 it.forEachRemaining(ele -&gt; System.out.printf("\t%d\n", ele));
 Assert.assertFalse(it.hasNext());
 }
}
</code></pre>

Output:

<pre><code>iterate:
 0
 1
 2
 3
 4
 5
 6
 7
 8
 9

consume elements:
 0
 1
 2
 3
 4
 5
 6
 7
 8
 9
</code></pre>

blocks|key|3792500|text|您可以做的是，使用while循环将迭代器设置为第一个位置。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3792501|while(iter.hasPrevious())
{
++++iter.previous();
}|code-block|syntax|javascript|3792502|当您走出循环时，您的迭代器将位于位置0|3792503|抱歉，有任何语法错误|3792504|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]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

What you can do is, set the iterator to the first position mannualy with a while loop.
<pre><code>while(iter.hasPrevious())
{
 iter.previous();
}
</code></pre>
when you get out of the loop you will have your iterator at the position 0
<h6 id="sorry-any-grammar-mistakes-ydm0">sorry any grammar mistakes</h6>

I need to know how to "reset" LinkedList iterator to its first element.

For example:

<code>LinkedList&lt;String&gt; list;</code>

<code>Iterator iter=list.listIterator;</code>

<code>iter.next();</code>

<code>iter.next();</code>

Over and over again and after many moves of the iterator I need to "reset" the position of iterator.
`

I want to ask how I can "reset" my iterator to first element

I know that I can get list iterator of the first element in this way:

<code>iter= list.listIterator(1);</code>

Is this the best solution? or maybe I missed something in Oracle docs?

java- reset list iterator to first element of the list

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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我需要知道如何“重置”LinkedList迭代器的第一个元素。例如：LinkedList<String> list;Iterator iter=list.listIterator;iter.next();iter.next();一遍又一遍，在迭代器的多次移动之后，我需要“重置”迭代器的位置。`我想问一下如何将迭代器“重置”为第一个元素我知道我可以用这种方式得到第一个元素的列表迭代器：iter= l

问java-将列表迭代器重置为列表的第一个元素
EN

回答 7

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