Question

有没有简单的方法来定义两个数字对的重叠范围？例如， Pair1 = [360,780] Pair2 = [420,800]

答案是[420,780]

Answer 1

确保您的间隔重叠。

然后取两个对的下限的最大值和上限的最小值。

int lower = Math.max(myPair1.x1, myPair2.x1);
int upper = Math.min(myPair1.x2, myPair2.x2);

if(lower < upper)
  return new Pair(lower, upper)
else 
  throw new CustomException("Intervals not overlap")

Answer 2

我会创建一个类似

的类

class Range {
    final int start;
    final int end;
    private Range(int start; int end) { this.start = start; this.end = end; }
    public static Range of(int start, int end) { return new Range(start, end); }
    public Range and(Range r) {
        return new Range(Math.max(start, r.start), Math.min(end, r.end);
    }
}

你可以写

Range one = Range.of(360, 780);
Range two = Range.of(420, 800);
Range both = one.and(two);

Answer 3

这个怎么样：

int from = Math.max(pair1.getX(), pair2.getX());
int to = Math.min(pair1.getY(), pair2.getY());

if (from <= to) {
   return new Pair(from, to);
} else {
    return null;
}

Answer 4

有点滥用java.awt.geom，请尝试：

public static double[] createIntersection(double[] a1, double[] a2) {
    Rectangle2D r1 = new Rectangle2D.Double(), r2 = new Rectangle2D.Double();
    r1.setFrameFromDiagonal(a1[0], 0, a1[1], 1);
    r2.setFrameFromDiagonal(a2[0], 0, a2[1], 1);
    Rectangle2D r3 = r1.createIntersection(r2);
    return new double[] { r3.getMinX(), r3.getMaxX() };
}

两个数字对的重叠范围

4 个答案: