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

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());
}

其他回答

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());
}

除了劳伦斯的答案,如果你也想让它为零安全:

private static <T> boolean listEqualsIgnoreOrder(List<T> list1, List<T> list2) {
    if (list1 == null)
        return list2==null;
    if (list2 == null)
        return list1 == null;
    return new HashSet<>(list1).equals(new HashSet<>(list2));
}

下面是一种比较两个集合的方法,它考虑了其中的重复。因此,集合大小不一定相同。因此,它会在'actual'中寻找'expected':

    private static <T> boolean containsAllExpected(Collection<T> actual, Collection<T> expected) {
        if (actual == null && expected == null) {
            return true;
        }
        if (actual == null || expected == null) {
            return false;
        }
        Collection<T> a = new ArrayList<>(actual);
        Collection<T> e = new ArrayList<>(expected);

        Iterator<T> ei = e.iterator();
        while (ei.hasNext()) {
            T item = ei.next();
            if (a.contains(item)) {
                ei.remove();
                a.remove(item);
            } else {
                return false;
            }
        }

        return true;
    }

享受:)

示例代码:

public static '<'T'>' boolean isListDifferent(List'<'T'>' previousList,
        List'<'T'>' newList) {

    int sizePrevoisList = -1;
    int sizeNewList = -1;

    if (previousList != null && !previousList.isEmpty()) {
        sizePrevoisList = previousList.size();
    }
    if (newList != null && !newList.isEmpty()) {
        sizeNewList = newList.size();
    }

    if ((sizePrevoisList == -1) && (sizeNewList == -1)) {
        return false;
    }

    if (sizeNewList != sizePrevoisList) {
        return true;
    }

    List n_prevois = new ArrayList(previousList);
    List n_new = new ArrayList(newList);

    try {
        Collections.sort(n_prevois);
        Collections.sort(n_new);
    } catch (ClassCastException exp) {
        return true;
    }

    for (int i = 0; i < sizeNewList; i++) {
        Object obj_prevois = n_prevois.get(i);
        Object obj_new = n_new.get(i);
        if (obj_new.equals(obj_prevois)) {
            // Object are same
        } else {
            return true;
        }
    }

    return false;
}

!集合。disjoint(Collection1, Collection2)如果它们有相同的元素,则返回true