Scan the binary form of the number from bit 1 to bit N-1, find count of numbers ends with 0 and 1

The ugly part is variable `c1`

, `pre`

and `current`

, this is to deal with the case of 10, from 4(100) to 8(1000), number ends with 0 in 4 contains 100, to number ends with 1 in 8 contains 1001, which is greater than 8, so needs to deduce 1. But if consecutive 1 appears, like 6(110) ->12(1100), then no need to worry about -1 since `e0`

of 6 does not contain 110

```
class Solution(object):
def findIntegers(self, num):
"""
:type num: int
:rtype: int
"""
if num <= 1:
return num + 1
e0 = 1
e1 = 1
#c1: consecutive 1 appears in the number
c1 = False
n = len(bin(num)) - 2
for i in range(n-2, -1, -1):
# pre: previous bit, curr: current bit
pre = (1 << i) & num
curr = (1 << (i + 1)) & num
e0, e1 = e1 + e0, e0
if not pre and not curr and not c1:
e1 -= 1
if not c1 and pre and curr:
c1 = True
return e0 + e1
```