将一个数组列表的元素复制到另一个数组列表，使用Java消除第二个数组列表中的冗余

Question

我想创建一个代码，帮助我将元素从“allWords”ArrayList复制到“差异词”。但他在不同单词ArrayList中的时间我希望一个单词只出现一个，消除“差异词”ArrayList中的冗余。

这是我想出的代码，我没有得到任何输出。

导入java。util。ArrayList；

公共类复制元素{

 static ArrayList<String> distinctWords = new ArrayList<String> ();
 static ArrayList<String> allWords = new ArrayList<String> ();
 static int allWordsCount = 0;
 static int distinctWordsCount;
 static int tracker;


public static void main(String[] args) {


    allWords.add("you");
    allWords.add("want");
    allWords.add("to");
    allWords.add("go");
    allWords.add("to");
    allWords.add("dubai");
    allWords.add("and");
    allWords.add("you");
    allWords.add("also");
    allWords.add("want");
    allWords.add("to");
    allWords.add("go");
    allWords.add("to");
    allWords.add("seychelles");




    //printing all items in the 'allwords' arraylist
    //System.out.println("CONTENTS OF 'ALL WORDS' ARRAYLIST : ");

    distinctWords.add(0,allWords.get(0));
    distinctWordsCount = 1;

    int i = 1;
        for(int j = 1; j <= distinctWords.size(); j++){
            if(i < allWords.size()){
                tracker = 0;
                if (tracker == j){
                    distinctWords.add(j, allWords.get(i));
                    System.out.println("\t\t\t" + distinctWords.get(j));
                    distinctWordsCount ++;
                    i++;
                } else
                    if (tracker != j){
                        if(allWords.get(i) !=  distinctWords.get(tracker)){
                            //distinctWords.add(allWords.get(i));
                            //distinctWordsCount ++;
                            tracker++;
                        }
                    else
                        if(allWords.get(i) == distinctWords.get(tracker)){
                              i++; 
                               }
                //System.out.println("CONTENTS OF 'ALL WORDS' ARRAYLIST : ");
                //System.out.println("\t\t\t" + distinctWords.get(j));

          }
        }
        //System.out.println("\t\t\t" + distinctWords.get(j));
    }

    //System.out.println("\n\nTHE NUMBER OF ITEMS IN THE 'ALLWORDS' ARRAYLIST IS : " + allWords.size());
    //System.out.println("\n\nTHE NUMBER OF ITEMS IN THE 'DISTINCTWORDS' ARRAYLIST IS : " + allWords.size());
    System.out.println("\n\nITEMS IN THE 'DISTINCTWORDS' ARRAYLIST ARE : " + distinctWords.size());

}

}

Answer 1

我想这可以给你一套所有不同的单词。

Set<String> distinctWords = new HashSet<>(allWords);

如果你确实需要一个数组，你可以简单地这样做：

ArrayList<String> distinctWordsArr = new ArrayList<>(distinctWords);

Answer 2

//al1 and al2 are ArrayList<Object>
for(Object obj:al1){
    if(!al2.contains(obj))
        al2.add(obj);
}

Answer 3

以下代码如何：

for (int i = 0, size = allWords.size(); i < size; i++)
{
    String word = allWords.get(i);

    // add the word if it is not in distinctWords
    if (!distinctWords.contains(word))
    {
        distinctWords.add(word);
    }
 }

如果你不在乎单词的顺序，那么你可以按照建议使用一套。

Set<String> distinctSet = new HashSet<String>(allWords);

// if you want the set put into your other arrayList
distinctWords.addAll(distinctSet);