Question

让我们说：

a = ["s", "i", "n", "e", "d"];
b = ["s", "e", "n", "d"];

（a和b的类型为List<String>）

如何确定b中是否包含a中的所有字母？ - 不一定按顺序排列（在这种情况下，[s,e,n,d]位于a和b之内，它是真的）

使用a.containsAll(b)并不总是有效！

另一个例子：

a=["b", "a", "z", "z", "z"]
b=["a", "a", "b", "b"]

此处我希望结果为false，因为[a,a,b,b] a中的任何排序都不会显示a.containsAll(b)，但使用true会返回{{1}} ！

Answer 1

只需将每个的添加到新的String变量中即可通过使用 .contains()
找到一个String变量值是否包含另一个String

List<String> a = ["b","a","n"]; List<String> b = ["b","a","n","a","n","a"]; String newA = null; String newB = null; for(String strA : a) { newA += strA; } for(String strB : b) { newB += strB; } if(newA.contains(newB)) return True; else return False;

Reference for String .contains()

Answer 2

试试这个：

private boolean containsAll(List<?> a, List<?> b) {
    // List doesn't support remove(), use ArrayList instead
    ArrayList<Object> x = new ArrayList<Object>();
    ArrayList<Object> y = new ArrayList<Object>();

    x.addAll(a);
    y.addAll(b);
    for (Object o : y) {
        if (!x.remove(o)) // an element in B is not in A!
            return false;
    }
    return true;          // all elements in B are also in A
}

我们的目的是从b中移除a中的每个字母。当您尝试删除不在a中的字母时，它会确认a不包含b中的所有字母。

（如果元素存在，remove()将返回true，否则false）

Answer 3

从较大的列表中删除未出现在较小列表中并等于它们的所有元素。如果它们相等，则较小的列表包含在较大的列表中：

static List<String> list1 = new ArrayList<String>(){{
    add("b");
    add("a");
    add("n");
    add("z");
    add("z");
    add("z");
    }};
static List<String> list2 = new ArrayList<String>(){{
    add("b");
    add("a");
    add("n");
    add("a");
    add("n");
    add("a");
}};

public static void main(String[] args) {

    if(deepContains(list1, list2)) 
        System.out.println("List2 is contained in List1"); 
}

public static boolean deepContains(List<String> one, List<String> two){     
    if (one == null && two == null){
        return true;
    }

    if((one == null && two != null) 
      || one != null && two == null){
        return false;
    }

    //to avoid messing the order and elements of the lists we will use a copy
    one = new ArrayList<String>(one);
    two = new ArrayList<String>(two);
    //This removes from one all the elements not contained in two
    one.retainAll(two); 
    int a = one.size();
    int b = two.size();

    //one has lesser elements than two, for sure two is not contained in one
    if(a < b) return false;

    //one has the same number of elements of two, check if they are the same
    if(a == b){
        Collections.sort(one);
        Collections.sort(two);      
        return one.equals(two); 
    }

    //one has more elements than two. Remove duplicate elements
    //and check for equality
    Set<String> set1 = new HashSet<String>(one);
    Set<String> set2 = new HashSet<String>(two);

    if(set1.size() == set2.size()){
        one = new ArrayList<String>(set1);
        two = new ArrayList<String>(set2);
        Collections.sort(one);
        Collections.sort(two);      
        return one.equals(two); 
    }
    return false;
}

Answer 4

这是适用于任何类型的任何集合的版本：

private <E> boolean containsAllIncludingDuplicates(Collection<E> container,
        Collection<E> items) {

    Set<E> checkedItems = new HashSet<>();
    for (E item : items) {
        if (checkedItems.add(item)
                && Collections.frequency(container, item) < Collections
                        .frequency(items, item)) {
            return false;
        }
    }
    return true;
}

使用Set确保当项目中存在重复时，频次检查不会重复多次。

ArrayList containsAll返回错误的值

4 个答案: