Java-解析具有2个迭代器和可怕的ConcurrentModificationException的ArrayList

Question

I am searching through a ArrayList and doing a compare with 2 iterators. I am writing out values to a String buffer that will eventually be a XML output. As I parse through the values I am checking for matching itemIds. The matches are normally parts and drawings. There can be many drawings to a part. For my XML I have to know the type and name of all the matches and append the values together.

Using this ArrayList:

itemId type name

1000 part hammer
1001 part nail
1000 dwg semantic
1002 part ruler

My sample XML output would roughly look like :

<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>

This is my first loop

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
  }

I written a method for the second loop and compare

private void compareEntriesofArray(Object currentEntry, Object currentType, Object currentDatasetName, ListIterator<ImportedTableObjects> compareInterator)
object nextEntry;   
while (compareInterator.hasNext()) {
    ImportedTableObjects next = compareInterator.next();
    nextEntry = next.identiferId;
    if(nextEntry.equals(currentEntry)) { 
    compareInterator.remove();
    }   

When it finds a match I am trying to remove the matching entry from the list. There is no need to re-check entries that has been matched. So when the 1st loop continues down the list - it does not have to check that entry again.

But of course I am getting the ConcurrentModificationException. I fully understand why. Is there a way that instead of trying to remove the entry, I can some how mark it with a boolean or something, so when the first loop gets to that entry in the list it knows to skip it and go to the next?

Answer 1

Add all elements you want to remove into a new List.

After iterating, call:

coll1.removeAll (coll2);

Not with iterators, and their hasNext/next, but with Lists, you can iterate with a for-loop from top to bottom. removing element (7) bevore visiting element (6) and so on has never been a problem for me, but I haven't seen it being recommended.

Here complete code

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 ();
    }   
}

Output:

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 ! 

Answer 2

最简单的方法可能是创建另一个列表，在该列表中放置“匹配的”条目，然后对照该列表进行检查。