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
什么新东西。只是可读性更强。
def overlap(event_1, event_2):
start_time_1 = event_1[0]
end_time_1 = event_1[1]
start_time_2 = event_2[0]
end_time_2 = event_2[1]
start_late = max(start_time_1, start_time_2)
end_early = min(end_time_1, end_time_2)
# The event that starts late should only be after the event ending early.
if start_late > end_early:
print("Absoloutly No overlap!")
else:
print("Events do overlap!")
其他回答
return x2 >= y1 && x1 <= y2;
为什么这样做有效: 范围不重叠的唯一情况是当一个范围的结束在另一个范围的开始之前。所以我们想要!(x2 < y1 || x1 > y2)它等价于上面。
如果你正在处理，给定两个范围[x1:x2]和[y1:y2]，自然/反自然顺序范围同时存在:
自然顺序:x1 <= x2 && y1 <= y2或 反自然顺序:x1 >= x2 && y1 >= y2
然后你可能想用这个来检查:
它们重叠<=> (y2 - x1) * (x2 - y1) >= 0
其中只涉及四个操作:
2倍 一个乘法 一个比较
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)
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)。
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
