java—如何比较(x,y)点进行排序,其中x总是第一个

t98cgbkg  于 2021-06-30  发布在  Java
关注(0)|答案(1)|浏览(362)

我使用javacomparator将位置对象与点(x,y)进行比较。
我需要能够比较两个点,以检索一个正整数或负整数,这将允许我排序(x,y)点,其中x值排序第一,然后y值第二(如果这有意义……)例如:

(3,4) (2,5) (1,1) (1,3) (3,3)

变成这样:

(1,1) (1,3) (2,5) (3,3) (3,4)

我认为有一种方法可以做到这一点,就是给x值一个大的优先级,把它乘以一个大的数字,比如1000。这样:比较(3,3)和(1,1):

int x_multiplier = 1000;
int value1 = (p1.x * x_multiplier ) + p1.y; // = 3 * 1000 + 3 = 3003
int value2 = (p2.x * x_multiplier ) + p2.y; // = 1 * 1000 + 1 = 1001
return value1-value2; // = 2002. Value1 is greater, thus p1 be later in list.

这是可行的,但问题是如果y值应该等于或大于x\u乘数,那么这就失败了(因为y值现在等于1 x值。。。再说一次,如果这有道理的话。)

// Comparing p1 = (2,0) & p2 = (1,18)
int x_multiplier = 10;
int value1 = (p1.x * x_multiplier ) + p1.y; // = 2 * 10 + 0  = 20
int value2 = (p2.x * x_multiplier ) + p2.y; // = 1 * 10 + 18 = 28
return value1-value2; // = -8, value2 is greater, and thus p2 will be later in the list. However, we know by looking at the points that p2 should come first.

我甚至不知道如何去寻找这个,所以如果有答案的话我就找不到了。

ckocjqey

ckocjqey1#

import java.util.*;

class Tuple {
    public int x, y;

    Tuple(int x, int y) {
        this.x = x;
        this.y = y;
    }

    @Override
    public String toString() {
        return "(" + x + ", " + y + ") ";
    }
}

public class coord {
    public static void main(String[] args) {
        LinkedList<Tuple> list = new LinkedList<Tuple>();
        list.add(new Tuple(3,4));
        list.add(new Tuple(2,5));
        list.add(new Tuple(1,1));
        list.add(new Tuple(1,3));
        list.add(new Tuple(3,3));
        for (Tuple t: list) {
            System.out.print(t);
        }
        Collections.sort(list, (Tuple t1, Tuple t2) -> {
            int result = Integer.compare(t1.x, t2.x);
            if (result == 0 ) result = Integer.compare(t1.y, t2.y);
            return result;
        });
        System.out.println("Sorted List: ");
        for (Tuple t: list) {
            System.out.print(t);
        }
    }
}

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