Rosalind Java|Introduction to Set Operations

Rosalind编程问题之集合操作。
题不难，但是格式处理繁琐。

Problem：
If A and B are sets, then their union A∪B is the set comprising any elements in either A or B; their intersection A∩B is the set of elements in both A and B; and their set difference A−B is the set of elements in A but not in B.

Furthermore, if A is a subset of another set U, then the set complement of A with respect to U is defined as the set Ac=U−A. See the Sample sections below for examples.

Given: A positive integer n (n≤20,000) and two subsets A and B of {1,2,…,n}.
Sample input：

10
{1, 2, 3, 4, 5}
{2, 8, 5, 10}

Return: Six sets: A∪B, A∩B, A−B, B−A, Ac, and Bc (where set complements are taken with respect to {1,2,…,n}).
Sample output：

{1, 2, 3, 4, 5, 8, 10}
{2, 5}
{1, 3, 4}
{8, 10}
{8, 9, 10, 6, 7}
{1, 3, 4, 6, 7, 9}

集合的交集，并集和补集是高中第一节课就学习的知识。Java语言对于构建集合有两类体系：collection单列集合和map双列集合。二者区别是前者每个元素只包含一个值，后者每个元素包含一个键值对（两个值）。本文使用了Set集合解题，Set集合基本的特点是添加的元素无序的，不重复，无索引。

主要知识点： 集合交并补处理核心公式：set.contains。

代码如下：
注意！！本文的代码需要在输入数据时手动去掉首尾括号。输出时手动更换 [ ] 括号为 { }。 否则都会被系统判定为错误。

例如本题示例数据：

10
{1, 2, 3, 4, 5}
{2, 8, 5, 10}

要以这个形式逐行输入：

10
1, 2, 3, 4, 5
2, 8, 5, 10

得到的答案是下述结果，需要手动更换[ ] 括号为 { }。

[1, 2, 3, 4, 5, 8, 10]
[2, 5]
[1, 3, 4]
[8, 10]
[6, 7, 8, 9, 10]
[1, 3, 4, 6, 7, 9]

public class Introduction_to_Set_Operations {
    public static void main(String[] args) {
        //1.初始化三个集合
        Scanner sc = new Scanner(System.in);
        System.out.println("请输入集合总元素数n：");
        int n  = sc.nextInt();
        List total = new ArrayList<>();
        for (int i = 1; i <= n; i++) {
            total.add(i);
        }


        //构建n个元素组成的总集合
        Scanner OA = new Scanner(System.in);
        System.out.println("请输入集合A(去掉首位括号)：");
        String[] strnumA  = OA.nextLine().split(", ");
        Scanner OB = new Scanner(System.in);
        System.out.println("请输入集合B(去掉首位括号)：");
        String[] strnumB  = OB.nextLine().split(", ");


        int[] numA = new int[strnumA.length];
        for (int i = 0; i < strnumA.length; i++) {
            numA[i] = Integer.parseInt(strnumA[i]);
        }

        int[] numB = new int[strnumB.length];
        for (int i = 0; i < strnumB.length; i++) {
            numB[i] = Integer.parseInt(strnumB[i]);
        }


        //将题目给出的两个子集转化成Set集合
        Set setA = new HashSet<>();
        Set setB = new HashSet<>();

        for (int i : numA) {
            setA.add(i);
        }

        for (int i : numB) {
            setB.add(i);
        }

        System.out.println("————答案如下：————");
        //2.并集
        Set Union = new HashSet<>();
        Union.addAll(setA);
        Union.addAll(setB);
        System.out.println(Union);


        //3.交集
        Set intersection = new HashSet<>();
        for (Integer eleA : setA) {
            if (setB.contains(eleA)){//如果集合A和B共有这个元素，则为交集元素
                intersection.add(eleA);
            }
        }
        System.out.println(intersection);

        //4.相对补集A-B
        Set A_B = new HashSet<>();
        for (Integer eleA : setA) {
            if (!setB.contains(eleA)){//如果B不包含A的元素，则该元素是A独有
                A_B.add(eleA);
            }
        }
        System.out.println(A_B);

        //5.相对补集B-A
        Set B_A = new HashSet<>();
        for (Integer eleB : setB) {
            if (!setA.contains(eleB)){//如果B不包含A的元素，则该元素是A独有
                B_A.add(eleB);
            }
        }
        System.out.println(B_A);

        //6.绝对补集A和B
        Set complementaryA = new HashSet<>();
        Set complementaryB = new HashSet<>();
        for (Integer eleC : total) {
            if (!setA.contains(eleC)){
                complementaryA.add(eleC);
            }
            if (!setB.contains(eleC)){
                complementaryB.add(eleC);
            }
        }
        System.out.println(complementaryA);
        System.out.println(complementaryB);
    }
}

集合处理的过程和逻辑非常简单，但是生成网站规范的答案需要对结果进一步处理。本文主要关注集合处理过程，格式处理并未深入撰写。有大佬处理的话欢迎交流哈~