[LeetCode] Find Minimum in Rotated Sorted Array II

Follow up for "Find Minimum in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

Find the minimum element.

The array may contain duplicates.

跟上一题一样，注意相等的情况要特殊考虑。

class Solution {
public:
    int findMinHelper(vector<int> &num, int left, int right) {
        int mid = (left + right) / 2;
        if (mid == left || mid == right) {
            return min(num[left], num[right]);
        }
        if (num[mid] < num[right]) {
            return findMinHelper(num, left, mid);
        } else if (num[mid] > num[right]) {
            return findMinHelper(num, mid, right);
        } else {
            return findMinHelper(num, left, right - 1);
        }
    }
    int findMin(vector<int> &num) {
       return findMinHelper(num, 0, num.size() - 1);
    }
};

或者：

class Solution {
public:
    int findMinHelper(vector<int> &num, int left, int right) {
        int mid = (left + right) / 2;
        if (mid == left || mid == right) {
            return min(num[left], num[right]);
        }
        if (num[mid] < num[right]) {
            return findMinHelper(num, left, mid);
        } else if (num[mid] > num[right]) {
            return findMinHelper(num, mid, right);
        } else {
            int res = num[left];
            for (int i = left; i <= right; ++i) {
                res = min(res, num[i]);
            }
            return res;
        }
    }
    int findMin(vector<int> &num) {
       return findMinHelper(num, 0, num.size() - 1);
    }
};