1390. Four Divisors (M)

Four Divisors (M)

Given an integer array nums, return the sum of divisors of the integers in that array that have exactly four divisors.

If there is no such integer in the array, return 0.

Example 1:

Input: nums = [21,4,7]
Output: 32
Explanation:
21 has 4 divisors: 1, 3, 7, 21
4 has 3 divisors: 1, 2, 4
7 has 2 divisors: 1, 7
The answer is the sum of divisors of 21 only.

Constraints:

• 1 <= nums.length <= 10^4
• 1 <= nums[i] <= 10^5

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题意

返回所有因数个数为4的数的因数之和。

思路

因数不重复，所以如果一个数是完全平方数，则它的因数个数一定是奇数，可直接排除。其它数因数都成对出现，当找到第三对因数时可直接跳过该数。

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代码实现

class Solution {
    public int sumFourDivisors(int[] nums) {
        int ans = 0;
        
        for (int num : nums) {
            int sqrt = (int) Math.sqrt(num);
            // 完全平方数直接跳过
            if (sqrt * sqrt == num) {
                continue;
            }
            int sum = 1 + num;
            int count = 1;
            for (int i = 2; i <= sqrt; i++) {
                if (num % i == 0) {
                    count++;
                    sum += i + sum / i;
                    if (count > 2) {
                        break;
                    }
                }
            }
            // 因数对数恰好为2时才计入答案
            if (count == 2) {
                ans += sum;
            }
        }
        
        return ans;
    }
}