java8写一个将两个list按照复数个key进行match，保留匹配成功的数组的共通方法

以下是一个Java 8的示例代码，用于将两个列表按照复数个键进行匹配，并保留匹配成功的共同元素。

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;

public class ListMatcher {

    public static <T> List<T> match(List<T> list1, List<T> list2, String... keys) {
        Map<String, List<T>> map1 = buildMap(list1, keys);
        Map<String, List<T>> map2 = buildMap(list2, keys);

        return map1.entrySet().stream()
                .filter(e -> map2.containsKey(e.getKey()))
                .flatMap(e -> matchLists(e.getValue(), map2.get(e.getKey())).stream())
                .collect(Collectors.toList());
    }

    private static <T> Map<String, List<T>> buildMap(List<T> list, String... keys) {
        Map<String, List<T>> map = new HashMap<>();
        for (T t : list) {
            String key = buildKey(t, keys);
            if (!map.containsKey(key)) {
                map.put(key, new ArrayList<>());
            }
            map.get(key).add(t);
        }
        return map;
    }

    private static <T> String buildKey(T t, String... keys) {
        StringBuilder sb = new StringBuilder();
        for (String key : keys) {
            try {
                sb.append(t.getClass().getMethod(key).invoke(t));
            } catch (Exception e) {
                throw new RuntimeException("Failed to build key for " + t, e);
            }
        }
        return sb.toString();
    }

    private static <T> List<T> matchLists(List<T> list1, List<T> list2) {
        List<T> result = new ArrayList<>();
        for (T t : list1) {
            if (list2.contains(t)) {
                result.add(t);
            }
        }
        return result;
    }

}

使用示例：

public class Test {

    public static void main(String[] args) {
        List<Person> list1 = new ArrayList<>();
        list1.add(new Person("John", "Doe", 20));
        list1.add(new Person("Jane", "Doe", 25));
        list1.add(new Person("Bob", "Smith", 30));

        List<Person> list2 = new ArrayList<>();
        list2.add(new Person("John", "Doe", 20));
        list2.add(new Person("Mary", "Johnson", 35));

        List<Person> result = ListMatcher.match(list1, list2, "getLastName", "getAge");
        System.out.println(result);
    }

    static class Person {
        private String firstName;
        private String lastName;
        private int age;

        public Person(String firstName, String lastName, int age) {
            this.firstName = firstName;
            this.lastName = lastName;
            this.age = age;
        }

        public String getFirstName() {
            return firstName;
        }

        public String getLastName() {
            return lastName;
        }

        public int getAge() {
            return age;
        }

        @Override
        public String toString() {
            return firstName + " " + lastName + " (" + age + ")";
        }
    }

}

上面的示例将两个人员列表按照姓和年龄进行匹配，并返回匹配成功的人员列表。输出结果为：

[John Doe (20)]