[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.

用了一个很愚笨的方式，先快速排序，然后返回第一个元素，即最小值。

class Solution {public:  void QuickSort(vector<int>& num, int l, int r){        if (l<r){            int i = l, j = r;            int x = num[l];            while (i<j){                while (i<j&&num[j] >= x)                    j--;                if (i<j)                    num[i++] = num[j];                while (i < j&&num[i]<x)                    i++;                if (i<j)                    num[j--] = num[i];            }            num[i] = x;            QuickSort(num, l, i - 1);            QuickSort(num, i + 1, r);        }    }    int findMin(vector<int> &num) {        if(num.size()>0){            QuickSort(num,0,num.size()-1);            return num[0];        }else{            exit(0);        }    }};