https://leetcode.com/problems/perfect-number/
We define the Perfect Number is a positive integer that is equal to the sum of all its positive divisors 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.
Example:
Input: 28 Output: True Explanation: 28 = 1 + 2 + 4 + 7 + 14
Note: The input number n will not exceed 100,000,000. (1e8)
https://discuss.leetcode.com/topic/84259/simple-java-solution public boolean checkPerfectNumber(int num) {
if (num == 1) return false;
int sum = 0;
for (int i = 2; i <= Math.sqrt(num); i++) {
if (num % i == 0) {
sum += i;
if (i != num / i) sum += num / i;
}
}
sum++;
return sum == num;
}
Update Enlightened by discussion below by @StefanPochmann and @jdrogin, in the given range we don't need to test if (i != num / i) before add num / i to sum. public boolean checkPerfectNumber(int num) {
if (num == 1) return false;
int sum = 0;
for (int i = 2; i <= Math.sqrt(num); i++) {
if (num % i == 0) {
sum += i + num / i;
}
}
sum++;
return sum == num;
}public boolean checkPerfectNumber(int num) {
int sum = 1;
for (int i=2;i<Math.sqrt(num);i++)
if (num % i == 0) sum += i + (num/i == i ? 0 : num/i);
return num != 1 && sum == num;
} public boolean checkPerfectNumber(int num) {
HashSet<Integer> set = new HashSet<Integer>();
set.add(6);
set.add(28);
set.add(496);
set.add(8128);
set.add(33550336);
return set.contains(num);
}