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
題解：
新的值需要有兩個(gè)因子，一個(gè)為prime, 一個(gè)存在ugly數(shù)組中。
創(chuàng)建一個(gè)數(shù)組遍希，idx[j]绽媒， 長度為prime的長度蜻底，記錄當(dāng)前prime對應(yīng)的ugly number
每次通過prime[j]*ugly[idx[j]碍庵，掃描j來獲得當(dāng)前的最小值了赌，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];
}
}
如果把兩個(gè)掃描j的方法合并為一個(gè)巩那？
