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### 题目描述

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.

### 例子

#### 例子 1

Input: `nums = [3,4,5,1,2]`
Output: `1`
Explanation: `The original array was [1,2,3,4,5] rotated 3 times.`

#### 例子 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.`

#### 例子 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.

### 解题思路

• `Left < Right`：由于数组不包括重复数字，显然这种情况下数组本身有序，返回第一个元素即可
• `Mid > Left`: 可以确定 `Mid``A` 中，因此收缩左边界：`Left = Mid + 1`
• `Mid <= Left`：可以确定 `Mid``B` 中，因此收缩右边界，但是这里注意我们要获取的是 `B` 中第一个元素，所以 `Mid` 本身还有合理的搜索范围中，因此： `Right = Mid`

• 时间复杂度: `O(log n)`
• 空间复杂度: `O(1)`

GitHub 代码同步地址： 153.FindMinimumInRotatedSortedArray.cpp

GitHub: Leetcode-C++-Solution