[LintCode] Super Ugly Number 超级丑陋数

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.

Notice:
1 is a super ugly number for any given primes.
The given numbers in primes are in ascending order.
0 < k ≤ 100, 0 < n ≤ 10^6, 0 < primes[i] < 1000

Example
Given n = 6, primes = [2, 7, 13, 19] return 13

LeetCode上的原题，请参见我之前的博客Super Ugly Number。

解法一：

class Solution {
public:
    /**
     * @param n a positive integer
     * @param primes the given prime list
     * @return the nth super ugly number
     */
    int nthSuperUglyNumber(int n, vector<int>& primes) {
        vector<int> res(1, 1), pos(primes.size(), 0);
        while (res.size() < n) {
            vector<int> t;
            for (int i = 0; i < primes.size(); ++i) {
                t.push_back(res[pos[i]] * primes[i]);
            }
            int mn = INT_MAX;
            for (int i = 0; i < primes.size(); ++i) {
                mn = min(mn, t[i]);
            }
            for (int i = 0; i < primes.size(); ++i) {
                if (t[i] == mn) ++pos[i];
            }
            res.push_back(mn);
        }
        return res.back();
    }
};

解法二：

class Solution {
public:
    /**
     * @param n a positive integer
     * @param primes the given prime list
     * @return the nth super ugly number
     */
    int nthSuperUglyNumber(int n, vector<int>& primes) {
        vector<int> dp(n, 1), pos(primes.size(), 0);
        for (int i = 1; i < n; ++i) {
            dp[i] = INT_MAX;
            for (int j = 0; j < primes.size(); ++j) {
                dp[i] = min(dp[i], dp[pos[j]] * primes[j]);
            }
            for (int j = 0; j < primes.size(); ++j) {
                if (dp[i] == dp[pos[j]] * primes[j]) {
                    ++pos[j];
                }
            }
        }
        return dp.back();
    }
};