第一种无重复的

title

link

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.

You may assume no duplicate exists in the array.

Example 1:

Input: [3,4,5,1,2]

Output: 1

Example 2:

Input: [4,5,6,7,0,1,2]

Output: 0

solution

class Solution {
public:
    int findMin(vector<int>& nums) {
        int i=0,j=nums.size()-1;
        while(i<=j){
            int mid=(i+j)/2;
            if(i==j) return nums[mid];
            if(nums[mid]>nums[j]){
                i=mid+1;
            
            }else{
                j=mid;
            }
        }
        
    }
};

第二种有重复的

title

problem link

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.

Example 1:

Input: [1,3,5]
Output: 1
Example 2:

Input: [2,2,2,0,1]
Output: 0
Note:

This is a follow up problem to Find Minimum in Rotated Sorted Array.

Would allow duplicates affect the run-time complexity? How and why?

solution

class Solution {
public:
	int findMin(vector<int>& nums) {
		int l = 0, r = nums.size() - 1;
		while (l <= r) {
			int mid = (l+r)/2;
			if (nums[mid] == nums[r]) {
				r--;
			}
			else if (nums[mid] > nums[r]) {
				l = mid + 1;
			}
			else {
				r = mid;
			}
		}
		return nums[l];
		//return min(nums[l], nums[r]);
	}
};