LeetCode Find Minimum In Rotated Sorted Array II

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

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.

第二轮：

class Solution {
public:
    /**
     * @param num: the rotated sorted array
     * @return: the minimum number in the array
     */
    int findMin(vector<int> &num) {
        // write your code here
        int lo = 0;
        int hi = num.size() - 1;
        
        while (lo < hi) {
            int mid = (lo + hi) / 2;
            if (num[mid] < num[hi]) {
                hi = mid;
            } else if (num[mid] > num[hi]) {
                lo = mid + 1;
            } else {
                hi--;
            }
        }
        return num[lo];
    }
};