LeetCode题解(1175)：质数排列(Python)

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)$O(N)$$O(N)$44ms (57.83%)Ans 2 (Python)$O(N)$$O(N)$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