leetcode Super Ugly Number

题目地址：https://leetcode.com/problems/super-ugly-number/

Super Ugly Number

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Total Accepted: 2812  Total Submissions: 9622  Difficulty: Medium

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.

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

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这道题其实是上一道题 Ugly Number II 的加强版而已，上一道题给出的素数只有2,3,5，而这道题，要求给定任意的素数集合，然后求出第n个ugly number。

Ugly Number II 题目地址：https://leetcode.com/problems/ugly-number-ii/

对于ugly Number2 这道题，我采用的做法分别维护三个数组ugly1，ugly2，ugly3分别对应由2,3,5作为因子的数的集合。如

minv = min(ugly1[i], ugly2[i], ugly3[i])

ugly1[i + 1]  =  minv  * 2

ugly2[i + 1]  = minv  * 3

ugly3[i + 1]  =  minv * 5

迭代n次，可以得到第n大的ugly NUmber。

同理，这道题也可以用这样的方法来做。不过由于prime数组的大小未知，开一个二维数组比较浪费空间。所以我们可以开一个跟prime数组大小一样的pointer数组，表示因子prime[i] 乘积的次数（这边这点比较抽象，需要看代码仔细理解哦）。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <limits.h>
using namespace std;

typedef unsigned long long ll;
class Solution {
public:
    int nthSuperUglyNumber(int n, vector<int>& primes) {
        int len = primes.size();
        vector<int> pointer(len, 0);
        vector<int> ugly(n + 1, INT_MAX);
        ugly[0] = 1;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < len; ++j) {
                int tmp = ugly[pointer[j]] * primes[j];
                ugly[i] = min(ugly[i], tmp);
            }
            for (int j = 0; j < len; ++j) {
                int tmp = ugly[pointer[j]] * primes[j];
                if (ugly[i] == tmp)
                    pointer[j]++;
            }
        }
        return ugly[n - 1];
    }
};

int main(int argc, char* argv[]) {

    Solution solution;
    vector<int> primes;
    primes.push_back(2);
    primes.push_back(3);
    primes.push_back(5);

    int n;
    while (cin >> n) {
        cout << solution.nthSuperUglyNumber(n, primes) << endl;
    }
    return 0;
}