313. 超级丑数

解法一

思想其实和丑数是一样的，参见：剑指 Offer 49. 丑数

class Solution {
    public int nthSuperUglyNumber(int n, int[] primes) {
        final int len = primes.length;
        // dp[i]表示第i+1个丑数
        int[] dp = new int[n];
        dp[0] = 1;
        int[] pointer = new int[len];
        int[] num = new int[len];
        for (int i = 0; i < len; ++i) {
            num[i] = dp[pointer[i]] * primes[i];
        }
        for (int i = 1; i < n; ++i) {
            dp[i] = min(num);
            for (int j = 0; j < len; ++j) {
                if (num[j] == dp[i]) {
                    ++pointer[j];
                    num[j] = dp[pointer[j]] * primes[j];
                }
            }
        }
        return dp[n - 1];
    }
    private int min(int[] arr) {
        int min = Integer.MAX_VALUE;
        for (int i : arr) {
            if (i < min) {
                min = i;
            }
        }
        return min;
    }
}