Question

我的allWordsList类似于：[aaa, bbb, ccc, ddd, eee]

如何在

中制作tempWordsList的副本（allWordsList）

for (String aWord : aWordsList)

没有迭代项目（即实现[bbb, ccc, ddd, eee]，然后[aaa, ccc, ddd, eee]等等）？

public class Anagrams {

    List<String> allWordsList = new ArrayList<String>();
    List<List<String>> myList = new ArrayList<List<String>>();
    List<String> tempWordsList = new ArrayList<String>();

    public Anagrams(String allWords) {
        getWordsList(allWordsList, allWords); // getting List to copy here
        getAnagramWordsList(allWordsList);
    }

    private void getAnagramWordsList(List<String> aWordsList) {
        for (String aWord : aWordsList){
            //tempWordsList.clear();
            tempWordsList = aWordsList;
            for (Iterator<String> iterator = tempWordsList.iterator(); iterator.hasNext();) {
                String string = iterator.next();
                if (string == aWord) {
                    // Remove the current element from the iterator and the list.
                    iterator.remove();
                }
            }
            myList.add(tempWordsList);
            System.out.println(aWordsList);
            System.out.println(tempWordsList); //before error both lists are without first item...
        }
    }

}

我经历了几次类似的案件，但仍然不太了解。

Answer 1

代码中最大的问题是tempWordsList和aWordsList引用同一个对象。您对tempWordsList所做的任何更改都会在同一时间aWordsList发生：

tempWordsList = aWordsList;

因此，myList将拥有aWordList的最后修改的多个副本：

myList.add(tempWordsList);

在循环的每次迭代中将相同的对象添加到myList。

要制作aWordsList和tempWordsList，您需要使用副本替换作业，如下所示：

tempWordsList = new List<String>(aWordsList);

Answer 2

应该是这样的：

Iterator<String> it = tempWordsList.iterator();
while(it.hasNext()){
    String value = it.next();
   // System.out.println("List Value:"+value);
   if (value.equals(aWord)) {
      it.remove();
        }
    }

Answer 3

解决问题的方法如下，但这不是一个好的解决方法。

//解决

private void getAnagramWordsList(List<String> aWordsList) {

    List<Integer> toRemove = new ArrayList<Integer>();

    for (String aWord : aWordsList){
        //tempWordsList.clear();
        tempWordsList = aWordsList;

        for (int i = tempWordsList.size()-1; i > 0; i--) {
            if (string.equals(tempWordsList.get(i)) {
                toRemove.add(i);
            }
        }
        for(int idx = 0; idx < toRemove.size(); idx++)
          tempWordsList.remove(idx);

        myList.add(tempWordsList);

        System.out.println(aWordsList);
        System.out.println(tempWordsList); //before error both lists are without first item...
    }

如何在不使用ConcurrentModificationException的情况下删除List迭代

3 个答案: