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.

一样，注意处理相同值的情况

1. int getMin(vector<int> &num, int start,int end) {
2. int min = num[start];
3. for(int i=start+1;i<end;i++) {
4. if(num[i] < min) min = num[i];
5. }
6. return min;
7. }
8. int findMin(vector<int> &num) {
10. int nSize = num.size();
11. if(nSize == 0) return 0;
12. if(nSize == 1) return num[0];
14. int start = 0;
15. int end = nSize - 1;
16. while(num[start]>=num[end]) {
17. if(end - start == 1) {
18. return num[end];
19. }
20. if(num[start] == num[end]) return getMin(num,start,end);
21. int mid = start + ((end - start) >> 1);
22. if(num[mid] > num[end]) {
23. start = mid;
24. } else {
25. end = mid;
26. }
27. }
28. return num[start];
29. }