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[0], nums[1], ..., 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?
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
}