# 81. Search in Rotated Sorted Array II ⭐

Medium
There is an integer array `nums` sorted in non-decreasing order (not necessarily with distinct values).
Before being passed to your function, `nums` is rotated at an unknown pivot index `k` (`0 <= 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, `[0,1,2,4,4,4,5,6,6,7]` might be rotated at pivot index `5` and become `[4,5,6,6,7,0,1,2,4,4]`.
Given the array `nums` after the rotation and an integer `target`, return `true` if `target` is in `nums`, or `false` if it is not in `nums`.
You must decrease the overall operation steps as much as possible.
Example 1:
Input: nums = [2,5,6,0,0,1,2], target = 0
Output:
true
Example 2:
Input: nums = [2,5,6,0,0,1,2], target = 3
Output:
false
Constraints:
• `1 <= nums.length <= 5000`
• `-104 <= nums[i] <= 104`
• `nums` is guaranteed to be rotated at some pivot.
• `-104 <= target <= 104`
Follow up: This problem is similar to Search in Rotated Sorted Array, but `nums` may contain duplicates. Would this affect the runtime complexity? How and why?

### 解題

func search(nums []int, target int) bool {
left := 0
right := len(nums) - 1
for left <= right {
mid := (left + right) / 2
if nums[mid] == target {
return true
}
if nums[left] == target || nums[right] == target {
return true
}
if nums[left] == nums[right] {
right--
left++
continue
}
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 false
}