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.

Example 2:

Input: nums = [21,21]

Output: 64

Example 3:

Input: nums = [1,2,3,4,5]

Output: 0

?

Constraints:

• 1 <= nums.length <= 104

• 1 <= nums[i] <= 105

主要就是計算有4個divisor的數(shù)字和，那么我們用set去儲存，遍歷從1到這個數(shù)的平方根，

如果能除盡，那么2個因子都放到set中，最后判斷set的大小為4，則返回每個數(shù)字的和，

如果不是4，返回0；

下面是代碼：

Runtime:?126 ms, faster than?11.96%?of?Java?online submissions for?Four Divisors.

Memory Usage:?43.2 MB, less than?31.58%?of?Java?online submissions for?Four Divisors.