154. Find Minimum in Rotated Sorted Array II

题目：

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 an array sorted in ascending order 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.

链接：https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/#/description

5/2/2017

有duplicate，O(N)时间复杂度

 1 public class Solution {
 2     public int findMin(int[] nums) {
 3         if (nums == null || nums.length == 0) {
 4             return -1;
 5         }
 6         for (int i = 1; i < nums.length; i++) {
 7             if (nums[i] < nums[i - 1]) {
 8                 return nums[i];
 9             }
10         }
11         return nums[0];
12     }
13 }

别人的答案

只让mid与end比较，但是时间复杂度还是O(N)

https://discuss.leetcode.com/topic/6468/my-pretty-simple-code-to-solve-it

 1 class Solution {
 2 public:
 3     int findMin(vector<int> &num) {
 4         int lo = 0;
 5         int hi = num.size() - 1;
 6         int mid = 0;
 7         
 8         while(lo < hi) {
 9             mid = lo + (hi - lo) / 2;
10             
11             if (num[mid] > num[hi]) {
12                 lo = mid + 1;
13             }
14             else if (num[mid] < num[hi]) {
15                 hi = mid;
16             }
17             else { // when num[mid] and num[hi] are same
18                 hi--;
19             }
20         }
21         return num[lo];
22     }
23 };