# 33. Search in Rotated Sorted Array

Medium
There is an integer array `nums` sorted in ascending order (with distinct values).
Prior to being passed to your function, `nums` is possibly rotated at an unknown pivot index `k` (`1 <= k < nums.length`) such that the resulting array is `[nums[k], nums[k+1], ..., nums[n-1], nums, nums, ..., nums[k-1]]` (0-indexed). For example, `[0,1,2,4,5,6,7]` might be rotated at pivot index `3` and become `[4,5,6,7,0,1,2]`.
Given the array `nums` after the possible rotation and an integer `target`, return the index of `target` if it is in `nums`, or `-1` if it is not in `nums`.
You must write an algorithm with `O(log n)` runtime complexity.
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
Example 3:
Input: nums = , target = 0
Output:
-1
Constraints:
• `1 <= nums.length <= 5000`
• `-104 <= nums[i] <= 104`
• All values of `nums` are unique.
• `nums` is an ascending array that is possibly rotated.
• `-104 <= target <= 104`

### 解題

Runtime: 0 ms, faster than 100%
Memory Usage: 2.5 MB, less than 100%
func search(nums []int, target int) int {
left := 0
right := len(nums) - 1
for left <= right {
mid := (left + right)/2
if nums[mid] == target {
return mid
}
if nums[mid] >= nums[left] {
if target <= nums[mid] && target >= nums[left] {
right = mid - 1
} else {
left = mid + 1
}
} else {
if target >= nums[mid] && target <= nums[right] {
left = mid + 1
} else {
right = mid - 1
}
}
}
return -1
}