背景:
一个东西的执行有多个入参和出参, 一个东西的出参又可以是别的东西的入参, 因此执行的依赖关系.

草图里a b c d e f为三个东西, 上面的数字是入参,下面的数字是出参
当前已知这6个东西, 和他们的入参出参
求他们的运行顺序.
要求同样执行顺序的东西可以并行执行.

代码如下:

import org.apache.logging.log4j.LogManager;
import org.apache.logging.log4j.Logger;
import org.junit.Test;

import java.util.LinkedList;

public class TestAOV {
    private static final Logger LOGGER = LogManager.getLogger(TestAOV.class);

    @Test
    public void test() {
        LOGGER.info("zzzzzzzzzz");
        Entity a = new Entity();
        a.input = new int[]{1, 3, 5};
        a.output = new int[]{7, 9};
        a.name = "a";

        Entity b = new Entity();
        b.input = new int[]{7};
        b.output = new int[]{11};
        b.name = "b";

        Entity c = new Entity();
        c.input = new int[]{9, 10};
        c.output = new int[]{13, 18};
        c.name = "c";

        Entity d = new Entity();
        d.input = new int[]{13, 14};
        d.output = new int[]{19, 7};
        d.name = "d";

        Entity e = new Entity();
        e.input = new int[]{13, 15};
        e.output = new int[]{21};
        e.name = "e";

        Entity f = new Entity();
        f.input = new int[]{11, 19};
        f.output = new int[]{20};
        f.name = "f";

        //如果一个entity的输入不是任何entity的输出,第一顺位 放入大集合.  //不需要mid-in mid-out//代码里选外部参数,再参数池的顺序找值.
        //大集合的结果参数加输入参数够 .第二顺位, 放大集合
        //同理一直到所有entity都排序好

        LinkedList<Entity> list = new LinkedList<>();
        list.add(a);
        list.add(b);
        list.add(c);
        list.add(d);
        list.add(e);
        list.add(f);

        LinkedList<LinkedList> result = new LinkedList<>();

        LinkedList<Integer> necessaryInput = new LinkedList<>();
//        for (Entity entity: list){
//            necessaryInput =
//        }
//        所有的input的集合减去output的集合 //代码不写了直接赋值
        necessaryInput.add(1);
        necessaryInput.add(3);
        necessaryInput.add(5);
        necessaryInput.add(10);
        necessaryInput.add(18);
        necessaryInput.add(14);
        necessaryInput.add(15);

        LinkedList<Integer> poolParams = new LinkedList<>();

        for (Entity entityX : list) {

            LinkedList<Entity> entities = new LinkedList<>();

            for (Entity entityY : list) {
                //if (entityY.input 是 ( necessaryInput 并 poolParams )的子集 ){
                if (is子集(entityY.input, necessaryInput, poolParams) && !在结果里(entityY, result)) {
                    entities.add(entityY);
                }
            }
            //如果有,放入结果
            if (entities.size() != 0) {
                //entities 的output都放进poolParams;
                for (Entity es :
                        entities) {
                    poolParams.addAll(makeList(es.output));
                }
                result.add(entities);
            }
        }


        //LOGGER.info(result);
        for (LinkedList<Entity> aa : result
        ) {
            LOGGER.info("-------------------------");
            for (Entity zzz : aa) {
                LOGGER.info(zzz.name);
            }
        }
    }

    public static LinkedList<Integer> makeList(int[] a) {
        LinkedList<Integer> x = new LinkedList<>();
        for (int i = 0; i < a.length; i++) {
            x.add(a[i]);
        }
        return x;
    }

    // a是(b并c)的子集
    public static boolean is子集(int[] a, LinkedList<Integer> b, LinkedList<Integer> c) {
        LinkedList<Integer> d = new LinkedList<>();
        d.addAll(b);
        d.addAll(c);
        //TODO 去重
        for (int i = 0; i < a.length; i++) {
            if (d.contains(a[i])) {

            } else {
                return false;
            }
        }
        return true;
    }

    // a在 result里
    public static boolean 在结果里(Entity a, LinkedList<LinkedList> result) {

        for (LinkedList<Entity> aa : result
        ) {


            if (aa.contains(a))
                return true;

        }
        return false;
    }

}

class Entity {
    int[] input;
    int[] output;
    String name;
}

执行结果: