lintcode-medium-Super Ugly Number

最新推荐文章于 2020-12-14 09:14:59 发布

转载最新推荐文章于 2020-12-14 09:14:59 发布 · 83 阅读

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原文链接：http://www.cnblogs.com/goblinengineer/p/5362008.html

本文介绍了一种算法，用于寻找第N个超丑数。超丑数是指所有质因数都在给定质数列表中的正整数。文中提供了一个Java实现的例子，并通过一个具体实例展示了如何使用该算法。

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.

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

public class Solution {
    /**
     * @param n a positive integer
     * @param primes the given prime list
     * @return the nth super ugly number
     */
    public int nthSuperUglyNumber(int n, int[] primes) {
        // Write your code here
        
        if(primes == null || primes.length == 0)
            return 1;
        
        if(n == 1)
            return 1;
        
        int curr = 1;
        int[] p = new int[primes.length];
        ArrayList<Integer> res = new ArrayList<Integer>();
        
        res.add(1);
        
        while(res.size() < n){
            int next = Integer.MAX_VALUE;
            
            for(int i = 0; i < p.length; i++){
                while(res.get(p[i]) * primes[i] <= curr)
                    p[i]++;
                
                next = Math.min(next, res.get(p[i]) * primes[i]);
            }
            
            curr = next;
            res.add(next);
        }
        
        return res.get(n - 1);
    }
}

转载于:https://www.cnblogs.com/goblinengineer/p/5362008.html