我可以从字符串中提取数字而不计算该字符串中的数字量吗？

Question

我需要将数字从字符串提取到int数组，但在Matcher类中找不到合适的函数。有什么办法比我做得更好吗？例如，这样的split（）函数按分隔符分割字符串并返回准备好的数组。

import java.util.regex.*;

public class Test
{
    public static void main(String[] args)
    {
        String stringWithNumbers = "1111llalbhahbah2222albhlahbahb3333lHVvlVLH4444ABHABHB";
        int[] numbersFromString;
        
        // find amount of numbers in stringWithNumbers
        Pattern toFindNumbers = Pattern.compile("\\d+");
        Matcher m1 = toFindNumbers.matcher(stringWithNumbers);
        int amountOfNumbers = 0;
        while (m1.find()) amountOfNumbers++;
        
        // fill array with numbers
        numbersFromString = new int[amountOfNumbers];
        m1.reset();
        for (int i = 0; m1.find(); i++)
        {
            numbersFromString[i] = Integer.parseInt(m1.group());
        }
        
        // print array (test)
        for (int i: numbersFromString)
        {
            System.out.println(i);
        }
    }
}

输出：
1111
2222
3333
4444

Answer 1

试试这个：

String stringWithNumbers = "1111llalbhahbah2222albhlahbahb3333lHVvlVLH4444ABHABHB";
int[] numbersFromString = Stream.of(stringWithNumbers.split("[a-zA-Z]+")).mapToInt(Integer::parseInt).toArray();

for (int i: numbersFromString)
{
     System.out.println(i);
}

输出：

1111
2222
3333
4444

Answer 2

您可以将String方法拆分与正则表达式\\D一起使用，如下所示：

String stringWithNumbers = "1111llalbhahbah2222albhlahbahb3333lHVvlVLH4444ABHABHB";
String[] s = stringWithNumbers.split("\\D+");
System.out.println(Arrays.toString(s)); //it will print [1111, 2222, 3333, 4444]