LeetCode 1390. Four Divisors四因数【Medium】【Python】【数学】
Problem
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 <= 10^41 <= nums[i] <= 10^5
问题
给你一个整数数组 nums，请你返回该数组中恰有四个因数的这些整数的各因数之和。
如果数组中不存在满足题意的整数，则返回 0 。
示例：
输入：nums = [21,4,7]
输出：32
解释：
21 有 4 个因数：1, 3, 7, 21
4 有 3 个因数：1, 2, 4
7 有 2 个因数：1, 7
答案仅为 21 的所有因数的和。
提示：
1 <= nums.length <= 10^41 <= nums[i] <= 10^5
思路
数学
暴力计算每个数的因数个数。
满足四个，就因数相加。
注意：因数不能重复。
时间复杂度: O(n*max(int(sqrt(x)), 4))，n 为 nums 个数。
空间复杂度: O(1)
Python3代码
from typing import List
class Solution:
def sumFourDivisors(self, nums: List[int]) -> int:
sum = 0
for x in nums:
if x == 1 or x == 2 or x == 3:
continue
num = 2
temp = [1, x]
# 计算因数
while num ** 2 <= x: # 用 num^2 <= x 比 num <= sqrt(x) 好
if len(temp) > 4:
break
if not x % num:
if num not in temp:
temp.append(num)
if int(x/num) not in temp:
temp.append(int(x/num))
num += 1
# print(temp)
if len(temp) == 4:
for _ in temp:
# print(_)
sum += _
return int(sum)
