Java-使用2个迭代器解析ArrayList和可怕的ConcurrentModificationException

我正在搜索ArrayList，并与2个迭代器进行比较。我将值写入一个字符串缓冲区，该缓冲区最终将成为XML输出。当我解析这些值时，我正在检查匹配的itemIds。匹配通常是零件和图纸。一个零件可以有许多图纸。对于XML，我必须知道所有匹配项的类型和名称，并将值附加在一起

使用此ArrayList：

itemId类型名称

1000件锤
1001件指甲
1000 dwg语义
1002零件尺

我的示例XML输出大致如下所示：

<Master itemId=1000 type=part name=hammer>
  <Extra type=dwg name=semantic>
</Master>
<Master itemId=1001 type=part name=nail>
</Master>
<Master itemId=1002 type=part name=ruler>
</Master>

这是我的第一个循环

while (theBaseInterator.hasNext()){
     ImportedTableObjects next = theBaseInterator.next(); 
     currentEntry = next.identiferId;
     currentType = next.typeId;
     currentDatasetName = next.nameId;
     compareInterator = tArray.listIterator(theBaseInterator.nextIndex());
     compareEntriesofArray(currentEntry, currentType, currentDatasetName, compareInterator); <======= calling method for 2nd loop and compare check
  }

while（baseInterator.hasNext（））{
ImportedTableObjects next=baseInterator.next（）；
currentEntry=next.identiferId；
currentType=next.typeId；
currentDatasetName=next.nameId；
compareInterator=tArray.listIterator（baseInterator.nextIndex（））；
compareEntriesofArray（currentEntry、currentType、currentDatasetName、compareInterator）；最简单的方法可能是创建另一个列表，在其中放置“匹配”的条目，然后只需对照该列表进行检查。
将所有要删除的元素添加到新列表中

迭代后，调用：

coll1.removeAll (coll2);

不是使用迭代器和它们的hasNext/next，而是使用列表，您可以使用for循环从上到下进行迭代。删除元素（7）bevore访问元素（6）等等对我来说从来都不是问题，但我没有看到有人推荐它

这里有完整的代码

import java.util.*;

public class GuessGame 
{
    public static void main ( String [] args )
    {
        char [] ca = "This is a test!".toCharArray ();
        List <Character> ls = new ArrayList <Character> ();
        for (char c: ca)
            ls.add (c);

        show (ls);
        // first method: remove from top/end and step backwise:
        for (int i = ls.size () - 1; i >= 0; --i)
        {
            char c = ls.get (i); 
            if (c == 'i' || c == 'a' || c == 'e')
                ls.remove (i); 
        }
        show (ls);

        // second approach: collect elements to remove ...
        ls = new ArrayList <Character> ();
        for (char c: ca)
            ls.add (c);
        show (ls);
        // ... in a separate list and 
        List <Character> toRemove = new ArrayList <Character> ();
        for (char c: ls)
        {
            if (c == 'i' || c == 'a' || c == 'e')
                toRemove.add (c); 
        }
        // ... remove them all in one go:
        ls.removeAll (toRemove);
        show (ls);
    }

    private static void show (List <Character> ls)
    {
        for (char c: ls)
            System.out.print (c + " ");
        System.out.println ();
    }   
}

您是否可以将已经处理过的项目存储在二级结构中，例如
树形图
，然后在那里进行检查？您是否可以在此建议上进行更多扩展
import java.util.*;

public class GuessGame 
{
    public static void main ( String [] args )
    {
        char [] ca = "This is a test!".toCharArray ();
        List <Character> ls = new ArrayList <Character> ();
        for (char c: ca)
            ls.add (c);

        show (ls);
        // first method: remove from top/end and step backwise:
        for (int i = ls.size () - 1; i >= 0; --i)
        {
            char c = ls.get (i); 
            if (c == 'i' || c == 'a' || c == 'e')
                ls.remove (i); 
        }
        show (ls);

        // second approach: collect elements to remove ...
        ls = new ArrayList <Character> ();
        for (char c: ca)
            ls.add (c);
        show (ls);
        // ... in a separate list and 
        List <Character> toRemove = new ArrayList <Character> ();
        for (char c: ls)
        {
            if (c == 'i' || c == 'a' || c == 'e')
                toRemove.add (c); 
        }
        // ... remove them all in one go:
        ls.removeAll (toRemove);
        show (ls);
    }

    private static void show (List <Character> ls)
    {
        for (char c: ls)
            System.out.print (c + " ");
        System.out.println ();
    }   
}

T h i s   i s   a   t e s t ! 
T h s   s     t s t ! 
T h i s   i s   a   t e s t ! 
T h s   s     t s t !