33. Search in Rotated Sorted Array
Đề Bài
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
Thuật Toán & Kỹ Thuật
⏱️ Thời gian
O(log n)
💾 Không gian
O(n)
Lời Giải
Python
0033-search-in-rotated-sorted-array.py
class Solution:
def searchMinimum(self, nums, left, right):
while left <= right:
mid = left + (right - left) // 2
if (mid == left or nums[mid - 1] > nums[mid]) and (mid == right or nums[mid + 1] > nums[mid]):
return mid
elif nums[mid] > nums[right]:
left = mid + 1
else:
right = mid - 1
def binarySearch(self, nums, left, right, target):
while left <= right:
mid = left + (right - left) // 2
if nums[mid] == target:
return mid
elif nums[mid] < target:
left = mid + 1
else:
right = mid - 1
return -1
def search(self, nums: List[int], target: int) -> int:
n = len(nums)
pivot = self.searchMinimum(nums, 0, n - 1)
res = self.binarySearch(nums, 0, pivot - 1, target)
if res != -1:
return res
return self.binarySearch(nums, pivot, n - 1, target)