It is not binary search, cost O(n) time. But it's very easy to understand.

```
int findMin(vector<int>& nums) {
int n = nums.size();
for(int i=1; i<=n/2; ++i)
{
if(nums[n-i]<nums[n-i-1]) return nums[n-i];
if(nums[i]<nums[i-1]) return nums[i];
}
return nums[0]; //no rotation or all the same numbers in the nums
}
```