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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.

Solution

和Find Minimum in Rotated Sorted Array I 类似，不过需要考虑a[mid] == a[right]的情况,

此时只需要把right--即可

Code

int findMin(vector<int> &nums) {
	int n = nums.size();
	if (n == 0)
		return INT_MIN;
	int s = 0, t = n - 1;
	while (s < t) {
		if (nums[s] < nums[t])
			return nums[s];
		int mid = (s + t) >> 1;
		if (nums[mid] > nums[t]) {
			s = mid + 1;
		} else if (nums[mid] < nums[t]) {
			t = mid;
		} else {
			t--;
		}
	}
	return nums[s];
}

扩展

Search in Rotated Sorted Array, 从旋转列表中查找某个值

Find Minimum in Rotated Sorted Array