## [Leetcode]154. Find Minimum in Rotated Sorted Array II(C++)

### 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 = [0,1,4,4,5,6,7]` 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, a, a, ..., a[n-1]]` 1 time results in the array `[a[n-1], a, a, a, ..., a[n-2]]`.

Given the sorted rotated array `nums` that may contain duplicates, return the minimum element of this array.

### 例子

#### 例子 1

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

#### 例子 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.

This is the same as `Find Minimum in Rotated Sorted Array` but with duplicates. Would allow duplicates affect the run-time complexity? How and why?

### 解题思路

``````class Solution {
public:
int findMin(vector<int>& nums) {
int left = 0;
int right = nums.size() - 1;
while (left < right) {
// move the left boundary if its equal to right boundary
while (left < right && nums[left] == nums[right]) {
left++;
}

// if the array is already sorted or there's only 1 element, break
if (left == right || nums[left] < nums[right]) {
break;
}

int mid = (left + right) / 2;
if (nums[mid] >= nums[left]) {
left = mid + 1;
} else {
right = mid;
}
}

return nums[left];
}
};
``````
• 时间复杂度: `O(log n) -> O(n)`
• 空间复杂度: `O(1)`

GitHub 代码同步地址： 154.FindMinimumInRotatedSortedArrayIi.cpp

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