# 313. Super-Ugly-Number

difficulty: Medium

section pre{ background-color: #eee; border: 1px solid #ddd; padding:10px; border-radius: 5px; }

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.

## Method One

```
class Solution {
    public int nthSuperUglyNumber(int n, int[] primes) {
        if( n < 2 ){
            return n;
        } 
        int[] pointers = new int[primes.length];
        int[] uglyNumbers = new int[n];
        uglyNumbers[0] = 1;
        for(int i = 1; i < n; i++){
            int min = Integer.MAX_VALUE;
            for(int j = 0; j < primes.length; j++ ){
                min = Math.min( min, primes[j]*uglyNumbers[ pointers[j] ] );
            }
            uglyNumbers[i] = min;
            for(int j = 0; j < primes.length; j++ ){
                if( min == primes[j]*uglyNumbers[ pointers[j] ] ){
                    pointers[j]++;
                }
            }
        }
        return uglyNumbers[ n - 1];
    }
}
​
```
