I know, I know, this is not the O(logn) solution :) But this is definitely the most intuitive one

```
class Solution:
# @return a float
def findMedianSortedArrays(self, A, B):
med1 = med2 = i = j = 0
n = len(A) + len(B)
while (i + j) <= n / 2:
if i < len(A) and j < len(B):
med2 = med1
if A[i] < B[j]:
med1 = A[i]
i += 1
else:
med1 = B[j]
j += 1
elif i < len(A):
med2 = med1
med1 = A[i]
i += 1
elif j < len(B):
med2 = med1
med1 = B[j]
j += 1
if n % 2 == 0:
return (med1 + med2) / 2.0
else:
return med1
```