题目：原题链接（简单）
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) |
|
| 44ms (57.83%) |
Ans 2 (Python) |
|
| 40ms (83.48%) |
Ans 3 (Python) |
LeetCode的Python执行用时随缘，只要时间复杂度没有明显差异，执行用时一般都在同一个量级，仅作参考意义。
解法一：
def numPrimeArrangements(self, n: int) -> int:
def countPrimes(k: int) -> int:
if k < 2:
return 0
num_list = [True] * k
num_list[0], num_list[1] = False, False
for i in range(2, int(pow(k, 0.5)) + 1):
if num_list[i]:
num_list[i * i::i] = [False] * ((k - i * i - 1) // i + 1)
return sum(num_list)
prime_num = countPrimes(n + 1)
other_num = n - prime_num
ans = 1
for i in range(1, prime_num + 1):
ans *= i
ans = ans % (10 ** 9 + 7)
for i in range(1, other_num + 1):
ans *= i
ans = ans % (10 ** 9 + 7)
return ans解法二（不但维护了质数表，还二分查找…）：
def numPrimeArrangements(self, n: int) -> int:
primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
prime_num = bisect.bisect_left(primes, n)
other_num = n - prime_num
ans = 1
for i in range(1, prime_num + 1):
ans *= i
ans = ans % (10 ** 9 + 7)
for i in range(1, other_num + 1):
ans *= i
ans = ans % (10 ** 9 + 7)
return ans
