[Swift]LeetCode1175. 質數排列 | Prime Arrangements

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Return the number of permutations of 1 to n so that prime numbers are at prime indices (1-indexed.)

(Recall that an integer is prime if and only if it is greater than 1, and cannot be written as a product of two positive integers both smaller than it.)

Since the answer may be large, return the answer modulo 10^9 + 7.

 

Example 1:

Input: n = 5
Output: 12
Explanation: For example [1,2,5,4,3] is a valid permutation, but [5,2,3,4,1] is not because the prime number 5 is at index 1.

Example 2:

Input: n = 100
Output: 682289015

 

Constraints:

• 1 <= n <= 100

---

 請你幫忙給從 1 到 n 的數設計排列方案，使得全部的「質數」都應該被放在「質數索引」（索引從 1 開始）上；你須要返回可能的方案總數。

讓咱們一塊兒來回顧一下「質數」：質數必定是大於 1 的，而且不能用兩個小於它的正整數的乘積來表示。

因爲答案可能會很大，因此請你返回答案 模 mod 10^9 + 7 以後的結果便可。

 

示例 1：

輸入：n = 5
輸出：12
解釋：舉個例子，[1,2,5,4,3] 是一個有效的排列，但 [5,2,3,4,1] 不是，由於在第二種狀況裏質數 5 被錯誤地放在索引爲 1 的位置上。

示例 2：

輸入：n = 100
輸出：682289015

 

提示：

• 1 <= n <= 100

---

0ms

 1 class Solution {
 2     private var count = 0
 3     private let modulo: UInt64 = 1_000_000_000 + 7
 4     
 5     func numPrimeArrangements(_ n: Int) -> Int {
 6         var primeCount = 0
 7         for i in 1 ... n {
 8             if isPrime(i) {
 9                 primeCount += 1
10             }
11         }
12         
13         
14         let primePermutations = permutationCount(primeCount)
15         let compositePermutations = permutationCount(n - primeCount)
16         
17         return Int((primePermutations * compositePermutations) % modulo)
18     }
19     
20     private func isPrime(_ n: Int) -> Bool {
21         guard n > 1 else { return false }
22         if n == 2 { return true }
23         if n == 3 { return true }
24         for k in 2 ... Int(sqrt(Float(n))) {
25             if n % k == 0 {
26                 return false
27             }
28         }
29         return true
30     }
31     
32     private func permutationCount(_ n: Int) -> UInt64 {
33         guard n > 0 else { return 1 }
34         var count: UInt64 = 1
35         for i in 1 ... n {
36             count = ( count * UInt64(i) ) % modulo
37         }
38         return count
39     }
40 }

---

Runtime: 8 ms

Memory Usage: 20.9 MB

 1 class Solution {
 2     func numPrimeArrangements(_ n: Int) -> Int {
 3         var p:Set<Int> = [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]
 4         let MOD:Int = 1000000007
 5         var ca:Int = 0
 6         var cb:Int = 0
 7         for i in 1...n
 8         {
 9             if p.contains(i)
10             {
11                 ca += 1
12             }
13             else
14             {
15                 cb += 1
16             }
17         }
18         var ans:Int = 1
19         if ca >= 1
20         {
21             for i in 1...ca
22             {
23                 ans = ans * i % MOD
24             }
25         }
26         if cb >= 1
27         {
28             for i in 1...cb
29             {
30                 ans = ans * i % MOD
31             }
32         }
33         return ans
34     }
35 }