def diap_to_prefix(a, b):
lst = ['%0*d'%(max(len(str(a)), len(str(b))), x) for x in range(int(a), int(b)+1)]
new_lst = []
while len(lst) != len(new_lst):
lst = new_lst or lst
new_lst = []
c = lst[0]
tmp_lst = [c]
for i in lst[1:]:
if c[:-1] == i[:-1]:
c = i
tmp_lst.append(c)
else:
if len(tmp_lst) == 10:
new_lst.append(c[:-1])
else:
new_lst.extend(tmp_lst)
c = i
tmp_lst = [c]
if len(tmp_lst) == 10:
new_lst.append(c[:-1])
else:
new_lst.extend(tmp_lst)
return lst
def diap_to_prefix(a, b):
def inner(aa, bb, p):
if p == 1:
if a <= aa <= b:
yield aa
return
for d in range(aa, bb + 1, p):
if a <= d and d + p - 1 <= b:
yield d // p
elif not (bb < a or aa > b):
for i in range(10):
yield from inner(d + i * p // 10, d + (i + 1) * p // 10 - 1, p // 10)
a, b = int(a), int(b)
p = 10**(max(len(str(x)) for x in (a, b)) - 1)
yield from inner(a // p * p, b // p * p + p - 1, p)
phone_len = 7
R = "33312345-3332345".split("-")
prefix = ""
for i in range(len(R[0])):
if R[0][i] == R[1][i]:
prefix += R[0][i]
else:
break
m = int(R[0]+"0"*(phone_len-len(R[0])))
M = int(R[1]+"9"*(phone_len-len(R[0])))
phones = [str(n) for n in range(m, M+1)]
我的工作代码。也不是很快。欢迎优化。在
我的新的更优化的解决方案(py3.4)
您需要获取以“-”分隔的值的公共前缀,因此:
.split
来获取这些内容,并对其进行迭代,直到找到不同之处phone_len
位,并对最大值执行相同的操作(使用9)这里是:
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