给定两个包含范围[x1:x2]和[y1:y2]，其中x1≤x2和y1≤y2，测试这两个范围是否有重叠的最有效方法是什么?
一个简单的实现如下:
bool testOverlap(int x1, int x2, int y1, int y2) {
return (x1 >= y1 && x1 <= y2) ||
(x2 >= y1 && x2 <= y2) ||
(y1 >= x1 && y1 <= x2) ||
(y2 >= x1 && y2 <= x2);
}
但是我希望有更有效的方法来计算这个。
就最少的操作而言，哪种方法是最有效的?
当前回答
反过来思考:如何使这两个范围不重叠?给定[x1, x2]，则[y1, y2]应在[x1, x2]之外，即y1 < y2 < x1或x2 < y1 < y2，这等价于y2 < x1或x2 < y1。
因此，使两个范围重叠的条件是:不(y2 < x1或x2 < y1)，这相当于y2 >= x1和x2 >= y1(与Simon接受的答案相同)。
其他回答
值域重叠是什么意思?这意味着存在一个在两个范围内的数C，即。
x1 <= C <= x2
and
y1 <= C <= y2
为了避免混淆，考虑范围为: [x1:x2]和[y1:y2]
现在，如果我们可以假设范围是构造良好的(因此x1 <= x2和y1 <= y2)，那么就足以进行测试
x1 <= y2 && y1 <= x2
OR
(StartA <= EndB)和(EndA >= StartB)
我的情况不同。我要检查两个时间范围是否重叠。不应该有单位时间的重叠。这里是Go的实现。
func CheckRange(as, ae, bs, be int) bool {
return (as >= be) != (ae > bs)
}
测试用例
if CheckRange(2, 8, 2, 4) != true {
t.Error("Expected 2,8,2,4 to equal TRUE")
}
if CheckRange(2, 8, 2, 4) != true {
t.Error("Expected 2,8,2,4 to equal TRUE")
}
if CheckRange(2, 8, 6, 9) != true {
t.Error("Expected 2,8,6,9 to equal TRUE")
}
if CheckRange(2, 8, 8, 9) != false {
t.Error("Expected 2,8,8,9 to equal FALSE")
}
if CheckRange(2, 8, 4, 6) != true {
t.Error("Expected 2,8,4,6 to equal TRUE")
}
if CheckRange(2, 8, 1, 9) != true {
t.Error("Expected 2,8,1,9 to equal TRUE")
}
if CheckRange(4, 8, 1, 3) != false {
t.Error("Expected 4,8,1,3 to equal FALSE")
}
if CheckRange(4, 8, 1, 4) != false {
t.Error("Expected 4,8,1,4 to equal FALSE")
}
if CheckRange(2, 5, 6, 9) != false {
t.Error("Expected 2,5,6,9 to equal FALSE")
}
if CheckRange(2, 5, 5, 9) != false {
t.Error("Expected 2,5,5,9 to equal FALSE")
}
你可以在边界比较中看到异或模式
给定两个范围[x1,x2]， [y1,y2]
def is_overlapping(x1,x2,y1,y2):
return max(x1,y1) <= min(x2,y2)
重叠(X, Y):= if (X1 <= Y1) then (Y1 <= X2) else (X1 <= Y2)。
证明:
考虑X在Y之前或与Y左对齐的情况，即X1 <= Y1。那么Y要么在X内部开始，要么在X的末尾开始，即Y1 <= X2;或者Y远离x，第一个条件是重叠;第二个，不是。
在互补的情况下，Y在X之前，同样的逻辑适用于交换的实体。
So,
重叠(X, Y):= if (X1 <= Y) then (Y1 <= X2) else重叠(Y, X)。
但这似乎并不完全正确。在递归调用中，第一个测试是多余的，因为我们已经从第一个调用的第一个测试中知道了实体的相对位置。因此，我们实际上只需要测试第二个条件，即交换后(X1 <= Y2)。所以,
重叠(X, Y):= if (X1 <= Y1) then (Y1 <= X2) else (X1 <= Y2)。
QED.
Ada的实现:
type Range_T is array (1 .. 2) of Integer;
function Overlap (X, Y: Range_T) return Boolean is
(if X(1) <= Y(1) then Y(1) <= X(2) else X(1) <= Y(2));
测试程序:
with Ada.Text_IO; use Ada.Text_IO;
procedure Main is
type Range_T is array (1 .. 2) of Integer;
function Overlap (X, Y: Range_T) return Boolean is
(if X(1) <= Y(1) then Y(1) <= X(2) else X(1) <= Y(2));
function Img (X: Range_T) return String is
(" [" & X(1)'Img & X(2)'Img & " ] ");
procedure Test (X, Y: Range_T; Expect: Boolean) is
B: Boolean := Overlap (X, Y);
begin
Put_Line
(Img (X) & " and " & Img (Y) &
(if B then " overlap .......... "
else " do not overlap ... ") &
(if B = Expect then "PASS" else "FAIL"));
end;
begin
Test ( (1, 2), (2, 3), True); -- chained
Test ( (2, 3), (1, 2), True);
Test ( (4, 9), (5, 7), True); -- inside
Test ( (5, 7), (4, 9), True);
Test ( (1, 5), (3, 7), True); -- proper overlap
Test ( (3, 7), (1, 5), True);
Test ( (1, 2), (3, 4), False); -- back to back
Test ( (3, 4), (1, 2), False);
Test ( (1, 2), (5, 7), False); -- disjoint
Test ( (5, 7), (1, 2), False);
end;
以上程序输出:
[ 1 2 ] and [ 2 3 ] overlap .......... PASS
[ 2 3 ] and [ 1 2 ] overlap .......... PASS
[ 4 9 ] and [ 5 7 ] overlap .......... PASS
[ 5 7 ] and [ 4 9 ] overlap .......... PASS
[ 1 5 ] and [ 3 7 ] overlap .......... PASS
[ 3 7 ] and [ 1 5 ] overlap .......... PASS
[ 1 2 ] and [ 3 4 ] do not overlap ... PASS
[ 3 4 ] and [ 1 2 ] do not overlap ... PASS
[ 1 2 ] and [ 5 7 ] do not overlap ... PASS
[ 5 7 ] and [ 1 2 ] do not overlap ... PASS
反过来思考:如何使这两个范围不重叠?给定[x1, x2]，则[y1, y2]应在[x1, x2]之外，即y1 < y2 < x1或x2 < y1 < y2，这等价于y2 < x1或x2 < y1。
因此，使两个范围重叠的条件是:不(y2 < x1或x2 < y1)，这相当于y2 >= x1和x2 >= y1(与Simon接受的答案相同)。
