[LeetCode] 507. Perfect Number

轉自LeetCode

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)

<Solution>

想法如下

• 其實就是暴力解，但要會懂得減枝。當找到一個數 n1 可以整除 num，你也同時找到一個數 n2 = num / n1 可以整除 num。所以，只要檢查到 sqrt(num) 就可以了，因為 sqrt(num) * sqrt(num) = num

code 如下

C++

	class Solution {
	public:
	bool checkPerfectNumber(int num) {
	if(num <= 1) {
	return false;
	}
	const int upper_bound = sqrt(num);
	int sum = 1;
	for(int i = 2; i <= upper_bound; i++) {
	if(num % i == 0) {
	sum += i;
	sum += num / i;
	}
	}
	return sum == num;
	}
	};

view raw perfectNumber.cpp hosted with ❤ by GitHub

Java

	class Solution {
	public boolean checkPerfectNumber(int num) {
	if(num <= 1) {
	return false;
	}
	final int upper_bound = (int)Math.sqrt(num);
	int sum = 1;
	for(int i = 2; i <= upper_bound; i++) {
	if(num % i == 0) {
	sum += i;
	sum += num / i;
	}
	}
	return sum == num;
	}
	}

view raw perfectNumber.java hosted with ❤ by GitHub