两个集合的并集，操作数组 Java

Question

思路是计算两个集合集合的并集，程序开始时会要求用户输入想要集合的长度，然后提示他分配集合中的元素，最后的思路是计算Union。我已经到了最后，编译我的程序后只打印了第一组的元素，我真的不知道为什么。

所以我的问题是如何计算集合并集，按照我开始的想法。

我的输出：

Desired array lengths: 3 3 First set elements: 1 2 3 Second set elements: 4 5 6 UNION of our sets: 1.002.003

public class Union{

public static void main(String[] args) {
    System.out.println("Desired array lengths: ");
    Scanner scan = new Scanner(System.in);

    //Infinite loop for reading input,if negative number entered break the loop!

    while (true) {
        int n1 = scan.nextInt();
        int n2 = scan.nextInt();
        if (n1 < 0 || n2 < 0)
            break;

        // Assigning elements to the first set.
        double[] s1 = new double[n1];
        System.out.println("First set elements: ");
        //We enter elements as long as the number of the elements added is less than the length of an array we assigned.
        for (int i = 0; i < n1; s1[i++] = scan.nextInt());

            if (n1 == 0)
            System.out.println();//If we do not enter any element go to new line

        //Assigning elements to the second set.
        double[] s2 = new double[n2];
        System.out.println("Second set elements: ");
        for (int i = 0; i < n2; s2[i++] = scan.nextInt());

        if (n2 == 0)
            System.out.println();//Same as before.

        // Calculating union

        double[] s3 = new double[n1 + n2];//We reserve memory space for the s3 array with the size equal to both n1 and n2 arrays.
        int n3 = 0; // Variable used to save number of elements,after reaching the end of the loop n3 WILL HAVE THE SIZE OF N1.
        while (n3 < n1) 
            s3[n3] = s1[n3++];
        for (int j = 0; j < n2; j++) {  //HERE WE ARE CHECKING IF THE ELEMENTS FROM N2 SET ARE PRESENT IN THE N1 
            int i = 0;
            while (i < n1 && s2[j] == s1[i])
                i++;
            if (i == n1)
                s3[n3++] = s2[j];
        }

        double[] pom = new double[n3];
        for (int i = 0; i < n3; pom[i] = s3[i++]);

        s3 = pom;
        pom = null;

        System.out.println("UNION of our sets: ");
        for (int i = 0; i < n3; System.out.printf("%.2f", s3[i++]))
            ;
        System.out.print("\n\n");
    }

}

Answer 1

错误在于你检查sets2的哪些元素的代码，你需要输入sets3.

基本上你需要检查 set s2 的元素是否存在于 set s1.

所以更改以下代码：

for (int j = 0; j < n2; j++) {  
       int i = 0;
       while (i < n1 && s2[j] == s1[i])
         i++;
       if (i == n1)
         s3[n3++] = s2[j];
}

此代码：

for (int j = 0; j < n2; j++) {  
       int i = 0;
       while (i < n1 && s2[j] != s1[i])
         i++;
       if (i == n1)
         s3[n3++] = s2[j];
}

只有当 set s1 中的元素 none 与元素 [=24= 匹配时，循环 while (i < n1 && s2[j] != s1[i]) 才会以 i = n1 结束] 这就是我们想要在 2 组的 UNION 中的元素。

Answer 2

你可以使用 Java Collections Framework? The Set interface 可以很容易地做到这一点。

Set<Double> union = new HashSet<>();
for (double e : s1)
    union.add(e);
for (double e : s2)
    union.add(e);
System.out.println(union);

如果您不能使用集合框架而必须自己滚动，那么您可以考虑滚动自己的二叉树或堆。

Answer 3

这是因为以下块：

while (i < n1 && s2[j] == s1[i])
    i++;
if (i == n1)

它正在尝试比较具有相似索引的两个数组的元素，例如如果 s1 和 s2 的第一个元素不相等，控制将跳出 while 循环，因此，i 永远不会是 n1，导致元素在 s2 的索引 j 处被跳过。

关于并集的计算，用Java中的Set就可以轻松完成，例如：

Set<Double> elements = new LinkedHashSet<>();
for(double number : s1){
    elements.add(number);
}
for(double number : s2){
    elements.add(number);
}
double[] union = new double[elements.size()];
int i = 0;
Iterator<Double> iterator = elements.iterator();
while(iterator.hasNext()){
    union[i++] = iterator.next();
}

两个集合的并集，操作数组 Java

Union of two sets,manipulating arrays Java

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