leetcode笔记:Super Ugly Number

一. 题目描述

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.

二. 题目分析

题目的大意是,编写程序寻找第n个“超级丑陋数“,以下给出超级丑数的定义:

超级丑数是指只包含给定的k个质因子的正数。例如,给定长度为4的质数序列primes = [2, 7, 13, 19],前12个超级丑陋数序列为:[1, 2, 4, 7, 8, 13, 14, 16, 19, 26, 28, 32]

注意:

  1. 1被认为是超级丑数,无论给定怎样的质数列表。
  2. 给定的质数列表以升序排列。
  3. 0 < k ≤ 100, 0 < n ≤ 106, 0 < primes[i] < 1000。

一种方法是使用一个数组用于记录每个primes内质数乘积的次数,用另一个数组存储解第1k个丑数的值。

三. 示例代码

class Solution {  
public:  
    int nthSuperUglyNumber(int n, vector<int>& primes) {  
        int len = primes.size();  
        vector<int> index(len, 0);
        vector<int> uglyNum(n, INT_MAX);
        vector<int> temp(len);
        uglyNum[0] = 1;
        for (int i = 1; i < n; ++i)
        {
            int minj = -1;
            int minNum = INT_MAX;

            for (int j = 0; j < len; ++j)
            {
                temp[j] = primes[j] * uglyNum[index[j]];
                if (temp[j] < uglyNum[i])
                {
                    minNum = temp[j];
                    uglyNum[i] = temp[j];
                    minj = j;
                }
            }
            for (int j = minj; j < len; ++j)
            {
                if (minNum == temp[j])
                ++index[j];
            }
        }
        return uglyNum[n - 1];
    }  
};

四. 小结

事实上,使用这种方法最少只需不到10行代码,为了省略一些运算,便用了更多的辅助内存。

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