如何比较整数列表然后对其进行升序排序？

Question

我正在学习如何对 ArrayList 中的正方形进行排序和比较。正方形的尺寸首先按长度排序。如果两个方格的长度相同，则要求宽度较短的方格在前。

public class Square {
private int length;
private int width;

public Square(int length, int width) {
    this.length = length;
    this.width = width;
}
}

public class Dimension {
private int length;
private int width;
ArrayList<Square> Dimension = new ArrayList<>();

public void addSquare(int length, int width) {
    Dimension.add(new Square(length, width));
}

public void sortDimension() {
    Collections.sort(Dimension, (length, width) -> length.compareTo(width));
}

public static void main(String[] args) {
    Square square;
    Dimension dimension = new Dimension();
    dimension.addSquare(10, 5);
    dimension.addSquare(8, 8);
    dimension.addSquare(10, 2);
    dimension.addSquare(12, 10);
    dimension.addSquare(8, 5);
    dimension.sortDimension();

    for(int i=0; i<Dimension.size();i++ ) {
        System.out.println(Dimension.get(i));
    }
}
}

我在互联网上找到了 Collections.sort(List, (obj1, obj2) -> obj1.compareTo(obj2))，但它不适用于我的代码。编译器说 cannot convert from Obj to int.

我需要你对这件事的建议。谢谢。

Answer 1

不要处处使用Dimension这个词，对于class，变量命名，这里有一些建议

ArrayList<Square> dims = new ArrayList<>(); // for the class attribut

Dimension value = new Dimension(); // in the main

当提供自定义Comparator时，你得到一对对象，需要确定第一个，你没有得到它的属性

Collections.sort(dims, (o1, o2) -> ); // Both o1 and o2 are Square instances

// You'd get for something like
Collections.sort(dims, (o1, o2) -> o1.getLength() == o2.getLength() ?
                                   o1.getWidth() - o2.getWidth() :
                                   o1.getLength() - o2.getLength());

但是你可以使用 Comparator 接口，它提供了很好的方法

dims.sort(Comparator.comparing(Square::getLength).thenComparing(Square::getWidth));

然后在主要访问您需要从实例中访问的列表，如value.dims.size()。在 Square class 中有一个很好的 toString 没关系

class Square {
    private int length, width;
    public Square(int length, int width) {
        this.length = length;
        this.width = width;
    }
    public int getLength() {return length;}
    public int getWidth() {return width; }
    @Override
    public String toString() {return "Square{" + "length=" + length + ", width=" + width + '}';}
}

class Dimension {

    private ArrayList<Square> dims = new ArrayList<>();

    public static void main(String[] args) {
        Dimension value = new Dimension();
        value.addSquare(10, 5);
        value.addSquare(8, 8);
        value.addSquare(10, 2);
        value.addSquare(12, 10);
        value.addSquare(8, 5);
        value.sortDimension();

        for (int i = 0; i < value.dims.size(); i++) {
            System.out.println(value.dims.get(i));
        }
    }

    public void addSquare(int length, int width) {
        dims.add(new Square(length, width));
    }
    public void sortDimension() {
        dims.sort(Comparator.comparing(Square::getLength).thenComparing(Square::getWidth));
    }
}

如何比较整数列表然后对其进行升序排序？

How to compare a list of integers and then sort it ascending?

java

sorting

collections

arraylist

hashmap