[leetcode] 1390. Four Divisors

[leetcode] 1390. Four Divisors

Description

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: 32Explanation:21 has 4 divisors: 1, 3, 7, 214 has 3 divisors: 1, 2, 47 has 2 divisors: 1, 7The answer is the sum of divisors of 21 only.

Constraints:

1 <= nums.length <= 10^41 <= nums[i] <= 10^5

分析

题目的意思是：找出数组中的四因数，然后因子相加求和。这道题直接的思路就是一个一个的求，我封装了count_divisor来计算一个数的因数的个数，计算完后，去除重复的因数，判断剩下的因数的个数满足4的话，就找到了一个满足的数，然后继续遍历其他的数。

代码

class Solution: def count_divisor(self,num): ans=[1,num] for i in range(2,int(sqrt(num))+1,1): if(num%i==0): ans.append(i) ans.append(num//i) return ans def sumFourDivisors(self, nums: List[int]) -> int: res=0 for num in nums: ans=self.count_divisor(num) ans=list(set(ans)) if(len(ans)==4): res+=sum(ans) return res

参考文献

​​【leetcode】1390. Four Divisors​​

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