leetcode Super Ugly Number

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.

新博客原文： leetcode-super-ugly-number
更多题解  https://www.hrwhisper.me/leetcode-algorithm-solution/

题意：

给定因子Primes , 让你求第n个super ugly number。

super ugly number定义为：整数，且因子全部都在primes中。 注意1为特殊的super ugly number。

思路：

和 leetcode Ugly Number  II 思路一样，要使得super ugly number 不漏掉，那么用每个因子去乘第一个，当前因子乘积是最小后，乘以下一个…..以此类推。

Java 130ms

// http://www.hrwhisper.me/leetcode-super-ugly-number/
public class Solution {
    public int nthSuperUglyNumber(int n, int[] primes) {
        int[] ugly_number = new int[n];
        ugly_number[0] = 1;
        PriorityQueue q = new PriorityQueue();
        for (int i = 0; i < primes.length; i++)
            q.add(new Node(0, primes[i], primes[i]));
        for (int i = 1; i < n; i++) {
            Node cur = q.peek();
            ugly_number[i] = cur.val;
            do {
                cur = q.poll();
                cur.val = ugly_number[++cur.index] * cur.prime;
                q.add(cur);
            } while (!q.isEmpty() && q.peek().val == ugly_number[i]);
        }
        return ugly_number[n - 1];
    }
}

class Node implements Comparable {
    int index;
    int val;
    int prime;

    Node(int index, int val, int prime) {
        this.val = val;
        this.index = index;
        this.prime = prime;
    }

    public int compareTo(Node x) {
        return this.val > x.val ? 1 : -1;
    }
}

本文地址：http://www.hrwhisper.me/leetcode-super-ugly-number/
本文由 hrwhisper 原创发布，转载请点名出处