153. Find Minimum in Rotated Sorted Array
Medium
Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:
[4,5,6,7,0,1,2]if it was rotated4times.[0,1,2,4,5,6,7]if it was rotated7times.
Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].
Given the sorted rotated array nums of unique elements, return the minimum element of this array.
You must write an algorithm that runs in O(log n) time.
Example 1:
Input: nums = [3,4,5,1,2] Output: 1 Explanation: The original array was [1,2,3,4,5] rotated 3 times.
Example 2:
Input: nums = [4,5,6,7,0,1,2] Output: 0 Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.
Example 3:
Input: nums = [11,13,15,17] Output: 11 Explanation: The original array was [11,13,15,17] and it was rotated 4 times.
Constraints:
n == nums.length1 <= n <= 5000-5000 <= nums[i] <= 5000- All the integers of
numsare unique. numsis sorted and rotated between1andntimes.
class Solution:
def findMin(self, nums: List[int]) -> int:
# if nums[0] < nums[-1]:
# return nums[0]
# left, right = 0, len(nums)-1
# while left < right:
# mid = (left+right)//2
# if nums[mid] > nums[mid+1]:
# return nums[mid+1]
# elif nums[mid] > nums[0]:
# left = mid+1
# else:
# right = mid
# return nums[left]
left = 0
right = len(nums)-1
while left < right:
mid = (left + right)//2
if nums[right] < nums[mid]:
left = mid+1
elif nums[right] > nums[mid]:
right = mid
else:
right -= 1
return nums[left]