1362. Closest Divisors (M)

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 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

题意

给定一个数num，求一组差最小的两个数，且这两个数之积等于num+1或num+2。

思路

从开根处往前搜索，最后比较得到的两组值。

代码实现

class Solution {
    public int[] closestDivisors(int num) {
        int[] p1 = solve(num + 1);
        int[] p2 = solve(num + 2);
        return p1[1] - p1[0] < p2[1] - p2[0] ? p1 : p2;
    }
    private int[] solve(int num) {
        int[] pair = new int[2];
        int mid = (int) Math.sqrt(num);
        for (int i = mid; i >= 1; i--) {
            if (num % i == 0) {
                pair[0] = i;
                pair[1] = num / i;
                break;
            }
        }
        return pair;
    }
}