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.

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.

和ugly number ii很像，解觉得方法就是设置两个数组，一个存数，一个存factors的index

所谓index，就是指下一个factor应该与结果数组里那个数相乘的index

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

每个数的运算都是先把当前最小的找出来，然后如果是哪个factor得来的，就把那个factor的index++

Keep track the index, use the minimum one.