Write a program to find the nth super ugly number.

Super ugly numbers are positive numbers whose all prime factors are in the given prime list primes of size k.

Example:

Input: n = 12, primes = [2,7,13,19]
Output: 32 
Explanation: [1,2,4,7,8,13,14,16,19,26,28,32] is the sequence of the first 12 
             super ugly numbers given primes
 
 = [2,7,13,19] of size 4.

Note:

• 1 is a super ugly number for any given primes.
• The given numbers in primes are in ascending order.
• 0 < k ≤ 100, 0 < n ≤ 106, 0 < primes[i] < 1000.
• The nth super ugly number is guaranteed to fit in a 32-bit signed integer.

編寫一段程式來查詢第 n 個超級醜數。

超級醜數是指其所有質因數都是長度為 k 的質數列表 primes 中的正整數。

示例:

輸入: n = 12, primes = [2,7,13,19]
輸出: 32 
解釋: 給定長度為 4 的質數列表 primes = [2,7,13,19]，前 12 個超級醜數序列為：[1,2,4,7,8,13,14,16,19,26,28,32] 。

說明:

• 1 是任何給定 primes 的超級醜數。
•  給定 primes
   中的數字以升序排列。
• 0 < k ≤ 100, 0 < n ≤ 106, 0 < primes[i] < 1000 。
• 第 n 個超級醜數確保在 32 位有符整數範圍內。

116 ms

 1 class Solution {
 2     func nthSuperUglyNumber(_ n: Int, _ primes: [Int]) -> Int {
 3         let count = primes.count
 4         
 5         var index = Array(repeatElement(0, count: count))
 6         var value = primes
 7         var temp = 0
 8 
 9         var ugly = [1]
10         for _ in 0..<n - 1 {
11             temp = Int.max
12             for j in 0..<count {
13                 temp = min(temp, value[j])
14             }
15             ugly.append(temp)
16             for j in 0..<count {
17                 if temp == value[j] {
18                     index[j] += 1
19                     value[j] = ugly[index[j]] * primes[j]
20                 }
21             }
22         }
23         return ugly[n - 1]
24     }
25 }

556ms

 1 class Solution {
 2     func nthSuperUglyNumber(_ n: Int, _ primes: [Int]) -> Int {
 3         var res = [1] 
 4         var c = Array(repeating: 0, count: primes.count)
 5         
 6         for _ in 0 ..< n-1 {
 7             var comp = [Int]()
 8             for i in 0 ..< primes.count {
 9                 comp.append(res[c[i]] * primes[i])
10             }
11             
12             let minP = comp.min()!
13             res.append(minP)
14             
15             for j in 0 ..< comp.count where comp[j] == minP {
16                 c[j] += 1
17             }
18         }
19         
20         return res.last!
21     }
22 }