313. 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. 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.

解法1：

和Ugly Number II的解法一样，只是这里要用一个数组来维护每一个对应的primes相乘的ugly number的位置。要注意这里用到的去重的方法是一旦当前的ugly number确定了就把可能产生的小于等于他的所有备选答案跳过（也就是对应的和prime相乘的index要++)

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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];
    }
}