I got an interview and the question is how to get the length of repeating decimal?

For example

```
1/3=0.3333..., it returns 1,
5/7=0.7142857142857143, it returns 6, since 714285 is the repeating decimal.
1/15=0.066666666666666, it returns 1.
17/150=0.11333333333333333, it returns 1. since 3 is the repeating decimal.
```

And I have tried to write a code

```
def solution(a, b):
n = a % b
if n == 0:
return 0
mem = []
n *= 10
while True:
n = n % b
if n == 0:
return 0
if n in mem:
i = mem.index(n)
return len(mem[i:])
else:
mem.append(n)
n *= 10
```

However, my code can't pass all tests. And it's time complexity is O(n*logn). How can I improve that?