LeetCode 154. Find Minimum in Rotated Sorted Array II

Binary Search

Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,4,4,5,6,7] might become:

• [4,5,6,7,0,1,4] if it was rotated 4 times.

• [0,1,4,4,5,6,7] if it was rotated 7 times.

Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].

Given the sorted rotated array nums that may contain duplicates, return the minimum element of this array.

Example 1:

Input: nums = [1,3,5]
Output: 1

Example 2:

Input: nums = [2,2,2,0,1]
Output: 0

Constraints:

• n == nums.length

• 1 <= n <= 5000

• -5000 <= nums[i] <= 5000

• nums is sorted and rotated between 1 and n times.

Follow up: This is the same as Find Minimum in Rotated Sorted Array but with duplicates. Would allow duplicates affect the run-time complexity? How and why?

Solution

English Version in Youtube

中文版解答Youtube Link

中文版解答Bilibili Link

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