Java 将多个 BigInteger 值添加到 ArrayList
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Adding multiple BigInteger values to an ArrayList
提问by aaa
I want to add multiple BigIntegervalues to an ArrayList. All I have found is examples that repeatedly add single values, each expressed on their own line of code. I'm looking for something like
我想将多个BigInteger值添加到ArrayList. 我发现的只是重复添加单个值的示例，每个值都用自己的代码行表示。我正在寻找类似的东西
ArrayList<BigInteger> array = {bigInt1, bigInt2, bigInt3};
and instead it's:
而是：
ArrayList<BigInteger> array = new ArrayList<BigInteger>();
array.add(bigInt1);
array.add(bigInt2);
array.add(bigInt3);
Can it be done, without adding one element/line or using a for loop?
可以在不添加一个元素/行或使用 for 循环的情况下完成吗？
采纳答案by cletus
I'm not really sure what you're after. You have four alternatives:
我不确定你在追求什么。你有四种选择：
1. Add items individually
1.单独添加项目
Instantiate a concrete Listtype and then call add()for each item:
实例化一个具体List类型，然后调用add()每个项目：
List<BigInteger> list = new ArrayList<BigInteger>();
list.add(new BigInteger("12345"));
list.add(new BigInteger("23456"));
2. Subclass a concrete Listtype (double brace initialization)
2. 子类化一个具体List类型（双括号初始化）
Some might suggest double brace initialization like this:
有些人可能会建议像这样进行双括号初始化：
List<BigInteger> list = new ArrayList<BigInteger>() {{
add(new BigInteger("12345"));
add(new BigInteger("23456"));
}};
I recommend not doing this. What you're actually doing here is subclassing ArrayList, which (imho) is not a good idea. That sort of thing can break Comparators, equals()methods and so on.
我建议不要这样做。你在这里实际做的是 subclassing ArrayList，这（恕我直言）不是一个好主意。那种东西可以破坏Comparators、equals()方法等等。
3. Using Arrays.asList()
3. 使用 Arrays.asList()
Another approach:
另一种方法：
List<BigInteger> list = new ArrayList<BigInteger>(Arrays.asList(
new BigInteger("12345"),
new BigInteger("23456")
));
or, if you don't need an ArrayList, simply as:
或者，如果您不需要ArrayList，只需如下：
List<BigInteger> list = Arrays.asList(
new BigInteger("12345"),
new BigInteger("23456")
);
I prefer one of the above two methods.
我更喜欢上述两种方法之一。
4. Collectionliterals (Java 7+)
4.Collection文字（Java 7+）
Assuming Collection literalsgo ahead in Java 7, you will be able to do this:
假设在 Java 7 中继续使用Collection 字面量，您将能够做到这一点：
List<BigInteger> list = [new BigInteger("12345"), new BigInteger("23456")];
As it currently stands, I don't believe this feature has been confirmed yet.
就目前而言，我认为此功能尚未得到确认。
That's it. Those are your choices. Pick one.
就是这样。这些都是你的选择。选一个。
回答by Chris Dennett
Arrays.asList(new BigInteger("1"), new BigInteger("2"), new BigInteger("3"), new BigInteger("4"));
You could probably make a method that returns a new BigInteger given a String, called something like bi(..) to reduce the size of this line.
您可能会创建一个方法，该方法在给定字符串的情况下返回一个新的 BigInteger，称为 bi(..) 之类的东西以减少该行的大小。
回答by Jatin
BigIntegerArrays.asList(1, 2, 3, 4);
Where BigIntegerArrays is a custom class which does what you need it to do. This helps if you are doing this often. No rocket science here - ArrayList BigIntegerArrays.asList(Integer... args) will use a FOR loop.
其中 BigIntegerArrays 是一个自定义类，它可以完成您需要做的事情。如果您经常这样做，这会有所帮助。这里没有火箭科学 - ArrayList BigIntegerArrays.asList(Integer... args) 将使用 FOR 循环。
回答by Pascal Thivent
If using a third party library is an option, then I suggest using Lists.newArrayList(E... elements)from Google's Guava:
如果使用第三方库是一种选择，那么我建议使用Lists.newArrayList(E... elements)Google 的Guava：
List<BigInteger> of = Lists.newArrayList(bigInt1, bigInt2, bigInt3);
And if mutability isn't required, then use an overload of ImmutableList.of():
如果不需要可变性，则使用以下重载ImmutableList.of()：
final List<BigInteger> of = ImmutableList.of(bigInt1, bigInt2, bigInt3);
This is IMO a very elegant solution.
这是 IMO 一个非常优雅的解决方案。
回答by Boann
This is easily accomplished with a helper function or two:
这可以通过一两个辅助函数轻松完成：
import java.util.*;
import java.math.BigInteger;
class Example {
public static void main(String[] args) {
ArrayList<BigInteger> array = newBigIntList(
1, 2, 3, 4, 5,
0xF,
"1039842034890394",
6L,
new BigInteger("ffff", 16)
);
// [1, 2, 3, 4, 5, 15, 1039842034890394, 6, 65535]
System.out.println(array);
}
public static void fillBigIntList(List<BigInteger> list, Object... numbers) {
for (Object n : numbers) {
if (n instanceof BigInteger) list.add((BigInteger)n);
else if (n instanceof String) list.add(new BigInteger((String)n));
else if (n instanceof Long || n instanceof Integer)
list.add(BigInteger.valueOf(((Number)n).longValue()));
else throw new IllegalArgumentException();
}
}
public static ArrayList<BigInteger> newBigIntList(Object... numbers) {
ArrayList<BigInteger> list = new ArrayList<>(numbers.length);
fillBigIntList(list, numbers);
return list;
}
}
