LeetCode——Find Minimum in Rotated Sorted Array II

Question

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.

Solution

抓住数组的特点。

Code

class Solution {
public:int findMin(vector<int>& nums) {int size = nums.size();if (size == 1)return nums[0];int low = 0;int high = size - 1;while (low < high) {// 没用递归，所以注意求中间节点的方法int mid = low + (high - low) / 2;// 如果更小，那么会比右边的都小if (nums[mid] < nums[high])high = mid;// 如果更大，那最小的肯定在右边else if (nums[mid] > nums[high])low = mid + 1;// 如果相等，不知道最大值在左边还是在右边elsehigh--;}return nums[low];}
};

转载于:https://www.cnblogs.com/zhonghuasong/p/7620023.html

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LeetCode——Find Minimum in Rotated Sorted Array II

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