如何在 java 中生成给定字符串列表的所有可能排序排列

Question

流程、资源、请求和发布列表：

A req R
A req S
A rel R
A rel S
B req S
B req R
B rel S
B rel R

无法重新安排流程中的顺序（在上面的示例中，A 必须始终在请求 S 之前先请求 R），但是可以重新安排跨流程的顺序（B 可以在 A 请求 R 之前请求 S 和 R）

package Test;

import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;
import java.util.Stack;

public class SortStr {
    public static void main(String[] args) {

        Set<String> s = new HashSet<String>();

        List<String> A = List.of("A req R", "A req S", "A rel R", "A rel S");
        List<String> B = List.of("B req S", "B req R", "B rel S", "B rel R");
        s.addAll(A);
        s.addAll(B);

        /* List<String> A = List.of("A");
        List<String> B = List.of("1");
        List<String> C = List.of("X");

        s.addAll(A);
        s.addAll(B);
        s.addAll(C); */

        Map<Integer, List<String>> result = new HashMap<>();
        permutations(s, new Stack<String>(), s.size(), result);

        result.forEach((key, value) -> 
            System.out.println("ORDER " + key + " : " + value)
        );
    }

    public static void permutations(Set<String> items, Stack<String> permutation, int size,
            Map<Integer, List<String>> result) {

        /* permutation stack has become equal to size that we require */
        if (permutation.size() == size) {
            /* print the permutation */
            // System.out.println(Arrays.toString(permutation.toArray(new String[0])));

            /* add the permutation in the result */
            Integer key = (result.size() > 0) ? result.size() + 1 : 0;
            result.put(key, Arrays.asList(permutation.toArray(new String[0])));
        }

        /* items available for permutation */
        String[] availableItems = items.toArray(new String[0]);
        for (String i : availableItems) {
            /* add current item */
            permutation.push(i);

            /* remove item from available item set */
            items.remove(i);

            /* pass it on for next permutation */
            permutations(items, permutation, size, result);

            /* pop and put the removed item back */
            items.add(permutation.pop());
        }
    }
}

以上程序生成所有可能的排序排列。但我的条件标准是

 a process can't be rearranged (A must always request R before it can request S in the example above), however orderings across processes can be rearranged (B can request S and R before A requests R)

那么，我怎样才能对所有可能的资源请求/释放顺序进行排序？

输出我想要的结果：-

ORDER 1: A req R, A req S, A rel R, A rel S, B req S, B req R, B rel S, B rel R
ORDER 2: A req R, A req S, A rel R, B req S, A rel S, B req R, B rel S, B rel R
...
ORDER X: A req R, B req S, A req S, B req R ...

---

测试用例输入：-

案例一：-

A req R A req S A rel R A rel S B req S B req T B rel S B rel T C req T 
C req R C rel T C rel R

案例2：-

A req R A req S A rel R A rel S B req T B rel T C req S C rel S D req U 
D req S D req T D rel U D rel S D rel T E req T E req V E rel T E rel V 
F req W F req S F rel W F rel S G req V G req U G rel V G rel U

我试了很多但我做不到 所以，请指导我实现这种排列生成。 谢谢你。