Leetcode: Find Minimum in Rotated Sorted Array II
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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.
如果有重复,时间复杂度可能退化到O(n),因为不能确定最小元素在中点的哪一侧。
class Solution {public: int findMin(vector<int> &num) { int left = 0; int right = num.size() - 1; int mid; while (left <= right) { mid = left + (right - left) / 2; if (num[mid] > num[left]) { if (num[left] >= num[right]) { left = mid + 1; } else { right = mid; } } else if (num[mid] == num[left]) { if (num[mid] > num[right]) { left = mid + 1; } else if (num[mid] == num[right]) { ++left; } else { return num[mid]; } } else { right = mid; } } return num[right]; }};
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