Question

我有两个类型为RemoveTest的ArrayLists。我把它们放在地图上。此外，我必须迭代地图，并为每个我需要检查的arraylist 如果任何元素的mid与任何其他元素相同然后rmid从列表中删除这两个元素。

我运行下面的代码..但我只需要确认我的方法。虽然我不是直接从hashmap中删除..而是从arraylist中删除在该arraylist的迭代完成之后。据我所知 ConcurrentModificationException不应该来......并且方法似乎是正确的只要但有人可以确认是否存在任何机会或情况 ConcurrentModificationException可以用于我的代码因为我最终得到了在HashMap中被修改的列表。

    List<RemoveTest> eleList = new ArrayList<RemoveTest>();
    List<RemoveTest> eleList2 = new ArrayList<RemoveTest>();

    Map<String, List<RemoveTest>> testMap =new HashMap<String,List<RemoveTest();
    RemoveTest obj1 = new RemoveTest();
    obj1.setmId("m1");
    obj1.setRmId("");

    RemoveTest obj2 = new RemoveTest();
    obj2.setmId("m2");
    obj2.setRmId("m1");

    RemoveTest obj3 = new RemoveTest();
    obj3.setmId("m3");
    obj3.setRmId("");

    RemoveTest obj4 = new RemoveTest();
    obj4.setmId("m4");
    obj4.setRmId("m3");

    RemoveTest obj5 = new RemoveTest();
    obj5.setmId("m5");
    obj5.setRmId("m4");

    RemoveTest obj = new RemoveTest();
    obj.setmId("m");
    obj.setRmId("");

    eleList.add(obj1);
    eleList.add(obj2);
    eleList.add(obj3);
    eleList.add(obj4);
    eleList.add(obj5);
    eleList.add(obj);

    //new
    RemoveTest obj6 = new RemoveTest();
    obj6.setmId("m6");
    obj6.setRmId("");

    RemoveTest obj7 = new RemoveTest();
    obj7.setmId("m7");
    obj7.setRmId("m6");

    eleList2.add(obj6);
    eleList2.add(obj7);
    testMap.put("t1", eleList);
    testMap.put("21", eleList2);


            Set<RemoveTest> removals = new HashSet<RemoveTest>();
    Set<String> matchingFwdRevMIds = new HashSet<String>();
    for(Map.Entry<String, List<RemoveTest>> removeTestMap : testMap.entrySet()){
    List<RemoveTest> grp = removeTestMap.getValue();
    String key = removeTestMap.getKey();
    for (int i=0;i<grp.size();i++){
        RemoveTest fwdM = grp.get(i); 
        String fwdMId = fwdM.getmId();
        if(matchingFwdRevMIds.contains(fwdMId)){
            System.out.println("continue becoz "+ fwdMId );
            continue;
        }
        for (int j=i+1; j<grp.size();j++){
            RemoveTest nextM = grp.get(j);
            String mId = nextM.getmId();
            String revMId = nextM.getRmId();
            if(revMId!=null){
                if(matchingFwdRevMIds.contains(revMId)){
                    System.out.println("adding "+ mId);
                    matchingFwdRevMIds.add(mId);
                    removals.add(nextM);

                }
                    if(revMId.equals(fwdMId)){
                        matchingFwdRevMIds.add(fwdMId);
                        matchingFwdRevMIds.add(mId);
                        removals.add(fwdM);
                        removals.add(nextM);
                    }   
                }
            }
        }
    grp.removeAll(removals);
    } 

    for(Map.Entry<String, List<RemoveTest>> removeTestMap : testMap.entrySet()){
        List<RemoveTest> grp = removeTestMap.getValue();
        for(RemoveTest test : grp){
            System.out.println("final mid "+ test.getmId());
        }
    }

Answer 1

不，当您在HashMap上进行迭代时，修改在HashMap中用作值的对象是完全正常的。只有当您在迭代时向ConcurrentModificationException添加或删除条目时才会出现HashMap。

顺便说一句，像你这样改变价值是好的。但是永远不会修改HashMap条目中HashMap条目的键对象。（确切地说，永远不要修改它的hashCode被改变和/或它的equals方法产生不同的结果），否则你可能无法找到你的条目。

修改放在HashMap中的ArrayList的内容

1 个答案: