153. Find Minimum in Rotated Sorted Array

153. Find Minimum in Rotated Sorted Array

Medium

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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 = [0,1,2,4,5,6,7] 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.

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

        All the integers of nums are unique.

        解题思路：若旋转过，则旋转处后面的数字递增且均小于第一个数。且题目保证数字不重复，用二分查找
        若 nums[l] < nums[mid] ，则[l,mid] 递增，nums[l]是这个区间内最小的数，和右区间比较
        右区间可能是个一直递增的数组，则nums[l]是最小的数，也可能有旋转，则旋转处是最小的数
        同理 nums[l] > nums[mid], 证明[l,mid] 之间出现了旋转处，nums[mid]可能是最小的，也可能不是
        """

        # 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]:
                # [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)