313. Super Ugly Number-优快云博客

本文介绍了一种高效算法来寻找第N个超级丑数。超级丑数是指其所有质因数都在给定质数列表中的正整数。文章详细解释了算法思路并提供了完整的Java实现代码。

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问题链接：https://leetcode.com/problems/super-ugly-number/?tab=Description

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.
(4) The nth super ugly number is guaranteed to fit in a 32-bit signed integer.

超级丑数：有限个给定质数的乘积，即将给定质数变成数组的方式，然后给定质数分别与超级丑数的数组元素相乘，则每个元素都是超级丑数。做法如下：

1.新建指针数组，指针数组在下标i位置的值表示超级丑数数组的i元素和给定质数数组i元素的乘积本次可能是最小的，两个数组超级丑数数组i-1元素已用过；

2.判断最小发生在哪里，则将指针数组对应位置+1

参考答案：https://my.oschina.net/Tsybius2014/blog/547766

public class Solution {
    public int nthSuperUglyNumber(int n, int[] primes) {
        int[] superUglyNums = new int[n];
        int[] indexs = new int[primes.length];
        Arrays.fill(indexs,0);
        int count = 1;
        superUglyNums[0] = 1;
        
        while(count<n) {
            int min=Integer.MAX_VALUE;
            for(int i=0;i<indexs.length;i++) {
                int temp=superUglyNums[indexs[i]]*primes[i];
                min= min<temp?min:temp;
            }
            
            for(int i=0;i<indexs.length;i++) {
                if(min==superUglyNums[indexs[i]]*primes[i])
                    indexs[i]++;
                
            }
            
            superUglyNums[count++]=min;
        }
        
        return superUglyNums[n-1];
    }
}