Java集合合并案例如何实操

wen java案例 2026-07-02 180

本文目录导读：

Java集合合并案例如何实操

基础集合合并案例
实际业务场景案例
高级合并技巧
最佳实践建议

我来通过几个实际案例,手把手教你Java集合合并的操作。

基础集合合并案例

案例1：两个List合并（去重）

import java.util.*;
import java.util.stream.Collectors;
public class ListMergeExample {
    public static void main(String[] args) {
        // 准备数据
        List<String> list1 = Arrays.asList("Java", "Python", "C++", "Java");
        List<String> list2 = Arrays.asList("Python", "Go", "Rust", "JavaScript");
        // 方法1：使用HashSet去重合并
        Set<String> mergedSet = new HashSet<>(list1);
        mergedSet.addAll(list2);
        List<String> mergedList = new ArrayList<>(mergedSet);
        System.out.println("方法1 - HashSet合并: " + mergedList);
        // 方法2：使用Stream API
        List<String> streamMerged = Stream.concat(list1.stream(), list2.stream())
                .distinct()
                .collect(Collectors.toList());
        System.out.println("方法2 - Stream合并: " + streamMerged);
    }
}

案例2：两个Map合并

import java.util.*;
import java.util.stream.Collectors;
import java.util.stream.Stream;
public class MapMergeExample {
    public static void main(String[] args) {
        Map<String, Integer> map1 = new HashMap<>();
        map1.put("A", 10);
        map1.put("B", 20);
        map1.put("C", 30);
        Map<String, Integer> map2 = new HashMap<>();
        map2.put("B", 25);  // 相同key
        map2.put("D", 40);
        map2.put("E", 50);
        // 方法1：使用putAll（简单覆盖）
        Map<String, Integer> merged1 = new HashMap<>(map1);
        merged1.putAll(map2);
        System.out.println("简单覆盖合并: " + merged1);
        // 方法2：使用merge（自定义合并逻辑）
        Map<String, Integer> merged2 = new HashMap<>(map1);
        map2.forEach((key, value) -> 
            merged2.merge(key, value, Integer::sum)  // 相同key时值相加
        );
        System.out.println("值相加合并: " + merged2);
        // 方法3：使用Stream
        Map<String, Integer> merged3 = Stream.concat(
                map1.entrySet().stream(),
                map2.entrySet().stream()
            )
            .collect(Collectors.toMap(
                Map.Entry::getKey,
                Map.Entry::getValue,
                (v1, v2) -> v1 + v2  // 冲突时相加
            ));
        System.out.println("Stream合并: " + merged3);
    }
}

实际业务场景案例

案例3：用户信息合并

import java.util.*;
import java.util.stream.Collectors;
class User {
    private Long id;
    private String name;
    private String email;
    private String phone;
    public User(Long id, String name, String email, String phone) {
        this.id = id;
        this.name = name;
        this.email = email;
        this.phone = phone;
    }
    // getters and setters
    public Long getId() { return id; }
    public String getName() { return name; }
    public String getEmail() { return email; }
    public String getPhone() { return phone; }
    public void setName(String name) { this.name = name; }
    public void setEmail(String email) { this.email = email; }
    public void setPhone(String phone) { this.phone = phone; }
    @Override
    public String toString() {
        return "User{id=" + id + ", name='" + name + "', email='" + email + "', phone='" + phone + "'}";
    }
}
public class UserMergeExample {
    public static void main(String[] args) {
        // 模拟从不同数据源获取的用户信息
        List<User> usersFromDB = Arrays.asList(
            new User(1L, "张三", "zhangsan@email.com", null),
            new User(2L, "李四", null, "13800138000"),
            new User(3L, "王五", "wangwu@email.com", "13900139000")
        );
        List<User> usersFromCache = Arrays.asList(
            new User(1L, null, null, "13600136000"),  // 补充电话
            new User(2L, null, "lisi@email.com", null), // 补充邮箱
            new User(4L, "赵六", "zhaoliu@email.com", "13700137000") // 新用户
        );
        // 合并用户信息（以DB数据为主，补充缓存数据）
        Map<Long, User> userMap = usersFromDB.stream()
            .collect(Collectors.toMap(User::getId, user -> user));
        for (User cacheUser : usersFromCache) {
            Long userId = cacheUser.getId();
            if (userMap.containsKey(userId)) {
                // 已存在用户，补充信息
                User existingUser = userMap.get(userId);
                if (existingUser.getName() == null && cacheUser.getName() != null) {
                    existingUser.setName(cacheUser.getName());
                }
                if (existingUser.getEmail() == null && cacheUser.getEmail() != null) {
                    existingUser.setEmail(cacheUser.getEmail());
                }
                if (existingUser.getPhone() == null && cacheUser.getPhone() != null) {
                    existingUser.setPhone(cacheUser.getPhone());
                }
            } else {
                // 新用户，直接添加
                userMap.put(userId, cacheUser);
            }
        }
        System.out.println("合并后的用户信息:");
        userMap.values().forEach(System.out::println);
    }
}

案例4：订单数据合并统计分析

import java.time.LocalDate;
import java.util.*;
import java.util.stream.Collectors;
class Order {
    private Long orderId;
    private LocalDate date;
    private Double amount;
    private String category;
    public Order(Long orderId, LocalDate date, Double amount, String category) {
        this.orderId = orderId;
        this.date = date;
        this.amount = amount;
        this.category = category;
    }
    public Long getOrderId() { return orderId; }
    public LocalDate getDate() { return date; }
    public Double getAmount() { return amount; }
    public String getCategory() { return category; }
}
public class OrderMergeExample {
    public static void main(String[] args) {
        // 分页查询的订单数据
        List<Order> page1 = Arrays.asList(
            new Order(1L, LocalDate.now(), 100.0, "电子产品"),
            new Order(2L, LocalDate.now(), 200.0, "图书"),
            new Order(3L, LocalDate.now().minusDays(1), 150.0, "电子产品")
        );
        List<Order> page2 = Arrays.asList(
            new Order(4L, LocalDate.now().minusDays(1), 300.0, "服装"),
            new Order(5L, LocalDate.now().minusDays(2), 250.0, "图书"),
            new Order(6L, LocalDate.now().minusDays(2), 180.0, "电子产品")
        );
        // 合并所有订单
        List<Order> allOrders = new ArrayList<>();
        allOrders.addAll(page1);
        allOrders.addAll(page2);
        // 按类别统计总销售额
        Map<String, Double> salesByCategory = allOrders.stream()
            .collect(Collectors.groupingBy(
                Order::getCategory,
                Collectors.summingDouble(Order::getAmount)
            ));
        System.out.println("按类别销售统计:");
        salesByCategory.forEach((category, total) -> 
            System.out.println(category + ": " + total + "元")
        );
        // 按日期统计销售额
        Map<LocalDate, Double> salesByDate = allOrders.stream()
            .collect(Collectors.groupingBy(
                Order::getDate,
                Collectors.summingDouble(Order::getAmount)
            ));
        System.out.println("\n按日期销售统计:");
        salesByDate.forEach((date, total) -> 
            System.out.println(date + ": " + total + "元")
        );
    }
}

高级合并技巧

案例5：高效合并大量数据

import java.util.*;
import java.util.concurrent.ConcurrentHashMap;
import java.util.stream.Collectors;
public class LargeDataMergeExample {
    // 批量合并优化
    public static <T> List<T> batchMerge(List<List<T>> batches) {
        if (batches == null || batches.isEmpty()) {
            return Collections.emptyList();
        }
        // 计算总大小预分配容量
        int totalSize = batches.stream().mapToInt(List::size).sum();
        List<T> result = new ArrayList<>(totalSize);
        // 批量添加
        for (List<T> batch : batches) {
            result.addAll(batch);
        }
        return result;
    }
    // 并行合并（大数据量时使用）
    public static Map<Long, String> parallelMerge(Map<Long, String> map1, Map<Long, String> map2) {
        Map<Long, String> result = new ConcurrentHashMap<>(map1);
        map2.entrySet().parallelStream().forEach(entry -> 
            result.merge(entry.getKey(), entry.getValue(), (v1, v2) -> v1 + " | " + v2)
        );
        return result;
    }
    public static void main(String[] args) {
        // 模拟多批数据
        List<List<String>> batches = new ArrayList<>();
        batches.add(Arrays.asList("A", "B", "C"));
        batches.add(Arrays.asList("D", "E", "F"));
        batches.add(Arrays.asList("G", "H", "I"));
        // 批量合并
        List<String> merged = batchMerge(batches);
        System.out.println("批量合并结果: " + merged);
        // 并行合并Map
        Map<Long, String> map1 = new HashMap<>();
        map1.put(1L, "张三");
        map1.put(2L, "李四");
        Map<Long, String> map2 = new HashMap<>();
        map2.put(2L, "李四2");
        map2.put(3L, "王五");
        Map<Long, String> result = parallelMerge(map1, map2);
        System.out.println("并行合并结果: " + result);
    }
}

最佳实践建议

性能优化原则：

// 不好的做法
List<String> result = new ArrayList<>();
for (List<String> batch : batches) {
    result.addAll(batch);  // 多次扩容
}
// 好的做法
int totalSize = batches.stream().mapToInt(List::size).sum();
List<String> result = new ArrayList<>(totalSize);  // 预分配容量
for (List<String> batch : batches) {
    result.addAll(batch);
}

线程安全合并：

import java.util.concurrent.ConcurrentHashMap;
// 多线程环境下的安全合并
public class ThreadSafeMerge {
    private ConcurrentHashMap<String, Integer> counter = new ConcurrentHashMap<>();
    public void mergeData(Map<String, Integer> data) {
        data.forEach((key, value) -> 
            counter.merge(key, value, Integer::sum)
        );
    }
}

这些案例覆盖了大多数常见的集合合并场景,根据实际业务需求选择合适的合并策略，注意性能优化和线程安全。

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