# Question

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

Your algorithm’s runtime complexity must be in the order of O(log n).

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4


Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1


# Solution

Use binary search.

class Solution:
def search(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
if len(nums) == 0:
return -1

left, right = 0, len(nums) - 1

if nums[left] == target:
return left
elif nums[right] == target:
return right

while True:
if right - left <= 1:
return -1

mid = (left + right) // 2
if nums[mid] == target:
return mid
if nums[left] < target < nums[mid]:
right = mid
elif nums[mid] < target < nums[right]:
left = mid
elif nums[mid] > nums[left]:
left = mid
elif nums[right] > nums[mid]:
right = mid
else:
return -1