5171-最接近的因数
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 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
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/closest-divisors
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O(n2)超时
class Solution {
public int[] closestDivisors(int num) {
int[] res = new int[2];
int diff = Integer.MAX_VALUE;
int sqrt = (int) Math.sqrt(num);
for(int i = sqrt; i <= num; i++) {
for(int j = sqrt + 1; j > 0; j--) {
int mul = i * j;
if((mul == num + 1 || mul == num + 2)) {
int abs = Math.abs(i - j);
if(abs < diff) {
res[0] = j;
res[1] = i;
diff = abs;
if(diff == 0) {
return res;
}
}
}
}
}
return res;
}
改进
public int[] closestDivisors(int num) {
int[] res = new int[2];
int num1 = num + 1;
int num2 = num + 2;
int sqrt1 = (int) Math.sqrt(num1);
if(Math.pow(sqrt1 + 1, 2) == num2) {
return new int[]{sqrt1 + 1, sqrt1 + 1};
}
for(int i = sqrt1; i <= num2; i++) {
int j = num2 % i;
int k = num1 % i;
if(j == 0 || k == 0) {
if(k == 0) {
j = num1 / i;
} else {
j = num2 / i;
}
res[0] = Math.min(i, j);
res[1] = Math.max(i, j);
return res;
}
}
return res;
}
继续改进:从1~sqrt显然比sqrt~num+2数更少
public int[] closestDivisors(int num) {
int[] r = new int[2];
int sqrt = (int) Math.sqrt(num + 2);
for (int i = sqrt; i >= 1; i--) {
if ((num + 1) % i == 0) {
r[0] = i;
r[1] = (num + 1) / i;
return r;
}
if ((num + 2) % i == 0) {
r[0] = i;
r[1] = (num + 2) / i;
return r;
}
}
return r;
}