求二维数组的并、交和差

Question

我正在尝试创建一个程序，给出由用户输入创建的两个数组的联合、交叉和差异。因此，如果用户希望第一个数组的大小为4，元素为1 2 3 4，第二数组的大小为5，而元素为3 4 5 6 7，则输出应返回；集A与集B的并为:1 2 3 4 5 6 7，集A与集B的交集为:3 4集合A与集B的差为:1.现在求交，但差不起作用。差异作为第一个数组的所有元素打印出来。我不知道从哪里开始工会。这是我的代码：

package rhc91310a13sets;

import java.util.*;

public class rhc91310a13sets {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int[] setA = null;
        int[] setB = null;

        System.out.print("Enter size for set A (must be less than or equal to 20): ");
        setA = new int[sc.nextInt()];

        System.out.print("Enter elements for set A between 1-20: ");
        for (int i = 0; i < setA.length; i++) {
            setA[i] = sc.nextInt();
        }

        System.out.print('\n');

        System.out.print("Enter size for set B (must be less than or equal to 20): ");
        setB = new int[sc.nextInt()];

        System.out.print("Enter elements for set B between 1-20: ");
        for (int i = 0; i < setB.length; i++) {
            setB[i] = sc.nextInt();
        }

        System.out.print('\n');

        System.out.print("The union of sets A and B are: ");
        for(int i = 0; i < setA.length; i++) {
            for(int j = 0; j < setB.length; j++) {
                }
            }       

        System.out.print('\n');

        System.out.print("The intersection of sets A and B are: ");
        for (int i = 0; i < setA.length; i++) {
            for (int j = 0; j < setB.length; j++) {
                if (setA[i] == setB[j]) {
                    System.out.print(setA[i] + " ");
                }
            }
        }

        System.out.print('\n');

        System.out.print("The difference of sets A and B are: ");
        for (int i = 0; i < setA.length; i++) {
            for (int j = 0; j < setB.length; j++) {
                if (!(setA[i] == setB[j])) 
                    System.out.print(setA[i] + " ");        
            }
        }
    }
  }

Answer 1

你意想不到的结果

The difference of sets A and B are: 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4

您的逻辑在problems.you和set B之间有一些检查差异，从setA中提取一个元素，并检查它是否不同于setB的元素，但是problem.for示例从setA获取1并检查它是否不同于3 ,4, 5, 6 ,7 .so --它不同于所有5个元素，因此，1将打印5 times.the，最坏的问题发生在您从setA获取3并检查setB.3与所有元素不同时，预期setB.so中的第一个元素，3会打印4 times .but等，这给了setA打印less than 5 times的一个概念元素，这意味着它不是不同的元素。您所要做的是检查是否所有5个元素都不同，而不是单一的，我将计数变量和检查与setB.but的5倍不同--我认为简单的方法是使用HashSet

int[] setA = {1,2,3,4};
int[] setB = {3 ,4, 5, 6 ,7};
int count=0;
System.out.print("The difference of sets A and B are: \n");
for (int i = 0; i < setA.length; i++) {
    count=0;
    for (int j = 0; j < setB.length ; j++) {
        if ((setA[i] != setB[j])){ 
            count++;

        } 
        if(count==setB.length){
            System.out.println(setA[i]);
        }
    }
}

output>>

The difference of sets A and B are: 
1
2

***update***

如果您想在数组中放置这2的不同元素，最好的方法是arraylist，.you可以这样做

int[] setA = {1, 2, 3, 4};
int[] setB = {3, 4, 5, 6, 7};
int count = 0;
ArrayList<Integer> arl = new ArrayList<Integer>();

System.out.print("The difference of sets A and B are: \n");
for (int i = 0; i < setA.length; i++) {
    count = 0;
    for (int j = 0; j < setB.length; j++) {
        if ((setA[i] != setB[j])) {
            count++;

        }
        if (count == setB.length) {
           // System.out.println(setA[i]);
            arl.add(i);
        }
    }
}
System.out.println(arl);

output>>

The difference of sets A and B are: 
1
2
[0, 1]

查找联合values.you可以使用HashSet。

ArrayList list = new ArrayList();
for(int i=0;i<setA.length;i++){
    list.add(setA[i]);
}
for(int i=0;i<setB.length;i++){
    list.add(setB[i]);
}
HashSet h = new HashSet();
h.addAll(list);
list.clear();
list .addAll(h);
System.out.println(list);