计算两个一维线段的重叠距离

Question

尝试编写一个函数，该函数将返回两条线段之间的重叠距离总和，用开始和结束整数表示。

目前我有这个：我从互联网上得到的。

def overlap(min1, max1, min2, max2):
    """returns the overlap between two lines that are 1D"""
    result = None
    if min1 >= max2 or min2 >= max1: result =  0
    elif min1 <= min2:
        result = max1 - min2
    else: result = max2 - min1
    return result

对于0 100和0,20的情况，这种方法可以工作但返回的结果是100。很明显这是错误的。 有没有一种简单的计算方法可以返回正确的值?

- Tonis F. Piip

Answer 1

def overlap(min1, max1, min2, max2):
    return max(0, min(max1, max2) - max(min1, min2))

>>> overlap(0, 10, 80, 90)
0
>>> overlap(0, 50, 40, 90)
10
>>> overlap(0, 50, 40, 45)
5
>>> overlap(0, 100, 0, 20)
20

- jamylak

---

我喜欢这样简单的答案。 - Noctis Skytower

Answer 2

尚未完全测试，但这个方案如何 -

def overlap(min1,max1,min2,max2):
    start = max(min1,min2)
    end = min(max1,max2)
    d = end - start
    if d < 0:
        return 0
    else:
        return d

#some tests
print overlap(0,100,0,20)
print overlap(5,10,15,20)
print overlap(1,3,0,5)
print overlap(-5,5,-2,10)

>>> 
20
0
2
7

- Brad

Answer 3

在一维中，检查重叠的前提很简单（我认为）。找到最小值和最大值中较大的一个，以及最小值和最大值中较小的一个。然后将两者相减。

def overlap(min1, max1, min2, max2):
    #Find out the bigger minimum
    if min1 >= min2: 
        bigger_min = min1
    else: 
        bigger_min = min2
    if max1 >= max2: 
        smaller_max = max2
    else: 
        smaller_max = max1
    if smaller_max <= bigger_min: 
        return 0
    else: 
        return smaller_max - bigger_min

结果

>>> overlap(20,40,30,70)
10
>>> overlap(0,100,200,300)
0
>>> overlap(0,100,0,30)
30
>>> overlap(0,100,30,60)
30
>>> overlap(0,100,30,70)
40
>>> overlap(20,100,30,70)
40
>>> overlap(20,30,30,70)
0
>>> overlap(0,50,0,50)
50

- RedBaron