java - 按数字升序对字符串列表进行排序，并保留多个小数

Question

我正在尝试对以下列表进行排序:

2.1.0.0
2.1.1.0
2.1.11.0
2.1.12.0
2.1.13.0
2.1.14.0
2.1.15.0
2.1.2.0
2.1.3.0
2.1.4.0
2.1.5.0
2.1.6.0
2.1.7.0
2.1.8.0
2.1.9.0

实现如下列表:

2.1.0.0
2.1.1.0
2.1.2.0
2.1.3.0
2.1.4.0
2.1.5.0
2.1.6.0
2.1.7.0
2.1.8.0
2.1.9.0
2.1.11.0
2.1.12.0
2.1.13.0
2.1.14.0
2.1.15.0

我正在尝试按第二个小数之后的顺序对此进行排序。我尝试过 Arrays.sort 但没有成功。这是我的代码

import java.util.Arrays;

public class NewSort {

    public static void main(String[] args) {
        String[] numbers = { "2.1.0.0", "2.1.1.0", "2.1.11.0", "2.1.12.0",
                "2.1.13.0", "2.1.14.0", "2.1.15.0", "2.1.2.0", "2.1.3.0",
                "2.1.4.0", "2.1.5.0", "2.1.6.0", "2.1.7.0", "2.1.8.0",
                "2.1.9.0" };

        Arrays.sort(numbers);

        for (String number : numbers) {
            System.out.println(number);
        }
    }

}

Answer 1

I am trying to sort this by what is after the second decimal by that order.

你可以自己写Comparator :

enum MyStringComparator implements Comparator<String> {
    INSTANCE;

    @Override
    public int compare(String s1, String s2) {
        int n1 = Integer.parseInt(s1.split("\\.")[2]);
        int n2 = Integer.parseInt(s2.split("\\.")[2]);
        return Integer.compare(n1, n2);
    }
}

现在:

String[] numbers = { "2.1.0.0", "2.1.1.0", "2.1.11.0", "2.1.12.0",
        "2.1.13.0", "2.1.14.0", "2.1.15.0", "2.1.2.0", "2.1.3.0",
        "2.1.4.0", "2.1.5.0", "2.1.6.0", "2.1.7.0", "2.1.8.0",
        "2.1.9.0" };

Arrays.sort(numbers, MyStringComparator.INSTANCE);

for (String number : numbers) {
    System.out.println(number);
}

2.1.0.02.1.1.02.1.2.02.1.3.02.1.4.02.1.5.02.1.6.02.1.7.02.1.8.02.1.9.02.1.11.02.1.12.02.1.13.02.1.14.02.1.15.0

I'll leave it to you to incorporate any error-checking mechanism that you may need.

---

In Java 8 it's a lot easier:

Arrays.sort(numbers,
            Comparator.comparingInt(s -> Integer.parseInt(s.split("\\.")[2])));