Java 通过替换特殊字符来转换一组字符串

Question

public static void main(String[] args) {
    String str = "astv*12atthh124ggh*dhr1234sfff123*dgdfg1234*mnaoj";
    Pattern p = Pattern.compile("\\d+");
    Matcher m = p.matcher(str);
    List<String> strings = new ArrayList<String>();
    List<Integer> nums = new ArrayList<Integer>();
    while (m.find()) {
        nums.add(Integer.parseInt(m.group()));
    }
    p = Pattern.compile("[a-z]+");
    m = p.matcher(str);
    while (m.find()) {
        strings.add(m.group());
    }
    System.out.println(nums);
    System.out.println(strings);
}

输出:

[12, 124, 1234, 123, 1234]
[astv, atthh, ggh, dhr, sfff, dgdfg, mnaoj]

但我想要这样的输出:

[12124, 1234123, 1234]
[astv, atthhggh, dhrsfff, dgdfg, mnaoj]

Answer 1

您可以将拆分与 * 一起使用，然后您可以处理每个元素，例如:

public static void main(String[] args) {
    String str = "astv*12atthh124ggh*dhr1234sfff123*dgdfg1234*mnaoj";
    String[] spl = str.split("\\*");//[astv, 12atthh124ggh, dhr1234sfff123, dgdfg1234, mnaoj]
    List<String> strings = new ArrayList<>();
    List<Integer> nums = new ArrayList<>();
    for (String s : spl) {

        String tmp = s.replaceAll("\\d+", "");//replace all the digits with empty
        if (!tmp.trim().isEmpty()) {
            strings.add(tmp);
        }

        tmp = s.replaceAll("[a-z]+", "");//replace all the character with empty
        if (!tmp.trim().isEmpty()) {
            nums.add(Integer.parseInt(tmp));
        }
    }

    System.out.println(nums);
    System.out.println(strings);
}

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输出

[12124, 1234123, 1234]
[astv, atthhggh, dhrsfff, dgdfg, mnaoj]

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Ideone demo