[LeetCode] 0153. find minimum in rotated sorted array

find minimum in rotated sorted array

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內容目錄

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 rotated 4 times.
  • [0,1,2,4,5,6,7] if it was rotated 7 times.

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.length
  • 1 <= n <= 5000
  • -5000 <= nums[i] <= 5000
  • All the integers of nums are unique.
  • nums is sorted and rotated between 1 and n times.

Python

				
					# time complexity: O(logn)
# space complexity: O(1)
from typing import List


class Solution:
    def findMin(self, nums: List[int]) -> int:
        if len(nums) == 1:
            return nums[0]
        left, right = 0, len(nums) - 1
        if nums[right] > nums[0]:
            return nums[0]
        while left <= right:
            mid = left + (right - left) // 2
            if nums[mid] > nums[mid + 1]:
                return nums[mid+1]
            if nums[mid] < nums[mid - 1]:
                return nums[mid]
            if nums[mid] > nums[-1]:
                left = mid + 1
            else:
                right = mid - 1
        return nums[mid]


nums = [11, 13, 15, 17]

print(Solution().findMin(nums))
				
			
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