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[0], nums[1], ..., 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 = [1], target = 0 Output: -1
Constraints:
1 <= nums.length <= 5000-104 <= nums[i] <= 104- All values of
numsare unique. numsis an ascending array that is possibly rotated.-104 <= target <= 104
class Solution:
def intersection(self, nums, left, right):
if nums[right] > nums[left]:
return 0
while left <= right:
mid = left + (right-left)//2
if nums[mid] > nums[mid+1]:
return mid + 1
else:
if nums[mid] < nums[left]:
right = mid - 1
else:
left = mid + 1
def binary_search(self, nums, target, left, right):
while left <= right:
mid = left + (right-left)//2
if nums[mid] == target:
return mid
elif nums[mid] > target:
right = mid - 1
else:
left = mid + 1
return -1
def search(self, nums: List[int], target: int) -> int:
n = len(nums)
if n == 1:
return 0 if nums[0] == target else -1
ins_point = self.intersection(nums, 0, n-1)
if target == nums[0]:
return 0
if ins_point == 0:
return self.binary_search(nums, target, 0, n-1 )
elif target > nums[0]:
return self.binary_search(nums, target, 0, ins_point)
else:
return self.binary_search(nums, target, ins_point, n-1)