Given an integer array nums
, return the greatest common divisor of the smallest number and largest number in nums
.
The greatest common divisor of two numbers is the largest positive integer that evenly divides both numbers.
Example 1:
Input: nums = [2,5,6,9,10] Output: 2 Explanation: The smallest number in nums is 2. The largest number in nums is 10. The greatest common divisor of 2 and 10 is 2.
Example 2:
Input: nums = [7,5,6,8,3] Output: 1 Explanation: The smallest number in nums is 3. The largest number in nums is 8. The greatest common divisor of 3 and 8 is 1.
Example 3:
Input: nums = [3,3] Output: 3 Explanation: The smallest number in nums is 3. The largest number in nums is 3. The greatest common divisor of 3 and 3 is 3.
Constraints:
2 <= nums.length <= 1000
1 <= nums[i] <= 1000
Solution:
Use std::minmax_element and std::gcd
Time complexity: O(n)
Space complexity: O(1)
C++
1 2 3 4 5 6 7 8 |
// Author: Huahua class Solution { public: int findGCD(vector<int>& nums) { auto p = minmax_element(begin(nums), end(nums)); return gcd(*p.first, *p.second); } }; |
请尊重作者的劳动成果,转载请注明出处!花花保留对文章/视频的所有权利。
如果您喜欢这篇文章/视频,欢迎您捐赠花花。
If you like my articles / videos, donations are welcome.
Be First to Comment