英文:
How to compare (x,y) points for sorting, where x is always first
问题
使用Java中的Comparator来比较具有坐标点（x，y）的位置对象。
我需要能够比较两个点，以获取一个正整数或负整数，使我能够对（x，y）点进行排序，其中首先对x值进行排序，然后对y值进行排序。（如果这样说得通的话...）
例如，这个：
(3,4) (2,5) (1,1) (1,3) (3,3)
变成了这个：
(1,1) (1,3) (2,5) (3,3) (3,4)
我思考过的一种方法实际上是通过将x值乘以一个大数值（比如1000）来赋予x值更高的优先级。像这样：
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较大，因此p1会在列表中位于p2之后。
这种方法是有效的，但问题在于，如果y值等于或大于x_multiplier，那么这种方法会失效（因为该y值现在等于1个x值...如果这样说得通的话）。
// 比较 p1 =（2,0）和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较大，因此p2将会在列表中位于p1之后。然而，通过查看这些点，我们知道p2应该在p1之前。
我实际上甚至不知道如何搜索这个问题，所以如果有答案，我也找不到。
英文:
I'm using Java Comparator to compare location objects with points (x,y).
I need to be able to compare two points to retrieve a positive or negative integer that will allow me to sort (x,y) points, where x-values are sorted first, then y-values secondly. (If that makes sense...)
For example, this:
(3,4) (2,5) (1,1) (1,3) (3,3)
Becomes this:
(1,1) (1,3) (2,5) (3,3) (3,4)
One way I've thought to do it is essentially by giving the x-value large precedence by multiplying it by a large number like 1000. Like so:
Comparing (3,3) and (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.
This works, but the issue with this is if the Y-value should ever be equal to or greater than the x_multiplier, then this breaks down (because that y-value is now equal to 1 x-value... again, if that makes sense.)
// 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.
I don't really even know how to search for this, so if there are answers out there I couldn't find them.
答案1
得分: 0
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);
}
}
}
英文:
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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