# LeetCode: 153. Find Minimum in Rotated Sorted Array

## 题目¶

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.

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.

## 解法¶

• 二分查找，如果中间元素的值比右边尾部元素的值要小，说明旋转没有超过一半，最小值在左边，在左边元素中继续进行二分操作
• 如果中间元素的值比右边尾部元素的值要大，说明旋转已经超过一半了，最小值在右边，在右边元素中继续进行二分查找
• 二分到最后的那个元素就是最小的那个元素

```class Solution:
def findMin(self, nums):
if len(nums) <= 2:
return min(nums)

left = 0
right = len(nums) - 1
while left < right:
mid = left + (right - left) // 2
if nums[mid] < nums[right]:
right = mid
else:
left = mid + 1

return nums[right]
```