【LeetCode】1362. 最接近的因数

思路一

我找出所有的相乘等于num+1或者num+2的 两个数。更新他们的距离。从而获得最终的结果。

/**
 * @Auther: Mason
 * @Date: 2020/07/20/15:24
 * @Description:
 */
public class Solution {
    
    
    public int[] closestDivisors(int num) {
    
    
        int[] res = new int[2];
        double range = Math.sqrt(num + 2);
        int min = Integer.MAX_VALUE;
        int tmp;
        for (int i = 1; i < range + 1; i++) {
    
    
            if ((num + 1) % i == 0) {
    
    
                tmp = (num + 1) / i;
                if (Math.abs(tmp - i) < min) {
    
    
                    res[0] = i;
                    res[1] = tmp;
                    min = Math.abs(tmp - i);
                }
            }
            if ((num + 2) % i == 0) {
    
    
                tmp = (num + 2) / i;
                if (Math.abs(tmp - i) < min) {
    
    
                    res[0] = i;
                    res[1] = tmp;
                    min = Math.abs(tmp - i);
                }
            }
        }
        return res;
    }

}

思路二

在思路一的基础上思考，其实越靠近平方根，其实距离是最小的。

/**
 * @Auther: Mason
 * @Date: 2020/07/20
 * @Description:
 */
public class Solution {
    
    
    public int[] closestDivisors(int num) {
    
    
        int[] res = new int[2];
        int sqrt1 = (int) Math.sqrt(num + 1);
        int sqrt2 = (int) Math.sqrt(num + 2);
        while (((num + 1) % sqrt1) != 0) {
    
    
            sqrt1--;
        }
        res[0] = sqrt1;
        res[1] = (num + 1) / sqrt1;
        while (((num + 2) % sqrt2) != 0) {
    
    
            sqrt2--;
        }
        if (Math.abs(sqrt2 - (num + 2) / sqrt2) < Math.abs(res[0] - res[1])) {
    
    
            res[0] = sqrt2;
            res[1] = (num + 2) / sqrt2;
        }
        return res;
    }


}