[Swift]LeetCode313. 超级丑数 | Super Ugly Number

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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.

---

编写一段程序来查找第 n 个超级丑数。

超级丑数是指其所有质因数都是长度为 k 的质数列表 primes 中的正整数。

示例:

输入: n = 12, primes = [2,7,13,19]
输出: 32 
解释: 给定长度为 4 的质数列表 primes = [2,7,13,19]，前 12 个超级丑数序列为：[1,2,4,7,8,13,14,16,19,26,28,32] 。

说明:

• 1 是任何给定 primes 的超级丑数。
•  给定 primes 中的数字以升序排列。
• 0 < k ≤ 100, 0 < n ≤ 106, 0 < primes[i] < 1000 。
• 第 n 个超级丑数确保在 32 位有符整数范围内。

---

116 ms

class Solution {
    func nthSuperUglyNumber(_ n: Int, _ primes: [Int]) -> Int {
        let count = primes.count
        
        var index = Array(repeatElement(0, count: count))
        var value = primes
        var temp = 0

        var ugly = [1]
        for _ in 0..<n - 1 {
            temp = Int.max
            for j in 0..<count {
                temp = min(temp, value[j])
            }
            ugly.append(temp)
            for j in 0..<count {
                if temp == value[j] {
                    index[j] += 1
                    value[j] = ugly[index[j]] * primes[j]
                }
            }
        }
        return ugly[n - 1]
    }
}

---

556ms

class Solution {
    func nthSuperUglyNumber(_ n: Int, _ primes: [Int]) -> Int {
        var res = [1] 
        var c = Array(repeating: 0, count: primes.count)
        
        for _ in 0 ..< n-1 {
            var comp = [Int]()
            for i in 0 ..< primes.count {
                comp.append(res[c[i]] * primes[i])
            }
            
            let minP = comp.min()!
            res.append(minP)
            
            for j in 0 ..< comp.count where comp[j] == minP {
                c[j] += 1
            }
        }
        
        return res.last!
    }
}