对数组 Integer 的排序数组，其中要排序的元素在 Java 中是 int

Question

我有这两个整数数组：

1. [[2021,06,24],[2021,06,22],[2021,04,19]]
2. [[2021,04,19],[2021,06,22],[2021,06,24]]

我想要两个数组的顺序相同，我如何在 Java 中做到这一点？ 所以我想要两个数组都按这个顺序：

1. [[2021,06,24],[2021,06,22],[2021,04,19]]
2. [[2021,06,24],[2021,06,22],[2021,04,19]]

Answer 1

按日期排序

由于您的日期代表日期，因此更容易比较它们的是使用LocalDate 构建日期对象。

从List<String> o 到LocalDate：

List<String> o = List.of("2021", "04", "19");
LocalDate.of(Integer.parseInt(o.get(0)), Integer.parseInt(o.get(1)), Integer.parseInt(o.get(2)))

然后用List.sort应用这个逻辑

List<List<String>> values = Arrays.asList(List.of("2021", "04", "19"),
        List.of("2021", "06", "22"), List.of("2021", "06", "24"));

values.sort((o1, o2) -> LocalDate.of(Integer.parseInt(o2.get(0)), Integer.parseInt(o2.get(1)), Integer.parseInt(o2.get(2)))
        .compareTo(LocalDate.of(Integer.parseInt(o1.get(0)), Integer.parseInt(o1.get(1)), Integer.parseInt(o1.get(2)))));

System.out.println(values); // [[2021, 06, 24], [2021, 06, 22], [2021, 04, 19]]

---

您可以将解析器提取到方法中并使用Comparator.comparing

class DateSorter {
    public static void main(String[] args) {
        List<List<String>> values = Arrays.asList(List.of("2021", "04", "19"),
                List.of("2021", "06", "22"), List.of("2021", "06", "24"));
        values.sort(Comparator.comparing(DateSorter::toDate, Comparator.reverseOrder()));

        System.out.println(values); // [[2021, 06, 24], [2021, 06, 22], [2021, 04, 19]]
    }

    static LocalDate toDate(List<String> o) {
        return LocalDate.of(Integer.parseInt(o.get(0)), Integer.parseInt(o.get(1)), Integer.parseInt(o.get(2)));
    }
}

---

按字符串排序

您也可以只对每个列表排序一个字符串，以防您的字符串是像 04 而不是 4 这样的双位数字符串

List<List<String>> values = Arrays.asList(List.of("2021", "04", "19"),
        List.of("2021", "06", "22"), List.of("2021", "06", "24"));
values.sort(Comparator.comparing(List::toString, Comparator.reverseOrder()));
System.out.println(values); // [[2021, 06, 24], [2021, 06, 22], [2021, 04, 19]]

Answer 2

[可以像这样使用Comparator.comparing和Comparator.reverseOrder()对输入列表进行排序：

List<List<String>> list = Arrays.asList(
    Arrays.asList("2021", "04", "19"),
    Arrays.asList("2021", "06", "22"),
    Arrays.asList("2021", "06", "24")
    );
list.sort(Comparator.comparing(List::toString, Comparator.reverseOrder()));

System.out.println("Reversed = " + list);

输出：

Reversed = [[2021, 06, 24], [2021, 06, 22], [2021, 04, 19]]

---

另一种方法是在Comparator.comparing 之后立即应用方法Comparator::reversed，如下所示：

list.sort(Comparator.comparing(List<String>::toString).reversed());

---

更新

如果应该从某个现有列表中检索新的排序列表，这可以使用 Stream API 实现：Stream::sorted 和 Stream::collect 和 Collectors.toList() 收集器：

List<List<String>> sortedPartitionsValues = partitionValues
        .stream()
        .sorted(Comparator.comparing(List<String>::toString).reversed())
        .collect(Collectors.toList());

Answer 3

这似乎解决了问题：

public class Tester {

    public static void main(String[] args) {
        int [][] ar = new int[][] {{2021, 04, 19}, {2021, 06, 22}, {2021, 06, 24}};
        List<int[]> list = Arrays.asList(ar);

        System.out.println("original:");
        list.forEach(subAr -> {
            System.out.println("[" + subAr[0] + ", " + subAr[1] + ", " + subAr[2] + "]");
        });

        list.sort((lhs, rhs) -> {
            LocalDate dlhs = LocalDate.of(lhs[0], lhs[1], lhs[2]);
            LocalDate drhs = LocalDate.of(rhs[0], rhs[1], rhs[2]);
            return drhs.compareTo(dlhs);
        });
        System.out.println("sorted:");
        list.forEach(subAr -> {
            System.out.println("[" + subAr[0] + ", " + subAr[1] + ", " + subAr[2] + "]");
        });
    }
}

输出是：

original:
[2021, 4, 19]
[2021, 6, 22]
[2021, 6, 24]
sorted:
[2021, 6, 24]
[2021, 6, 22]
[2021, 4, 19]