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
西蒙的回答很好，但对我来说，相反的情况更容易思考。
什么时候两个范围不重叠?当其中一个开始后另一个结束时，它们不会重叠:
dont_overlap = x2 < y1 || x1 > y2
当它们重叠时，很容易表示:
overlap = !dont_overlap = !(x2 < y1 || x1 > y2) = (x2 >= y1 && x1 <= y2)
其他回答
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
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
以下是我的看法:
int xmin = min(x1,x2)
, xmax = max(x1,x2)
, ymin = min(y1,y2)
, ymax = max(y1,y2);
for (int i = xmin; i < xmax; ++i)
if (ymin <= i && i <= ymax)
return true;
return false;
除非您正在对数十亿个宽间距整数运行一些高性能的范围检查器，否则我们的版本应该执行类似的操作。我的观点是，这是微观优化。
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)。
