leetcode Super Ugly Number

最新推荐文章于 2024-06-04 11:57:06 发布

转载最新推荐文章于 2024-06-04 11:57:06 发布 · 3.5k 阅读

本文详细阐述了如何使用优先队列优化技术来解决寻找超丑数的问题，超丑数是所有质因数仅限于一组给定质数的正整数。文章深入分析了解决该问题的核心思想，并提供了Java实现代码，帮助读者理解和实现这一经典算法。

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<Node> q = new PriorityQueue<Node>();
        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<Node> {
    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;
    }
}

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