313. Super Ugly Number

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. For example, [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) 1 is a super ugly number for any given primes.
(2) The given numbers in primes are in ascending order.
(3) 0 < k ≤ 100, 0 < n ≤ 106, 0 < primes[i] < 1000.
(4) The nth super ugly number is guaranteed to fit in a 32-bit signed integer.

一刷
题解：
新的值需要有两个因子，一个为prime, 一个存在ugly数组中。
创建一个数组，idx[j]， 长度为prime的长度，记录当前prime对应的ugly number
每次通过prime[j]*ugly[idx[j]，扫描j来获得当前的最小值，并update idx[j]

class Solution {
    public int nthSuperUglyNumber(int n, int[] primes) {
        int[] ugly = new int[n];
        int[] idx = new int[primes.length];

        ugly[0] = 1;
        for (int i = 1; i < n; i++) {
            //find next
            ugly[i] = Integer.MAX_VALUE;
            for (int j = 0; j < primes.length; j++)
                ugly[i] = Math.min(ugly[i], primes[j] * ugly[idx[j]]);
            
            //slip duplicate
            for (int j = 0; j < primes.length; j++) {
                while (primes[j] * ugly[idx[j]] <= ugly[i]) idx[j]++;
            }
        }

        return ugly[n - 1];
    }
}

如果把两个扫描j的方法合并为一个？