我想我找到了一个非常简单的方法来创建一个深度复制数组列表。假设你想复制一个字符串数组列表arrayA。

ArrayList<String>arrayB = new ArrayList<String>();
arrayB.addAll(arrayA);

如果对你不起作用请告诉我。

我认为目前的绿色答案很糟糕，为什么你会问?

它可能需要添加大量代码 它要求你列出所有要复制的列表并这样做

序列化的方式在我看来也是不好的，你可能不得不到处添加Serializable。

那么解决方案是什么呢?

Java深度克隆库 克隆库是一个小型的开源(apache许可)java库，它对对象进行深度克隆。对象不必实现克隆接口。实际上，这个库可以克隆任何java对象。它可以用在缓存实现中，如果你不想修改缓存对象，或者当你想创建对象的深度副本时。

Cloner cloner=new Cloner();
XX clone = cloner.deepClone(someObjectOfTypeXX);

请登录https://github.com/kostaskougios/cloning查看

其他的海报是正确的:你需要迭代列表并复制到一个新的列表。

然而…… 如果列表中的对象是不可变的-你不需要克隆它们。如果你的对象有一个复杂的对象图，它们也需要是不可变的。

不可变性的另一个好处是它们也是线程安全的。

一些其他用于将ArrayList复制为深度复制的替代方法

Alernative 1 -使用外部包common -lang3，方法SerializationUtils.clone():

SerializationUtils.clone()

假设我们有一个类dog，其中类的字段是可变的，并且至少有一个字段是String类型和mutable类型的对象——而不是基本数据类型(否则浅拷贝就足够了)。

浅拷贝的例子:

List<Dog> dogs = getDogs(); // We assume it returns a list of Dogs
List<Dog> clonedDogs = new ArrayList<>(dogs);

现在回到狗的深度复制。

Dog类只有可变字段。

狗类:

public class Dog implements Serializable {
    private String name;
    private int age;

    public Dog() {
        // Class with only mutable fields!
        this.name = "NO_NAME";
        this.age = -1;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public int getAge() {
        return age;
    }

    public void setAge(int age) {
        this.age = age;
    }

    @Override
    public String toString() {
        return "Dog{" +
                "name='" + name + '\'' +
                ", age=" + age +
                '}';
    }
}

注意，类Dog实现了Serializable!这使得可以使用方法“SerializationUtils.clone(dog)”

阅读main方法中的注释以理解结果。这表明我们已经成功地对ArrayList()进行了深度复制。看到 在“SerializationUtils.clone(dog)”下面:

public static void main(String[] args) {
    Dog dog1 = new Dog();
    dog1.setName("Buddy");
    dog1.setAge(1);

    Dog dog2 = new Dog();
    dog2.setName("Milo");
    dog2.setAge(2);

    List<Dog> dogs = new ArrayList<>(Arrays.asList(dog1,dog2));

    // Output: 'List dogs: [Dog{name='Buddy', age=1}, Dog{name='Milo', age=2}]'
    System.out.println("List dogs: " + dogs);

    // Let's clone and make a deep copy of the dogs' ArrayList with external package commons-lang3:
    List<Dog> clonedDogs = dogs.stream().map(dog -> SerializationUtils.clone(dog)).collect(Collectors.toList());
    // Output: 'Now list dogs are deep copied into list clonedDogs.'
    System.out.println("Now list dogs are deep copied into list clonedDogs.");

    // A change on dog1 or dog2 can not impact a deep copy.
    // Let's make a change on dog1 and dog2, and test this
    // statement.
    dog1.setName("Bella");
    dog1.setAge(3);
    dog2.setName("Molly");
    dog2.setAge(4);

    // The change is made on list dogs!
    // Output: 'List dogs after change: [Dog{name='Bella', age=3}, Dog{name='Molly', age=4}]'
    System.out.println("List dogs after change: " + dogs);

    // There is no impact on list clonedDogs's inner objects after the deep copy.
    // The deep copy of list clonedDogs was successful!
    // If clonedDogs would be a shallow copy we would see the change on the field
    // "private String name", the change made in list dogs, when setting the names
    // Bella and Molly.
    // Output clonedDogs:
    // 'After change in list dogs, no impact/change in list clonedDogs:\n'
    // '[Dog{name='Buddy', age=1}, Dog{name='Milo', age=2}]\n'
    System.out.println("After change in list dogs, no impact/change in list clonedDogs: \n" + clonedDogs);
}

输出:

List dogs: [Dog{name='Buddy', age=1}, Dog{name='Milo', age=2}]
Now list dogs are deep copied into list clonedDogs.
List dogs after change: [Dog{name='Bella', age=3}, Dog{name='Molly', age=4}]
After change in list dogs, no impact/change in list clonedDogs:
[Dog{name='Buddy', age=1}, Dog{name='Milo', age=2}]

备注: 因为改变列表狗后对列表克隆狗没有影响/改变， 那么ArrayList的深度复制成功!

Alernative 2 -不使用外部包:

Dog类中引入了一个新方法“clone()”，与替代方案1相比，“implements Serializable”被删除了。

clone()

狗类:

public class Dog {
    private String name;
    private int age;

    public Dog() {
        // Class with only mutable fields!
        this.name = "NO_NAME";
        this.age = -1;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public int getAge() {
        return age;
    }

    public void setAge(int age) {
        this.age = age;
    }

    /**
     * Returns a deep copy of the Dog
     * @return new instance of {@link Dog}
     */
    public Dog clone() {
        Dog newDog = new Dog();
        newDog.setName(this.name);
        newDog.setAge(this.age);
        return newDog;
    }

    @Override
    public String toString() {
        return "Dog{" +
                "name='" + name + '\'' +
                ", age=" + age +
                '}';
    }
}

阅读下面主要方法中的评论以理解结果。这表明我们已经成功地对ArrayList()进行了深度复制。看到 下面是上下文中的“clone()”方法:

public static void main(String[] args) {
    Dog dog1 = new Dog();
    dog1.setName("Buddy");
    dog1.setAge(1);

    Dog dog2 = new Dog();
    dog2.setName("Milo");
    dog2.setAge(2);

    List<Dog> dogs = new ArrayList<>(Arrays.asList(dog1,dog2));

    // Output: 'List dogs: [Dog{name='Buddy', age=1}, Dog{name='Milo', age=2}]'
    System.out.println("List dogs: " + dogs);

    // Let's clone and make a deep copy of the dogs' ArrayList:
    List<Dog> clonedDogs = dogs.stream().map(dog -> dog.clone()).collect(Collectors.toList());
    // Output: 'Now list dogs are deep copied into list clonedDogs.'
    System.out.println("Now list dogs are deep copied into list clonedDogs.");

    // A change on dog1 or dog2 can not impact a deep copy.
    // Let's make a change on dog1 and dog2, and test this
    // statement.
    dog1.setName("Bella");
    dog1.setAge(3);
    dog2.setName("Molly");
    dog2.setAge(4);

    // The change is made on list dogs!
    // Output: 'List dogs after change: [Dog{name='Bella', age=3}, Dog{name='Molly', age=4}]'
    System.out.println("List dogs after change: " + dogs);

    // There is no impact on list clonedDogs's inner objects after the deep copy.
    // The deep copy of list clonedDogs was successful!
    // If clonedDogs would be a shallow copy we would see the change on the field
    // "private String name", the change made in list dogs, when setting the names
    // Bella and Molly.
    // Output clonedDogs:
    // 'After change in list dogs, no impact/change in list clonedDogs:\n'
    // '[Dog{name='Buddy', age=1}, Dog{name='Milo', age=2}]\n'
    System.out.println("After change in list dogs, no impact/change in list clonedDogs: \n" + clonedDogs);
}

输出:

List dogs: [Dog{name='Buddy', age=1}, Dog{name='Milo', age=2}]
Now list dogs are deep copied into list clonedDogs.
List dogs after change: [Dog{name='Bella', age=3}, Dog{name='Molly', age=4}]
After change in list dogs, no impact/change in list clonedDogs:
[Dog{name='Buddy', age=1}, Dog{name='Milo', age=2}]

备注: 因为改变列表狗后对列表克隆狗没有影响/改变， 那么ArrayList的深度复制成功!

注一: 方案1比方案2慢得多， 但更容易维护，因为您不需要 更新任何方法，如clone()。

注2:对于替代方案1，以下maven依赖项用于方法“SerializationUtils.clone()””:

<!-- https://mvnrepository.com/artifact/org.apache.commons/commons-lang3 -->
<dependency>
    <groupId>org.apache.commons</groupId>
    <artifactId>commons-lang3</artifactId>
    <version>3.9</version>
</dependency>

更多common-lang3版本请访问:

https://mvnrepository.com/artifact/org.apache.commons/commons-lang3

下面是一个使用泛型模板类型的解决方案:

public static <T> List<T> copyList(List<T> source) {
    List<T> dest = new ArrayList<T>();
    for (T item : source) { dest.add(item); }
    return dest;
}

基本上有三种不需要手动迭代的方法，

1使用构造函数

ArrayList<Dog> dogs = getDogs();
ArrayList<Dog> clonedList = new ArrayList<Dog>(dogs);

2使用addAll(Collection<?c)扩展;

ArrayList<Dog> dogs = getDogs();
ArrayList<Dog> clonedList = new ArrayList<Dog>();
clonedList.addAll(dogs);

3使用addAll(int index, Collection<?用int形参扩展E> c)方法

ArrayList<Dog> dogs = getDogs();
ArrayList<Dog> clonedList = new ArrayList<Dog>();
clonedList.addAll(0, dogs);

注意:如果指定的集合在操作进行时被修改，这些操作的行为将是未定义的。