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)。

 1 class Solution {
 2 public:
 3     int min(vector<int>& primes, vector<int>& pointers, vector<int>& superUglyNumber) {
 4         int res = INT_MAX;
 5         for (int i = 0, n = primes.size(); i < n; i++)
 6             if (res > superUglyNumber[pointers[i]] * primes[i])
 7                 res = superUglyNumber[pointers[i]] * primes[i];
 8         for (int i = 0, n = primes.size(); i < n; i++)
 9             if (res == superUglyNumber[pointers[i]] * primes[i])
10                 pointers[i]++;
11         return res;
12     }
13     int nthSuperUglyNumber(int n, vector<int>& primes) {
14         vector<int> superUglyNumber;
15         superUglyNumber.push_back(1);
16         vector<int> pointers(primes.size());
17         for (int i = 1; i < n; i++)
18             superUglyNumber.push_back(min(primes, pointers, superUglyNumber));
19         return superUglyNumber.back();
20     }
21 };