java - 查找2数组的不同元素

Question

我尝试打印 2 个数组 A 和 B 之间的集合的非交集。但是，我遇到了如何打印 A 不同 B 上的元素的问题。这是我的示例代码:

public class Array {

    public static void main(String[] args) {
        for (int i = 0; i <= arrA.length - 1; i++) {
            arrA[i] = sc.nextInt();
        }
        for (int i = 0; i <= arrB.length - 1; i++) {
            arrB[i] = sc.nextInt();
        }

        boolean x = true;
        int y = 0;
        for (int i = 0; i < arrA.length; i++) {
            for (int j = 0; j < arrB.length; j++) {
                if (arrA[i] == arrB[j]) {
                    arrTestA[i] = true;
                }else y = arrA[i];
            }
        }

        for (int i = 0; i < arrA.length; i++) {
            x = x && arrTestA[i];
        }

        if (x) {
            System.out.println("All the elements of A contained in B.");
        }else {
            System.out.println("There are elements on A different B.");
            System.out.println("The elements of A which is not in B = "); //My Problem
        }

    }
}

Answer 1

要实现这一点，您可以使用 Collections 和 retainAll 方法。例如:

List<Integer> arrTestA = new ArrayList<>();
List<Integer> arrTestB = new ArrayList<>();

[...]

List<Integer> common = new ArrayList<>(arrTestA);
common.retainAll(arrTestB);

List<Integer> diff = new ArrayList<>();
for(Integer element : arrTestA) 
    if(!common.contains(element))
        diff.add(element);

[here you print out elements of diff as The elements of A which is not in B]

预计到达时间:非收集尝试:

int[] arr1 = { 1, 11, 5, 9, 4, 3, 4, 8 };
int[] arr2 = { 1, 7, 5, 3, 4, 8 };

Arrays.sort(arr1);
Arrays.sort(arr2);

for (int i : arr1) {
    boolean contains = false;
    for (int j : arr2) {
        if (i == j) {
            contains = true;
            break;
        }
    }
    if (!contains)
        System.out.println("Arr2 doesn't contain number: " + i);
}

...或者循环看起来像这样:

outer: for (int i : arr1) {
    for (int j : arr2) {
        if (i == j) {
            continue outer;
        }
    }
    System.out.println("Arr2 doesn't contain number: " + i);
}

这只是一种方法，但希望您明白这一点。

预计到达时间 2:在我的做法中，其实这些Array是不需要排序的。您可以简单地删除负责排序的代码行。