## # 33. Search in Rotated Sorted Array

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

Approach 1: binary search -- there's a case where the array is rotated less than a half, and another case where it's rotated more than a half.

### # Code (Python)

Approach 1:

``````class Solution:
def search(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
if not nums:
return -1
left, right = 0, len(nums) - 1
while left < right:
mid = (left + right) // 2
for index in (left, mid, right):
if target == nums[index]:
return index
if nums[mid] > nums[right]:
# example: [4,5,6,7,0,1,2]
if nums[left] < target < nums[mid]:
right = mid - 1
else:
left = mid + 1
else:
# example: [5,0,1,2,3]
if nums[mid] < target < nums[right]:
left = mid + 1
else:
right = mid - 1
return left if nums[left] == target else -1
``````

### # Code (C++)

Approach 1:

``````class Solution {
public:
int search(vector<int>& nums, int target) {
int tail = nums.size() - 1;
{
if (nums[mid] == target)
return mid;
{
if (target < nums[mid])
tail = mid - 1;
else
}
else
{
if (target < nums[mid])
{
if (nums[mid] <= nums[tail])
tail = mid - 1;
else if (nums[mid] > nums[tail] && target <= nums[tail])
else
tail = mid - 1;
}
else
{
if (nums[mid] > nums[tail])