1362 Closest Divisors 最接近的因数
Description:
Given an integer num, find the closest two integers in absolute difference whose product equals num + 1 or num + 2.
Return the two integers in any order.
Example:
Example 1:
Input: num = 8
Output: [3,3]
Explanation: For num + 1 = 9, the closest divisors are 3 & 3, for num + 2 = 10, the closest divisors are 2 & 5, hence 3 & 3 is chosen.
Example 2:
Input: num = 123
Output: [5,25]
Example 3:
Input: num = 999
Output: [40,25]
Constraints:
1 <= num <= 10^9
题目描述：
给你一个整数 num，请你找出同时满足下面全部要求的两个整数：
两数乘积等于 num + 1 或 num + 2
以绝对差进行度量，两数大小最接近
你可以按任意顺序返回这两个整数。
示例：
示例 1：
输入：num = 8
输出：[3,3]
解释：对于 num + 1 = 9，最接近的两个因数是 3 & 3；对于 num + 2 = 10, 最接近的两个因数是 2 & 5，因此返回 3 & 3 。
示例 2：
输入：num = 123
输出：[5,25]
示例 3：
输入：num = 999
输出：[40,25]
提示：
1 <= num <= 10^9
思路：
数学
两个最近的因数在 n ^ 1 / 2 两侧
从 n ^ 1 / 2 开始向下搜索直到找到 n 的因子
时间复杂度为 O(n ^ 1 / 2), 空间复杂度为 O(1)
代码:
C++:
class Solution
{
public:
vector<int> closestDivisors(int num)
{
vector<int> result1 = helper(num + 1), result2 = helper(num + 2);
return abs(result1.front() - result1.back()) < abs(result2.front() - result2.back()) ? result1 : result2;
}
private:
vector<int> helper(int num)
{
vector<int> result(2);
for (int i = (int)sqrt(num); i > 0; i--)
{
if (!(num % i))
{
result.front() = i;
result.back() = num / i;
break;
}
}
return result;
}
};
Java:
class Solution {
public int[] closestDivisors(int num) {
int[] result1 = helper(num + 1), result2 = helper(num + 2);
return Math.abs(result1[0] - result1[1]) < Math.abs(result2[0] - result2[1]) ? result1 : result2;
}
private int[] helper(int num) {
int result[] = new int[2];
for (int i = (int)Math.sqrt(num); i > 0; i--) {
if (num % i == 0) {
result[0] = i;
result[1] = num / i;
break;
}
}
return result;
}
}
Python:
class Solution:
def closestDivisors(self, num: int) -> List[int]:
def helper(num: int) -> List[int]:
for x in range(int(num ** 0.5), 0, -1):
if not num % x:
return [x, num // x]
return result1 if abs((result1 := helper(num + 1))[0] - result1[1]) < abs((result2 := helper(num + 2))[0] - result2[1]) else result2
