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
你已经有了最有效的表示——这是需要检查的最小值，除非你确定x1 < x2等，然后使用其他人提供的解决方案。
你可能应该注意到，一些编译器实际上会为你优化它——只要这4个表达式中的任何一个返回true就返回。如果其中一个返回true，那么最终结果也会返回true——因此可以跳过其他检查。
其他回答
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
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!")
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。
如果有人正在寻找计算实际重叠的一行程序:
int overlap = ( x2 > y1 || y2 < x1 ) ? 0 : (y2 >= y1 && x2 <= y1 ? y1 : y2) - ( x2 <= x1 && y2 >= x1 ? x1 : x2) + 1; //max 11 operations
如果你想要少一些操作，但多一些变量:
bool b1 = x2 <= y1;
bool b2 = y2 >= x1;
int overlap = ( !b1 || !b2 ) ? 0 : (y2 >= y1 && b1 ? y1 : y2) - ( x2 <= x1 && b2 ? x1 : x2) + 1; // max 9 operations
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)
