这是霍格沃兹学院的一条完全笔直的街道，路两边有完全相同的房子。当然，你想知道街道另一边人们的门牌号。街道看起来是这样的:
1| |6 3| |4 5| |2右边的偶数递减，左边的奇数递增。房屋数量从1开始增加，没有间隔。例如，当 n = 3时，1与6相对，3与4相对，5与2相对。
请编写一个函数
over_the_road(x, n)，给你门牌号x和街道长度 n，返回街道对面的门牌号。难度：中等
来源：codewars: OVER THE ROAD
def over_the_road(x, n):
pass
assert 6 == over_the_road(1, 3)
assert 4 == over_the_road(3, 3)
assert 5 == over_the_road(2, 3)
assert 8 == over_the_road(3, 5)
def over_the_load(x, n):
list1 = []
list2 = []
for i in range(1, n+1):
list1.append(2 * i)
list2.append(2 * i - 1)
list1 = list1[::-1]
if x in list1:
num = list1.index(x)
return list2[num]
else:
num = list2.index(x)
return list1[num]
assert 6 == over_the_load(1, 3)
assert 4 == over_the_load(3, 3)
assert 5 == over_the_load(2, 3)
assert 8 == over_the_load(3, 5)
def over_the_road(x, n):
return (n * 2 + 1) - x
assert 6 == over_the_road(1, 3)
assert 4 == over_the_road(3, 3)
assert 5 == over_the_road(2, 3)
assert 8 == over_the_road(3, 5)
def over_the_road(x, n):
odds = list(filter(lambda x:x % 2 == 1, range(1, n*2+1)))
evens = sorted(list(filter(lambda x:x % 2 == 0, range(1, n*2+1))), reverse=True)
return evens[odds.index(x)] if x in odds else odds[evens.index(x)]
assert 6 == over_the_road(1, 3)
assert 4 == over_the_road(3, 3)
assert 5 == over_the_road(2, 3)
assert 8 == over_the_road(3, 5)
#conding=utf-8
def over_the_road(x, n):
"""
这是霍格沃兹学院的一条完全笔直的街道，路两边有完全相同的房子。当然，你想知道街道另一边人们的门牌号。街道看起来是这样的:
1| |6
3| |4
5| |2
右边的偶数递减，左边的奇数递增。房屋数量从1开始增加，没有间隔。例如，当 n = 3时，1与6相对，3与4相对，5与2相对。
请编写一个函数over_the_road(x, n)，给你门牌号x和街道长度 n，返回街道对面的门牌号
:return:
"""
a=[]
#创建空列表，把所有门牌号存进去
for i in range(1,n*2+1):
a.append(i)
#判断左边还是右边，两边计算方法不一样
# if x%2==0:
# print(f"{a[2*n-x]}| |{a[x-1]}")
# elif x%2==1:
# print(f"{a[x-1]}| |{a[2*n-x]}")
# if x%2==0:
# return(a[2*n-x])
# elif x%2==1:
# return (a[2*n-x])
return a[2*n-x]
if __name__ == '__main__':
# over_the_road(3, 6)
# over_the_road(4,4)
assert 6 == over_the_road(1, 3)
assert 4 == over_the_road(3, 3)
assert 5 == over_the_road(2, 3)
assert 8 == over_the_road(3, 5)
def over_the_road(x, n):
result_a = {}
result_b = {}
list_a = [i for i in range(1,2*n,2)] # 先将左侧一列的值存入列表a中
list_b = [j for j in range(2*n,1,-2)] # 再将右侧一列的值存入列表b中
# 结果1，左-右
for m in range(n):
result_a[list_a[m]] = list_b[m]
# 结果2，右-左
for o in range(n):
result_b[list_b[o]] = list_a[o]
# 取另一边人们的门牌号
if x in result_a.keys():
return result_a[x]
elif x in result_b.keys():
return result_b[x]
assert 6 == over_the_road(1, 3)
assert 4 == over_the_road(3, 3)
assert 5 == over_the_road(2, 3)
assert 8 == over_the_road(3, 5)
def over_the_road(x, n):
""" 返回街道门牌号"""
left_list = [i for i in range(1, 2*n + 1, 2)]
right_list = [j for j in range(2*n, -1, -2)]
if x in left_list:
return right_list[left_list.index(x)]
else:
return left_list[right_list.index(x)]