在标准Java库中,找出两个list是否包含完全相同的元素的最简单方法是什么?

这两个list是否为相同实例并不重要,这两个list的类型参数是否不同也不重要。

e.g.

List list1
List<String> list2; 
// ... construct etc

list1.add("A");
list2.add("A"); 
// the function, given these two lists, should return true

我知道可能有什么东西在盯着我的脸:-)


编辑:为了澄清,我正在寻找完全相同的元素和元素的数量,按顺序。


当前回答

我的解决方案适用于不关心列表中的顺序的情况——换句话说:具有相同元素但顺序不同的列表将被认为具有相同的内容。

示例:["word1", "word2"]和["word2", "word1"]被认为内容相同。

我已经谈到了订购,我还需要说一些关于副本的事情。列表需要具有相同数量的元素才能被认为是相等的。

例如:["word1"]和["word1", "word1"]被认为不具有相同的内容。

我的解决方案:

public class ListUtil {

    public static <T> boolean hasSameContents(List<T> firstList, List<T> secondList) {      
        if (firstList == secondList) { // same object
            return true;
        }
        if (firstList != null && secondList != null) {
            if (firstList.isEmpty() && secondList.isEmpty()) {
                return true;
            }
            if (firstList.size() != secondList.size()) {
                return false;
            }
            List<T> tmpSecondList = new ArrayList<>(secondList);
            Object currFirstObject = null;
            for (int i=1 ; i<=firstList.size() ; i++) {
                currFirstObject = firstList.get(i-1);
                boolean removed = tmpSecondList.remove(currFirstObject);
                if (!removed) {
                    return false;
                }
                if (i != firstList.size()) { // Not the last element
                    if (tmpSecondList.isEmpty()) {
                        return false;
                    }
                }
            }
            if (tmpSecondList.isEmpty()) {
                return true;
            }
        }
        return false;
    }
}

我用Strings进行了测试,如下所示:

@Test public void testHasSameContents() throws Exception { // comparing with same list => no duplicate elements Assert.isTrue(ListUtil.hasSameContents(List.of("one", "two", "three"), List.of("one", "two", "three"))); // comparing with same list => duplicate elements Assert.isTrue(ListUtil.hasSameContents(List.of("one", "two", "three", "one"), List.of("one", "two", "three", "one"))); // compare with disordered list => no duplicate elements Assert.isTrue(ListUtil.hasSameContents(List.of("one", "two", "three"), List.of("three", "two", "one"))); // compare with disordered list => duplicate elements Assert.isTrue(ListUtil.hasSameContents(List.of("one", "two", "three", "one"), List.of("three", "two", "one", "one"))); // comparing with different list => same size, no duplicate elements Assert.isFalse(ListUtil.hasSameContents(List.of("one", "two", "three"), List.of("four", "five", "six"))); // comparing with different list => same size, duplicate elements Assert.isFalse(ListUtil.hasSameContents(List.of("one", "two", "two"), List.of("one", "two", "three"))); Assert.isFalse(ListUtil.hasSameContents(List.of("one", "two", "three"), List.of("one", "two", "two"))); // comparing with different list => different size, no duplicate elements Assert.isFalse(ListUtil.hasSameContents(List.of("one", "two", "three", "four"), List.of("one", "two", "three"))); Assert.isFalse(ListUtil.hasSameContents(List.of("one", "two", "three"), List.of("one", "two", "three", "four"))); // comparing with different list => different sizes, duplicate elements Assert.isFalse(ListUtil.hasSameContents(List.of("one", "two", "three", "one"), List.of("one", "two", "three"))); Assert.isFalse(ListUtil.hasSameContents(List.of("one", "two", "three"), List.of("one", "two", "three", "one"))); }

其他回答

我的解决方案适用于不关心列表中的顺序的情况——换句话说:具有相同元素但顺序不同的列表将被认为具有相同的内容。

示例:["word1", "word2"]和["word2", "word1"]被认为内容相同。

我已经谈到了订购,我还需要说一些关于副本的事情。列表需要具有相同数量的元素才能被认为是相等的。

例如:["word1"]和["word1", "word1"]被认为不具有相同的内容。

我的解决方案:

public class ListUtil {

    public static <T> boolean hasSameContents(List<T> firstList, List<T> secondList) {      
        if (firstList == secondList) { // same object
            return true;
        }
        if (firstList != null && secondList != null) {
            if (firstList.isEmpty() && secondList.isEmpty()) {
                return true;
            }
            if (firstList.size() != secondList.size()) {
                return false;
            }
            List<T> tmpSecondList = new ArrayList<>(secondList);
            Object currFirstObject = null;
            for (int i=1 ; i<=firstList.size() ; i++) {
                currFirstObject = firstList.get(i-1);
                boolean removed = tmpSecondList.remove(currFirstObject);
                if (!removed) {
                    return false;
                }
                if (i != firstList.size()) { // Not the last element
                    if (tmpSecondList.isEmpty()) {
                        return false;
                    }
                }
            }
            if (tmpSecondList.isEmpty()) {
                return true;
            }
        }
        return false;
    }
}

我用Strings进行了测试,如下所示:

@Test public void testHasSameContents() throws Exception { // comparing with same list => no duplicate elements Assert.isTrue(ListUtil.hasSameContents(List.of("one", "two", "three"), List.of("one", "two", "three"))); // comparing with same list => duplicate elements Assert.isTrue(ListUtil.hasSameContents(List.of("one", "two", "three", "one"), List.of("one", "two", "three", "one"))); // compare with disordered list => no duplicate elements Assert.isTrue(ListUtil.hasSameContents(List.of("one", "two", "three"), List.of("three", "two", "one"))); // compare with disordered list => duplicate elements Assert.isTrue(ListUtil.hasSameContents(List.of("one", "two", "three", "one"), List.of("three", "two", "one", "one"))); // comparing with different list => same size, no duplicate elements Assert.isFalse(ListUtil.hasSameContents(List.of("one", "two", "three"), List.of("four", "five", "six"))); // comparing with different list => same size, duplicate elements Assert.isFalse(ListUtil.hasSameContents(List.of("one", "two", "two"), List.of("one", "two", "three"))); Assert.isFalse(ListUtil.hasSameContents(List.of("one", "two", "three"), List.of("one", "two", "two"))); // comparing with different list => different size, no duplicate elements Assert.isFalse(ListUtil.hasSameContents(List.of("one", "two", "three", "four"), List.of("one", "two", "three"))); Assert.isFalse(ListUtil.hasSameContents(List.of("one", "two", "three"), List.of("one", "two", "three", "four"))); // comparing with different list => different sizes, duplicate elements Assert.isFalse(ListUtil.hasSameContents(List.of("one", "two", "three", "one"), List.of("one", "two", "three"))); Assert.isFalse(ListUtil.hasSameContents(List.of("one", "two", "three"), List.of("one", "two", "three", "one"))); }

private boolean listHaveEqualObjects(List<?> list1, List<?> list2){
    return list1.containsAll(list2) && list2.containsAll(list1);

很晚才加入这个派对,但想添加这个空安全检查:

Objects.equals(list1, list2)

如果您正在使用(或乐于使用)Apache Commons Collections,您可以使用CollectionUtils。isEqualCollection“如果给定集合包含完全相同的元素和完全相同的基数,则返回true。”

当两个列表具有相同的元素,但顺序不同时的解决方案:

public boolean isDifferentLists(List<Integer> listOne, List<Integer> listTwo) {
    if(isNullLists(listOne, listTwo)) {
        return false;
    }

    if (hasDifferentSize(listOne, listTwo)) {
        return true;
    }

    List<Integer> listOneCopy = Lists.newArrayList(listOne);
    List<Integer> listTwoCopy = Lists.newArrayList(listTwo);
    listOneCopy.removeAll(listTwoCopy);

    return CollectionUtils.isNotEmpty(listOneCopy);
}

private boolean isNullLists(List<Integer> listOne, List<Integer> listTwo) {
    return listOne == null && listTwo == null;
}

private boolean hasDifferentSize(List<Integer> listOne, List<Integer> listTwo) {
    return (listOne == null && listTwo != null) || (listOne != null && listTwo == null) || (listOne.size() != listTwo.size());
}