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.
You must decrease the overall operation steps as much as possible.
Example 1:
Input: nums = [1,3,5] Output: 1
Example 2:
Input: nums = [2,2,2,0,1] Output: 0
Constraints:
n == nums.length1 <= n <= 5000-5000 <= nums[i] <= 5000nums is sorted and rotated between 1 and n times.
Follow up: This problem is similar to Find Minimum in Rotated Sorted Array, but nums may contain duplicates. Would this affect the runtime complexity? How and why?
class Solution {
public int findMin(int[] nums) {
int n = nums.length, lt = 0, rt = n - 1;
while (lt != rt) {
int mid = ((lt + rt + (rt < lt ? n : 0)) / 2) % n;
if (nums[mid] < nums[rt])
rt = mid;
else if (nums[mid] == nums[rt])
rt--;
else
lt = mid + 1;
}
return nums[lt];
}
}