Formatted question description: https://leetcode.ca/all/507.html
507. Perfect Number (Easy)
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)
Related Topics:
Math
Similar Questions:
Solution 1.
// OJ: https://leetcode.com/problems/perfect-number/
// Time: O(sqrt(N))
// Space: O(1)
class Solution {
public:
bool checkPerfectNumber(int num) {
if (num <= 1) return false;
int sum = 0, end = sqrt(num);
for (int i = 1; i <= end; ++i) {
if (num % i) continue;
int j = num / i;
sum += i;
if (j != i && j != num) sum += j;
}
return sum == num;
}
};
Java
class Solution {
public boolean checkPerfectNumber(int num) {
if (num <= 0)
return false;
int[][] primeFactorization = primeFactorize(num);
long factorsSum = 1;
for (int[] primePower : primeFactorization) {
int prime = primePower[0], power = primePower[1];
long primeSum = 0;
long factor = 1;
for (int i = 0; i <= power; i++) {
primeSum += factor;
factor *= prime;
}
factorsSum *= primeSum;
}
return factorsSum == (long) (2 * num);
}
public int[][] primeFactorize(int num) {
List<Integer> primes = new ArrayList<Integer>();
List<Integer> powers = new ArrayList<Integer>();
if (num % 2 == 0) {
primes.add(2);
int power2 = 0;
while (num % 2 == 0) {
num /= 2;
power2++;
}
powers.add(power2);
}
int prime = 3;
while (num > 1) {
if (num % prime == 0) {
primes.add(prime);
int power = 0;
while (num % prime == 0) {
num /= prime;
power++;
}
powers.add(power);
}
prime += 2;
}
int length = primes.size();
int[][] primeFactorization = new int[length][2];
for (int i = 0; i < length; i++) {
primeFactorization[i][0] = primes.get(i);
primeFactorization[i][1] = powers.get(i);
}
return primeFactorization;
}
}