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？
