如何使用Java8流将字符串数组转换为大整数数组

我有一个字符串数组，我想用Java8流将它转换成BigInteger数组

String[] output = bigSorting(new String[]{"31415926535897932384626433832795", "1", "4900146572543628830293235422623540449026979", "10", "57500297590012603652986133599394871645776460", "5", 
                    "497010206818067722087306230802257700034825862515267073569769100385728461314", "57500297590012603652986133599394871645776460497010206818067722087306230802257700034825862515267073569769100385728461314497010206818067722087306230802257700034825862515267073569769100385728461314497010206818067722087306230802257700034825862515267073569769100385728461314497010206818067722087306230802257700034825862515267073569769100385728461314497010206818067722087306230802257700034825862515267073569769100385728461314"});
        Object[] unsortedBigIntegerArr = convertFromStringArrayToBigIntegerArray(output);

这是我尝试过的，但是我不能得到BigInteger数组，但是我可以得到对象数组

private static Object[] convertFromStringArrayToBigIntegerArray(String[] unsorted) {
        return Arrays.stream(unsorted).map(BigSorting2::convertFromStringToBigInteger).toArray();
    }

    private static BigInteger convertFromStringToBigInteger(String unsorted) {
        return new BigInteger(unsorted);
    }

是否有任何方法可以完全使用Java 8 streams完成此操作。

简单到：

String[] output = bigSorting(new String[]{"31415926535897932384626433832795", "1", "4900146572543628830293235422623540449026979", "10", "57500297590012603652986133599394871645776460", "5", 
                    "497010206818067722087306230802257700034825862515267073569769100385728461314", "57500297590012603652986133599394871645776460497010206818067722087306230802257700034825862515267073569769100385728461314497010206818067722087306230802257700034825862515267073569769100385728461314497010206818067722087306230802257700034825862515267073569769100385728461314497010206818067722087306230802257700034825862515267073569769100385728461314497010206818067722087306230802257700034825862515267073569769100385728461314"});
        Object[] unsortedBigIntegerArr = convertFromStringArrayToBigIntegerArray(output);

return Arrays.stream(unsorted)
             .map(BigSorting2::convertFromStringToBigInteger)
             .toArray(BigInteger[]::new);

简单到：

return Arrays.stream(unsorted)
             .map(BigSorting2::convertFromStringToBigInteger)
             .toArray(BigInteger[]::new);

---

非常简单的事情，比如：

public void test(String[] args) {
    String[] output = new String[]{"31415926535897932384626433832795", "1", "4900146572543628830293235422623540449026979", "10", "57500297590012603652986133599394871645776460", "5",
            "497010206818067722087306230802257700034825862515267073569769100385728461314", "57500297590012603652986133599394871645776460497010206818067722087306230802257700034825862515267073569769100385728461314497010206818067722087306230802257700034825862515267073569769100385728461314497010206818067722087306230802257700034825862515267073569769100385728461314497010206818067722087306230802257700034825862515267073569769100385728461314497010206818067722087306230802257700034825862515267073569769100385728461314"};
    Object[] bigIntegers = Arrays.stream(output)
            .map(BigInteger::new)
            .toArray();
    System.out.println(Arrays.toString(bigIntegers));
}

---

非常简单的事情，比如：

public void test(String[] args) {
    String[] output = new String[]{"31415926535897932384626433832795", "1", "4900146572543628830293235422623540449026979", "10", "57500297590012603652986133599394871645776460", "5",
            "497010206818067722087306230802257700034825862515267073569769100385728461314", "57500297590012603652986133599394871645776460497010206818067722087306230802257700034825862515267073569769100385728461314497010206818067722087306230802257700034825862515267073569769100385728461314497010206818067722087306230802257700034825862515267073569769100385728461314497010206818067722087306230802257700034825862515267073569769100385728461314497010206818067722087306230802257700034825862515267073569769100385728461314"};
    Object[] bigIntegers = Arrays.stream(output)
            .map(BigInteger::new)
            .toArray();
    System.out.println(Arrays.toString(bigIntegers));
}

---

天哪，这太酷了，谢谢你。你能给我建议一下Java8流的最佳学习路径吗。@pjj我从Oracle教程和官方文档开始……你甚至可以做

.map（BigInteger:：new）

@pjj我从堆栈溢出问题和答案中学到了很多东西。例如，搜索[java stream]标记。@pjj老实说，大多数不熟悉java类型系统的人都希望您尝试

Arrays.stream（未排序）.map（BigSorting2:：ConvertFromStringToBiginger）.toArray（）

生成

流

，但显然这是不可能的。由于java的限制，必须显式指定数组构造函数引用以检索所需类型的数组，因此为

.toArray（BigInteger[]：：new）天哪，这太酷了，谢谢你。你能给我建议一下Java8流的最佳学习路径吗。@pjj我从Oracle教程和官方文档开始……你甚至可以做
.map（BigInteger:：new）
@pjj我从堆栈溢出问题和答案中学到了很多东西。例如，搜索[java stream]标记。@pjj老实说，大多数不熟悉java类型系统的人都希望您尝试
Arrays.stream（未排序）.map（BigSorting2:：ConvertFromStringToBiginger）.toArray（）
生成
流
，但显然这是不可能的。由于java的限制，必须显式指定数组构造函数引用以检索所需类型的数组，因此为
.toArray（BigInteger[]：：new）我想你误解了我的问题，请看我的问题或@Eugene-answer，谢谢你的回答。我理解这个问题的原因是，是否有可能从
字符串[]
中获得
biginger[]
而不是
对象[]
。我想你误解了我的问题，请查看我的问题或@Eugene answer，谢谢您的回答。根据我的理解，问题是是否可以从
字符串[]
获取
BigInteger[]
而不是
对象[]
。