[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。

 1 class Solution {

 2 public:

 3     int findMinHelper(vector<int> &num, int left, int right) {

 4         int mid = (left + right) / 2;

 5         if (mid == left || mid == right) {

 6             return min(num[left], num[right]);

 7         }

 8         if (num[mid] < num[right]) {

 9             return findMinHelper(num, left, mid);

10         } else if (num[mid] > num[right]) {

11             return findMinHelper(num, mid, right);

12         } else {

13             return findMinHelper(num, left, right - 1);

14         }

15     }

16     int findMin(vector<int> &num) {

17        return findMinHelper(num, 0, num.size() - 1);

18     }

19 };

或者：

 1 class Solution {

 2 public:

 3     int findMinHelper(vector<int> &num, int left, int right) {

 4         int mid = (left + right) / 2;

 5         if (mid == left || mid == right) {

 6             return min(num[left], num[right]);

 7         }

 8         if (num[mid] < num[right]) {

 9             return findMinHelper(num, left, mid);

10         } else if (num[mid] > num[right]) {

11             return findMinHelper(num, mid, right);

12         } else {

13             int res = num[left];

14             for (int i = left; i <= right; ++i) {

15                 res = min(res, num[i]);

16             }

17             return res;

18         }

19     }

20     int findMin(vector<int> &num) {

21        return findMinHelper(num, 0, num.size() - 1);

22     }

23 };