```
class Solution(object):
def findMin(self, nums):
l = 0; r = len(nums)-1
while(l<r):
mid = l +(r-l)/2
if (nums[mid]<nums[-1]):
r = mid
else:
l = mid+1
return nums[l]
```

If there's a rotation, then the list consists of two increasing sequences: a1,a2,...an, then b1,b2,bn, where

bn <a1 since bn was in front of a1 before rotation.

So the problem could be converted to find the left most element which is smaller than the last element bn.

This could be solved by using standard binary search techniques. The testing function is f(x) = x<bn, and is a 00001111 style problem.