154. Find Minimum in Rotated Sorted Array II

154. Find Minimum in Rotated Sorted Array II

Hard

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Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,4,4,5,6,7] might become:

[4,5,6,7,0,1,4] if it was rotated 4 times. [0,1,4,4,5,6,7] 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 that may contain duplicates, return the minimum element of this array.

You must decrease the overall operation steps as much as possible.

Example 1:

Input: nums = [1,3,5]
Output: 1

Example 2:

Input: nums = [2,2,2,0,1]
Output: 0

Constraints:

n == nums.length 1 <= n <= 5000 -5000 <= nums[i] <= 5000 nums is sorted and rotated between 1 and n times.

Follow up: This problem is similar to , but nums may contain duplicates. Would this affect the runtime complexity? How and why?

class Solution:
    def findMin(self, nums: List[int]) -> int:
        """
        assert Solution().findMin([5, 1, 2, 3, 4]) == 1
        assert Solution().findMin([2, 1]) == 1
        assert Solution().findMin([3, 4, 5, 1, 2]) == 1
        assert Solution().findMin([1]) == 1
        assert Solution().findMin([4, 5, 6, 7, 0, 1, 2]) == 0
        assert Solution().findMin([11, 13, 15, 17]) == 11
        assert Solution().findMin([2, 2, 2, 0, 1]) == 0
        assert Solution().findMin([2, 2, 2]) == 2
        
        解题思路：参考题目 153. Find Minimum in Rotated Sorted Array
        这题会出现重复的数，即当 nums[l] == nums[mid]时，不知道最小值在左边还是右边，则l++
        如 [2, 2, 2, 0, 1] 和 [1, 1, 1, 2, 3]
        时间复杂度：最坏O(n)
        """

        # 二分查找
        def find(l: int, r: int) -> int:
            if l >= r:
                return nums[l]
            if r - l <= 1:
                # 解决[l,r]间隔1，mid总是等于l,忽略了r
                return min(nums[l], nums[r])
            mid = int((l + r) / 2)
            if nums[l] == nums[mid]:
                return find(l + 1, r)
            elif nums[l] < nums[mid]:
                # [l,mid] 递增
                num = find(mid + 1, r)
                return min(nums[l], num)
            else:
                # 肯定在[l,mid]之间
                num = find(l, mid - 1)
                return min(nums[mid], num)

        return find(0, len(nums) - 1)