Super Ugly Number -- LeetCode

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.

思路：解法和ugly number II一样。复杂度O(kn)。

这个算法有一个改进的地方，min函数可以用最小堆来实现。这样复杂度是O(log(k)n)。

 class Solution {
 public:
     int min(vector<int>& primes, vector<int>& pointers, vector<int>& superUglyNumber) {
         int res = INT_MAX;
         for (int i = , n = primes.size(); i < n; i++)
             if (res > superUglyNumber[pointers[i]] * primes[i])
                 res = superUglyNumber[pointers[i]] * primes[i];
         for (int i = , n = primes.size(); i < n; i++)
             if (res == superUglyNumber[pointers[i]] * primes[i])
                 pointers[i]++;
         return res;
     }
     int nthSuperUglyNumber(int n, vector<int>& primes) {
         vector<int> superUglyNumber;
         superUglyNumber.push_back();
         vector<int> pointers(primes.size());
         for (int i = ; i < n; i++)
             superUglyNumber.push_back(min(primes, pointers, superUglyNumber));
         return superUglyNumber.back();
     }
 };