Perfect Number
https://leetcode.com/problems/perfect-number/description/
We define the Perfect Number is a positive integer that is equal to the sum of all itspositivedivisors except itself.
Now, given an integer n, write a function that returns true when it is a perfect number and false when it is not.
Thoughts
依次检查从2到sqrt(num)看是否能被num整除, 能就把i和对应的num / i 加进去. 因为已经考虑过了num/ i, 所以只到sqrt(num)即可.
Code
class Solution {
public boolean checkPerfectNumber(int num) {
if (num <= 1) {
return false;
}
int sum = 0;
sum++;
for (int i = 2; i <= Math.sqrt(num); i++) {
if (num % i == 0) {
sum += i + num / i;
//System.out.println(i + ", " + num / i);
}
}
return sum == num;
}
}
Analysis
时间复杂度O(sqrt(N)).
Last updated
Was this helpful?