[LeetCode] Find Minimum in Rotated Sorted Array II

This problem is more or less the same as Find Minimum in Rotated Sorted Array. And one key difference is as stated in the solution tag. That is, due to duplicates, we may not be able to throw one half sometimes. And in this case, we could just apply linear search and the time complexity will become O(n).

The idea to solve this problem is still to use invariants. We set l to be the left pointer and r to be the right pointer. Since duplicates exist, the invatiant is nums[l] >= nums[r] (if it does not hold, then nums[l] will simply be the minimum). We then begin binary search by comparingnums[l], nums[r] with nums[mid].

1. If nums[l] = nums[r] = nums[mid], simply apply linear search within nums[l..r].
2. If nums[mid] <= nums[r], then the mininum cannot appear right to mid, so set r = mid;
3. If nums[mid] > nums[r], then mid is in the first larger half and r is in the second smaller half, so the minimum is to the right of mid: set l = mid + 1.

The code is as follows.

 1 class Solution {
 2 public:
 3     int findMin(vector<int>& nums) {
 4         int l = 0, r = nums.size() - 1;
 5         while (nums[l] >= nums[r]) {
 6             int mid = (l & r) + ((l ^ r) >> 1);
 7             if (nums[l] == nums[r] && nums[mid] == nums[l])
 8                 return findMinLinear(nums, l, r);
 9             if (nums[mid] <= nums[r]) r = mid;
10             else l = mid + 1;
11         }
12         return nums[l];
13     } 
14 private:
15     int findMinLinear(vector<int>& nums, int l, int r) {
16         int minnum = nums[l];
17         for (int p = l + 1; p <= r; p++)
18             minnum = min(minnum, nums[p]);
19         return minnum;
20     }
21 };