Java8根据ID移除List元素并收集被移除元素到另一列表

需求梳理 #

• 输入为两个存储Employee类型对象的列表list1、list2
• 匹配规则：以Employee对象的id属性为匹配依据，匹配时忽略大小写
• 处理规则：从list1中移除所有id未在list2中存在的元素，所有被移除的元素存入新列表list3

原有代码问题 #

你写的实现逻辑存在3个核心错误，所以结果不符合预期：

• 遍历方向错误：以list2为遍历基准，会把list2独有的id纳入无效判断，逻辑完全颠倒
• 判断逻辑错误：每次遍历list2元素时，只要找到任意一个id和当前list2元素id不等的list1元素就判定为待移除，判断条件完全不成立
• 性能差且有风险：遍历过程中反复全表扫描list1、多次调用removeIf修改列表，时间复杂度高，还可能触发并发修改异常

正确实现思路 #

1. 先提取list2中所有员工的id，统一转为小写后存入HashSet，后续id匹配的时间复杂度可以从O(n)降到O(1)
2. 直接调用list1的removeIf方法做批量处理：对每个list1元素判断id是否在提前准备的id集合中，不在的话就先把元素加入list3，再返回true将该元素从list1中移除
3. 提前做空判断，避免空指针异常，不需要额外引入第三方工具类

完整可运行代码 #

import java.util.*;
import java.util.stream.Collectors;
import java.util.stream.Stream;

// Employee实体类，没有用Lombok的话手动补全getter、全参构造、toString方法即可
class Employee {
    private String id;
    private String city;
    private String state;

    public Employee(String id, String city, String state) {
        this.id = id;
        this.city = city;
        this.state = state;
    }

    public String getId() {
        return id;
    }

    @Override
    public String toString() {
        return "Employee{" +
                "id='" + id + '\'' +
                ", city='" + city + '\'' +
                ", state='" + state + '\'' +
                '}';
    }
}

public class EmployeeListProcess {
    public static void main(String[] args) {
        // 示例输入list1
        List<Employee> list1 = Stream.of(
                        new Employee("100","Boston","Massachusetts"),
                        new Employee("400","Atlanta","Georgia"),
                        new Employee("300","pleasanton","California"),
                        new Employee("200","Decatur","Texas"),
                        new Employee("500","Cumming","Atlanta"),
                        new Employee("98","sula","Maine"),
                        new Employee("156","Duluth","Ohio"))
                .collect(Collectors.toList());
        // 匹配基准list2
        List<Employee> list2 = Stream.of(
                        new Employee("100","Boston","Massachusetts"),
                        new Employee("800","pleasanton","California"),
                        new Employee("400","Atlanta","Georgia"),
                        new Employee("10","Decatur","Texas"),
                        new Employee("500","Cumming","Atlanta"),
                        new Employee("50","sula","Maine"),
                        new Employee("156","Duluth","Ohio"))
                .collect(Collectors.toList());

        // 核心处理逻辑
        List<Employee> list3 = new ArrayList<>();
        if (list1 != null && !list1.isEmpty() && list2 != null && !list2.isEmpty()) {
            // 预收集list2的所有id，统一转小写实现忽略大小写匹配
            Set<String> existIds = list2.stream()
                    .map(emp -> emp.getId().toLowerCase(Locale.ROOT))
                    .collect(Collectors.toSet());
            // 批量移除不匹配元素，移除前同步收集到list3
            list1.removeIf(emp -> {
                boolean needRemove = !existIds.contains(emp.getId().toLowerCase(Locale.ROOT));
                if (needRemove) {
                    list3.add(emp);
                }
                return needRemove;
            });
        }

        // 结果输出验证
        System.out.println("处理后list1：");
        list1.forEach(System.out::println);
        System.out.println("\n被移除的元素list3：");
        list3.forEach(System.out::println);
    }
}

运行结果说明 #

代码运行后完全符合预期输出：

• 处理后的list1保留id为100、400、500、156的4条员工数据
• list3存储被移除的id为300、200、98的3条员工数据

内容的提问来源于stack exchange，提问作者Musa