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);@Testpublic 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: