[Java] ArrayList 类_arraylist list = new arraylist<>();

java.util.ArrayList<E>

从书中各种代码来看，java.util.ArrayList<E> 是非常重要的一个类，在代码中广泛使用，E表示泛型，ArrayList是一个泛型类。
ArrayList相当于C++ 的vector，用于存储对象。与数组不同，数组一旦创建，长度固定，但是ArrayList的长度是动态的，不受限制，可以存储任意多的对象，但是只能存储对象，不能存储原生数据类型例如int。

数组和ArrayList的比较

创建一个存储字符串的ArrayList对象：

ArrayList<String> cities = new ArrayList<String>();
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创建一个存储日期的ArrayList对象：

ArrayList<java.util.Date> dates = new ArrayList<java.util.Date> ();
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JDK 7 之后，下述表达式

ArrayList<AConcreteType> list = new ArrayList<AConcreteType>();
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可以简化为：

ArrayList<AConcreteType> list = new ArrayList<>();
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因为编译器有一个新的feature叫做类型推断(type inference)， 能够从变量声明推断类型。

例1

import java.util.ArrayList;
public class TestArrayList {
    public static void main(String[] args) {
        // Create a list to store cities
        ArrayList<String> cityList = new ArrayList<String>();
        
        // Add some cities in the list
        cityList.add("London");
        // cityList now contains [London]
        
        cityList.add("Denver");
        // cityList now contains [London, Denver]
        
        cityList.add("Paris");
        // cityList now contains [London, Denver, Paris]
        
        cityList.add("Miami");
        // cityList now contains [London, Denver, Paris, Miami]
        
	    cityList.add("Seoul");
        // Contains [London, Denver, Paris, Miami, Seoul]
        
        cityList.add("Tokyo");
        // Contains [London, Denver, Paris, Miami, Seoul, Tokyo]

        System.out.println("List size? " + cityList.size());  // 6
        System.out.println("Is Miami in the list? " + cityList.contains("Miami"));  // true
        System.out.println("The location of Denver in the list? " + cityList.indexOf("Denver")); // 1 返回索引，如果不在list中，返回-1
        System.out.println("Is the list empty? " + cityList.isEmpty()); // Print false

        // Insert a new city at index 2
        cityList.add(2, "Xian");
        // Contains [London, Denver, Xian, Paris, Miami, Seoul, Tokyo]

        // Remove a city from the list
        cityList.remove("Miami");
        // Contains [London, Denver, Xian, Paris, Seoul, Tokyo]

        // Remove a city at index 1
        cityList.remove(1);
        // Contains [London, Xian, Paris, Seoul, Tokyo]

        // Display the contents in the list
        System.out.println(cityList.toString());

        // Display the contents in the list in reverse order
        for (int i = cityList.size() - 1; i >= 0; i--)
            System.out.print(cityList.get(i) + " ");
        System.out.println();

        // Create a list to store two circles
        ArrayList<CircleFromSimpleGeometricObject> list = new ArrayList<CircleFromSimpleGeometricObject>();

        // Add two circles
        list.add(new CircleFromSimpleGeometricObject(2));
        list.add(new CircleFromSimpleGeometricObject(3));

        // Display the area of the first circle in the list
        System.out.println("The area of the circle? " + list.get(0).getArea());
    }
}
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其中System.out.println(cityList.toString()); 等同于 System.out.println(cityList);

toString()方法返回列表的字符串表示，形式为 [e0.toString(), e1.toString(), ..., ek.toString()]，e0, e1, . . . ,和 ek 都是列表中的元素。

由于ArrayList只能存储对象，不能存储原生数据类型数据，下面的代码是错误的：
ArrayList<int> list = new ArrayList<>(); 错误！！！！！
只能写成：
ArrayList<Integer> list = new ArrayList<>();

例2

用户输入一个数字序列，假定输入以0结尾，且0不计入数字序列，打印序列中不重复的数字：

package TEST_ALL;
import java.util.ArrayList;
import java.util.Scanner;

public class DistinctNumbers {
    public static void main(String[] args) {
        ArrayList<Integer> list = new ArrayList<Integer>();
        Scanner input = new Scanner(System.in);
        System.out.print("Enter integers (input ends with 0): ");
        int value;
        do {
            value = input.nextInt(); // Read a value from the input
            if (!list.contains(value) && value != 0)
            list.add(value); // Add the value if it is not in the list
        } while (value != 0);
        // Display the distinct numbers
        for (int i = 0; i < list.size(); i++)
            System.out.print(list.get(i) + " ");
    }
}
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运行结果：

Enter integers (input ends with 0): 2 32 3 1 2 3 2 9 0
2 32 3 1 9 
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可以用下列 foreach 循环遍历一个array list：

for (elementType element: arrayList) {
	// Process the element 
}
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例如上例中的语句：

for (int i = 0; i < list.size(); i++)
    System.out.print(list.get(i) + " ");
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可以改写为：

for (int number: list)
    System.out.print(number + " ");
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书后的练习

我做的答案：

11.30 How do you do the following?

a. Create an ArrayList for storing double values?

ArrayList<Double> list_double = new ArrayList<Double>();`
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b. Append an object to a list?

Double o = new Double(11);
list_double.add(o);
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c. Insert an object at the beginning of a list?

Double o = new Double(11);
list_double.add(0, o);
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d. Find the number of objects in a list?

list_double.size();
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e. Remove a given object from a list?

Double o = new Double(11);
while(list_double.contains(o))
    list_double.remove(o);
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f. Remove the last object from the list?

list_double.remove(list_double.size() - 1);
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g. Check whether a given object is in a list?

Double o = new Double(11);
if (list_double.contains(o))
    System.out.print("Y");
else 
    System.out.print("N");
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h. Retrieve an object at a specified index from a list?

int index = 111;
list_double.get(index);
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11.31 Identify the errors in the following code.

ArrayList<String> list = new ArrayList<>();
list.add("Denver");
list.add("Austin");
list.add(new java.util.Date()); # 错误，元素数据类型必须一致，不能改成Date
String city = list.get(0);
list.set(3, "Dallas"); #
System.out.println(list.get(3)); # 错误，index 超出 size
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list.set(3, "Dallas"); 和list.get(3)都是错的，You cannot use the get(index) and set(index, element) methods if the element is not in the list. 看源代码，get 和 set 都有RangeCheck(index); 这个函数：

private void RangeCheck(int index) {
    if (index >= size)
        throw new IndexOutOfBoundsException("Index: "+index+", Size: "+size);
}
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11.32 Suppose the ArrayList list contains {"Dallas", "Dallas", "Houston", "Dallas"}.What is the list after invoking list.remove("Dallas")one time?
Does the following code correctly remove all elements with value "Dallas" from
the list? If not, correct the code.

for (int i = 0; i < list.size(); i++)
    list.remove("Dallas");
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调用一次list.remove("Dallas")， list 变为 { "Dallas", "Houston", "Dallas"}，
code不能完全删除 "Dallas"，因为最后结果是{"House", "Dallas"}， 应该改成：

while(list.contains("Dallas"))
    list.remove("Dallas");
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11.33 Explain why the following code displays [1, 3] rather than [2, 3].

ArrayList<Integer> list = new ArrayList<>();
list.add(1);
list.add(2);
list.add(3);
list.remove(1);
System.out.println(list);
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因为remove方法有两种签名：list.remove(Object)，list.remove(index); 此处调用的是第二种，如果要删除第一个元素，必须改成list.remove((Integer)1).
(这里不是特别理解)

11.34 Explain why the following code is wrong.

ArrayList<Double> list = new ArrayList<>();
list.add(1);
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错误是因为， 1 为整型，与 Double 类型不符，必须改成 1.0.
也不能改为list.add((Double)1), 否则会显示 inconvertible type 的错误.

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[1] Introduction to Java Programming 10th edition Chapter 11.11