動機

這題可以與33一起看，會有新的理解

Problem

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: 1Explanation: The original array was [1,2,3,4,5] rotated 3 times.

Example 2:

Input: nums = [4,5,6,7,0,1,2]Output: 0Explanation: 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: 11Explanation: 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.

Sol

對rotate的長度做二分

一直捨棄有序的部分，留下的就是最小!!

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