Question

我有2个字符串对象的arraylists。

List<String> sourceList = new ArrayList<String>();
List<String> destinationList = new ArrayList<String>();

我有一些逻辑，我需要处理源列表，并最终得到目标列表。目标列表将添加一些其他元素添加到源列表或从源列表中删除。

我的预期输出是2个字符串ArrayList，其中第一个列表应该从源中删除所有字符串，第二个列表应该包含新添加到源的所有字符串。

任何更简单的方法来实现这一目标？

Answer 1

将列表转换为Collection并使用removeAll

    Collection<String> listOne = new ArrayList(Arrays.asList("a","b", "c", "d", "e", "f", "g"));
    Collection<String> listTwo = new ArrayList(Arrays.asList("a","b",  "d", "e", "f", "gg", "h"));


    List<String> sourceList = new ArrayList<String>(listOne);
    List<String> destinationList = new ArrayList<String>(listTwo);


    sourceList.removeAll( listTwo );
    destinationList.removeAll( listOne );



    System.out.println( sourceList );
    System.out.println( destinationList );

输出：

[c, g]
[gg, h]

<强> [编辑]

其他方式（更清楚）

  Collection<String> list = new ArrayList(Arrays.asList("a","b", "c", "d", "e", "f", "g"));

    List<String> sourceList = new ArrayList<String>(list);
    List<String> destinationList = new ArrayList<String>(list);

    list.add("boo");
    list.remove("b");

    sourceList.removeAll( list );
    list.removeAll( destinationList );


    System.out.println( sourceList );
    System.out.println( list );

输出：

[b]
[boo]

Answer 2

这应检查两个列表是否相等，它首先进行一些基本检查（即空值和长度），然后排序并使用collections.equals方法检查它们是否相等。

public  boolean equalLists(List<String> a, List<String> b){     
    // Check for sizes and nulls

    if (a == null && b == null) return true;


    if ((a == null && b!= null) || (a != null && b== null) || (a.size() != b.size()))
    {
        return false;
    }

    // Sort and compare the two lists          
    Collections.sort(a);
    Collections.sort(b);      
    return a.equals(b);
}

Answer 3

将List转换为String并检查字符串是否相同

import java.util.ArrayList;
import java.util.List;



/**
 * @author Rakesh KR
 *
 */
public class ListCompare {

    public static boolean compareList(List ls1,List ls2){
        return ls1.toString().contentEquals(ls2.toString())?true:false;
    }
    public static void main(String[] args) {

        ArrayList<String> one  = new ArrayList<String>();
        ArrayList<String> two  = new ArrayList<String>();

        one.add("one");
        one.add("two");
        one.add("six");

        two.add("one");
        two.add("two");
        two.add("six");

        System.out.println("Output1 :: "+compareList(one,two));

        two.add("ten");

        System.out.println("Output2 :: "+compareList(one,two));
    }
}

Answer 4

答案在@dku-rajkumar帖子中给出。

ss

Answer 5

最简单的方法是逐个遍历源列表和目标列表像这样：

List<String> newAddedElementsList = new ArrayList<String>();
List<String> removedElementsList = new ArrayList<String>();
for(String ele : sourceList){
    if(destinationList.contains(ele)){
        continue;
    }else{
        removedElementsList.add(ele);
    }
}
for(String ele : destinationList){
    if(sourceList.contains(ele)){
        continue;
    }else{
        newAddedElementsList.add(ele);
    }
}

如果您的源列表和目标列表包含许多元素，但它确实更简单，它可能效率不高。

Answer 6

如果您的要求是维持插入顺序并检查两个arraylist的内容，那么您应该执行以下操作：

List<String> listOne = new ArrayList<String>();
List<String> listTwo = new ArrayList<String>();

listOne.add("stack");
listOne.add("overflow");

listTwo.add("stack");
listTwo.add("overflow");

boolean result = Arrays.equals(listOne.toArray(),listTwo.toArray());

这将返回true。

但是，如果您更改顺序，例如：

listOne.add("stack");
listOne.add("overflow");

listTwo.add("overflow");
listTwo.add("stack");

boolean result = Arrays.equals(listOne.toArray(),listTwo.toArray());

将返回false，因为顺序不同。

Answer 7

据我所知，我认为使用4个列表最简单： - 你的sourceList - 你的目的地名单 - removedItemsList --NEnewAddedItemsList

Answer 8

private int compareLists(List<String> list1, List<String> list2){
    Collections.sort(list1);
    Collections.sort(list2);

    int maxIteration = 0;
    if(list1.size() == list2.size() || list1.size() < list2.size()){
        maxIteration = list1.size();
    } else {
        maxIteration = list2.size();
    }

    for (int index = 0; index < maxIteration; index++) {
        int result = list1.get(index).compareTo(list2.get(index));
        if (result == 0) {
            continue;
        } else {
            return result;
        }
    }
    return list1.size() - list2.size();
}

Answer 9

boolean isEquals(List<String> firstList, List<String> secondList){
    ArrayList<String> commons = new ArrayList<>();

    for (String s2 : secondList) {
        for (String s1 : firstList) {
           if(s2.contains(s1)){
               commons.add(s2);
           }
        }
    }

    firstList.removeAll(commons);
    secondList.removeAll(commons);
    return !(firstList.size() > 0 || secondList.size() > 0) ;
}

Answer 10

List<String> oldList = Arrays.asList("a", "b", "c", "d", "e", "f");
List<String> modifiedList = Arrays.asList("a", "b", "c", "d", "e", "g");

List<String> added = new HashSet<>(modifiedList);
List<String> removed = new HashSet<>(oldList);

modifiedList.stream().filter(removed::remove).forEach(added::remove);

// added items
System.out.println(added);
// removed items
System.out.println(removed);

比较2个ArrayLists的简单方法

10 个答案: